Applied Linear Statistical S：Instructor's Solutions Manual(5th,2005)
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Instructor
Solutions Manual
to accompany
Applied Linear
Statistical Models
Fifth Edition
Michael H. Kutner
Emory University
Christopher J. Nachtsheim
University of Minnesota
John Neter
University of Georgia
William Li
University of Minnesota
2005
McGraw-Hill/Irwin
Chicago, IL
Boston, MA
PREFACE
This Solutions Manual gives intermediate and final numerical results for all end-of-chapter
Problems, Exercises, and Projects with computational elements contained in Applied Linear
Statistical M odels, 5th edition. This Solutions Manual also contains proofs for all Exercises
that require derivations. No solutions are provided for the Case Studies.
In presenting calculational results we frequently show, for ease in checking, more digits
than are significant for the original data. Students and other users may obtain slightly
different answers than those presented here, because of different rounding procedures. When
a problem requires a percentile (e.g. of the t or F distributions) not included in the Appendix
B Tables, users may either interpolate in the table or employ an available computer program
for finding the needed value. Again, slightly different values may be obtained than the ones
shown here.
We have included many more Problems, Exercises, and Projects at the ends of chapters
than can be used in a term, in order to provide choice and flexibility to instructors in assigning
problem material. For all major topics, three or more problem settings are presented, and the
instructor can select different ones from term to term. Another option is to supply students
with a computer printout for one of the problem settings for study and class discussion and to
select one or more of the other problem settings for individual computation and solution. By
drawing on the basic numerical results in this Manual, the instructor also can easily design
additional questions to supplement those given in the text for a given problem setting.
The data sets for all Problems, Exercises, Projects and Case Studies are contained in the
compact disk provided with the text to facilitate data entry. It is expected that the student
will use a computer or have access to computer output for all but the simplest data sets,
where use of a basic calculator would be adequate. For most students, hands-on experience
in obtaining the computations by computer will be an important part of the educational
experience in the course.
While we have checked the solutions very carefully, it is possible that some errors are
still present. We would be most grateful to have any errors called to our attention. Errata
can be reported via the website for the book: http://www.mhhe.com/KutnerALSM5e. We
acknowledge with thanks the assistance of Lexin Li and Yingwen Dong in the checking of
Chapters 1-14 of this manual. We, of course, are responsible for any errors or omissions that
remain.
Michael H. Kutner
Christopher J. Nachtsheim
John Neter
William Li
i
ii
Contents
1 LINEAR REGRESSION WITH ONE PREDICTOR VARIABLE
1-1
2 INFERENCES IN REGRESSION AND CORRELATION ANALYSIS
2-1
3 DIAGNOSTICS AND REMEDIAL MEASURES
3-1
4 SIMULTANEOUS INFERENCES AND OTHER TOPICS IN REGRESSION ANALYSIS
4-1
5 MATRIX APPROACH TO SIMPLE LINEAR REGRESSION ANALYSIS
5-1
6 MULTIPLE REGRESSION – I
6-1
7 MULTIPLE REGRESSION – II
7-1
8 MODELS FOR QUANTITATIVE AND QUALITATIVE PREDICTORS 8-1
9 BUILDING THE REGRESSION MODEL I: MODEL SELECTION AND
VALIDATION
9-1
10 BUILDING THE REGRESSION MODEL II: DIAGNOSTICS
10-1
11 BUILDING THE REGRESSION MODEL III: REMEDIAL MEASURES11-1
12 AUTOCORRELATION IN TIME SERIES DATA
12-1
13 INTRODUCTION TO NONLINEAR REGRESSION AND NEURAL NETWORKS
13-1
14 LOGISTIC REGRESSION, POISSON REGRESSION,AND GENERALIZED LINEAR MODELS
14-1
15 INTRODUCTION TO THE DESIGN OF EXPERIMENTAL AND OBSERVATIONAL STUDIES
15-1
16 SINGLE-FACTOR STUDIES
16-1
17 ANALYSIS OF FACTOR LEVEL MEANS
17-1
iii
18 ANOVA DIAGNOSTICS AND REMEDIAL MEASURES
18-1
19 TWO-FACTOR ANALYSIS OF VARIANCE WITH EQUAL SAMPLE
SIZES
19-1
20 TWO-FACTOR STUDIES – ONE CASE PER TREATMENT
20-1
21 RANDOMIZED COMPLETE BLOCK DESIGNS
21-1
22 ANALYSIS OF COVARIANCE
22-1
23 TWO-FACTOR STUDIES WITH UNEQUAL SAMPLE SIZES
23-1
24 MULTIFACTOR STUDIES
24-1
25 RANDOM AND MIXED EFFECTS MODELS
25-1
26 NESTED DESIGNS, SUBSAMPLING, AND PARTIALLY NESTED DESIGNS
26-1
27 REPEATED MEASURES AND RELATED DESIGNS
27-1
28 BALANCED INCOMPLETE BLOCK, LATIN SQUARE, AND RELATED
DESIGNS
28-1
29 EXPLORATORY EXPERIMENTS – TWO-LEVEL FACTORIAL AND
FRACTIONAL FACTORIAL DESIGNS
29-1
30 RESPONSE SURFACE METHODOLOGY
30-1
Appendix D: RULES FOR DEVELOPING ANOVA MODELS AND TABLES
FOR BALANCED DESIGNS
D.1
iv
Chapter 1
LINEAR REGRESSION WITH ONE
PREDICTOR VARIABLE
1.1.
No
1.2.
Y = 300 + 2X, functional
1.5.
No
1.7.
a.
No
b.
Yes, .68
1.8.
Yes, no
1.10. No
1.12. a. Observational
1.13. a. Observational
1.18. No
1.19. a.
β0 = 2.11405, β1 = 0.03883, Ŷ = 2.11405 + .03883X
c.
Ŷh = 3.27895
d.
β1 = 0.03883
1.20. a.
d.
1.21. a.
Ŷ = −0.5802 + 15.0352X
Ŷh = 74.5958
Ŷ = 10.20 + 4.00X
b.
Ŷh = 14.2
c.
4.0
d.
(X̄, Ȳ ) = (1, 14.2)
1.22. a.
Ŷ = 168.600000 + 2.034375X
1-1
b.
Ŷh = 249.975
c.
β1 = 2.034375
1.23. a.
i:
1
2
ei : 0.9676 1.2274
Yes
b.
1.24. a.
M SE = 0.388,
√
...
...
119
-0.8753
120
-0.2532
M SE = 0.623, grade points
i:
1
2
...
ei : -9.4903 0.4392 . . .
44
1.4392
45
2.4039
P 2
ei = 3416.377
P 2
Min Q =
b.
1.25. a.
b.
ei
M SE = 79.45063,
√
M SE = 8.913508, minutes
e1 = 1.8000
P 2
ei = 17.6000, M SE = 2.2000, σ 2
1.26. a.
i:
1
ei : -2.150
2
3
3.850 -5.150
i:
7
ei : -2.425
8
9
10
5.575 3.300 .300
i: 13
ei : .025
1.27. a.
b.
6
2.575
11
12
1.300 -3.700
14
15
16
-1.975 3.025 -3.975
Yes
b.
4
5
-1.150 .575
M SE = 10.459,
√
M SE = 3.234, Brinell units
Ŷ = 156.35 − 1.19X
(1) b1 = −1.19, (2) Ŷh = 84.95, (3) e8 = 4.4433,
(4) M SE = 66.8
1.28. a.
b.
Ŷ = 20517.6 − 170.575X
(1) b1 = −170.575, (2) Ŷh = 6871.6, (3) e10 = 1401.566,
(4) M SE = 5552112
1.31. No, no
1.32. Solving (1.9a) and (1.9b) for b0 and equating the results:
P
Yi − b1
n
P
Xi
P
=
1-2
X i Yi − b 1
P
Xi
P
Xi2
and then solving for b1 yields:
b1 =
1.33. Q =
n
P
P
Xi Yi − Xi Yi
=
P
n Xi2 − ( Xi )2
P
P
P
X i Yi
X i Yi −
P n
( Xi )2
P 2
Xi −
n
P
P
P
(Yi − β0 )2
X
dQ
= −2 (Yi − β0 )
dβ0
Setting the derivative equal to zero, simplifying, and substituting the least squares
estimator b0 yields:
P
(Yi − b0 ) = 0 or b0 = Ȳ
1.34. E{b0 } = E{Ȳ } =
1X
1X
E{Yi } =
β0 = β0
n
n
1.35. From the first normal equation (1.9a):
P
1.36.
Yi = nb0 + b1
P
Xi =
P
(b0 + b1 Xi ) =
P
Ŷi from (1.13)
P
P
P
P
P
Ŷ e = (b0 + b1 Xi )ei = b0 ei + b1 Xi ei = 0 because ei = 0 from (1.17) and
P i i
Xi ei = 0 from (1.19).
1.38. (1) 76, yes; (2) 60, yes
1.39. a.
Applying (1.10a) and (1.10b) to (5, Ȳ1 ), (10, Ȳ2 ) and (15, Ȳ3 ), we
obtain:
Ȳ3 − Ȳ1
4Ȳ1 + Ȳ2 − 2Ȳ3
b0 =
10
3
Using (1.10a) and (1.10b) with the six original points yields the same results.
b1 =
b.
Yes
1.40. No
1.41. a.
P
Q = (Yi − β1 Xi )2
X
dQ
= −2 (Yi − β1 Xi )Xi
dβ1
Setting the derivative equal to zero, simplifying, and substituting the least squares
estimator b1 yields:
P
Yi Xi
b1 = P 2
Xi
b.
L=
·
1
1
exp − 2 (Yi − β1 Xi )2
2
1/2
2σ
i=1 (2πσ )
n
Q
1-3
¸
It is more convenient to work with loge L :
n
1 X
loge L = − loge (2πσ 2 ) − 2
(Yi − β1 Xi )2
2
2σ
d loge L
1 X
=
(Yi − β1 Xi )Xi
dβ1
σ2
c.
1.42. a.
b.
Setting the derivative equal to zero, simplifying, and substituting the maximum
likelihood estimator b1 yields:
P
Yi Xi
P
(Yi − b1 Xi )Xi = 0 or b1 = P 2
Xi
Yes
(P
)
Yi Xi
1 X
E{b1 } = E P 2 = P 2
Xi E{Yi }
Xi
Xi
1 X
=P 2
Xi (β1 Xi ) = β1
Xi
1
1
exp[− (Yi − β1 Xi )2 ]
32
i=1
32π
−30
L(17) = 9.45 × 10 , L(18) = 2.65 × 10−7 , L(19) = 3.05 × 10−37
L(β1 ) =
6
Q
√
β1 = 18
c.
b1 = 17.928, yes
d.
Yes
1.43. a.
Total population: Ŷ = −110.635 + 0.0027954X
Number of hospital beds: Ŷ = −95.9322 + 0.743116X
Total personal income: Ŷ = −48.3948 + .131701X
c.
Total population: M SE = 372, 203.5
Number of hospital beds: M SE = 310, 191.9
Total personal income: M SE = 324, 539.4
1.44. a.
Region 1: Ŷ = −1723.0 + 480.0X
Region 2: Ŷ = 916.4 + 299.3X
Region 3: Ŷ = 401.56 + 272.22X
Region 4: Ŷ = 396.1 + 508.0X
c.
Region 1: M SE = 64, 444, 465
Region 2: M SE = 141, 479, 673
Region 3: M SE = 50, 242, 464
Region 4: M SE = 514, 289, 367
1.45. a.
Infection risk: Ŷ = 6.3368 + .7604X
Facilities: Ŷ = 7.7188 + .0447X
1-4
X-ray: Ŷ = 6.5664 + .0378X
c.
Infection risk: M SE = 2.638
Facilities: M SE = 3.221
X-ray: M SE = 3.147
1.46. a.
Region 1: Ŷ = 4.5379 + 1.3478X
Region 2: Ŷ = 7.5605 + .4832X
Region 3: Ŷ = 7.1293 + .5251X
Region 4: Ŷ = 8.0381 + .0173X
c.
Region 1: M SE = 4.353
Region 2: M SE = 1.038
Region 3: M SE = .940
Region 4: M SE = 1.078
b.
1
1
exp[− (Yi − β0 − β1 Xi )2 ]
32
i=1
32π
b0 = 1.5969, b1 = 17.8524
c.
Yes
1.47. a.
L(β0 , β1 ) =
6
Q
√
1-5
1-6
Chapter 2
INFERENCES IN REGRESSION
AND CORRELATION ANALYSIS
2.1.
a. Yes, α = .05
2.2.
No
2.4.
a.
t(.995; 118) = 2.61814, .03883 ± 2.61814(.01277), .00540 ≤ β1 ≤ .07226
b.
H0 : β1 = 0, Ha : β1 6= 0. t∗ = (.03883 − 0)/.01277 = 3.04072. If |t∗ | ≤ 2.61814,
conclude H0 , otherwise Ha . Conclude Ha .
c.
0.00291
a.
t(.95; 43) = 1.6811, 15.0352 ± 1.6811(.4831), 14.2231 ≤ β1 ≤ 15.8473
b.
H0 : β1 = 0, Ha : β1 6= 0. t∗ = (15.0352 − 0)/.4831 = 31.122. If |t∗ | ≤ 1.681
conclude H0 , otherwise Ha . Conclude Ha . P -value= 0+
c.
Yes
d.
H0 : β1 ≤ 14, Ha : β1 > 14. t∗ = (15.0352 − 14)/.4831 = 2.1428. If t∗ ≤ 1.681
conclude H0 , otherwise Ha . Conclude Ha . P -value= .0189
a.
t(.975; 8) = 2.306, b1 = 4.0, s{b1 } = .469, 4.0 ± 2.306(.469),
2.5.
2.6.
2.918 ≤ β1 ≤ 5.082
b.
H0 : β1 = 0, Ha : β1 6= 0. t∗ = (4.0 − 0)/.469 = 8.529. If |t∗ | ≤ 2.306 conclude H0 ,
otherwise Ha . Conclude Ha . P -value= .00003
c.
b0 = 10.20, s{b0 } = .663, 10.20 ± 2.306(.663), 8.671 ≤ β0 ≤ 11.729
d.
H0 : β0 ≤ 9, Ha : β0 > 9. t∗ = (10.20 − 9)/.663 = 1.810. If t∗ ≤ 2.306 conclude H0 ,
otherwise Ha . Conclude H0 . P -value= .053
e.
H0 : β1 = 0: δ = |2 − 0|/.5 = 4, power = .93
H0 : β0 ≤ 9: δ = |11 − 9|/.75 = 2.67, power = .78
2.7.
a.
t(.995; 14) = 2.977, b1 = 2.0344, s{b1 } = .0904, 2.0344 ± 2.977(.0904),
1.765 ≤ β1 ≤ 2.304
2-1
2.8.
b.
H0 : β1 = 2, Ha : β1 6= 2. t∗ = (2.0344 − 2)/.0904 = .381. If |t∗ | ≤ 2.977 conclude
H0 , otherwise Ha . Conclude H0 . P -value= .71
c.
δ = |.3|/.1 = 3, power = .50
a.
H0 : β1 = 3.0, Ha : β1 6= 3.0. t∗ = (3.57 − 3.0)/.3470 = 1.643,
t(.975; 23) = 2.069. If |t∗ | ≤ 2.069 conclude H0 , otherwise Ha . Conclude H0 .
b.
2.10. a.
δ = |.5|/.35 = 1.43, power = .30 (by linear interpolation)
Prediction
b.
Mean response
c.
Prediction
2.12. No, no
2.13. a.
Ŷh = 3.2012, s{Ŷh } = .0706, t(.975; 118) = 1.9803, 3.2012 ± 1.9803(.0706),
3.0614 ≤ E{Yh } ≤ 3.3410
b.
s{pred} = .6271, 3.2012 ± 1.9803(.6271), 1.9594 ≤ Yh( new) ≤ 4.4430
c.
Yes, yes
d.
W 2 = 2F (.95; 2, 118) = 2(3.0731) = 6.1462, W = 2.4792, 3.2012 ± 2.4792(.0706),
3.0262 ≤ β0 + β1 Xh ≤ 3.3762, yes, yes
2.14. a.
Ŷh = 89.6313, s{Ŷh } = 1.3964, t(.95; 43) = 1.6811, 89.6313 ± 1.6811(1.3964),
87.2838 ≤ E{Yh } ≤ 91.9788
b.
s{pred} = 9.0222, 89.6313 ± 1.6811(9.0222), 74.4641 ≤ Yh(new) ≤ 104.7985, yes,
yes
c.
87.2838/6 = 14.5473, 91.9788/6 = 15.3298, 14.5473 ≤ Mean time per machine
≤ 15.3298
d.
W 2 = 2F (.90; 2, 43) = 2(2.4304) = 4.8608, W = 2.2047, 89.6313±2.2047(1.3964),
86.5527 ≤ β0 + β1 Xh ≤ 92.7099, yes, yes
2.15. a.
Xh = 2: Ŷh = 18.2, s{Ŷh } = .663, t(.995; 8) = 3.355, 18.2 ± 3.355(.663), 15.976 ≤
E{Yh } ≤ 20.424
Xh = 4: Ŷh = 26.2, s{Ŷh } = 1.483, 26.2 ± 3.355(1.483), 21.225 ≤ E{Yh } ≤ 31.175
b.
s{pred} = 1.625, 18.2 ± 3.355(1.625), 12.748 ≤ Yh(new) ≤ 23.652
c.
s{predmean} = 1.083, 18.2 ± 3.355(1.083), 14.567 ≤ Ȳh(new) ≤ 21.833, 44 =
3(14.567) ≤ Total number of broken ampules ≤ 3(21.833) = 65
d.
W 2 = 2F (.99; 2, 8) = 2(8.649) = 17.298, W = 4.159
Xh = 2: 18.2 ± 4.159(.663), 15.443 ≤ β0 + β1 Xh ≤ 20.957
Xh = 4: 26.2 ± 4.159(1.483), 20.032 ≤ β0 + β1 Xh ≤ 32.368
yes, yes
2.16. a.
Ŷh = 229.631, s{Ŷh } = .8285, t(.99; 14) = 2.624, 229.631±2.624(.8285), 227.457 ≤
E{Yh } ≤ 231.805
2-2
b.
s{pred} = 3.338, 229.631 ± 2.624(3.338), 220.872 ≤ Yh(new) ≤ 238.390
c.
s{predmean} = 1.316, 229.631 ± 2.624(1.316), 226.178 ≤ Ȳh(new) ≤ 233.084
d.
Yes, yes
e.
W 2 = 2F (.98; 2, 14) = 2(5.241) = 10.482, W = 3.238, 229.631 ± 3.238(.8285),
226.948 ≤ β0 + β1 Xh ≤ 232.314, yes, yes
2.17. Greater, H0 : β1 = 0
2.20. No
2.21. No
2.22. Yes, yes
2.23. a.
Source
SS
Regression 3.58785
Error
45.8176
Total
49.40545
df
1
118
119
MS
3.58785
0.388285
P
b.
σ 2 + β12
(Xi − X̄)2 , σ 2 , when β1 = 0
c.
H0 : β1 = 0, Ha : β1 6= 0. F ∗ = 3.58785/0.388285 = 9.24, F (.99; 1, 118) = 6.855. If
F ∗ ≤ 6.855 conclude H0 , otherwise Ha . Conclude Ha .
d.
SSR = 3.58785, 7.26% or 0.0726, coefficient of determination
e.
+0.2695
f.
R2
2.24. a.
Source
Regression
Error
Total
SS
df
MS
76,960.4 1 76,960.4
3,416.38 43 79.4506
80,376.78 44
Source
SS
df
MS
Regression
76,960.4 1 76,960.4
Error
3,416.38 43 79.4506
Total
80,376.78 44
Correction for mean 261,747.2 1
Total, uncorrected
342,124 45
b.
H0 : β1 = 0, Ha : β1 6= 0. F ∗ = 76, 960.4/79.4506 = 968.66, F (.90; 1, 43) = 2.826.
If F ∗ ≤ 2.826 conclude H0 , otherwise Ha . Conclude Ha .
c.
95.75% or 0.9575, coefficient of determination
d.
+.9785
e.
R2
2-3
2.25. a.
Source
Regression
Error
Total
SS
160.00
17.60
177.60
df
1
8
9
MS
160.00
2.20
b.
H0 : β1 = 0, Ha : β1 6= 0. F ∗ = 160.00/2.20 = 72.727, F (.95; 1, 8) = 5.32. If
F ∗ ≤ 5.32 conclude H0 , otherwise Ha . Conclude Ha .
c.
t∗ = (4.00 − 0)/.469 = 8.529, (t∗ )2 = (8.529)2 = 72.7 = F ∗
d.
R2 = .9009, r = .9492, 90.09%
2.26. a.
Source
SS
Regression 5,297.5125
Error
146.4250
Total
5,443.9375
b.
df
MS
1 5,297.5125
14
10.4589
15
H0 : β1 = 0, Ha : β1 6= 0, F ∗ = 5, 297.5125/10.4589 = 506.51, F (.99; 1, 14) =
8.86.If F ∗ ≤ 8.86 conclude H0 , otherwise Ha . Conclude Ha .
c.
d.
2.27. a.
i:
Yi − Ŷi :
Ŷi − Ȳ :
1
2
3
-2.150
3.850
-5.150
-24.4125 -24.4125 -24.4125
i:
Yi − Ŷi :
Ŷi − Ȳ :
7
8
-2.425
5.575
-8.1375 -8.1375
i:
Yi − Ŷi :
Ŷi − Ȳ :
13
14
.025
-1.975
24.4125 24.4125
4
-1.150
-24.4125
9
10
3.300
.300
8.1375 8.1375
15
3.025
24.4125
5
.575
-8.1375
11
1.300
8.1375
6
2.575
-8.1375
12
-3.700
8.1375
16
-3.975
24.4125
R2 = .9731, r = .9865
H0 : β1 ≥ 0, Ha : β1 < 0. s{b1 } = 0.090197,
t∗ = (−1.19 − 0)/.090197 = −13.193, t(.05; 58) = −1.67155.
If t∗ ≥ −1.67155 conclude H0 , otherwise Ha . Conclude Ha .
P -value= 0+
c.
2.28. a.
t(.975; 58) = 2.00172, −1.19 ± 2.00172(.090197), −1.3705 ≤ β1 ≤ −1.0095
Ŷh = 84.9468, s{Ŷh } = 1.05515, t(.975; 58) = 2.00172,
84.9468 ± 2.00172(1.05515), 82.835 ≤ E{Yh } ≤ 87.059
b.
s{Yh(new) } = 8.24101, 84.9468 ± 2.00172(8.24101), 68.451 ≤ Yh(new) ≤ 101.443
c.
W 2 = 2F (.95; 2, 58) = 2(3.15593) = 6.31186, W = 2.512342,
84.9468 ± 2.512342(1.05515), 82.296 ≤ β0 + β1 Xh ≤ 87.598, yes, yes
2.29. a.
2-4
i:
1
2
...
Yi − Ŷi : 0.823243 -1.55675 . . .
Ŷi − Ȳ : 20.2101 22.5901 . . .
59
-0.666887
-14.2998
60
8.09309
-19.0598
b.
Source
SS
Regression 11,627.5
Error
3,874.45
Total
15,501.95
df
MS
1 11,627.5
58 66.8008
59
c.
H0 : β1 = 0, Ha : β1 6= 0. F ∗ = 11, 627.5/66.8008 = 174.0623,
F (.90; 1, 58) = 2.79409. If F ∗ ≤ 2.79409 conclude H0 , otherwise Ha . Conclude
Ha .
d.
24.993% or .24993
e.
R2 = 0.750067, r = −0.866064
2.30. a.
H0 : β1 = 0, Ha : β1 6= 0. s{b1 } = 41.5743,
t∗ = (−170.575 − 0)/41.5743 = −4.1029, t(.995; 82) = 2.63712.
If |t∗ | ≤ 2.63712 conclude H0 , otherwise Ha . Conclude Ha .
P -value = 0.000096
b.
−170.575 ± 2.63712(41.5743), −280.2114 ≤ β1 ≤ −60.9386
2.31. a.
Source
SS
Regression 93,462,942
Error
455,273,165
Total
548,736,107
df
MS
1 93,462,942
82 5,552,112
83
b.
H0 : β1 = 0, Ha : β1 6= 0. F ∗ = 93, 462, 942/5, 552, 112 = 16.8338, F (.99; 1, 82) =
6.9544. If F ∗ ≤ 6.9544 conclude H0 , otherwise Ha. Conclude Ha . (t∗ )2 =
(−4.102895)2 = 16.8338 = F ∗ . [t(.995; 82)]2 = (2.63712)2 = 6.9544 = F (.99; 1, 82).
Yes.
c.
SSR = 93, 462, 942, 17.03% or 0.1703
d.
-0.4127
2.32. a.
Full: Yi = β0 + β1 Xi + εi , reduced: Yi = β0 + εi
b. (1) SSE(F ) = 455, 273, 165, (2) SSE(R) = 548, 736, 107,
(3) dfF = 82, (4) dfR = 83,
(5) F ∗ = [(548, 736, 107 − 455, 273, 165)/1] ÷[455, 273, 165/82] = 16.83376, (6) If
F ∗ ≤ F (.99; 1, 82) = 6.95442 conclude H0 , otherwise Ha .
c.
2.33. a.
Yes
H0 : β0 = 7.5, Ha : β0 6= 7.5
b.
Full: Yi = β0 + β1 Xi + εi , reduced: Yi − 7.5 = β1 Xi + εi
c.
Yes, dfR − dfF = (n − 1) − (n − 2) = 1
2-5
2.36
Regression model
2.38. No
2.39. a.
Normal, mean µ1 = 50, standard deviation σ1 = 3
b.
Normal, mean E{Y2 |Y1 = 55} = 105.33, standard deviation σ2|1 = 2.40
c.
Normal, mean E{Y1 |Y2 = 95} = 47, standard deviation σ1|2 = 1.80
2.40. (1) No, (2) no, (3) yes
2.41. No
2.42. b.
.95285, ρ12
c.
q
√
H0 : ρ12 = 0, Ha : ρ12 6= 0. t∗ = (.95285 13)/ 1 − (.95285)2 = 11.32194,
t(.995; 13) = 3.012. If |t∗ | ≤ 3.012 conclude H0 , otherwise Ha . Conclude Ha .
d.
No
2.43. a.
q
√
H0 : ρ12 = 0, Ha : ρ12 6= 0. t∗ = (.61 82)/ 1 − (.61)2 = 6.9709,
t(.975; 82) = 1.993. If |t∗ | ≤ 1.993 conclude H0 , otherwise Ha . Conclude Ha .
b.
z 0 = .70892, σ{z 0 } = .1111, z(.975) = 1.960, .70892 ± 1.960(.1111), .49116 ≤ ζ ≤
.92668, .455 ≤ ρ12 ≤ .729
c.
.207 ≤ ρ212 ≤ .531
2.44. a.
q
√
H0 : ρ12 = 0, Ha : ρ12 6= 0. t∗ = (.87 101)/ 1 − (.87)2 = 17.73321, t(.95; 101) =
1.663. If |t∗ | ≤ 1.663 conclude H0 , otherwise Ha . Conclude Ha .
b.
z 0 = 1.33308, σ{z 0 } = .1, z(.95) = 1.645, 1.33308 ± 1.645(.1), 1.16858 ≤ ζ ≤
1.49758, .824 ≤ ρ12 ≤ .905
c.
.679 ≤ ρ212 ≤ .819
2.45. a.
z 0 = 1.18814, σ{z 0 } = .0833, z(.995) = 2.576, 1.18814 ± 2.576(.0833),
.97356 ≤ ζ ≤ 1.40272, .750 ≤ ρ12 ≤ .886.
b.
2.46. a.
b.
2.47. a.
b.
.563 ≤ ρ212 ≤ .785
0.9454874
H0 : There is no association between Y1 and Y2
Ha : There is an association
between Y1 and Y2
√
0.9454874
13
= 10.46803. t(0.995, 13) = 3.012276. If |t∗ | ≤ 3.012276,
t∗ = q
2
1 − (0.9454874)
conclude H0 , otherwise, conclude Ha . Conclude Ha .
-0.866064,
q
√
H0 : ρ12 = 0, Ha : ρ12 6= 0. t∗ = (−0.866064 58)/ 1 − (−0.866064)2 =
−13.19326, t(.975; 58) = 2.00172. If |t∗ | ≤ 2.00172 conclude H0 , otherwise Ha .
Conclude Ha .
2-6
c.
-0.8657217
d.
H0 : There is no association between X and Y
Ha : There is an association
between X and Y
√
−0.8657217 58
t∗ = q
= −13.17243. t(0.975, 58) = 2.001717. If |t∗ | ≤
2
1 − (−0.8657217)
2.001717, conclude H0 , otherwise, conclude Ha . Conclude Ha .
2.48. a.
b.
2.49. a.
b.
2.50.
−0.4127033
q
√
H0 : ρ12 = 0, Ha : ρ12 6= 0. t∗ = (−0.4127033 82)/ 1 − (−0.4127033)2 =
−4.102897, t(.995; 82) = 2.637123. If |t∗ | ≤ 2.637123 conclude H0 , otherwise Ha .
Conclude Ha .
-0.4259324
H0 : There is no association between X and Y
Ha : There is an association
between X and Y
√
−0.4259324 58
t∗ = q
= −4.263013. t(0.995, 80) = 2.637123. If |t∗ | ≤
1 − (−0.4259324)2
2.637123, conclude H0 , otherwise, conclude Ha . Conclude Ha .
P
ki Xi =
=
X
Ã
!
Xi − X̄
Xi
P
(Xi − X̄)2
X (Xi − X̄)(Xi − X̄)
P
2
P
=P
(Xi − X̄)
because
X (Xi − X̄)X̄
=0
P
2
(Xi − X̄)
2
(Xi − X̄)
=1
(Xi − X̄)2
2.51. E{b0 } = E{Ȳ − b1 X̄}
=
1X
E{Yi } − X̄E{b1 }
n
=
1X
(β0 + β1 Xi ) − X̄β1
n
= β0 + β1 X̄ − X̄β1 = β0
2.52. σ 2 {b0 } = σ 2 {Ȳ − b1 X̄}
= σ 2 {Ȳ } + X̄ 2 σ 2 {b1 } − 2X̄σ{Ȳ , b1 }
=
σ2
σ2
−0
+ X̄ 2 P
(Xi − X̄)2
n
"
= σ2
2.53. a.
1
X̄ 2
+P
n
(Xi − X̄)2
n
Y
#
¸
·
1
1
√
L=
exp − 2 (Yi − β0 − β1 Xi )2 g(Xi )
2
2σ
2πσ
i=1
2-7
b.
Maximum likelihood estimators can be found more easily by working with loge L:
X
n
1 X
loge L = − loge (2πσ 2 ) − 2
(Yi − β0 − β1 Xi )2 +
loge g(Xi )
2
2σ
∂ loge L
1 X
= 2
(Yi − β0 − β1 Xi )
∂β0
σ
∂ loge L
1 X
= 2
(Yi − β0 − β1 Xi )(Xi )
∂β1
σ
µ ¶
µ ¶
∂ loge L
n 1
1X
1
2
(Yi − β0 − β1 Xi )
=−
+
2
2
∂σ
2 σ
2
σ4
Setting each derivative equal to zero, simplifying, and substituting the maximum
likelihood estimators b0 , b1 , and σ̂ 2 yields:
(1)
(2)
P
P
P
Yi − nb0 − b1
Yi Xi − b0
P
P
Xi = 0
Xi − b1
P
Xi2 = 0
(Yi − b0 − b1 Xi )2
= σ̂ 2
n
Equations (1) and (2) are the same as the least squares normal equations (1.9),
hence the maximum likelihood estimators b0 and b1 are the same as those in (1.27).
(3)
2.54. Yes, no
P
P
2.55. SSR = (Ŷi − Ȳ )2 = [(b0 + b1 Xi ) − Ȳ ]2
P
= [(Ȳ − b1 X̄) + b1 Xi − Ȳ ]2
P
= b21 (Xi − X̄)2
2.56. a.
b.
2.57. a.
b.
E{M SR} = 1, 026.36, E{M SE} = .36
E{M SR} = 90.36, E{M SE} = .36
Yi − 5Xi = β0 + εi , n − 1
Yi − 2 − 5Xi = εi , n
2.58. If ρ12 = 0, (2.74) becomes:
(
1
1
f (Y1 , Y2 ) =
exp −
2πσ1 σ2
2
"
µ
1
1 Y 1 − µ1
=√
exp −
2
σ1
2πσ1
"µ
¶2 #
Y 1 − µ1
σ1
¶2
µ
Y2 − µ 2
+
σ2
"
¶2 #)
µ
1
1 Y 2 − µ2
·√
exp −
2
σ2
2πσ2
¶2 #
= f1 (Y1 ) · f2 (Y2 )
2.59. a.
L=
n
Y
i=1
1
q
2πσ1 σ2 1 − ρ212
× exp{−
Yi1 − µ1 2
1
[(
)
2(1 − ρ212 )
σ1
Yi1 − µ1 Yi2 − µ2
Yi2 − µ2 2
)(
)+(
) ]}
σ1
σ2
σ2
Maximum likelihood estimators can be found more easily by working with loge L:
−2ρ12 (
2-8
n
loge (1 − ρ212 )
2
n
X
1
Yi1 − µ1 2
Yi1 − µ1 Yi2 − µ2
[(
−
) − 2ρ12 (
)(
)
2
2(1 − ρ12 ) i=1
σ1
σ1
σ2
loge L = −n loge 2π − n loge σ1 − n loge σ2 −
Yi2 − µ2 2
)]
σ2
X
X
ρ12
∂ loge L
1
= 2
(Y
−
µ
)
−
(Yi2 − µ2 )
i1
1
∂µ1
σ1 (1 − ρ212 )
σ1 σ2 (1 − ρ212 )
X
X
ρ12
∂ loge L
1
(Y
−
µ
)
−
= 2
(Yi1 − µ1 )
i2
2
∂µ2
σ2 (1 − ρ212 )
σ1 σ2 (1 − ρ212 )
+(
∂ loge L
n
1
=− +
∂σ1
σ1 (1 − ρ12 )2
"P
(Yi1 − µ1 )2
− ρ12
σ13
"P
P
(Yi1 − µ1 )(Yi2 − µ2 )
σ12 σ2
P
#
n
1
(Yi1 − µ1 )(Yi2 − µ2 )
∂ loge L
(Yi2 − µ2 )2
=− +
− ρ12
3
2
∂σ2
σ2 (1 − ρ12 )
σ2
σ1 σ22
X µ Yi1 − µ1 ¶ µ Yi2 − µ2 ¶
∂ loge L
nρ12
1
ρ12
=
+
−
2
2
∂ρ12
1 − ρ12 1 − ρ12
σ1
σ2
(1 − ρ212 )2
"µ
¶
µ
¶µ
¶
µ
#
¶ #
Yi1 − µ1 2
Yi2 − µ2 2
Yi1 − µ1
Yi2 − µ2
×
− 2ρ12
+
σ1
σ1
σ2
σ2
Setting the derivatives equal to zero, simplifying, and substituting the maximum
likelihood estimators µ̂1 , µ̂2 , σ̂ 1 , σ̂ 2 , and ρ̂12 yields:
1 X
ρ̂ X
(1)
(Yi1 − µ̂1 ) − 12
(Yi2 − µ̂2 ) = 0
σ̂ 1
σ̂ 2
1 X
ρ̂ X
(2)
(Yi2 − µ̂2 ) − 12
(Yi1 − µ̂1 ) = 0
σ̂ 2
σ̂ 1
P
P
(Yi1 − µ̂1 )2
(Yi1 − µ̂1 )(Yi2 − µ̂2 )
(3)
− n(1 − ρ̂212 ) = 0
− ρ̂12
2
σ̂ 1 σ̂ 2
σ̂ 1
P
P
(4)
(Yi2 − µ̂2 )2
− ρ̂12
σ̂ 22
(5)
ρ̂212 )
nρ̂12 (1 −
−ρ̂12
P
P
+ (1 +
(Yi1 − µ̂1 )(Yi2 − µ̂2 )
− n(1 − ρ̂212 ) = 0
σ̂ 1 σ̂ 2
P
ρ̂212 )
Ã
Yi1 − µ̂1
σ̂ 1
!Ã
Yi2 − µ̂2
σ̂ 2
!
Ã
!2 Ã
!2
Y
−
µ̂
Y
−
µ̂
i1
i2
1
2
=0
+
σ̂ 1
σ̂ 2
Solving equations (1) and (2) yields:
µ̂1 = Ȳ1
µ̂2 = Ȳ2
Using these results in equations (3), (4), and (5), it will be found that the maximum likelihood estimators are:
sP
µ̂1 = Ȳ1
sP
σ̂ 2 =
(Yi1 − Ȳ1 )2
n
P
(Yi1 − Ȳ1 )(Yi2 − Ȳ2 )
µ̂2 = Ȳ2
(Yi2 − Ȳ2 )2
n
σ̂ 1 =
ρ̂12 = P
1 P
1
[ (Yi1 − Ȳ1 )2 ] 2 [ (Yi2 − Ȳ2 )2 ] 2
2-9
b.
α̂1|2 = µ̂1 − µ̂2 ρ̂12
σ̂ 1
σ̂ 2
# qP
(Yi1 − Ȳ1 )2 /n
(Yi1 − Ȳ1 )(Yi2 − Ȳ2 )
q
= Ȳ1 − Ȳ2 P
1 P
1
P
2
[ (Yi1 − Ȳ1 )2 ] 2 [ (Yi2 − Ȳ2 )2 ] 2
(Yi2 − Ȳ2 ) /n
#
"P
"
P
(Yi1 − Ȳ1 )(Yi2 − Ȳ2 )
P
(Yi2 − Ȳ2 )2
= Ȳ1 − Ȳ2
β̂ 12 = ρ̂12
σ̂ 1
σ̂ 2
# qP
(Yi1 − Ȳ1 )2 /n
q
= P
1 P
1
P
2
2
2
2
2
[ (Yi1 − Ȳ1 ) ] [ (Yi2 − Ȳ2 ) ]
(Yi2 − Ȳ2 ) /n
P
"
=
P
(Yi1 − Ȳ1 )(Yi2 − Ȳ2 )
(Yi1 − Ȳ1 )(Yi2 − Ȳ2 )
P
(Yi2 − Ȳ2 )2
σ̂ 21|2 = σ̂ 21 (1 − ρ̂212 )
"
P
c.
#
P
(Yi1 − Ȳ1 )2
[ (Yi1 − Ȳ1 )(Yi2 − Ȳ2 )]2
=
1− P
P
n
(Yi1 − Ȳ1 )2 (Yi2 − Ȳ2 )2
P
P
(Yi1 − Ȳ1 )2 [ (Yi1 − Ȳ1 )(Yi2 − Ȳ2 )]2
=
−
P
n
n (Yi2 − Ȳ2 )2
The equivalence is shown by letting Yi1 and Yi2 in part (b) be Yi and Xi , respectively.
2.60. Using regression notation and letting
X
(Xi − X̄)2 = (n − 1)s2X
and
X
(Yi − Ȳ )2 = (n − 1)s2Y ,
we have from (2.84) with Yi1 = Yi and Yi2 = Xi
"P
(Yi − Ȳ )2
sY
b1 = r12
since b1 = P
sX
(Xi − X̄)2
SSE =
#1
2
r12
P
[ (Xi − X̄)(Yi − Ȳ )]2
(Yi − Ȳ ) −
P
(Xi − X̄)2
P
2
2
2
)
(n − 1)s2Y = (n − 1)s2Y (1 − r12
= (n − 1)s2Y − r12
s2 {b1 } =
Hence:
2
2
)
)
s2 (1 − r12
(n − 1)s2Y (1 − r12
÷ (n − 1)s2X = Y
2
n−2
(n − 2)sX
q
2
sY 1 − r12
b1
sY
´
= r12
÷ ³√
=
s{b1 }
sX
n − 2 sX
2-10
³√
´
n − 2 r12
q
1−
2
r12
= t∗
"
#2
i
Σ(Yi1 − Ȳ1 )(Yi2 − Ȳ2 ) h
2
Σ(Y
−
Ȳ
)
i1
1
Σ(Yi1 − Ȳ1 )2
SSR(Y1 )
=
SST O
Σ(Yi2 − Ȳ2 )2
2.61.
P
[ (Yi1 − Ȳ1 )(Yi2 − Ȳ2 )]2
=P
P
(Yi1 − Ȳ1 )2 (Yi2 − Ȳ2 )2
"P
SSR(Y2 )
=
SST O
P
#2
i
(Yi2 − Ȳ2 )(Yi1 − Ȳ1 ) hP
2
(Y
−
Ȳ
)
P
i2
2
(Yi2 − Ȳ2 )2
P
(Yi1 − Ȳ1 )2
[ (Yi2 − Ȳ2 )(Yi1 − Ȳ1 )]2
=P
P
(Yi1 − Ȳ1 )2 (Yi2 − Ȳ2 )2
Total population: R2 = 0.884067
2.62.
Number of hospital beds: R2 = 0.903383
Total personal income: R2 = 0.898914
2.63.
Region 1: 480.0 ± 1.66008(110.1), 297.2252 ≤ β1 ≤ 662.7748
Region 2: 299.3 ± 1.65936(154.2), 43.42669 ≤ β1 ≤ 555.1733
Region 3: 272.22 ± 1.65508(70.34), 155.8017 ≤ β1 ≤ 388.6383
Region 4: 508.0 ± 1.66543(359.0), −89.88937 ≤ β1 ≤ 1105.889
Infection rate: R2 = .2846
2.64.
Facilities: R2 = .1264
X-ray: R2 = .1463
2.65.
Region 1: 1.3478 ± 2.056(.316), .6981 ≤ β1 ≤ 1.9975
Region 2: .4832 ± 2.042(.137), .2034 ≤ β1 ≤ .7630
Region 3: .5251 ± 2.031(.111), .2997 ≤ β1 ≤ .7505
Region 4: .0173 ± 2.145(.306), −.6391 ≤ β1 ≤ .6737
2.66. a.
E{Yh } = 36 when Xh = 4, E{Yh } = 52 when Xh = 8, E{Yh } = 68 when Xh = 12,
E{Yh } = 84 when Xh = 16, E{Yh } = 100 when Xh = 20
s
c.
d.
25
= .3953
160
Expected proportion is .95
E{b1 } = 4, σ{b1 } =
2-11
Chapter 3
DIAGNOSTICS AND REMEDIAL
MEASURES
3.3.b.and c.
i:
Ŷi :
ei :
1
2
3
...
2.92942 2.65763 3.20121 . . .
0.967581 1.22737 0.57679 . . .
118
119
120
3.20121 2.73528 3.20121
0.71279 -0.87528 -0.25321
d.
Ascending order:
1
2
3
...
Ordered residual: -2.74004 -1.83169 -1.24373 . . .
Expected value: -1.59670 -1.37781 -1.25706 . . .
e.
119
120
0.99441 1.22737
1.37781 1.59670
H0 : Normal, Ha : not normal. r = 0.97373. If r ≥ .987 concludeH0 , otherwise Ha .
Conclude Ha .
n1 = 65, d¯1 = 0.43796,
n2 = 55, d¯2 = 0.50652, s = 0.417275, t∗BF = (0.43796 −
q
0.50652)/0.417275 (1/65) + (1/55) = −0.89674, t(.995; 18) = 2.61814. If |t∗BF | ≤
2.61814 conclude error variance constant, otherwise error variance not constant.
Conclude error variance constant.
3.4.c and d.
i:
Ŷi :
ei :
1
29.49034
-9.49034
2
...
59.56084 . . .
0.43916 . . .
44
45
59.56084 74.59608
1.43916 2.40392
e.
Ascending order:
1
2
...
Ordered residual: -22.77232 -19.70183 . . .
Expected value: -19.63272 -16.04643 . . .
44
45
14.40392 15.40392
16.04643 19.63272
H0 : Normal, Ha : not normal. r = 0.9891. If r ≥ .9785 conclude H0 , otherwise
Ha . Conclude H0 .
g.
3.5.
2
= (15, 155/2) ÷ (3416.38/45)2 = 1.314676,
SSR∗ = 15, 155, SSE = 3416.38, XBP
2
≤ 3.84 conclude error variance constant, otherwise error
χ2 (.95; 1) = 3.84. If XBP
variance not constant. Conclude error variance constant.
c.
3-1
i: 1
ei : 1.8
2
-1.2
3
4
5
6
7
-1.2 1.8 -.2 -1.2 -2.2
8
.8
9
.8
10
.8
e.
Ascending Order:
1
2
Ordered residual: -2.2 -1.2
Expected value: -2.3 -1.5
H0 : Normal, Ha : not normal. r
Conclude H0 .
g.
3
4
5 6
-1.2 -1.2 -.2 .8
-1.0 -.6 -.2 .2
= .961. If r ≥ .879
7
8
.8 .8
.6 1.0
conclude
9
1.8
1.5
H0 ,
10
1.8
2.3
otherwise Ha .
2
SSR∗ = 6.4, SSE = 17.6, XBP
= (6.4/2) ÷ (17.6/10)2 = 1.03, χ2 (.90; 1) = 2.71.
2
If XBP
≤ 2.71 conclude error variance constant, otherwise error variance not
constant. Conclude error variance constant.
Yes.
3.6.a and b.
i:
ei :
Ŷi :
1
2
3
4
-2.150
3.850
-5.150
-1.150
201.150 201.150 201.150 201.150
5
.575
217.425
6
2.575
217.425
i:
ei :
Ŷi :
7
8
9
10
11
-2.425
5.575
3.300
.300
1.300
217.425 217.425 233.700 233.700 233.700
12
-3.700
233.700
i:
ei :
Ŷi :
13
14
15
16
.025
-1.975
3.025
-3.975
249.975 249.975 249.975 249.975
c. and d.
Ascending order:
1
2
3
4
5
6
Ordered residual: -5.150 -3.975 -3.700 -2.425 -2.150 -1.975
Expected value -5.720 -4.145 -3.196 -2.464 -1.841 -1.280
e∗i : -1.592 -1.229 -1.144 -.750 -.665 -.611
Ascending order:
7
8
9
10
11
12
Ordered residual: -1.150 .025 .300 .575 1.300 2.575
Expected value: -.755 -.250 .250 .755 1.280 1.841
e∗i : -.356 .008 .093 .178 .402 .796
Ascending order:
13
14
15
16
Ordered residual: 3.025 3.300 3.850 5.575
Expected value: 2.464 3.196 4.145 5.720
e∗i : .935 1.020 1.190 1.724
H0 : Normal, Ha : not normal. r = .992. If r ≥ .941 conclude H0 , otherwise Ha .
Conclude H0 . t(.25; 14) = −.692, t(.50; 14) = 0, t(.75; 14) = .692
Actual:
e.
4/16
7/16
11/16
¯
¯
n1 = 8, d1 = 2.931, n2 = 8, dq
2 = 2.194, s = 1.724,
∗
tBF = (2.931 − 2.194)/1.724 (1/8) + (1/8) = .86, t(.975; 14) = 2.145. If |t∗BF | ≤
2.145 conclude error variance constant, otherwise error variance not constant.
Conclude error variance constant.
3-2
3.7.b and c.
i:
ei :
Ŷi :
1
2
...
0.82324
-1.55675 . . .
105.17676 107.55675 . . .
59
-0.66689
70.66689
60
8.09309
65.90691
d.
Ascending order:
1
2
...
Ordered residual: -16.13683 -13.80686 . . .
Expected value: -18.90095 -15.75218 . . .
59
60
13.95312 23.47309
15.75218 18.90095
H0 : Normal, Ha : not normal. r = 0.9897. If r ≥ 0.984 conclude H0 , otherwise
Ha . Conclude H0 .
e.
SSR∗ = 31, 833.4, SSE = 3, 874.45,
2
2
≤
= (31, 833.4/2) ÷ (3, 874.45/60)2 = 3.817116, χ2 (.99; 1) = 6.63. If XBP
XBP
6.63 conclude error variance constant, otherwise error variance not constant. Conclude error variance constant. Yes.
3.8.b and c.
i:
ei :
Ŷi :
1
2
...
591.964 1648.566 . . .
7895.036 6530.434 . . .
83
621.141
6359.859
84
28.114
7553.886
d.
Ascending order:
1
2
...
Ordered residual: -5278.310 -3285.062 . . .
Expected value: -5740.725 -4874.426 . . .
e.
83
84
4623.566 6803.265
4874.426 5740.725
H0 : Normal, Ha : not normal. r = 0.98876. If r ≥ 0.9854 conclude H0 , otherwise
Ha . Conclude H0 .
n1 = 8, d¯1 = 1751.872, n2 = 76, d¯2 = 1927.083, s = 1327.772,
q
t∗BF = (1751.872 − 1927.083)/1327.772 (1/8) + (1/76) = −0.35502, t(.975; 82) =
1.98932. If |t∗BF | ≤ 1.98932 conclude error variance constant, otherwise error
variance not constant. Conclude error variance constant.
3.10. b. 4, 4
3.11. b.
2
SSR∗ = 330.042, SSE = 59.960, XBP
= (330.042/2) ÷ (59.960/9)2 = 3.72,
2
2
χ (.95; 1) = 3.84. If XBP ≤ 3.84 conclude error variance constant, otherwise error
variance not constant. Conclude error variance constant.
3.13. a.
H0 : E{Y } = β0 + β1 X, Ha : E{Y } 6= β0 + β1 X
b.
3.14. a.
SSP E = 2797.66, SSLF = 618.719, F ∗ = (618.719/8)÷(2797.66/35) = 0.967557,
F (.95; 8, 35) = 2.21668. If F ∗ ≤ 2.21668 conclude H0 , otherwise Ha . Conclude
H0 .
H0 : E{Y } = β0 + β1 X, Ha : E{Y } 6= β0 + β1 X. SSP E = 128.750,
SSLF = 17.675, F ∗ = (17.675/2) ÷ (128.750/12) = .824, F (.99; 2, 12) = 6.93. If
F ∗ ≤ 6.93 conclude H0 , otherwise Ha . Conclude H0 .
3-3
3.15. a.
b.
Ŷ = 2.57533 − 0.32400X
H0 : E{Y } = β0 + β1 X, Ha : E{Y } 6= β0 + β1 X. SSP E = .1575, SSLF =
2.7675, F ∗ = (2.7675/3) ÷ (.1575/10) = 58.5714, F (.975; 3, 10) = 4.83. If F ∗ ≤
4.83 conclude H0 , otherwise Ha . Conclude Ha .
3.16. b.
λ:
-.2
-.1
SSE: .1235 .0651
c.
0
.1
.0390 .0440
.2
.0813
Ŷ 0 = .65488 − .19540X
e.
i:
1
2
3
4
5
6
7
8
ei : -.051
.058
.007 -.083 -.057 .035 .012 .086
0
Ŷi : -1.104 -1.104 -1.104 -.713 -.713 -.713 -.322 -.322
Expected value: -.047
.062
.000 -.086 -.062 .035 .008 .086
i:
9
10
11
12
13
14
15
ei : .046 .018 -.008 -.039 -.006 -.050 .032
0
Ŷi : -.322 .069 .069 .069 .459 .459 .459
Expected value: .047 .017 -.017 -.026 -.008 -.035 .026
f.
Ŷ =antilog10 (.65488 − .19540X) = 4.51731(.63768)X
3.17. b.
λ:
.3
.4
.5
SSE: 1099.7 967.9 916.4
c.
.6
942.4
.7
1044.2
Ŷ 0 = 10.26093 + 1.07629X
e.
f.
3.18. b.
i:
1
2
3
ei : -.36
.28
.31
0
Ŷi : 10.26 11.34 12.41
Expected value: -.24
.14
.36
4
-.15
13.49
-.14
5
.30
14.57
.24
i:
6
7
8
ei : -.41
.10
-.47
0
Ŷi : 15.64 16.72 17.79
Expected value: -.36
.04
-.56
9
.47
18.87
.56
10
-.07
19.95
-.04
Ŷ = (10.26093 + 1.07629X)2
Ŷ = 1.25470 − 3.62352X 0
d.
e.
i:
1
ei : -1.00853
Ŷi : 15.28853
Expected value: -0.97979
√
Ŷ = 1.25470 − 3.62352 X
2
3
...
-3.32526 1.64837 . . .
12.12526 10.84163 . . .
-3.10159 1.58857 . . .
3-4
110
-0.67526
12.12526
-0.59149
111
0.49147
15.28853
0.36067
3.21.
PP
=
(Yij − Ŷij )2 =
PP
PPh
(Yij − Ȳj )2 +
i2
(Yij − Ȳj ) + (Ȳj − Ŷij )
PP
(Ȳj − Ŷij )2 + 2
PP
(Yij − Ȳj )(Ȳj − Ŷij )
PP
Now,
(Yij − Ȳj )(Ȳj − Ŷij )
PP
PP 2
PP
PP
=
Yij Ȳj −
Ȳj −
Yij Ŷij +
Ȳj Ŷij
P
P
P
P
= nj Ȳj2 − nj Ȳj2 − Ŷij nj Ȳj + nj Ȳj Ŷij = 0
j
j
j
j
since Ŷij = b0 + b1 Xj is independent of i.
(P P
)
(Yij − Ȳj )2
1 X
3.22. E{M SP E} = E
=
E{(nj − 1)s2j }
n−c
n−c
1 X
σ2 X
=
E{σ 2 χ2 (nj − 1)} =
(nj − 1) = σ 2
n−c
n−c
3.23. Full: Yij = µj + εij , reduced: Yij = β1 Xj + εij
dfF = 20 − 10 = 10, dfR = 20 − 1 = 19
3.24. a.
Ŷ = 48.66667 + 2.33333X
i:
1
2
3
4
ei : 2.6667 -.3333 -.3333 -1.0000
5
-4.0000
6
-7.6667
7
13.3333
8
-2.6667
b.
Ŷ = 53.06796 + 1.62136X
c.
Ŷh = 72.52428, s{pred} = 3.0286, t(.995; 5) = 4.032, 72.52428 ± 4.032(3.0286),
60.31296 ≤ Yh(new) ≤ 84.73560, yes
3.27. b.
Ŷ = 6.84922 + .60975X
Xh = 6.5: Ŷh = 10.81260, s{pred} = 1.2583, t(.975; 109) = 1.982, 10.81260 ±
1.982(1.2583), 8.31865 ≤ Yh(new) ≤ 13.30655
Xh = 5.9: Ŷh = 10.44675, s{pred} = 1.2512, 10.44675 ± 1.982(1.2512), 7.96687 ≤
Yh(new) ≤ 12.92663
Yes
3.29. a.
Band
1
2
3
4
b.
Median
X
Y
2 23.5
4
57
5 81.5
7
111
F (.90; 2, 43) = 2.43041, W = 2.204727
Xh = 2: 29.4903 ± 2.204727(2.00609), 25.067 ≤ E{Yh } ≤ 33.913
Xh = 4: 59.5608 ± 2.204727(1.43307), 56.401 ≤ E{Yh } ≤ 62.720
Xh = 5: 74.5961 ± 2.204727(1.32983), 71.664 ≤ E{Yh } ≤ 77.528
3-5
Xh = 7: 104.667 ± 2.204727(1.6119), 101.113 ≤ E{Yh } ≤ 108.221
No
c.
Neighborhood
1
2
3
4
5
6
Xc
2
3
4
5
6
7
Ŷc
27.000
43.969
60.298
77.905
93.285
107.411
3.30. a.
Band
1
2
3
4
5
Median
X
Y
0.5 116.5
2.5 170.0
4.5 226.5
6.5 291.5
8.5 384.5
b.
Neighborhood
1
2
3
4
5
6
7
c.
Xc
1
2
3
4
5
6
7
Ŷc
131.67
158.33
187.00
210.33
245.33
271.67
319.00
F (.95; 2, 8) = 4.46, W = 2.987
Xh = 1: 124.061 ± 2.987(7.4756), 101.731 ≤ E{Yh } ≤ 146.391
Xh = 2: 156.558 ± 2.987(6.2872), 137.778 ≤ E{Yh } ≤ 175.338
Xh = 3: 189.055 ± 2.987(5.3501), 173.074 ≤ E{Yh } ≤ 205.036
Xh = 4: 221.552 ± 2.987(4.8137), 207.174 ≤ E{Yh } ≤ 235.931
Xh = 5: 254.049 ± 2.987(4.8137), 239.671 ≤ E{Yh } ≤ 268.428
Xh = 6: 286.546 ± 2.987(5.3501), 270.565 ≤ E{Yh } ≤ 302.527
Xh = 7: 319.043 ± 2.987(6.2872), 300.263 ≤ E{Yh } ≤ 337.823
Yes
3-6
Chapter 4
SIMULTANEOUS INFERENCES
AND OTHER TOPICS IN
REGRESSION ANALYSIS
4.1.
No, no
4.2.
90 percent
4.3.
a.
Opposite directions, negative tilt
b.
B = t(.9875; 43) = 2.32262, b0 = −0.580157, s{b0 } = 2.80394, b1 = 15.0352, s{b1 } =
0.483087
4.4.
4.5.
4.6.
−0.580157 ± 2.32262(2.80394)
−7.093 ≤ β0 ≤ 5.932
15.0352 ± 2.32262(0.483087)
13.913 ≤ β1 ≤ 16.157
c.
Yes
a.
Opposite directions, negative tilt
b.
B = t(.9975; 8) = 3.833, b0 = 10.2000, s{b0 } = .6633, b1 = 4.0000, s{b1 } = .4690
a.
10.2000 ± 3.833(.6633)
7.658 ≤ β0 ≤ 12.742
4.0000 ± 3.833(.4690)
2.202 ≤ β1 ≤ 5.798
B = t(.975; 14) = 2.145, b0 = 168.6000, s{b0 } = 2.6570, b1 = 2.0344, s{b1 } =
.0904
168.6000 ± 2.145(2.6570)
162.901 ≤ β0 ≤ 174.299
2.0344 ± 2.145(.0904)
1.840 ≤ β1 ≤ 2.228
b.
Negatively, no
a.
B = t(.9975; 14) = 2.91839, b0 = 156.347, s{b0 } = 5.51226, b1 = −1.190,s{b1 } =
0.0901973
156.347 ± 2.91839(5.51226)
−1.190 ± 2.91839(0.0901973)
b.
140.260 ≤ β0 ≤ 172.434
− 1.453 ≤ β1 ≤ −0.927
Opposite directions
4-1
4.7.
c.
No
a.
F (.90; 2, 43) = 2.43041, W = 2.204727
Xh = 3: 44.5256 ± 2.204727(1.67501)
40.833 ≤ E{Yh } ≤ 48.219
Xh = 5: 74.5961 ± 2.204727(1.32983) 71.664 ≤ E{Yh } ≤ 77.528
Xh = 7: 104.667 ± 2.204727(1.6119) 101.113 ≤ E{Yh } ≤ 108.221
b.
F (.90; 2, 43) = 2.43041, S = 2.204727; B = t(.975; 43) = 2.01669; Bonferroni
c.
Xh = 4: 59.5608 ± 2.01669(9.02797) 41.354 ≤ Yh(new) ≤ 77.767
Xh = 7: 104.667 ± 2.01669(9.05808) 86.3997 ≤ Yh(new) ≤ 122.934
4.8.
a.
F (.95; 2, 8) = 4.46, W = 2.987
Xh = 0: 10.2000 ± 2.987(.6633) 8.219 ≤ E{Yh } ≤ 12.181
Xh = 1: 14.2000 ± 2.987(.4690) 12.799 ≤ E{Yh } ≤ 15.601
Xh = 2: 18.2000 ± 2.987(.6633) 16.219 ≤ E{Yh } ≤ 20.181
b.
B = t(.99167; 8) = 3.016, yes
c.
F (.95; 3, 8) = 4.07, S = 3.494
Xh = 0: 10.2000 ± 3.494(1.6248) 4.523 ≤ Yh(new) ≤ 15.877
Xh = 1: 14.2000 ± 3.494(1.5556) 8.765 ≤ Yh(new) ≤ 19.635
Xh = 2: 18.2000 ± 3.494(1.6248) 12.523 ≤ Yh(new) ≤ 23.877
4.9.
d.
B = 3.016, yes
a.
B = t(.9833; 14) = 2.360
Xh = 20: 209.2875 ± 2.360(1.0847) 206.727 ≤ E{Yh } ≤ 211.847
Xh = 30: 229.6312 ± 2.360(0.8285) 227.676 ≤ E{Yh } ≤ 231.586
Xh = 40: 249.9750 ± 2.360(1.3529) 246.782 ≤ E{Yh } ≤ 253.168
b.
F (.90; 2, 14) = 2.737, W = 2.340, no
c.
F (.90; 2, 14) = 2.737, S = 2.340, B = t(.975; 14) = 2.145
Xh = 30: 229.6312 ± 2.145(3.3385)
222.470 ≤ Yh(new) ≤ 236.792
Xh = 40: 249.9750 ± 2.145(3.5056) 242.455 ≤ Yh(new) ≤ 257.495
4.10. a.
F (.95; 2, 58) = 3.15593, W = 2.512342
Xh = 45: 102.797 ± 2.512342(1.71458) 98.489 ≤ E{Yh } ≤ 107.105
Xh = 55: 90.8968 ± 2.512342(1.1469) 88.015 ≤ E{Yh } ≤ 93.778
Xh = 65: 78.9969 ± 2.512342(1.14808) 76.113 ≤ E{Yh } ≤ 81.881
b.
B = t(.99167; 58) = 2.46556, no
c.
B = 2.46556
Xh = 48: 99.2268 ± 2.46556(8.31158) 78.734 ≤ Yh(new) ≤ 119.720
Xh = 59: 86.1368 ± 2.46556(8.24148) 65.817 ≤ Yh(new) ≤ 106.457
4-2
Xh = 74: 68.2869 ± 2.46556(8.33742) 47.730 ≤ Yh(new) ≤ 88.843
d.
4.12. a.
Yes, yes
Ŷ = 18.0283X
c.
H0 : β1 = 17.50, Ha : β1 6= 17.50. M SE = 20.3113, s{b1 } = .07948, t∗ = (18.0283−
17.50)/.07948 = 6.65, t(.99; 11) = 2.718. If |t∗ | ≤ 2.718 conclude H0 , otherwise
Ha . Conclude Ha .
d.
Ŷh = 180.283, s{pred} = 4.576, 180.283 ± 2.718(4.576), 167.845 ≤ Yh(new) ≤
192.721
4.13. a.
i:
1
ei : 1.802
2
3
-3.340 10.717
i:
7
8
9
ei : -.849 6.292 -.510
4
-2.283
5
-2.396
10
11
-3.283 2.887
6
-4.708
12
-1.170
No
b.
4.14. a.
H0 : E{Y } = β1 X, Ha : E{Y } 6= β1 X. SSLF = 40.924, SSP E = 182.500,
F ∗ = (40.924/8) ÷ (182.500/3) = .084, F (.99; 8, 3) = 27.5. If F ∗ ≤ 27.5 conclude
H0 , otherwise Ha . Conclude H0 . P -value = .997
Ŷ = 0.121643X
b.
s{b1 } = 0.00263691, t(.975; 19) = 1.9801, 0.121643 ± 1.9801(0.00263691), 0.116 ≤
β1 ≤ 0.127
c.
Ŷh = 3.64929, s{Ŷh } = 0.0791074, 3.64929 ± 1.9801(0.0791074), 3.493 ≤ E{Yh } ≤
3.806
4.15. b.
i:
1
ei : 1.3425
2
2.1820
...
...
119
-0.0863
120
-0.4580
No
c.
4.16. a.
H0 : E{Y } = β1 X, Ha : E{Y } 6= β1 X. SSLF = 23.3378, SSP E = 39.3319,
F ∗ = (23.3378/20) ÷ (39.3319/99) = 2.93711, F (.995; 20, 99) = 2.22939. If F ∗ ≤
2.22939 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0.0002
Ŷ = 14.9472X
b.
s{b1 } = 0.226424, t(.95; 44) = 1.68023, 14.9472 ± 1.68023(0.226424), 14.567 ≤
β1 ≤ 15.328
c.
Ŷh = 89.6834, s{pred} = 8.92008, 89.6834 ± 1.68023(8.92008), 74.696 ≤ Yh(new) ≤
104.671
4.17. b.
i:
1
ei : -9.89445
2
...
0.21108 . . .
44
1.2111
No
4-3
45
2.2639
c.
H0 : E{Y } = β1 X, Ha : E{Y } 6= β1 X. SSLF = 622.12, SSP E = 2797.66, F ∗ =
(622.12/9) ÷ (2797.66/35) = 0.8647783, F (.99; 9, 35) = 2.96301. If F ∗ ≤ 2.96301
conclude H0 , otherwise Ha . Conclude H0 . P -value = 0.564
4.18. No
4.19. a.
X̂h(new) = 33.11991, t(.95; 118) = 1.657870, s{predX} = 16.35037,
33.11991 ± 1.657870(16.35037), 6.013 ≤ Xh(new) ) ≤ 60.227
b.
4.20. a.
No, 0.297453 > .1
X̂h(new) = 34.1137, t(.995; 14) = 2.977, s{predX} = 1.6610,
34.1137 ± 2.977(1.6610), 29.169 ≤ Xh(new) ≤ 39.058
b.
Yes, .0175 < .1
4.21. Yes, no
4.22. Let Ā3 denote the event that statement 3 is correct and B̄ the event Ā1 ∩ Ā2 . Then by
(4.2a):
P (B̄ ∩ Ā3 ) = P (Ā1 ∩ Ā2 ∩ Ā3 ) ≥ 1 − 2α − α = 1 − 3α
4.23. From (4.13) it follows at once that:
P
Xi (Yi − b1 Xi ) =
P
Xi ei = 0
4.24. From Exercise 1.41c, we have that E{b1 } = β1 . Hence:
E{Ŷ } = E{b1 X} = XE{b1 } = β1 X = E{Y }.
4.25. σ 2 {Ŷh } = σ 2 {b1 Xh } = Xh2 σ 2 {b1 } = Xh2 (σ 2 /
P
s2 {Ŷh } = Xh2 (M SE/ Xi2 ).
4.26. a.
P
Xi2 ); hence,
B = t(.9875; 438) = 2.24913, b0 = −110.635, s{b0 } = 34.7460,
b1 = 0.00279542, s{b1 } = 0.0000483694
−110.635 ± 2.24913(34.7460)
− 188.783 ≤ β0 ≤ −32.487
0.00279542 ± 2.24913(0.00004837) 0.00269 ≤ β1 ≤ 0.0029
b.
Yes
c.
F (.90; 2, 438) = 2.31473, W = 2.151618;
B = t(.9833; 438) = 2.13397;
Bonferroni
d.
Xh = 500: −109.237 ± 2.13397(34.7328) − 183.356 ≤ E{Yh } ≤ −35.118
4.27. a.
Xh = 1, 000: −107.839 ± 2.13397(34.7196)
− 181.930 ≤ E{Yh } ≤ −33.748
Xh = 5, 000: −96.6577 ± 2.13397(34.6143)
− 170.524 ≤ E{Yh } ≤ −22.792
B = t(.975; 111) = 1.982, b0 = 6.3368, s{b0 } = .5213, b1 = .7604, s{b1 } = .1144
4-4
6.3368 ± 1.982(.5213)
5.304 ≤ β0 ≤ 7.370
0.7604 ± 1.982(.1144)
0.534 ≤ β1 ≤ 0.987
b.
No
c.
F (.95; 2, 111) = 3.08, W = 2.482; B = t(.99375; 111) = 2.539; Working-Hotelling
d.
Xh = 2: 7.858 ± 2.482(.3098)
7.089 ≤ E{Yh } ≤ 8.627
Xh = 3: 8.618 ± 2.482(.2177)
8.078 ≤ E{Yh } ≤ 9.158
Xh = 4: 9.378 ± 2.482(.1581)
8.986 ≤ E{Yh } ≤ 9.770
Xh = 5: 10.139 ± 2.482(.1697)
9.718 ≤ E{Yh } ≤ 10.560
4-5
Chapter 5
MATRIX APPROACH TO SIMPLE
LINEAR REGRESSION ANALYSIS
5.1.
"
(5)
5.2.
(1)
(4)
5.3.
(1)
(2)
h
9 26
26 76
5 9
11 11
10 8
6 12
14
49
71
76
Y1
Y2
Y3
Y4
22
54
82
80
−
Ŷ1
Ŷ2
Ŷ3
Ŷ4
(2)
(1) 1,259
(2)
(1) 2,194
(2)
"
(5)
=
e1
e2
e3
e4
X1 X2 X3 X4
(1) 503.77
(3)
h
58 80
11 8
20 26
32 38
28 40
"
5.6.
−1 −7
−5 −1
0
6
2
4
(2)
"
5.5.
#
"
5.4.
2 7
0 1
23 24 1
13 17 22
(1) 3 10 (2) 1 2 (3) 36 40 2 (4) 20 26 34
5 13
1 3
49 56 3
27 35 46
i
5
0
0 160
6 17
17 55
10 10
10 20
63 94
55 73
#
e1
e2
e3
e4
h
i
= 0
#
"
(3)
#
"
(3)
#
"
(3)
49.7
−39.2
81
261
142
182
#
#
5-1
#
i
"
5.7.
(1) 819,499
5.8.
a.
Yes
b.
2
c.
0
a.
Yes
b.
Yes
c.
2
d.
0
5.9.
"
5.10. A−1 =
16
448
448 13, 824
(2)
"
(3)
#
−.1
.4
.3 −.2
#
B−1
.33088 −.15441 −.03676
.09559
.13971 −.19853
−.26471
.32353
.02941
"
5.12.
"
5.13.
"
5.14. a.
"
b.
"
5.15. a.
"
b.
5.16.
5.17. a.
.2
0
0 .00625
#
1.34146 −.41463
−.41463
.14634
4 7
2 3
y1
y2
#"
#
"
Ŷ1
Ŷ2
Ŷ3
Ŷ4
Ŷ5
#"
#
"
=
"
=
y1
y2
0
4
= Ȳ
#
#
25
12
#
#
4.5
1
=
5 2
23 7
y1
y2
y1
y2
#
"
=
8
28
#
#
1
1
1
1
1
+ b1
#
.10870 −.08696
.10870
.02174 −.15217
= .34783
−.23913
.14130
.01087
5.11.
3, 609
103, 656
X1 − X̄
X2 − X̄
X3 − X̄
X4 − X̄
X5 − X̄
1
1
1
Y1
W1
0
Y2
W2 = 1 −1
Y3
W3
1 −1 −1
5-2
b.
1
1
W1
E
W2 = 1 −1
W3
c.
1
E{Y1 }
E{Y1 } + E{Y2 } + E{Y3 }
0
E{Y2 } = E{Y1 } − E{Y2 }
1 −1 −1
E{Y3 }
E{Y1 } − E{Y2 } − E{Y3 }
1
1
1
σ 2 {Y1 } σ{Y1 , Y2 } σ{Y1 , Y3 }
2
0 σ{Y2 , Y1 } σ 2 {Y2 } σ{Y2 , Y3 }
σ {W} = 1 −1
1 −1 −1
σ{Y3 , Y1 } σ{Y3 , Y2 } σ 2 {Y3 }
1
1
1
× 1 −1 −1
1
0 −1
Using the notation σ12 for σ 2 {Y1 }, σ12 for σ{Y1 , Y2 }, etc., we obtain:
σ 2 {W1 } = σ12 + σ22 + σ32 + 2σ12 + 2σ13 + 2σ23
σ 2 {W2 } = σ12 + σ22 − 2σ12
σ 2 {W3 } = σ12 + σ22 + σ32 − 2σ12 − 2σ13 + 2σ23
σ{W1 , W2 } = σ12 − σ22 + σ13 − σ23
σ{W1 , W3 } = σ12 − σ22 − σ32 − 2σ23
σ{W2 , W3 } = σ12 + σ22 − 2σ12 − σ13 + σ23
"
5.18. a.
W1
W2
("
b.
E
#
"
=
W1
W2
#)
1
4
1
2
"
=
"
c.
1
4
1
2
σ 2 {W} =
1
4
1
2
1
4
1
2
1
4
− 12
1
4
− 12
1
[E{Y1 }
4
1
[E{Y1 }
2
1
4
− 12
1
4
Y1
Y2
Y3
Y4
+ E{Y2 } + E{Y3 } + E{Y4 }]
+ E{Y2 } − E{Y3 } − E{Y4 }]
1
4
− 12
1
4
1
4
× 1
4
#
#
1
2
1
2
− 21
− 21
2
#
σ 2 {Y1 } σ{Y1 , Y2 } σ{Y1 , Y3 } σ{Y1 , Y4 }
σ{Y2 , Y1 } σ 2 {Y2 } σ{Y2 , Y3 } σ{Y2 , Y4 }
σ{Y3 , Y1 } σ{Y3 , Y2 } σ 2 {Y3 } σ{Y3 , Y4 }
σ{Y4 , Y1 } σ{Y4 , Y2 } σ{Y4 , Y3 } σ 2 {Y4 }
Using the notation σ12 for σ {Y1 }, σ12 for σ{Y1 , Y2 }, etc., we obtain:
1
(σ12 + +σ22 + σ32 + σ42 + 2σ12 + 2σ13 + 2σ14 + 2σ23 + 2σ24 + 2σ34 )
16
σ 2 {W2 } = 14 (σ12 + σ22 + σ32 + σ42 + 2σ12 − 2σ13 − 2σ14 − 2σ23 − 2σ24 + 2σ34 )
σ{W1 , W2 } = 81 (σ12 + σ22 − σ32 − σ42 + 2σ12 − 2σ34 )
σ 2 {W1 } =
"
5.19.
"
5.20.
3 5
5 17
#
7 −4
−4
8
#
5-3
5.21. 5Y12 + 4Y1 Y2 + Y22
5.22. Y12 + 3Y22 + 9Y32 + 8Y1 Y3
"
5.23. a.
(1)
"
(5)
c.
d.
(2)
#
9.940
−.245
.00987
0
0
.000308
.6
.4
.2
0
−.2
.4
.3
.2
.1
0
(1)
"
(5)
c.
d.
(3) 9.604
(6) 11.41
(7) .02097
.43902
4.60976
(4) .148
.2
.2
.2
.2
.2
0 −.2
.1
0
.2
.2
.3
.4
.4
.6
#
(2)
6.8055 −2.1035
−2.1035
.7424
(1) −2.1035
.01973 −.01973 −.00987
.00000
.00987
−.01973
.03453 −.00987 −.00493
.00000
−.00987 −.00987
.03947 −.00987 −.00987
.00000 −.00493 −.00987
.03453 −.01973
.00987
.00000 −.00987 −.01973
.01973
"
b.
#
5.24. a.
−.18
.04
.26
.08
−.20
−2.8781
−.0488
.3415
.7317
−1.2683
3.1219
(3) 145.2073
#
(2) 6.8055
(6) 18.878
(7) 6.9290
(3) .8616
.366 −.146 .024 .195 .195
.366
−.146
.658 .390 .122 .122 −.146
.024
.390 .268 .146 .146
.024
.195
.122 .146 .171 .171
.195
.195
.122 .146 .171 .171
.195
.366 −.146 .024 .195 .195
.366
3.217
.742 −.124 −.990 −.990 −1.856
.742
1.732 −1.980 −.619 −.619
.742
−.124 −1.980
3.712 −.742 −.742 −.124
−.990 −.619 −.742 4.207 −.866 −.990
−.990 −.619 −.742 −.866 4.207 −.990
−1.856
.742 −.124 −.990 −.990
3.127
5-4
(4) 20.2927
"
5.25. a.
(1)
(4)
.2 −.1
−.1
.1
#
"
(2)
(1) .22
c.
5.26. a.
(6)
.44 −.22
−.22
.22
(2) −.22
#
(7) 18.2
(1)
.675000 −.021875
−.021875 .00078125
(8) .44
(3) .663
0
0
0
0
0 0
0 0
0
.1 −.1
.1 −.2 0
.1 0
0 −.1
.1 −.1
.2 0 −.1 0
0
.1 −.1
.1 −.2 0
.1 0
0 −.2
.2 −.2
.4 0 −.2 0
0
0
0
0
0 0
0 0
0
.1 −.1
.1 −.2 0
.1 0
0
0
0
0
0 0
0 0
0 −.1
.1 −.1
.2 0 −.1 0
0
.1 −.1
.1 −.2 0
.1 0
"
(5) 17.60
#
.1
.1 .1
.1
.1 .1
.1 .1 .1
.1
.1
.2 0
.2 −.1 .1
.2 .1 0
.2
.1
0 .2
0
.3 .1
0 .1 .2
0
.1
.2 0
.2 −.1 .1
.2 .1 0
.2
.1 −.1 .3 −.1
.5 .1 −.1 .1 .3 −.1
.1
.1 .1
.1
.1 .1
.1 .1 .1
.1
.1
.2 0
.2 −.1 .1
.2 .1 0
.2
.1
.1 .1
.1
.1 .1
.1 .1 .1
.1
.1
.0 .2
0
.3 .1
0 .1 .2
0
.1
.2 0
.2 −.1 .1
.2 .1 .0
.2
"
b.
10.2
4.0
(3)
1.8
−1.2
−1.2
1.8
−.2
−1.2
−2.2
.8
.8
.8
#
"
(2)
5-5
0
0
−.1
.1
.1 −.1
−.1
.1
.2 −.2
0
0
−.1
.1
0
0
.1 −.1
−.1
.1
168.600000
2.034375
#
(3)
(4)
201.150
201.150
201.150
201.150
217.425
217.425
217.425
217.425
233.700
233.700
233.700
233.700
249.975
249.975
249.975
249.975
.175
.175
.175
..
.
.175 · · · −.050 −.050 −.050
.175 · · · −.050 −.050 −.050
.175 · · · −.050 −.050 −.050
..
..
..
..
.
.
.
.
−.050 −.050 −.050 · · ·
.175
.175
.175
−.050 −.050 −.050 · · ·
.175
.175
.175
−.050 −.050 −.050 · · ·
.175
.175
.175
"
(5) 146.425
b.
(1) 7.0598
c.
5.27.
.175
.175
.175
..
.
(6)
7.0598 −.2288
−.2288 .008171
(2) −.2288
#
(7) 11.1453
(3) .0904
.825 −.175 −.175 · · ·
.050
.050
.050
−.175
.825 −.175 · · ·
.050
.050
.050
−.175 −.175
.825 · · ·
.050
.050
.050
..
..
..
..
..
..
.
.
.
.
.
.
.050
.050
.050 · · ·
.825 −.175 −.175
.050
.050
.050 · · · −.175
.825 −.175
.050
.050
.050 · · · −.175 −.175
.825
ε1
0
ε
0
2
E
=
=0
ε
0
3
ε4
0
5.28. Let
X=
X1
X2
..
.
Xn
5-6
Then by (5.60) b = (X0 X)−1 X0 Y =
5.29
P
Xi Yi /
P
Xi2 .
E{b} = E{(X0 X)−1 X0 Y} = (X0 X)−1 X0 E{Y}
= (X0 X)−1 X0 Xβ = β
5.30. Ŷh = X0h b is a scalar, hence it equals its transpose. By (5.32) then,
X0h b = (X0h b)0 = b0 Xh .
5.31. σ 2 {Ŷ} = Hσ 2 {Y}H0
= Hσ 2 IH
= σ2H
[by (5.46)]
(since H is symmetric)
(since HH = H)
5-7
5-8
Chapter 6
MULTIPLE REGRESSION – I
6.1.
a.
X=
b.
X=
6.2.
a.
X=
b.
6.5.
a.
X=
1
1
1
1
X11
X21
X31
X41
X11 X12
X21 X22
X31 X32
X41 X42
1
1
1
1
X11
X21
X31
X41
X12
X22
X32
X42
X11
X21
X31
X41
X51
1
1
1
1
1
X12
X22
X32
X42
X52
X11
X21
X31
X41
X51
β0
β = β1
β2
2
X11
2
X21
2
X31
2
X41
2
X51
β0
β = β1
β2
log10 X12
log10 X22
log10 X32
log10 X42
log10 X52
β1
β = β2
β3
β0
β = β1
β2
Y
1.000 .892 .395
1.000 .000
X1
1.000
X2
b.
b0 = 37.650, b1 = 4.425, b2 = 4.375, Ŷ = 37.650 + 4.425X1 + 4.375X2
c&d.
i:
1
2
ei : −.10 .15
Expected Val.: −.208 .208
3
4
−3.10 3.15
−3.452 2.661
5
6
−.95 −1.70
−.629 −1.533
i:
9
10
11
12
13
ei : 1.20 −1.55 4.20 2.45 −2.65
Expected Val.: 1.066 −1.066 4.764 2.052 −2.661
6-1
7
8
−1.95 1.30
−2.052 1.533
14
15
−4.40 3.35
−4.764 3.452
16
.60
.629
6.6.
6.7.
6.8.
e.
2
SSR∗ = 72.41, SSE = 94.30, XBP
= (72.41/2) ÷ (94.30/16)2 = 1.04, χ2 (.99; 2) =
2
9.21. If XBP ≤ 9.21 conclude error variance constant, otherwise error variance
not constant. Conclude error variance constant.
f.
H0 : E{Y } = β0 + β1 X1 + β2 X2 , Ha : E{Y } 6= β0 + β1 X1 + β2 X2 . M SLF = 7.46,
M SP E = 7.125, F ∗ = 7.46/7.125 = 1.047, F (.99; 5, 8) = 6.63. If F ∗ ≤ 6.63
conclude H0 , otherwise Ha . Conclude H0 .
a.
H0 : β1 = β2 = 0, Ha : not all βk = 0 (k = 1, 2). M SR = 936.350, M SE = 7.254,
F ∗ = 936.350/7.254 = 129.083, F (.99; 2, 13) = 6.70. If F ∗ ≤ 6.70 conclude H0 ,
otherwise Ha . Conclude Ha .
b.
P -value = 0+
c.
s{b1 } = .301, s{b2 } = .673, B = t(.9975; 13) = 3.372
4.425 ± 3.372(.301)
3.410 ≤ β1 ≤ 5.440
4.375 ± 3.372(.673)
2.106 ≤ β2 ≤ 6.644
a.
SSR = 1, 872.7, SST O = 1, 967.0, R2 = .952
b.
.952, yes.
a.
Ŷh = 77.275, s{Ŷh } = 1.127, t(.995; 13) = 3.012, 77.275 ± 3.012(1.127),
73.880 ≤ E{Yh } ≤ 80.670
b.
6.9.
c.
6.10. a.
s{pred} = 2.919, 77.275 ± 3.012(2.919), 68.483 ≤ Yh(new) ≤ 86.067
Y
X1
X2
X3
1.0000
.2077
1.0000
.0600
.0849
1.0000
.8106
.0457
.1134
1.0000
Ŷ = 4149.89 + 0.000787X1 − 13.166X2 + 623.554X3
b&c.
e.
i:
1
2
ei : −32.0635 169.2051
Expected Val.: −24.1737 151.0325
n1 = 26, d¯1 = 145.0, n2 = 26, d¯2 = 77.4,
b.
51
−184.8776
−212.1315
52
64.5168
75.5358
s = 81.7,
77.4)/[81.7 (1/26) + (1/26)] = 2.99, t(.995; 50) = 2.67779. If
conclude error variance constant, otherwise error variance not
constant. Conclude error variance not constant.
t∗BF = (145.0 −
|t∗BF | ≤ 2.67779
6.11. a.
...
...
...
q
H0 : β1 = β2 = β3 = 0, Ha : not all βk = 0 (k = 1, 2,3). M SR = 725, 535,
M SE = 20, 531.9, F ∗ = 725, 535/20, 531.9 = 35.337, F (.95; 3, 48) = 2.79806. If
F ∗ ≤ 2.79806 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+.
s{b1 } = .000365, s{b3 } = 62.6409, B = t(.9875; 48) = 2.3139
0.000787 ± 2.3139(.000365)
− .000058 ≤ β1 ≤ 0.00163
623.554 ± 2.3139(62.6409)
478.6092 ≤ β3 ≤ 768.4988
6-2
c.
6.12. a.
SSR = 2, 176, 606, SST O = 3, 162, 136, R2 = .6883
F (.95; 4, 48) = 2.56524, W = 3.2033; B = t(.995; 48) = 2.6822
Xh1
302, 000
245, 000
280, 000
350, 000
295, 000
Xh2
7.2
7.4
6.9
7.0
6.7
Xh3
0:
0:
0:
0:
1:
4292.79 ± 2.6822(21.3567)
4245.29 ± 2.6822(29.7021)
4279.42 ± 2.6822(24.4444)
4333.20 ± 2.6822(28.9293)
4917.42 ± 2.6822(62.4998)
4235.507 ≤ E{Yh } ≤ 4350.073
4165.623 ≤ E{Yh } ≤ 4324.957
4213.855 ≤ E{Yh } ≤ 4344.985
4255.606 ≤ E{Yh } ≤ 4410.794
4749.783 ≤ E{Yh } ≤ 5085.057
b.Yes, no
6.13.
F (.95; 4, 48) = 2.5652, S = 3.2033; B = t(.99375; 48) = 2.5953
Xh1
230, 000
250, 000
280, 000
340, 000
6.14. a.
Xh2
7.5
7.3
7.1
6.9
Xh3
0:
0:
0:
0:
4232.17 ± 2.5953(147.288)
4250.55 ± 2.5953(146.058)
4276.79 ± 2.5953(145.134)
4326.65 ± 2.5953(145.930)
3849.913 ≤ Yh(new)
3871.486 ≤ Yh(new)
3900.124 ≤ Yh(new)
3947.918 ≤ Yh(new)
≤ 4614.427
≤ 4629.614
≤ 4653.456
≤ 4705.382
Ŷh = 4278.37, s{predmean} = 85.82262, t(.975; 48) = 2.01063,
4278.37 ± 2.01063(85.82262), 4105.812 ≤ Ȳh(new) ≤ 4450.928
b.
12317.44 ≤ Total labor hours≤ 13352.78
Y
X1
X2
X3
6.15. b.
c.
1.000 −.7868 −.6029 −.6446
1.000
.5680
.5697
1.000
.6705
1.000
Ŷ = 158.491 − 1.1416X1 − 0.4420X2 − 13.4702X3
d&e.
i:
1
2 ...
ei : .1129
−9.0797 . . .
Expected Val.: −0.8186 −8.1772 . . .
45
−5.5380
−5.4314
46
10.0524
8.1772
f.
No
g.
2
SSR∗ = 21, 355.5, SSE = 4, 248.8, XBP
= (21, 355.5/2) ÷ (4, 248.8 /46)2 =
2
2
1.2516, χ (.99; 3) = 11.3449. If XBP ≤ 11.3449 conclude error variance constant,
otherwise error variance not constant. Conclude error variance constant.
6.16. a.
H0 : β1 = β2 = β3 = 0, Ha : not all βk = 0 (k = 1, 2, 3).
M SR = 3, 040.2, M SE = 101.2, F ∗ = 3, 040.2/101.2 = 30.05, F (.90; 3, 42) =
2.2191. If F ∗ ≤ 2.2191 conclude H0 , otherwise Ha . Conclude Ha . P -value =
0.4878
b.
s{b1 } = .2148, s{b2 } = .4920, s{b3 } = 7.0997, B = t(.9833; 42) = 2.1995
−1.1416 ± 2.1995(.2148)
− 1.6141 ≤ β1 ≤ −0.6691
−.4420 ± 2.1995(.4920)
− 1.5242 ≤ β2 ≤ 0.6402
−13.4702 ± 2.1995(7.0997)
− 29.0860 ≤ β3 ≤ 2.1456
6-3
c.
6.17. a.
b.
SSR = 9, 120.46, SST O = 13, 369.3, R = .8260
Ŷh = 69.0103, s{Ŷh } = 2.6646, t(.95; 42) = 1.6820, 69.0103 ± 1.6820(2.6646),
64.5284 ≤ E{Yh } ≤ 73.4922
s{pred} = 10.405, 69.0103 ± 1.6820(10.405), 51.5091 ≤ Yh(new) ≤ 86.5115
6.18. b.
Y
X1
X2
X3
X4
1.0000 −.2503
1.0000
.4138
.0665
.3888 −.2527
1.0000 −.3798
1.0000
.5353
.2886
.4407
.0806
1.0000
c.
Ŷ = 12.2006 − .1420X1 + .2820X2 + 0.6193X3 + 0.0000079X4
i:
1
2 ...
80
81
d&e.
ei : −1.0357 −1.5138 . . . −2.0302 −.9068
Expected Val.: −1.1524 −1.5857 . . . −1.9321 −1.0407
f. No
g. n1 = 40, d¯1 =q0.8696, n2 = 41, d¯2 = 0.7793, s = 0.7357, t∗BF = (0.8696 −
0.7793)/0.7357 (1/40) + (1/41) = 0.5523, t(.975; 79) = 1.9905. If |t∗BF | ≤ 1.9905
conclude error variance constant, otherwise error variance not constant. Conclude
error variance constant.
6.19. a.
H0 : β1 = β2 = β3 = β4 = 0, Ha : not all βk = 0 (k = 1, 2, 3, 4). M SR = 34.5817
M SE = 1.2925, F ∗ = 34.5817/1.2925 = 26.7557, F (.95; 4, 76) = 2.4920. If
F ∗ ≤ 2.4920 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
b.
s{b1 } = .02134, s{b2 } = .06317, s{b3 } = 1.08681, s{b4 } = .00000138, B =
t(.99375; 76) = 2.5585
−.1420 ± 2.5585(.02134)
−.1966 ≤ β1 ≤ −.0874
.2820 ± 2.5585(.06317)
.1204 ≤ β2 ≤ .4436
.6193 ± 2.5585(1.08681)
−2.1613 ≤ β3 ≤ 3.3999
.0000079 ± 2.5585(.00000138)
c.
.0000044 ≤ β1 ≤ .0000114
SSR = 138.327, SST O = 236.5576, R2 = .5847
6.20. F (.95; 5, 76) = 2.3349, W = 3.4168; B = t(.99375; 76) = 2.5585
Xh1
5
6
14
12
Xh2
8.25
8.50
11.50
10.25
Xh3
0
.23
.11
0
Xh4
250, 000:
270, 000:
300, 000:
310, 000:
15.7981 ± 2.5585(.2781)
16.0275 ± 2.5585(.2359)
15.9007 ± 2.5585(.2222)
15.8434 ± 2.5585(.2591)
15.087 ≤ E{Yh } ≤ 16.510
15.424 ≤ E{Yh } ≤ 16.631
15.332 ≤ E{Yh } ≤ 16.469
15.180 ≤ E{Yh } ≤ 16.506
15.1485 ± 1.9917(1.1528)
15.5425 ± 1.9917(1.1535)
16.9138 ± 1.9917(1.1946)
12.852 ≤ Yh(new) ≤ 17.445
13.245 ≤ Yh(new) ≤ 17.840
14.535 ≤ Yh(new) ≤ 19.293
6.21. t(.975; 76) = 1.9917
Xh1
4
6
12
Xh2
10.0
11.5
12.5
Xh3
0.10
0
.32
Xh4
80, 000:
120, 000:
340, 000:
6-4
85 percent
6.22. a.
Yes
b.
2
No, yes, Yi0 = loge Yi = β0 + β1 Xi1 + β2 Xi2
+ ε0i , where ε0i = loge εi
c.
Yes
d.
No, no
e.
No, yes, Yi0 = loge (Yi−1 − 1) = β0 + β1 Xi1 + εi
6.23. a.
P
Q = (Yi − β1 Xi1 − β2 Xi2 )2
X
∂Q
= −2 (Yi − β1 Xi1 − β2 Xi2 )Xi1
∂β1
X
∂Q
= −2 (Yi − β1 Xi1 − β2 Xi2 )Xi2
∂β2
Setting the derivatives equal to zero, simplifying, and substituting the least squares
estimators b1 and b2 yields:
P
P
Yi Xi1 − b1
Yi Xi2 − b1
and:
P
b1 =
P
b.
6.24. a.
P
P
2
− b2
Xi1
P
Xi1 Xi2 = 0
Xi1 Xi2 − b2
P
2
Xi2
=0
P
P
P
P
P
P
2
Yi Xi2 Xi1 Xi2 − Yi Xi1 Xi2
P
P
P
2
2
( Xi1 Xi2 )2 − Xi1
Xi2
2
Yi Xi1 Xi1 Xi2 − Yi Xi2 Xi1
b2 =
P
P
P
2
2
( Xi1 Xi2 )2 − Xi1
Xi2
·
¸
n
1
1
Q
√
L=
exp − 2 (Yi − β1 Xi1 − β2 Xi2 )2
2σ
i=1
2πσ 2
It is more convenient to work with loge L:
n
1 X
loge L = − loge (2πσ 2 ) − 2
(Yi − β1 Xi1 − β2 Xi2 )2
2
2σ
∂ loge L
1 X
= 2
(Yi − β1 Xi1 − β2 Xi2 )Xi1
∂β1
σ
∂ loge L
1 X
= 2
(Yi − β1 Xi1 − β2 Xi2 )Xi2
∂β2
σ
Setting the derivatives equal to zero, simplifying, and substituting the maximum
likelihood estimators b1 and b2 yields the same normal equations as in part (a),
and hence the same estimators.
P
2
− β3 Xi2 )2
Q = (Yi − β0 − β1 Xi1 − β2 Xi1
X
∂Q
2
− β3 Xi2 )
= −2 (Yi − β0 − β1 Xi1 − β2 Xi1
∂β0
X
∂Q
2
= −2 (Yi − β0 − β1 Xi1 − β2 Xi1
− β3 Xi2 )Xi1
∂β1
X
∂Q
2
2
= −2 (Yi − β0 − β1 Xi1 − β2 Xi1
− β3 Xi2 )Xi1
∂β2
6-5
X
∂Q
2
= −2 (Yi − β0 − β1 Xi1 − β2 Xi1
− β3 Xi2 )Xi2
∂β3
Setting the derivatives equal to zero, simplifying, and substituting the least squares
estimators b0 , b1 , b2 , and b3 yields the normal equations:
P
P
P
P
b.
Yi − nb0 − b1
Yi Xi1 − b0
2
Yi Xi1
− b0
P
P
Xi1 − b2
Xi1 − b1
P
2
Xi1
− b1
P
P
P
P
P
2
− b3
Xi1
2
Xi1
− b2
3
Xi1
− b2
P
P
P
Xi2 = 0
3
Xi1
− b3
4
Xi1
− b3
P
P
P
Xi1 Xi2 = 0
2
Xi1
Xi2 = 0
P
2
2
Yi Xi2 − b0 Xi2 − b1 Xi1 Xi2 − b2 Xi1
Xi2 − b3 Xi2
=0
n
1
1
Q
2
√
L=
exp[− 2 (Yi − β0 − β1 Xi1 − β2 Xi1
− β3 Xi2 )2 ]
2
2σ
i=1
2πσ
or
n
1 X
2
loge L = − loge (2πσ 2 ) − 2
(Yi − β0 − β1 Xi1 − β2 Xi1
− β3 Xi2 )2
2
2σ
6.25. Fit Yi0 = β0 + β1 Xi1 + β3 Xi3 + εi , where Yi0 = Yi − 4Xi2
6.26. For regression model (6.1), R2 = 1 −
SSE(X1 , X2 )
SST O
When regressing Yi on Ŷi , SST O remains unchanged and the fitted
regression equation:
Ŷi∗ = b∗0 + b∗1 Ŷi
has coefficients b∗0 = 0, b∗1 = 1 because:
P
b∗1
=
P
=
since
P
ei Ŷi = 0 and
(Ŷi − Ȳ )(Yi − Ȳ )
=
P
(Ŷi − Ȳ )2
P
(Ŷi − Ȳ )[(Yi − Ŷi ) + (Ŷi − Ȳ )]
P
(Ŷi − Ȳ )2
(Ŷi − Ȳ )[ei + (Ŷi − Ȳ )]
=1
P
(Ŷi − Ȳ )2
P
ei Ȳ = 0 by (1.20) and (1.17).
b∗0 = Ȳ − b∗1 Ȳ = 0
Hence Ŷi∗ = Ŷi and SSE(Ŷ ) =
r2 = 1 −
6.27. a.
P
(Yi − Ŷi∗ )2 =
P
(Yi − Ŷi )2 = SSE(X1 , X2 ), and:
SSE(Ŷ )
SSE(X1 , X2 )
=1−
= R2
SST O
SST O
33.93210
2.78476
−.26442
6-6
b.
c.
d.
e.
f.
g.
6.28. b.
c.
d.
6.29. a.
−2.6996
−1.2300
−1.6374
−1.3299
−.0900
6.9868
.2314
.2517
.2118
.1489 −.0548 .2110
.3124
.0944
.2663 −.1479 .2231
.2517
.2118
.0944
.7044 −.3192
.1045 .2041
.1489
.2663 −.3192
.6143
.1414 .1483
−.0548 −.1479
.1045
.1414
.9404 .0163
.2110
.2231
.2041
.1483
.0163 .1971
3,009.926
715.4711 −34.1589 −13.5949
1.6617
.6441
−34.1589
−13.5949
.6441
.2625
53.8471
5.4247
Model I:
Y
1.0000 0.9402 0.0781 0.9481
X1
1.0000 0.1731 0.9867
X2
1.0000 0.1271
X3
1.0000
Model II:
Y
1.0000 0.4064 −0.0031
0.9481
X1
1.0000
0.0292
0.3162
X2
1.0000 −0.0227
X3
1.0000
Model
Model
Model
Model
I: Ŷ = −13.3162 + 0.000836618X1 − 0.065523X2 + 0.094132X3
II: Ŷ = −170.574 + 0.0961589X1 + 6.33984X2 + 0.126566X3
I: 0.902643
II: 0.911749
Region 1:
Region 2:
Region 3:
Region 4:
Ŷ
Ŷ
Ŷ
Ŷ
= −26, 140 + 16.34X1 + 0.3834X2 + 291.1X3
= 63, 104.1209 + 2.5883X1 + 3.6022X2 − 854.5493X3
= 56, 929.3851 + 0.3065X1 + 4.8955X2 − 800.3958X3
= 37, 720 − 0.9915X1 + 3.627X2 − 489.0X3
c.
Region
Region
Region
Region
1:
2:
3:
4:
M SE
8.0728 × 108
1.4017 × 108
1.9707 × 108
2.1042 × 108
R2
0.831
0.9392
0.8692
0.9713
6-7
6.30. b.
Model I:
Y
1.000 .189 .533
.356
X1
1.000 .001 −.040
X2
1.000
.413
1.000
X3
Model II:
Y
1.000 .409 .533 .356
X1
1.000 .360 .795
X2
1.000 .413
X3
1.000
c.
Model I: Ŷ = 1.38646 + .08371X1 + .65845X2 + .02174X3
Model II: Ŷ = 6.46738 + .00302X1 + .64771X2 − .00929X3
d.
Model I: .3448
Model II: .3407
6.31. a.
Region 1: Ŷ = −3.34958 + .11695X1 + .05824X2 + .00151X3 + .00661X4
Region 2: Ŷ = 2.29154 + .00474X1 + .05803X2 + .00117X3 + .01502X4
Region 3: Ŷ = −.14386 + .03085X1 + .10228X2 + .00411X3 + .00804X4
Region 4: Ŷ = 1.56655 + .03524X1 + .04033X2 − .00066X3 + .01279X4
Region
Region
Region
Region
1:
2:
3:
4:
M SE
1.022
1.212
.937
.954
R2
.4613
.4115
.6088
.0896
6-8
Chapter 7
MULTIPLE REGRESSION – II
7.1.
(1) 1 (2) 1 (3) 2 (4) 3
7.3.
a.
SSR(X1 ) = 1, 566.45, SSR(X2 |X1 ) = 306.25, SSE(X1 , X2 ) = 94.30, df : 1, 1, 13.
b.
H0 : β2 = 0, Ha : β2 6= 0. SSR(X2 |X1 ) = 306.25, SSE(X1 , X2 ) = 94.30, F ∗ =
(306.25/1) ÷ (94.30/13) = 42.219, F (.99; 1, 13) = 9.07. If F ∗ ≤ 9.07 conclude H0 ,
otherwise Ha . Conclude Ha . P -value = 0+.
a.
SSR(X1 ) = 136, 366, SSR(X3 |X1 ) = 2, 033, 566, SSR(X2 |X1 , X3 ) = 6, 674,
SSE(X1 , X2 , X3 ) = 985, 530, df : 1, 1, 1,48.
b.
H0 : β2 = 0, Ha : β2 6= 0. SSR(X2 |X1 , X3 ) = 6, 674, SSE(X1 , X2 , X3 ) = 985, 530,
F ∗ = (6, 674/1) ÷ (985, 530/48) = 0.32491, F (.95; 1, 17) = 4.04265. If F ∗ ≤
4.04265 conclude H0 , otherwise Ha . Conclude H0 . P -value = 0.5713.
c.
Yes, SSR(X1 )+SSR(X2 |X1 ) = 136, 366+5, 726 = 142, 092, SSR(X2 )+SSR(X1 |X2 ) =
11, 394.9 + 130, 697.1 = 142, 092.
7.4.
Yes.
7.5.
a.
SSR(X2 ) = 4, 860.26, SSR(X1 |X2 ) = 3, 896.04, SSR(X3 |X2 , X1 ) = 364.16,
SSE(X1 , X2 , X3 ) = 4, 248.84, df : 1, 1, 1, 42
b.
H0 : β3 = 0, Ha : β3 6= 0. SSR(X3 |X1 , X2 ) = 364.16, SSE(X1 , X2 , X3 ) =4, 248.84,
F ∗ = (364.16/1)÷(4, 248.84/42) = 3.5997, F (.975; 1, 42) = 5.4039. If F ∗ ≤ 5.4039
conclude H0 , otherwise Ha . Conclude H0 . P -value = 0.065.
7.6.
H0 : β2 = β3 = 0, Ha : not both β2 and β3 = 0. SSR(X2 , X3 |X1 ) = 845.07,
SSE(X1 , X2 , X3 ) = 4, 248.84, F ∗ = (845.07/2)÷(4, 248.84/42) = 4.1768, F (.975; 2, 42) =
4.0327. If F ∗ ≤ 4.0327 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0.022.
7.7.
a.
SSR(X4 ) = 40.5033, SSR(X1 |X4 ) = 42.2746, SSR(X2 |X1 , X4 ) = 27.8575,
SSR(X3 |X1 , X2 , X4 ) = 0.4195, SSE(X1 , X2 , X3 , X4 ) = 98.2306, df : 1, 1, 1, 1, 76.
b.
H0 : β3 = 0, Ha : β3 6= 0. F ∗ = (0.42/1) ÷ (98.2306/76) = 0.3249, F (.99; 1, 76) =
6.9806. If F ∗ ≤ 6.9806 conclude H0 , otherwise Ha . Conclude H0 . P -value
= .5704.
7-1
7.8.
H0 : β2 = β3 = 0, Ha : not both β2 and β3 = 0. SSR(X2 , X3 |X1 , X4 ) = 28.277,
SSE(X1 , X2 , X3 , X4 ) = 98.2306, F ∗ = (28.277/2)÷(98.2306/76) = 10.9388, F (.99; 2, 20) =
4.8958. If F ∗ ≤ 4.8958 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+.
7.9.
H0 : β1 = −1.0, β2 = 0; Ha : not both equalities hold. Full model: Yi = β0 + β1 Xi1 +
β2 Xi2 + β3 Xi3 + εi , reduced model: Yi + Xi1 = β0 + β3 Xi3 + εi . SSE(F ) = 4, 248.84,
dfF = 42, SSE(R) = 4, 427.7, dfR = 44, F ∗ = [(4427.7−4248.84)/2]÷(4, 248.84/42) =
.8840, F (.975; 2, 42) = 4.0327. If F ∗ ≤ 4.0327 conclude H0 , otherwise Ha . Conclude
H0 .
7.10. H0 : β1 = −.1, β2 = .4; Ha : not both equalities hold. Full model: Yi = β0 + β1 Xi1 +
β2 Xi2 + β3 Xi3 + β4 Xi4 + εi , reduced model: Yi + .1Xi1 − .4Xi2 = β0 + β3 Xi3 + β4 Xi4 + εi .
SSE(F ) = 98.2306, dfF = 76, SSE(R) = 110.141, dfR = 78, F ∗ = [(110.141 −
98.2306)/2] ÷ (98.2306/76) = 4.607, F (.99; 2, 76) = 4.89584. If F ∗ ≤ 4.89584 conclude
H0 , otherwise Ha . Conclude H0 .
2
7.11. a. RY2 1 = .550, RY2 2 = .408, R12
= 0, RY2 1|2 = .929, RY2 2|1 = .907, R2 = .958
2
7.12. RY2 1 = .796, RY2 2 = .156, R12
= 0, RY2 1|2 = .943, RY2 2|1 = .765, R2 = .952
2
= .0072, RY2 1|2 = 0.0415, RY2 2|1 = 0.0019, RY2 2|13 =
7.13. RY2 1 = .0431, RY2 2 = .0036, R12
2
.0067 R = .6883
7.14. a.
RY2 1 = .6190, RY2 1|2 = .4579, RY2 1|23 = .4021
b.
RY2 2 = .3635, RY2 2|1 = .0944, RY2 2|13 = .0189
2
7.15. RY2 4 = .2865, RY2 1 = .0626, RY2 1|4 = .2505, R14
= .4652, RY2 2|14 = .2202, RY2 3|124 =
.0043, R2 = .5848
7.16. a.
c.
7.17. a.
b.
c.
Ŷ ∗ = .89239X1∗ + .39458X2∗
11.45135
sY = 11.45135, s1 = 2.30940, s2 = 1.03280, b1 =
(.89239) = 4.425,
2.30940
11.45135
b2 =
(.39458) = 4.375, b0 = 81.7500 − 4.425(7) − 4.375(3) = 37.650.
1.03280
Ŷ ∗ = .17472X1∗ − .04639X2∗ + .80786X3∗
2
2
2
R12
= .0072, R13
= .0021,R23
= .0129
249.003
sY = 249.003, s1 = 55274.6, s2 = .87738, s3 = .32260 b1 =
(.17472) =
55274.6
249.003
249.003
(−.04639) = −13.16562, b3 =
(.80786) = 623.5572,
.00079, b2 =
.87738
5.32260
b0 = 4363.04−.00079(302, 693) +13.16562(7.37058)−623.5572(0.115385) = 4149.002.
7.18. a.
Ŷ ∗ = −.59067X1∗ − .11062X2∗ − .23393X3∗
b.
2
2
2
= .44957
= .32456, R23
= .32262, R13
R12
c.
17.2365
sY = 17.2365, s1 = 8.91809, s2 = 4.31356, s3 = .29934, b1 =
(−.59067) =
8.91809
17.2365
17.2365
−1.14162, b2 =
(−.11062) = −.44203, b3 =
(−.23393) = −13.47008,
4.31356
.29934
b0 = 61.5652+1.14162(38.3913) +.44203(50.4348)+13.47008(2.28696) = 158.4927
7-2
7.19. a.
c.
Ŷ ∗ = −.547853X1∗ + .423647X2∗ + .0484614X3∗ + .502757X4∗
sY = 1.71958, s1 = 6.63278, s2 = 2.58317, s3 = .13455,s4 = 109099, b1 =
1.71958
1.71958
(−.547853) = −.14203, b2 =
(.423647) = .28202,
6.63278
2.58317
1.71958
1.71958
b3 =
(.0484614) = .61934, b4 =
(.502757) = 7.9243 × 10−6 ,
.13455
109099
b0 = 15.1389 + .14203(7.8642) − .28202(9.68815) − .61934(.08099) − 7.9243 ×
10−6 (160633) = 12.20054.
7.21. b. The line of fitted values when .5X1 − X2 = −5.
7.24. a.
Ŷ = 50.775 + 4.425X1
c.
Yes, SSR(X1 ) = 1, 566.45, SSR(X1 |X2 ) = 1, 566.45
d.
r12 = 0
7.25. a.
Ŷ = 4079.87 + 0.000935X2
c.
No, SSR(X1 ) = 136, 366, SSR(X1 |X2 ) = 130, 697
d.
r12 = .0849
7.26. a.
c.
Ŷ = 156.672 − 1.26765X1 − 0.920788X2
No, SSR(X1 ) = 8, 275.3, SSR(X1 |X3 ) = 3, 483.89
No, SSR(X2 ) = 4, 860.26, SSR(X2 |X3 ) = 708
d.
r12 = .5680, r13 = .5697, r23 = .6705
7.27. a.
Ŷ = 14.3613 − .11447X1 + .00001X4
c.
No, SSR(X4 ) = 67.7751, SSR(X4 |X3 ) = 66.8582
No, SSR(X1 ) = 14.8185, SSR(X1 |X3 ) = 13.7744
d.
7.28. a.
r12 = .4670, r13 = .3228, r23 = .2538
(1) SSR(X1 , X5 ) − SSR(X1 ) or SSE(X1 ) − SSE(X1 , X5 )
(2) SSR(X1 , X3 , X4 ) − SSR(X1 ) or SSE(X1 ) − SSE(X1 , X3 , X4 )
(3) SSR(X1 , X2 , X3 , X4 ) − SSR(X1 , X2 , X3 )
or SSE(X1 , X2 , X3 ) − SSE(X1 , X2 , X3 , X4 )
b.
7.29. a.
SSR(X5 |X1 , X2 , X3 , X4 ), SSR(X2 , X4 |X1 , X3 , X5 )
SSR(X1 ) + SSR(X2 , X3 |X1 ) + SSR(X4 |X1 , X2 , X3 )
= SSR(X1 ) + [SSR(X1 , X2 , X3 ) − SSR(X1 )]
+[SSR(X1 , X2 , X3 , X4 ) − SSR(X1 , X2 , X3 )]
= SSR(X1 , X2 , X3 , X4 )
b.
SSR(X2 , X3 ) + SSR(X1 |X2 , X3 ) + SSR(X4 |X1 , X2 , X3 )
= SSR(X2 , X3 ) + [SSR(X1 , X2 , X3 ) − SSR(X2 , X3 )]
+[SSR(X1 , X2 , X3 , X4 ) − SSR(X1 , X2 , X3 )] = SSR(X1 , X2 , X3 , X4 )
7-3
7.30. a.
b.
c.
Ŷ = 68.625 + 4.375X2
i:
1
2
3
ei : -13.3750 -13.1250 -16.3750
4
-10.1250
5
-5.3750
i:
7
ei : -6.3750
8
9
10
11
-3.1250 5.6250 2.8750 8.6250
i:
13
ei : 10.6250
14
8.8750
X̂1 = 7
i: 1 2
ei : -3 -3
6
-6.1250
12
6.8750
15
16
16.6250 13.8750
3 4 5 6
-3 -3 -1 -1
i: 9 10 11 12 13
ei : 1 1 1 1 3
r = .971 = rY 1|2
7
-1
14
3
8
-1
15 16
3 3
7.31. (1) Yi = β0 + β1 Xi1 + β2 Xi2 + εi
√
(2) Yi = β0 + β1 Xi1 + β2 Xi2 + β4 Xi3 + εi
√
(3) Yi0 = Yi − 5(Xi1 + Xi2 ) = β0 + β3 Xi1 Xi2 + β4 Xi3 + εi
√
(4) Yi0 = Yi − 7 Xi3 = β0 + β1 Xi1 + β2 Xi2 + β3 Xi1 Xi2 + εi
7.32. (1) Yi = β0 + β2 Xi2 + εi
2
(2) Yi = β1 Xi1 + β2 Xi2 + β3 Xi1
+ εi
2
(3) Yi0 = Yi − 5Xi1
= β0 + β1 Xi1 + β2 Xi2 + εi
2
+ εi
(4) Yi0 = Yi − 10 = β1 Xi1 + β2 Xi2 + β3 Xi1
2
+ εi , where βc = β1 = β2
(5) Yi = β0 + βc (Xi1 + Xi2 ) + β3 Xi1
7.33. Let:
yi = Yi − Ȳ
xi1 = Xi1 − X̄1
xi2 = Xi2 − X̄2
P
Then: SSR(X1 ) =
(
xi1 yi )2
P 2
x
=
i1
X
yi2 rY2 1 by (1.10a), (2.51) and (2.84)
P
( xi1 yi )2 X 2
P 2
=
yi (1 − rY2 1 )
SSE(X1 ) = yi − P 2
xi1
P
P
SSR(X1 , X2 ) = b1
xi1 yi + b2
xi2 yi
by (2.43) and
Further:
P
"P
yi2
xi1 yi
P 2 − P 2
xi1
xi1
b1 =
2
1 − r12
#1/2
rY 2 r12
by (7.56)
7-4
P
yi = 0
and similarly:
"P
P
xi2 yi
yi2
P 2 − P 2
xi2
xi2
b2 =
2
1 − r12
#1/2
rY 1 r12
Substituting these expressions for b1 and b2 into SSR(X1 , X2 ), we obtain after some
simplification:
SSR(X1 , X2 ) =
i
X
X
1 hX 2 2
2 2
2
y
r
+
y
r
−
2
y
r
r
r
i Y1
i Y2
i Y 1 Y 2 12
2
1 − r12
Now by (7.36) and (7.2b), we have:
rY2 2|1 =
SSR(X1 , X2 ) − SSR(X1 )
SSE(X1 )
Substituting the earlier expressions into the above, we obtain after some simplifying:
rY2 2|1 = P
X
1
2
yi2 (1 − rY2 1 )(1 − r12
)
[
yi2 rY2 1 +
2
− (1 − r12
)
X
yi2 rY2 2 − 2
X
yi2 rY 1 rY 2 r12
P 2 2
yi rY 1 ]
After some further simplifying, we obtain:
rY2 2|1 =
"
7.34. a.
(1)
"
(3)
b.
(rY 2 − r12 rY 1 )2
2
(1 − rY2 1 )(1 − r12
)
1 0
0 1
#
"
(2)
.7420
.6385
#
"
(4)
.7420
.6385
#
.0083
0
0
.0083
#
1.069
(5.375) = .742
7.745
.5345
b∗2 =
(9.250) = .638
7.745
From (7.53), b∗1 =
7.35. From (7.45), we have:
∗
∗
+ ε∗i
+ β2∗ Xi2
Yi∗ = β1∗ Xi1
1
√
n−1
Ã
Yi − Ȳ
sY
!
=
β1∗ √
1
n−1
Ã
Xi1 − X̄1
s1
!
+
β2∗ √
1
n−1
Ã
Xi2 − X̄2
s2
Simplifying, we obtain:
Yi = (Ȳ − β1∗
√
sY
sY
sY
sY
X̄1 − β2∗ X̄2 ) + β1∗ Xi1 + β2∗ Xi2 + n − 1sY ε∗i
s1
s2
s1
s2
7-5
!
+ ε∗i
Hence:
sY
β1∗
= β1
s1
β2∗
sY
= β2
s2
"
∗
7.36. X Y=
7.37. a.
∗ ∗
ΣXi1
Yi
∗ ∗
ΣXi2 Yi
#
=
Σ(Xi1 − X̄1 )(Yi − Ȳ )
(n − 1)s1 sY
Σ(Xi2 − X̄2 )(Yi − Ȳ )
(n − 1)s2 sY
=
"
rY 1
rY 2
#
RY2 3|12 = .02883, RY2 4|12 = .00384, RY2 5|12 = .55382, RY2 6|12 = .00732
b.
X5 , yes.
c.
Full model: Yi = β0 + β1 Xi1 + β2 Xi2 + β5 Xi5 + εi . H0 : β5 = 0, Ha : β5 6= 0.
SSR(X5 |X1 , X2 ) = 78, 070, 132, SSE(X1 , X2 , X5 ) = 62, 896, 949,
F ∗ = (78, 070, 132/1) ÷ (62, 896, 949/436) = 541.1801, F (.99; 1, 137) = 6.69336.
If F ∗ ≤ 6.69336 conclude H0 , otherwise Ha . Conclude Ha . No.
7.38. a.
RY2 3|12 = .01167, RY2 4|12 = .13620, RY2 5|12 = .03737, RY2 6|12 = .03639
b.
X4 , yes.
c.
Full model: Yi = β0 + β1 Xi1 + β2 Xi2 + β4 Xi4 + εi . H0 : β4 = 0, Ha : β4 6= 0.
SSR(X4 |X1 , X2 ) = 37.89858, SSE(X1 , X2 , X4 ) = 240.35163, F ∗ = (37.89858/1)÷
(240.35163/109) = 17.187, F (.95; 1, 109) = 3.93. If F ∗ ≤ 3.93 conclude H0 , otherwise Ha . Conclude Ha . No.
7-6
Chapter 8
MODELS FOR QUANTITATIVE
AND QUALITATIVE PREDICTORS
8.4.
8.5.
a.
Ŷ = 82.9357 − 1.18396x + .0148405x2 , R2 = .76317
b.
H0 : β1 = β11 = 0, Ha : not both β1 and β11 = 0. M SR = 5915.31, M SE = 64.409,
F ∗ = 5915.31/64.409 = 91.8398, F (.95; 2, 57) = 3.15884. If F ∗ ≤ 3.15884 conclude
H0 , otherwise Ha . Conclude Ha .
c.
Ŷh = 99.2546, s{Ŷh } = 1.4833, t(.975; 57) = 2.00247, 99.2546 ± 2.00247(1.4833),
96.2843 ≤ E{Yh } ≤ 102.2249
d.
s{pred} = 8.16144, 99.2546 ± 2.00247(8.16144), 82.91156 ≤ Yh(new) ≤ 115.5976
e.
H0 : β11 = 0, Ha : β11 6= 0. s{b11 } = .00836, t∗ = .0148405/.00836 = 1.7759,
t(.975; 57) = 2.00247. If |t∗ | ≤ 2.00247 conclude H0 , otherwise Ha . Conclude
H0 . Alternatively, SSR(x2 |x) = 203.1, SSE(x, x2 ) = 3671.31, F ∗ = (203.1/1) ÷
(3671.31/57) = 3.15329, F (.95; 1, 57) = 4.00987. If F ∗ ≤ 4.00987 conclude H0 ,
otherwise Ha . Conclude H0 .
f.
Ŷ = 207.350 − 2.96432X + .0148405X 2
g.
rX,X 2 = .9961, rx,x2 = −.0384
a.
i:
1
2
3
...
ei : -1.3238 -4.7592 -3.8091 . . .
58
-11.7798
59
-.8515
60
6.22023
b.
H0 : E{Y } = β0 + β1 x + β11 x2 , Ha : E{Y } 6= β0 + β1 x + β11 x2 . M SLF = 62.8154,
M SP E = 66.0595, F ∗ = 62.8154/66.0595 = 0.95, F (.95; 29, 28) = 1.87519. If
F ∗ ≤ 1.87519 conclude H0 , otherwise Ha . Conclude H0 .
c.
Ŷ = 82.92730 − 1.26789x + .01504x2 + .000337x3
H0 : β111 = 0, Ha : β111 6= 0. s{b111 } = .000933, t∗ = .000337/.000933 = .3612,
t(.975; 56) = 2.00324. If |t∗ | ≤ 2.00324 conclude H0 , otherwise Ha . Conclude H0 . Yes. Alternatively, SSR(x3 |x, x2 ) = 8.6, SSE(x, x2 , x3 ) = 3662.78,
F ∗ = (8.6/1) ÷ (3662.78/56) = .13148, F (.95; 1, 56) = 4.01297. If F ∗ ≤ 4.01297
conclude H0 , otherwise Ha . Conclude H0 . Yes.
8.6.
a.
Ŷ = 21.0942 + 1.13736x − .118401x2 , R2 = .81434
8-1
b.
H0 : β1 = β11 = 0, Ha : not all βk = 0 (k = 1, 11). M SR = 523.133, M SE =
9.9392, F ∗ = 523.133/9.9392 = 52.6333,F (.99; 2, 24) = 5.6136. If F ∗ ≤ 5.6136
conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
F (.99; 3, 24) = 5.04, W = 3.7622; B = t(.99833; 24) = 3.25756
X
10: 20.6276 ± 3.25756(1.8945)
8.7.
8.8.
8.9.
14.45615 ≤ E{Yh } ≤ 26.79905
15: 11.5142 ± 3.25756(4.56694)
− 3.36288 ≤ E{Yh } ≤ 26.39128
20: −3.5192 ± 3.25756(8.50084)
− 31.2112 ≤ E{Yh } ≤ 24.1728
d.
s{pred} = 5.54942, t(.995; 24) = 2.79694, 11.5142 ± 2.79694(5.54942), −4.0072 ≤
Yh(new) ≤ 27.0356
e.
H0 : β11 = 0, Ha : β11 6= 0. s{b11 } = .02347, t∗ = −.118401/.02347 = −5.04478,
t(.995; 24) = 2.79694. If |t∗ | ≤ 2.79694 conclude H0 , otherwise Ha . Conclude Ha .
Alternatively, SSR(x2 |x) = 252.989, SSE(x, x2 ) = 238.541, F ∗ = (252.989/1) ÷
(238.541/24) = 25.4536, F (.99; 1, 24) = 7.82287. If F ∗ ≤ 7.82287 conclude H0 ,
otherwise Ha . Conclude Ha .
f.
Ŷ = −26.3254 + 4.87357X − .118401X 2
a.
i:
1
2
ei : 3.96746 -1.42965
...
...
26
2.10202
27
-2.43692
b.
H0 : E{Y } = β0 + β1 x + β11 x2 , Ha : E{Y } 6= β0 + β1 x + β11 x2 . M SLF = 6.65396,
M SP E = 13.2244, F ∗ = 6.65396/13.2244 = 0.50316, F (.99; 12, 12) = 4.15526. If
F ∗ ≤ 4.15526 conclude H0 , otherwise Ha . Conclude H0 .
a.
Ŷ = 10.1893 − .181775x1 + .0141477x21 + .314031X2 + .000008X4
b.
.5927
c.
H0 : β11 = 0, Ha : β11 =
6 0. s{b11 } = .005821, t∗ = .0141477/.005821 = 2.43046,
t(.975; 76) = 1.99167. If |t∗ | ≤ 1.99167 conclude H0 , otherwise Ha . Conclude Ha .
d.
Ŷh = 17.2009, s{Ŷh } = .37345, t(.975; 76) = 1.99167, 17.2009 ± 1.99167(.37345),
16.45711 ≤ E{Yh } ≤ 17.94469
e.
Ŷ = 12.4938 − .404296x1 + .0141477x21 + .314031X2 + .000008X4
a.
X2 = 3: E{Y } = 37 + 7.5X1
X2 = 6: E{Y } = 49 + 12X1
8.10. a.
X1 = 1: E{Y } = 21 + X2
X1 = 4: E{Y } = 42 − 11X2
8.11. a.
b.
Ŷ = 27.150 + 5.925X1 + 7.875X2 − .500X1 X2
H0 : β3 = 0, Ha : β3 6= 0. M SR(X1 X2 |X1 , X2 ) = 20.0000, M SE = 6.1917,
F ∗ = 20.0000/6.1917 = 3.23, F (.95; 1, 12) = 4.75. If F ∗ ≤ 4.75 conclude H0 ,
otherwise Ha . Conclude H0 .
8.13. E{Y } = 25.3 + .20X1 for mutual firms,
8-2
E{Y } = 13.2 + .20X1 for stock firms.
8.15. b.
c.
Ŷ = −0.92247 + 15.0461X1 + .75872X2
s{b2 } = 2.77986, t(.975; 42) = 2.01808, .75872 ± 2.01808(2.77986), −4.85126 ≤
β2 ≤ 6.3687
e.
i:
1
2
...
Xi1 Xi2 :
2
0
...
ei : -9.92854 .73790 . . .
8.16. b.
c.
44
0
1.73790
45
0
2.69176
Ŷ = 2.19842 + .03789X1 − .09430X2
H0 : β2 = 0, Ha : β2 6= 0. s{b2 } = .11997, t∗ = −.09430/.11997 = −.786,
t(.995; 117) = 2.6185. If |t∗ | ≤ 2.6185 conclude H0 , otherwise Ha . Conclude H0 .
d.
i:
1
2
...
Xi1 Xi2 :
0
14
...
ei : .90281 1.25037 . . .
119
16
-.85042
120
0
-.31145
8.17. No
8.18. E{Y } = 25 + .30X1 for mutual firms,
E{Y } = 12.5 + .35X1 for stock firms.
8.19. a.
b.
8.20. a.
b.
8.21. a.
Ŷ = 2.81311 + 14.3394X1 − 8.14120X2 + 1.77739X1 X2
H0 : β3 = 0, Ha : β3 6= 0. s{b3 } = .97459, t∗ = 1.77739/.97459 = 1.8237,
t(.95; 41) = 1.68288. If |t∗ | ≤ 1.68288 conclude H0 , otherwise Ha . Conclude Ha .
Alternatively, SSR(X1 X2 |X1 , X2 ) = 255.9, SSE(X1 , X2 , X1 X2 ) = 3154.44, F ∗ =
(255.9/1)÷ (3154.44/ 41) = 3.32607, F (.90; 1, 41) = 2.83208. If F ∗ ≤ 2.83208
conclude H0 , otherwise Ha . Conclude Ha .
Ŷ = 3.22632 − .00276X1 − 1.64958X2 + .06224X1 X2
H0 : β3 = 0, Ha : β3 6= 0. s{b3 } = .02649, t∗ = .06224/.02649 = 2.3496,
t(.975; 116) = 1.9806. If |t∗ | ≤ 1.9806 conclude H0 , otherwise Ha . Conclude Ha .
Alternatively, SSR(X1 X2 |X1 , X2 ) = 2.07126, SSE(X1 , X2 , X1 X2 ) = 45.5769,
F ∗ = (2.07126/1) ÷ (45.5769/116) = 5.271665, F (.95; 1, 116) = 3.9229. If F ∗ ≤
3.9229 conclude H0 , otherwise Ha . Conclude Ha .
Hard hat: E{Y } = (β0 + β2 ) + β1 X1
Bump cap: E{Y } = (β0 + β3 ) + β1 X1
None: E{Y } = β0 + β1 X1
b. (1) H0 : β3 ≥ 0, Ha : β3 < 0; (2) H0 : β2 = β3 , Ha : β2 6= β3
8.22.
E{Y } = β0 + β1 X1
Tool models M1
E{Y } = (β0 + β2 ) + (β1 + β5 )X1
Tool models M2
8-3
E{Y } = (β0 + β3 ) + (β1 + β6 )X1
Tool models M3
E{Y } = (β0 + β4 ) + (β1 + β7 )X1
Tool models M4
8.24. b.
Ŷ = −126.905 + 2.7759X1 + 76.0215X2 − 1.10748X1 X2 ,
H0 : β2 = β3 = 0, Ha : not both β2 = 0 and β3 = 0. SSR(X2 , X1 X2 | X1 ) = 566.15,
SSE(X1 , X2 , X1 X2 ) = 909.105, F ∗ = (369.85/2) ÷ (909.105/60) = 12.2049,
F (.95; 2, 60) = 3.15041. If F ∗ ≤ 3.15041 conclude H0 , otherwise Ha . Conclude
Ha .
c.
Ŷ = −126.9052 + 2.7759X1
for noncorner lots
Ŷ = −50.8836 + 1.6684X1
for corner lots
8.25. a.
Ŷ = 4295.72+.000903x1 −(1.5767×10−9 )x21 +614.393X3 −.000188x1 X3 +(1.8076×
10−9 )x21 X3
b.
H0 : β2 = β4 = β5 = 0, Ha : not all β2 = 0,β4 = 0 and β5 = 0.
SSR(x21 , x1 X3 , x21 X3 |x1 , X3 ) = 1442, SSE(x1 , x21 , X3 , x1 X3 , x21 X3 ) = 990762, F ∗ =
(1442/3) ÷ (990762/46) = .02232, F (.95; 3, 46) = 2.8068. If F ∗ ≤ 2.80681 conclude H0 , otherwise Ha . Conclude H0 .
Set 1
8.29.
X
1 .990 .966
2
X
1
−
X3
1
x
1 .379 .904
2
x
1
−
x3
1
Set 2
X
1 .970 .929
X2
1
−
3
X
1
8.30.
dE{Y }
= β1 + 2β11 x
dx
d2 E{Y }
= 2β11
dx2
8.31. a.
x
1 .846 .89
x2
1
−
3
x
1
Ŷ = b0 + b1 x + b11 x2
= b0 + b1 (X − X̄) + b11 (X − X̄)2
= b0 + b1 X − b1 X̄ + b11 X 2 + b11 X̄ 2 − 2b11 X X̄
= (b0 − b1 X̄ + b11 X̄ 2 ) + (b1 − 2b11 X̄)X + b11 X 2
Hence:
b00 = b0 − b1 X̄ + b11 X̄ 2
b01 = b1 − 2b11 X̄
b011 = b11
8-4
σ02 σ01 σ02
σ 2 {b} =
σ01 σ12 σ12
σ02 σ12 σ22
1 −X̄ X̄ 2
1 −2X̄
b. A =
0
0 0
1
where σ02 = σ 2 {b0 }, σ01 = σ{b0 , b1 }, etc. for the regression coefficients in the
transformed x variables.
The variance-covariance matrix of the regression coefficients in the original X
variables, A [σ 2 {b}] A0 , then yields:
σ 2 {b00 } = σ02 − 2X̄σ01 + 2X̄ 2 σ02 + X̄ 2 σ12 − 2X̄ 3 σ12 + X̄ 4 σ22
σ 2 {b01 } = σ12 − 4X̄σ12 + 4X̄ 2 σ22
σ 2 {b02 } = σ22
σ{b00 , b01 } = σ01 − 2X̄σ02 + 3X̄ 2 σ12 − X̄σ12 − 2X̄ 3 σ22
σ{b00 , b02 } = σ02 − X̄σ12 + X̄ 2 σ22
σ{b01 , b02 } = σ12 − 2X̄σ22
8.32. When Xi are equally spaced,
P
P
Yi = nb0 + b11
P 3
xi = 0; hence (8.4) becomes:
P 2
x
i
P
xi Yi = b1 x2i
P 2
P
P
xi Yi = b0 x2i + b11 x4i
8.33. a.
b.
Yi = β0 +β1 xi1 +β2 x2i1 +β3 Xi2 +β4 xi1 Xi2 +β5 x2i1 Xi2 +β6 Xi3 +β7 xi1 Xi3 +β8 x2i1 Xi3 +εi
(1) H0 : β3 = β4 = β5 = β6 = β7 = β8 = 0
Ha : not all βk = 0 (k = 3, ..., 8)
SSE(R) = SSE(x1 , x21 )
SSE(R) − SSE(F ) SSE(F )
F∗ =
÷
6
n−9
If F ∗ ≤ F (.99; 6, n − 9) conclude H0 , otherwise Ha .
(2) H0 : β3 = β6 = 0, Ha : not both β3 = 0 and β6 = 0
SSE(R) = SSE(x1 , x21 , x1 X2 , x21 X2 , x1 X3 , x21 X3 )
SSE(R) − SSE(F ) SSE(F )
F∗ =
÷
2
n−9
∗
If F ≤ F (.99; 2, n − 9) conclude H0 , otherwise Ha .
(3) H0 : β4 = β5 = β7 = β8 = 0, Ha : not all βk = 0 (k = 4, 5, 7, 8)
SSE(R) = SSE(x1 , x21 , X2 , X3 )
SSE(R) − SSE(F ) SSE(F )
F∗ =
÷
4
n−9
If F ∗ ≤ F (.99; 4, n − 9) conclude H0 , otherwise Ha .
8.34. a.
b.
Yi = β0 + β1 Xi1 + β2 Xi2 + β3 Xi3 + εi
Commercial: E{Y } = (β0 + β2 ) + β1 X1
8-5
Mutual savings: E{Y } = (β0 + β3 ) + β1 X1
Savings and loan: E{Y } = (β0 − β2 − β3 ) + β1 X1
8.35. a.
Let n2 = n − n1 and define:
1 0
. .
.. ..
← n1
1 0
X=
1 1
. .
. .
. . ← n2
1 1
P
P
Yi1
Yi2
Ȳ1 =
Ȳ2 =
n1
n2
Then:
"
X0 X =
n n2
n2 n2
"
0
−1
(X X)
"
Ȳ1
Ȳ2 − Ȳ1
=
#
1
n1
− n11
b=
c.
SSR = n1 Ȳ12 + n2 Ȳ22 − nȲ 2
8.36. a.
PP
Yn 1 1
Y=
Y12
.
.
.
Yn 2 2
PP
Ȳ =
"
X0 Y =
− n11
1
+ n12
n1
Y11
.
..
#
b.
SSE =
#
Yij
n1 + n2
nȲ
n2 Ȳ2
#
Yij2 − n1 Ȳ12 − n2 Ȳ22
Ŷ = 999.912 + .00296x − 3.29518 × 10−11 x2
b.
R2 = .8855 for second-order model; R2 = .6711 for first-order model.
c.
H0 : β11 = 0, Ha : β11 6= 0. s{b11 } = 1.400396 × 10−11 , t∗ = −3.29518 ×
10−11 /1.400396 × 10−11 = −2.353, t(.975; 437) = 1.9654. If |t∗ | ≤ 1.9654 conclude H0 , otherwise Ha . Conclude Ha . Alternatively, SSR(x2 |x) = 2, 039, 681,
SSE(x, x2 ) = 160, 985, 454, F ∗ = (2, 039, 681/1) ÷ (160, 985, 454/437) = 5.5368,
F (.95; 1, 437) = 3.8628. If F ∗ ≤ 3.8628 conclude H0 , otherwise Ha . Conclude Ha .
8.37. a.
Ŷ = .056288 + 0.000004585x1 − .000088x3 + 2.6982 × 10−12 x21 + .00016293x23 +
8.3337 × 10−7 x1 x3 , R2 =.2485
b.
H0 : β11 = β33 = β13 = 0, Ha : not all βk = 0 (k = 11, 33, 13).
SSR(x21 , x23 , x1 x3 |x1 , x3 ) = .005477, SSE(x1 , x3 , x21 , x23 , x1 x3 ) = .246385, F ∗ =
(.005477/3) ÷ (.246385/437) = 3.2381, F (.99; 1, 437) = 3.8267. If F ∗ ≤ 3.8267
conclude H0 , otherwise Ha . Conclude H0 .
c.
Ŷ = .0584998+2.9419×10−8 x1 −5.5765×10−7 x2 +.00068244x3 −3.3559×10−15 x21 ,
R2 = .1444
8.38. a.
b.
Ŷ = 150.07921 + 7.06617x + .10116x2
R2 = .6569 for second-order model; R2 = .6139 for first-order model.
8-6
c.
8.39. a.
H0 : β11 = 0, Ha : β11 6= 0. s{b11 } = .02722, t∗ = .10116/.02722 = 3.716,
t(.995; 110) = 2.621. If |t∗ | ≤ 2.621 conclude H0 , otherwise Ha . Conclude Ha .
Alternatively, SSR(x2 |x) = 93, 533.252, SSE(x, x2 ) = 745, 203.642,
F ∗ = (93, 533.252/1) ÷ (745, 203.642/ 110) = 13.807, F (.99; 1, 110) = 6.871. If
F ∗ ≤ 6.871 conclude H0 , otherwise Ha . Conclude Ha .
Ŷ = −207.5 + .0005515X1 + .107X2 + 149.0X3 + 145.5X4 + 191.2X5
b.
b3 −b4 = 3.5, s{b3 −b4 } = 1.68, t(.95; 434) = 1.6484, 3.5±1.6484(1.68), 0.730688 ≤
β3 − β4 ≤ 6.2693
c.
H0 : β3 = β4 = β5 = 0, Ha : not all βk = 0 (k = 3, 4, 5). SSR(X3 , X4 , X5 |X1 , X2 ) =
1, 873, 626, SSE(X1 , X2 , X3 , X4 , X5 ) = 139, 093, 455,
F ∗ = (1, 873, 626/3) ÷ (139, 093, 455/434) = 1.9487, F (.90; 3, 434) = 2.09645. If
F ∗ ≤ 2.09645 conclude H0 , otherwise Ha . Conclude H0 . P −value=.121.
8.40. a.
Ŷ = .85738 + .28882X1 − .01805X2 + .01995X3 + .28782X4
b.
s{b4 } = .30668, t(.99; 108) = 2.361, .28782 ± 2.361(.30668), −.476 ≤ β4 ≤ 1.012
c.
Ŷ = .99413 + .26414X1 − .02283X2 + .02429X3 − 5.69520X4 + .15576X2 X4 −
.02406X3 X4
H0 : β5 = β6 = 0, Ha : not both β5 = 0 and β6 = 0. SSR(X2 X4 , X3 X4
|X1 , X2 , X3 , X4 ) = 5.1964, SSE(X1 , X2 , X3 , X4 , X2 X4 , X3 X4 ) = 122.0468, F ∗ =
(5.1964/2) ÷ (122.0468/106) = 2.257, F (.90; 2, 106) = 2.353. If F ∗ ≤ 2.353 conclude H0 , otherwise Ha . Conclude H0 .
8.41. a.
Ŷ = 2.0478+.10369X1 +.04030X2 +.00660X3 −.020761X4 +2.14999X5 +1.19033X6 +
.63348X7
b.
H0 : β2 = 0, Ha : β2 6= 0. s{b2 } = .01430, t∗ = .04030/.01430 = 2.818,
t(.975; 105) = 1.983. If |t∗ | ≤ 1.983 conclude H0 , otherwise Ha . Conclude Ha .
Alternatively, SSR(X2 |X1 , X3 , X4 , X5 , X6 , X7 ) = 15.52782,
SSE(X1 , X2 , X3 , X4 , X5 , X6 , X7 ) = 205.3634, F ∗ = (15.52782/1)÷(205.3634/105) =
7.9392, F (.95; 1, 105) = 3.932. If F ∗ ≤ 3.932 conclude H0 , otherwise Ha . Conclude Ha .
c.
s{b5 } = .46152, s{b6 } = .43706, s{b7 } = .42755, B = t(.99167; 105) = 2.433
8.42. a.
b.
2.14999 ± 2.443(.46152)
1.0225 ≤ β5 ≤ 3.2775
1.19033 ± 2.443(.43706)
.1226 ≤ β6 ≤ 2.2581
.63348 ± 2.443(.42755)
− .4110 ≤ β7 ≤ 1.6780
Ŷ = 3.0211−.247X1 −.000097X2 +.4093X3 +.124X4 −.01324X5 {1999} −.1088X5 {2001}−
.08306X5 {2002}
Ŷ = 2.38 − 0.453x1 − 0.000144x2 + 0.00016x1 x2 + 0.92x21 + 0.000001x22 + 0.394X3 +
0.115X4 + 0.012X5 {1999} − 0.101X5 {2001} − 0.0581X5 {2002}
H0 : β3 = β4 = β5 = 0, Ha : not all βk = 0 (k = 3, 4, 5). SSE(R) = .65424, dfR =
28, SSE(F ) = .62614, dfF = 25, M SE(F ) = .02505 F ∗ = .37392, F (.95; 3, 30) =
2.9223. If F ∗ ≤ 2.9223 conclude H0 , otherwise Ha . Conclude H0 .
8-7
c.
H0 : β2 = β5 = β6 = β7 = 0, Ha : not all βk = 0 (k = 2, 5, 6, 7). SSE(R) =
.71795, dfR = 32, SSE(F ) = .65424, dfF = 28, M SE(F ) = .02337, F ∗ = .68154,
F (.95; 4, 28) = 2.71408. If F ∗ ≤ 2.71408 conclude H0 , otherwise Ha . Conclude
H0 .
8-8
Chapter 9
BUILDING THE REGRESSION
MODEL I: MODEL SELECTION
AND VALIDATION
Variables in Model
None
X1
X2
X3
X1 , X2
X1 , X3
X2 , X3
X1 , X2 , X3
9.9.
9.10. b.
X1
X2
X3
X4
c.
Rp2
0
.6190
.3635
.4155
.6550
.6761
.4685
.6822
AICp
262.916
220.529
244.131
240.214
217.968
215.061
237.845
216.185
Cp
P RESSp
88.16 13,970.10
8.35 5,569.56
42.11 9,254.49
35.25 8,451.43
5.60 5,235.19
2.81 4,902.75
30.25 8,115.91
4.00 5,057.886
1 .102 .181 .327
1 .519 .397
1 .782
1
Ŷ = −124.3820 + .2957X1 + .0483X2 + 1.3060X3 + .5198X4
9.11. a.
Subset
X1 , X3 , X4
X1 , X2 , X3 , X4
X1 , X3
X1 , X2 , X3
9.12.
2
Ra,p
.9560
.9555
.9269
.9247
Subset
X3 , X5 , X6
X5
X3 , X5 , X6 , X8 = 2001
SBCp
−126.601
−124.970
−124.969
Note: Variable numbers for predictors are those in the appendix.
9-1
9.13. b.
X1
1 .653 −.046
X2
1 −.423
X3
1
c.
Ŷ = 87.1875 − .5645X1 − .5132X2 − .0720X3
9.14. a.
Subset
x1 , x2 , x21 , x22
x1 , x 2 , x 1 x2
x1 , x2 , x1 x2 , x22
9.15. b.
2
Ra,p
.75067
.75066
.74156
X1
1 .468 −.089
X2
1
.068
X3
1
c.
Ŷ = 120.0473 − 39.9393X1 − .7368X2 + .7764X3
9.16. a.
Subset
x1 , x2 , x3 , x23 , x1 x2
x1 , x2 , x3 , x22, x23, x1 x2
x1 , x2 , x3, x23 , x1 x2 , x1 x3
9.17. a.
2
Ra,p
.8668
.8652
.8638
X1 , X3
b.
.10
c.
X1 , X3
d.
X1 , X3
9.18. a. X1 , X3 , X4
9.19
a.
x1 , x2 , x3 , x1 x2
b.
2
Ra,p
= .8615
9.20. X3 , X5 , X6 in appendix.
9.21. P RESS = 760.974, SSE = 660.657
9.22. a.
X1
X2
X3
X4
1 .011 .177 .320
1 .344 .221
1 .871
1
b.
9-2
Model-building
data set
−127.596
12.685
.348
.054
1.823
.123
27.575
.933
b0 :
s{b0 }:
b1 :
s{b1 }:
b3 :
s{b3 }:
M SE:
R2 :
Validation
data set
−130.652
12.189
.347
.048
1.848
.122
21.446
.937
c.
M SP R = 486.519/25 = 19.461
d.
Ŷ = −129.664 + .349X1 + 1.840X3 , s{b0 } = 8.445, s{b1 } = .035, s{b3 } = .084
9.23. a. P RESS = 5, 102.494, SSE = 1, 680.465
9.24. Xi = 10: [E{Ŷ } − E{Y }]2 = (−55)2 , [Ŷ − E{Ŷ }]2 = (−47)2
Xi = 20: [E{Ŷ } − E{Y }]2 = (−705)2 , [Ŷ − E{Ŷ }]2 = (−97)2
9.25. b.
X3
X4
X5
X6
X7
X10
X11
X12
1 .025 −.101 .161 −.198 −.172 −.236 −.164
1
.448 .334 .490
.501
.530
.453
1
.195
.168
.204
.239
.240
1
.067
.086
.060
.128
1
.990
.909
.764
1
.904
.729
1
.707
1
Note: Variable numbers for predictor variables are those in the data set description.
Subset
X3 , X6 , X10
X3 , X6 , X10 , X11
X3 , X6 , X7 , X10
c.
9.26. b.
X4
X6
X7
X8
X9
X11
X12
X13
X14
X15
X16
1
−.063
1
.016
−.599
1
.040
.174
−.023
1
Cp
3.81
3.86
4.27
.021
.113
.054
.921
1
−.113
.245
−.240
.056
−.100
1
−.145
.486
−.359
.264
.059
.722
1
.169
−.025
−.003
.033
.173
−.753
−.466
1
.181
−.222
.189
−.056
.034
−.671
−.551
.513
1
−.190
.078
−.028
.312
.145
.585
.748
−.649
−.379
1
.078
.116
−.023
.934
.891
.087
.238
−.052
−.023
.347
1
Note: Variable numbers for predictor variables are those in the data set description.
9-3
Subset
X6 , X9 , X13 , X14
X6 , X8 , X9 , X13 , X14 , X15
X6 , X9 , X13 , X14 , X15
c.
SBCp
3407.16
3407.41
3408.09
9.27. a.
b0 :
s{b0 }:
b3 :
s{b3 }:
b6 :
s{b6 }:
b10 :
s{b10 }:
M SE:
R2 :
Model-building
data set
.6104
.0888
.00388
.00163
.00117
.000419
.000293
.0000456
.00305
.519
Validation
data set
.6189
.1248
.00399
.00211
.00152
.000437
.000157
.0000622
.00423
.293
b.
M SP R = .258271/56 = .00461
c.
Ŷ 0 = .6272 + .00353X3 + .00143X6 + .000236X10 , s{b0 } = .0738, s{b3 } = .00129,
s{b6 } = .000297, s{b10 } = .0000374, where Y 0 = log10 Y.
9.28. a.
b0 :
s{b0 }:
b6 :
s{b6 }:
b9 :
s{b9 }:
b13 :
s{b13 }:
b14 :
s{b14 }:
M SE:
R2 :
b.
M SP R =
Model-building Validation
data set
data set
243.680
3015.63
1322.82
1189.63
122.507
34.3137
41.1906
34.2984
.578662
.221509
.075844
.057344
296.117
269.557
34.3417
39.0049
-224.020
-128.343
77.1406
70.4556
4,816,124 4,484,316
.463
.284
2, 259, 424, 814
= 10, 270, 113
220
9-4
Chapter 10
BUILDING THE REGRESSION
MODEL II: DIAGNOSTICS
10.5.
a.
i:
e(Y | X1 ):
e(X2 | X1 ):
e(Y | X2 ):
e(X1 | X2y ):
e(Y
e(X2
e(Y
e(X1
c.
i:
| X1 ):
| X1 ):
| X2 ):
| X2 ):
1
−4.475
−1
−13.38
−3
9
−3.175
−1
5.625
1
2
4.525
1
−13.13
−3
10
2.825
1
2.875
1
3
−7.475
−1
−16.38
−3
11
−.175
−1
8.625
1
...
...
...
...
...
...
...
...
...
...
6
2.675
1
−6.125
−1
14
−.025
1
8.875
3
7
−6.325
−1
−6.375
−1
15
−1.025
−1
16.625
3
8
5.675
1
−3.125
−1
16
4.975
1
13.875
3
Ŷ (X1 ) = 50.775 + 4.425X1 , X̂2 (X1 ) = 3
[Y − Ŷ (X1 )] = 4.375[X2 − X̂2 (X1 )], Ŷ = 37.650 + 4.425X1 + 4.375X2
10.6.
a.
b.
d.
Ŷ = 3995.48 + .00091916X1 + 12.1205X2
e(Y
e(X2
e(Y
e(X1
i:
| X1 ):
| X1 ):
| X2 ):
| X2 ):
1
2
−101.811 108.842
−.205
−1.205
−95.621 152.904
4, 036.66 32, 043.7
...
...
...
...
...
51
−279.061
.312
−184.865
106, 600
52
−11.1165
.414
−27.843
−12, 742.4
Ŷ (X1 ) = 4079.87 + .000935X1 , X̂2 (X1 ) = 6.96268 + .00000135X1
[Y − Ŷ (X1 )] = 12.1205[X2 − X̂2 (X1 )], Ŷ = 3995.48 + .000919X1 + 12.1205X2
10.7.
a.
e(Y
e(X3
e(Y
e(X2
e(Y
e(X1
i:
1
| X1 , X2 ):
1.671
| X1 , X2 ):
−.116
| X1 , X3 ):
.589
| X1 , X3 ): −1.077
| X2 , X3 ): −12.537
| X2 , X3 ):
11.081
2
−11.680
.193
−7.218
−4.213
−9.734
.573
10-1
...
...
...
...
...
...
...
45
−1.967
−.265
−7.365
4.134
−4.086
−1.272
46
13.179
−.232
9.808
.554
12.696
−2.316
10.8.
a.
e(Y
e(X3
e(Y
e(X2
e(Y
e(X1
i:
1
| X1 , X2 ): −.630
| X1 , X2 ): −.039
| X1 , X3 ): −2.085
| X1 , X3 ): −3.491
| X2 , X3 ): −.242
| X2 , X3 ): −2.405
2
−1.768
.179
−2.75065
−1.155
−3.259
8.596
...
...
...
...
...
...
...
80
81
−.129 −1.068
−.016
.016
.316 −.240
.905
1.950
−.259 −1.490
1.437
3.198
10.9. a&g.
i:
1
2
ti : −.041 .061
Di : .0002 .0004
3
4
5
−1.361 1.386 −.367
.1804 .1863 .0077
i:
7
8
ti : −.767 .505
Di : .0323 .0144
9
.465
.0122
i:
13
14
ti : −1.140 −2.103
Di : .1318
.3634
10
11
−.604 1.823
.0204 .1498
15
1.490
.2107
6
−.665
.0245
12
.978
.0510
16
.246
.0068
t(.9969; 12) = 3.31. If |ti | ≤ 3.31 conclude no outliers, otherwise outliers. Conclude no outliers.
c.
2p/n = 2(3)/16 = .375, no
d.
X0new =
h
1 10 3
i
(X0 X)−1
1.2375 −.0875 −.1875
.0125
0
=
0
.0625
hnew,new = .175, no extrapolation
e.
Case 14:
f.
DF F IT S
−1.174
DF BET AS
b0
b1
b2
D
.839 −.808 −.602 .3634
.68%
10.10. a&f.
i:
1
2
...
ti : −.224 1.225 . . .
Di : .0003 .0245 . . .
51
52
−1.375 .453
.0531 .0015
t(.9995192; 47) = 3.523. If |ti | ≤ 3.523 conclude no outliers, otherwise outliers.
Conclude no outliers.
b.
2p/n = 2(4)/52 = .15385. Cases 3, 5, 16, 21, 22, 43, 44, and 48.
c.
X0new = [ 1 300, 000 7.2 0 ]
10-2
(X0 X)−1
1.8628 −.0000 −.1806
.0473
.0000 −.0000 −.0000
=
.0260 −.0078
.1911
hnew, new = .01829, no extrapolation
d.
Case
Case
Case
Case
Case
Case
Case
Case
e.
DF F IT S
−.554
.055
.562
−.147
.459
−.651
.386
.397
16:
22:
43:
48:
10:
32:
38:
40:
b0
−.2477
.0304
−.3578
.0450
.3641
.4095
−.0996
.0738
DF BET AS
b1
b2
−.0598
.3248
−.0253 −.0107
.1338
.3262
−.0938
.0090
−.1044 −.3142
.0913 −.5708
−.0827
.2084
−.2121
.0933
b3
−.4521
.0446
.3566
−.1022
−.0633
.1652
−.1270
−.1110
D
.0769
.0008
.0792
.0055
.0494
.0998
.0346
.0365
Case 16: .161%, case 22: .015%, case 43: .164%, case 48: .042%,
case 10: .167%, case 32: .227%, case 38: .152%, case 40: .157%.
10.11. a&f.
i:
1
ti : .0116
Di : .000003
2
...
−.9332 . . .
.015699 . . .
45
−.5671
.006400
46
1.0449
.024702
t(.998913; 41) = 3.27. If |ti | ≤ 3.27 conclude no outliers, otherwise outliers.
Conclude no outliers.
b.
2p/n = 2(4)/46 = .1739. Cases 9, 28, and 39.
c.
X0new = [ 1 30 58 2.0 ]
(X0 X)−1
3.24771 .00922 −.06793 −.06730
.00046 −.00032 −.00466
=
.00239 −.01771
.49826
hnew, new = .3267, extrapolation
d.
Case 11:
Case 17:
Case 27:
e.
10.12. a&f.
DF F IT S
b0
.5688
.0991
.6657 −.4491
−.6087 −.0172
DF BET AS
b1
b2
b3
D
−.3631 −.1900 .3900 .0766
−.4711
.4432 .0893 .1051
.4172 −.2499 .1614 .0867
Case 11: 1.10%, case 17: 1.32% , case 27: 1.12%.
i:
1
ti : −.9399
Di :
.0117
2 ...
−1.3926 . . .
.0308 . . .
80
81
−1.9232 −.8095
.0858
.0046
t(.999938; 75) = 4.05. If |ti | ≤ 4.05 conclude no outliers, otherwise outliers.
Conclude no outliers.
10-3
b.
2p/n = 2(5)/81 = .1235. Cases 3, 8, 53, 61, and 65.
c.
X0new = [ 1 10 12 .05 350, 000 ]
(X0 X)−1
.2584 −.0003 −.0251 −.2508
.0000
.0004
−.0002
.0031
−.0000
.0031
.0219 −.0000
=
.9139 −.0000
.0000
hnew, new = .0402, no extrapolation
d.
Case 61:
Case 8:
Case 3:
Case 53:
Case 6:
Case 62:
e.
DF F IT S
.639
.116
−.284
.525
−.873
.690
b0
−.0554
−.0142
−.2318
−.0196
.1951
.2758
DF BET AS
b1
b2
b3
.0242 −.0076
.5457
−.0072
.0030
.0955
−.1553
.2364
.1008
−.0240 −.0243
.4180
−.5649 −.1767 −.6182
−.3335 −.2595
.0627
b4
.0038
.0126
−.0115
.0490
.4482
.4051
D
.082
.003
.016
.055
.137
.088
Case 61: .300%, case 8: .054%, case 3: .192%, case 53: .235%,
case 6: .556%, case 62: .417%.
10.13. a.
Ŷ = 1.02325 + .96569X1 + .62916X2 + .67603X3
b.
H0 : β1 = β2 = β3 = 0, Ha : not all βk = 0 (k = 1, 2, 3). M SR = 127.553,
M SE = 3.33216, F ∗ = 127.553/3.33216 = 38.28, F (.95; 3, 10) = 2.84. If F ∗ ≤
2.84 conclude H0 , otherwise Ha . Conclude Ha .
c.
H0 : βk = 0, Ha : βk 6= 0. t(.975; 10) = 2.021. If |t∗ | ≤ 2.021 conclude H0 ,
otherwise Ha .
b1 = .96569, s{b1 } = .70922, t∗1 = 1.362, conclude H0
b2 = .62916, s{b2 } = .77830, t∗2 = .808, conclude H0
b3 = .67603, s{b3 } = .35574, t∗3 = 1.900, conclude H0
No
d.
X1
1 .9744 .3760
X2
1
.4099
X3
1
10.14. a.
(V IF )1 = (1 − .950179)−1 = 20.072
(V IF )2 = (1 − .951728)−1 = 20.716
(V IF )3 = (1 − .178964)−1 = 1.218
b.
Ŷ = 3.16277 + 1.65806X1
10.15. b.
(V IF )1 = 1, (V IF )2 = 1
10.16. b.
(V IF )1 = 1.0086, (V IF )2 = 1.0196, (V IF )3 = 1.0144.
10-4
10.17. b.
(V IF )1 = 1.6323, (V IF )2 = 2.0032, (V IF )3 = 2.0091
10.18. b.
(V IF )1 = (1 − .193775)−1 = 1.2403
(V IF )2 = (1 − .393287)−1 = 1.6482
(V IF )3 = (1 − .244458)−1 = 1.3236
(V IF )4 = (1 − .292147)−1 = 1.4127
10.19.a,b&c.
i:
1
2
ei : 3.308
5.494
e(Y | X1 ):
4.35
8.16
e(X3 | X1 ):
.57
1.46
e(Y | X3 ): −2.63 −8.66
e(X1 | X3 ): −17.03 −40.61
Exp. value: 4.744
5.590
3
−2.525
.53
1.68
−.53
5.73
−2.724
···
23
24
· · · −.202
.172
· · · −14.52 27.41
· · · −7.85
14.94
···
1.15
−5.14
···
3.87
−15.24
···
.522
1.050
25
2.035
−3.12
−2.83
−1.72
−10.79
2.724
H0 : normal, Ha : not normal. r = .983. If r ≥ .939 conclude H0 , otherwise
Ha . Conclude H0 .
d and e.
i:
hii :
ti :
1
2
.071 .214
.645 1.191
3
···
23
24
.046 · · ·
.074 .171
−.484 · · · −.039 .035
25
.060
.392
t(.999; 21) = 3.53. If |ti | ≤ 3.53 conclude no outliers, otherwise outliers.
Conclude no outliers.
f.
Case
7
16
18
g.
DF F IT S
−.340
.603
1.000
DF BET AS
b0
b1
b3
D
−.240 −.151 .303 .040
−.069
.152 .051 .092
−.464
.878 .115 .308
(V IF )1 = (V IF )2 = 1.034
10.20.a&b. Ŷ = 134.400 − 2.133X1 − 1.699X2 + .0333X1 X2
i:
1
2
3
···
17
ei : 17.740 4.161 −4.616 · · · −7.061
Exp. value: 14.564 5.789 −2.788 · · · −7.467
18
−.582
.000
r = .963
c.
(V IF )1 = 5.431, (V IF )2 = 11.640, (V IF )3 = 22.474
d&e.
i:
1
2
hii : .276 .083
ti : 2.210 .399
3
.539
−.629
···
···
···
17
.144
−.709
10-5
18
19
.139
.077
−.057 −.802
19
−8.256
−11.609
t(.9987; 14) = 3.65. If |ti | ≤ 3.65 conclude no outliers, otherwise outliers. Conclude no outliers.
f.
Case DF F IT S
3
−.680
7
1.749
8
−4.780
15
.175
10.21. a.
DF BET AS
b1
b2
.592
.433
−1.278 −.742
1.187 3.162
−.035
.077
b0
−.652
1.454
−1.547
−.016
b3
D
−.482 .121
.848 .459
−3.286 4.991
−.016 .008
(V IF )1 = 1.305, (V IF )2 = 1.300, (V IF )3 = 1.024
b&c.
i:
1
ei : 13.181
e(Y | X2 , X3 ): 26.368
e(X1 | X2 , X3 ): −.330
e(Y | X1 , X3 ): 18.734
e(X2 | X1 , X3 ): −7.537
e(Y | X1 , X2 ): 11.542
e(X3 | X1 , X2 ): −2.111
Exp. value: 11.926
10.22. a.
2
3
−4.042
3.060
−2.038 −31.111
−.050
.856
−17.470
8.212
18.226 −6.993
−7.756
15.022
−4.784
15.406
−4.812
1.886
···
32
· · · 14.335
···
6.310
···
.201
· · · 12.566
···
2.401
···
6.732
· · · −9.793
· · · 17.591
33
1.396
5.845
.111
−8.099
12.888
−15.100
−21.247
−.940
Ŷ 0 = −2.0427 − .7120X10 + .7474X20 + .7574X30 , where Y 0 = loge Y , X10 = loge X1 ,
X20 = loge (140 − X2 ), X30 = loge X3
b.
i:
ei :
Exp. value:
c.
1
−.0036
.0238
2
.0005
.0358
3
···
31
32
−.0316 · · · −.1487 .2863
−.0481 · · · −.1703 .2601
33
.1208
.1164
(V IF )1 = 1.339, (V IF )2 = 1.330, (V IF )3 = 1.016
d&e.
i:
1
hii :
.101
ti : −.024
2
3
.092
.176
.003 −.218
···
···
···
31
32
33
.058 .069 .149
−.975 1.983 .829
t(.9985; 28) = 3.25. If |ti | ≤ 3.25 conclude no outliers, otherwise outliers. Conclude no outliers.
f.
Case
28
29
DF F IT S
.739
−.719
b0
.530
−.197
DF BET AS
b1
b2
−.151 −.577
−.310 −.133
b3
−.187
.420
D
.120
.109
10.23. Ŷ = Xb, Ŷ(i) = Xb(i) . From (10.33a), we obtain:
Di =
(Xb − Xb(i) )0 (Xb − Xb(i) )
(b − b(i) )0 X0 X(b − b(i) )
=
pM SE
pM SE
10-6
10.24. H = X(X0 X)−1 X0 = XX−1 (X0 )−1 X0 = II = I
hii = 1, Ŷi = Yi
"
10.25. M SE(i) =
(n − p)SSE
e2i
−
n−p
1 − hii
#
÷ (n − p − 1)
from (10.25)
Substitution into (10.24a) yields (10.26):
"
ti = ei
n−p−1
SSE(1 − hii ) − e2i
# 1/2
10.26. From Exercise 5.31, σ 2 {Ŷ} = Hσ 2 or σ 2 {Ŷi } = σ 2 hii ; hence
P 2
P
σ {Ŷi } = σ 2 hii = σ 2 p by (10.27)
10.27.a&b.
i:
ei :
Exp. value:
57
−.086
−.077
58
59
−.064 −.004
−.057 −.005
···
111
112
· · · −.049 .086
· · · −.043 .084
113
.019
.010
H0 : normal, Ha : not normal. r = .990. If r ≥ .980 conclude H0 , otherwise Ha .
Conclude H0 .
c.
(V IF )3 = 1.065, (V IF )6 = 1.041, (V IF )10 = 1.045
X3
1 .161 −.172
X6
1
.086
X10
1
Note: Variable numbers for predictor variables are those in the data set description.
d&e.
i:
57
hii :
.055
ti : −1.617
58
59
···
111
112
.055
.069 · · ·
.042 .288
−1.201 −.079 · · · −.911 1.889
113
.067
.348
t(.9999; 52) = 4.00. If |ti | ≤ 4.00 conclude no outliers, otherwise outliers. Conclude no outliers.
f.
Case DF F IT S
62
.116
75
.254
87
−.411
106
.757
112
1.200
b0
.010
.222
.025
−.437
−.464
DF BET AS
b3
b6
.007 −.061
−.242
.069
−.031
.022
.626 −.400
.372
.051
10.28.a&b.
10-7
b10
.094
−.066
−.291
−.032
1.132
D
.003
.016
.040
.138
.343
i:
ei :
Exp. value:
2
4
6
−.794 .323 4.615
−1.011 .644 1.011
···
···
···
436
.078
.249
438
.007
.052
440
−.008
−.010
H0 : normal, Ha : not normal. r = .636. If r ≥ .982 conclude H0 , otherwise Ha .
Conclude Ha .
c.
(V IF )6 = 1.0093, (V IF )8 = 4.5906,(V IF )9 = 4.2859, (V IF )13 = 1.4728,(V IF )14 =
1.1056, (V IF )15 = 1.4357,
X6
X8
X9
X13
X14
X15
1 .174 .113 −.025 −.222 .078
1 .921 .033 −.056 .312
1
.173
.034
.145
1
.513 −.649
1
−.379
Note: Variable numbers for predictor variables are those in the data set description.
d&e.
i:
2
hii :
.514
ti : −3.182
4
6
···
.090
.095 · · ·
.926 31.797 · · ·
436
.063
.219
438
440
.041
.007
.020 −.021
t(.99989; 212) = 3.759. If |ti | ≤ 3.759 conclude no outliers, otherwise outliers.
Conclude case 6 is an outlier.
f.
Case
2
8
48
128
206
404
6
DF F IT S
−3.27
−.60
.31
−.10
.20
−.12
10.29
DF BET AS
b0
b6
b8
−2745.72
.403
−.479
595.06 −.052
.456
134.34 −.079
.290
170.25 −.003
−.014
−399.49 −.056
.030
220.73 −.019
−.011
−8536.94
.274 −4.236
10-8
b9
−.815
−.548
−.271
.023
.0005
.018
6.678
b13
1.184
.006
.088
−.093
.157
−.001
2.729
b14
−.825
−.488
−.215
−.222
−.142
−.654
5.196
b15
1.188
−.178
.030
−.038
−.275
−.028
2.110
D
1.467
.052
.014
.001
.006
.002
2.634
Chapter 11
BUILDING THE REGRESSION
MODEL III: REMEDIAL
MEASURES
11.6. a.
b.
Ŷ = 19.4727 + 3.2689X
i:
1
ei : 5.225
2
3
4.763 -6.389
i:
7
ei : 2.838
8
9
1.032 6.418
4
5
-2.162 -3.237
10
11
-1.700 2.687
6
-5.044
12
-4.431
n1 = 6, d¯1 = 2.821, n2 = 6, d¯2 = 4.833, s = 1.572,
³
q
´
t∗BF = (2.821 − 4.833)/ 1.572 1/6 + 1/6 = −2.218,
t(.975; 10) = 2.228. If |t∗BF | ≤ 2.228 conclude error variance constant, otherwise
error variance not constant. Conclude error variance constant.
d.
ŝ = −.905 + .3226X; smallest weight = .02607, case 3; largest weight = .18556,
cases 4 and 7.
e.
Ŷ = 17.3006 + 3.4211X
f.
s{b0 }:
s{b1 }:
g.
11.7. a.
Unweighted Weighted
5.5162
4.8277
.3651
.3703
Ŷ = 17.2697 + 3.4234X
Ŷ = −5.750 + .1875X
i:
1
2
ei : -3.75 5.75
i:
7
ei : -10.50
b.
3
-13.50
4
5
6
-16.25 -9.75 7.50
8
9
10
11
26.75 14.25 -17.25 -1.75
SSR∗ = 123, 753.125, SSE = 2, 316.500,
11-1
12
18.50
2
2
XBP
= (123, 753.125/2)/(2, 316.500/12)2 = 1.66, χ2 (.90; 1) = 2.71. If XBP
≤ 2.71
conclude error variance constant, otherwise error variance not constant. Conclude
error variance constant.
d.
v̂ = −180.1 + 1.2437X
i:
1
2
weight: .01456 .00315
i:
7
weight: .00518
e.
8
.00315
3
.00518
4
.00315
5
.01456
6
.00518
9
.01456
10
.00315
11
.01456
12
.00518
Ŷ = −6.2332 + .1891X
f.
s{b0 }:
s{b1 }:
g.
11.8. b.
c.
Unweighted Weighted
16.7305
13.1672
.0538
.0506
Ŷ = −6.2335 + .1891X
Ŷ = 31.4714 + 10.8120X1 + 22.6307X2 + 1.2581X3 + 1.8523X4
i:
1
2
3
···
63
64
65
ei : -3.2892 -3.2812 -.3274 · · · 36.9093 -18.6811 -5.3643
n1 = 33, d¯1 = 2.7595, n2 = 32, d¯2 = 10.1166, s = 6.3643,
q
³
´
t∗BF = (2.7595 − 10.1166)/ 6.3643 1/33 + 1/32 = −4.659,
t(.995; 63) = 2.656. If |t∗BF | ≤ 2.656 conclude error variance constant, otherwise
error variance not constant. Conclude error variance not constant.
e.
ŝ = 2.420 + .3996X3 + .2695X4
i:
1
2
3
weight: .0563 .0777 .0015
f.
···
···
63
.1484
64
.0941
65
.0035
Ŷ = 29.4255 + 10.8996X1 + 26.6849X2 + 1.4253X3 + 1.7239X4
g.
s{b0 }:
s{b1 }:
s{b2 }:
s{b3 }:
s{b4 }:
h.
11.9. b.
c.
Unweighted Weighted
2.8691
1.3617
3.2183
1.4918
3.4846
1.6686
.2273
.2002
.2276
.3206
Ŷ = 29.0832 + 11.0075X1 + 26.8142X2 + 1.4904X3 + 1.6922X4
c = .06
Yˆ∗ = .410X1∗ + .354X2∗ + .165X3∗
Ŷ = 21.7290 + 1.7380X1 + .1727X2 + .6929X3
11.10. a.
Ŷ = 3.32429 + 3.76811X1 + 5.07959X2
11-2
d.
c = .07
e.
Ŷ = 6.06599 + 3.84335X1 + 4.68044X2
11.11. a.
Ŷ = 1.88602 + 15.1094X (47 cases)
Ŷ = −.58016 + 15.0352X (45 cases)
i:
1
2
...
46
47
ui : -1.4123 -.2711 . . . 4.6045 10.3331
smallest weights: .13016 (case 47), .29217 (case 46)
b.
c.
Ŷ = −.9235 + 15.13552X
d.
2nd iteration: Ŷ = −1.535 + 15.425X
3rd iteration: Ŷ = −1.678 + 15.444X
smallest weights: .12629 (case 47), .27858 (case 46)
11.12. a.
Ŷ = −193.924 + 5.248X
b.
smallest weight: .5582 (case 2)
c.
Ŷ = −236.259 + 5.838X
d.
2nd iteration: Ŷ = −241.577 + 5.914X
3rd iteration: Ŷ = −242.606 + 5.928X
smallest weight: .5025 (case 2)
P
11.13. Qw =
1
(Yi − β0 − β1 Xi )2
kXi
X 1
∂Qw
= −2
(Yi − β0 − β1 Xi )
∂β0
kXi
X1
∂Qw
= −2
(Yi − β0 − β1 Xi )
∂β1
k
Setting the derivatives equal to zero, simplifying, and substituting the least squares
estimators b0 and b1 yields:
P Yi
Xi
P
− b0
X 1
Xi
Yi − nb0 − b1
P
11.14. bw1 =
since
P
− nb1 = 0
P
Xi = 0
wi (Xi − X̄w )(Yi − Ȳw )
P
wi (Xi − X̄w )2
wi (Xi − X̄w )(Yi − Ȳw ) =
=
and
P
wi (Xi − X̄w )2 =
P
P
P
wi Xi Yi − (
P
P
wi Xi Yi −
P
wi Xi2 − (
wi )X̄w2
11-3
wi )X̄w Ȳw
wi Xi
P
P
wi
w i Yi
=
11.17.
P
P
wi Xi2
−
(
wi Xi )2
wi
P
X1 /.3
0
0
0
0
X2 /.3
0
0
0
0
X3 /.3
0
0
0
0
X4 /.3
bw = (X0 WX)−1 X0 WY
11.18.
σ 2 {bw } = [(X0 WX)−1 X0 W](kW−1 )[(X0 WX)−1 X0 W]0
= k(X0 WX)−1 X0 WW−1 [W0 X(X0 WX)−1 ]
(since (X0 WX)−1 is symmetric)
= k(X0 WX)−1 X0 IWX(X0 WX)−1
(since W is symmetric)
= k(X0 WX)−1
11.19. E{bR − β}2 = E{bR − E{bR } + E{bR } − β}2
= E{bR − E{bR }}2 + 2E{bR − E{bR }}[E{bR } − β] + E{E{bR } − β}2
= σ 2 {bR } + 0 + [E{bR } − β]2
11.20. a.
b.
11.21. a.
c.
38.3666
38.5822, yes.
Xh : 10 20
E{Yh }: 120 220
30
320
40
420
50
520
Ordinary least squares: E{b1 } = 10, σ 2 {b1 } = .024
Weighted least squares: E{b1 } = 10, σ 2 {b1 } = .01975
11.22. a.
c
.000
.005
.010
.020
.030
.040
.050
(V IF )1
1.6323
1.6000
1.5687
1.5089
1.4527
1.3997
1.3497
(V IF )2
2.0032
1.9506
1.9002
1.8054
1.7181
1.6374
1.5626
(V IF )3
2.0091
1.9561
1.9054
1.8101
1.7222
1.6411
1.5659
R2
.68219
.68218
.68215
.68204
.68185
.68160
.68129
11-4
c
.000
.005
.010
.020
.030
.040
.050
11.23. a.
bR
1
−.5907
−.5868
−.5831
−.5758
−.5687
−.5619
−.5553
bR
2
−.1106
−.1123
−.1140
−.1171
−.1200
−.1228
−.1253
bR
3
−.2339
−.2338
−.2337
−.2334
−.2331
−.2329
−.2326
Ŷ = 62.4054 + 1.5511X1 + .5102X2 + .1019X3 − .1441X4
b.
c
.000
.002
.004
.006
.008
.020
.040
.060
.080
.010
(V IF )1
38.496
9.844
5.592
4.183
3.530
2.456
1.967
1.674
1.455
1.284
c
.000
.002
.004
.006
.008
.020
.040
.060
.080
.100
11.24. a.
b.
c.
bR
1
.6065
.5524
.5351
.5257
.5193
.4975
.4751
.4577
.4429
.4300
(V IF )2
254.423
51.695
21.903
12.253
7.957
2.108
.986
.715
.591
.516
bR
2
.5277
.3909
.3519
.3337
.3233
.3033
.2986
.2986
.2992
.2998
(V IF )3 (V IF )4
R2
46.868 282.513 .9824
11.346
57.092 .9823
6.089
23.971 .9822
4.359
13.248 .9822
3.566
8.478 .9821
2.323
2.015 .9819
1.833
.820 .9813
1.560
.560 .9805
1.360
.454 .9794
1.204
.396 .9783
bR
3
.0434
−.0159
−.0343
−.0439
−.0502
−.0694
−.0864
−.0984
−.1079
−.1157
bR
4
−.1603
−.3043
−.3452
−.3641
−.3748
−.3942
−.3958
−.3920
−.3873
−.3824
Ŷ = 12.2138 − 0.1462X1 + .2893X2 + 1.4277X3 + 0.0000X4
P
|Yi − Ŷi | = 64.8315
66.9736, yes.
11.25. a.
Ŷ = 50.3840 − .7620x1 − .5300x2 − .2929x21
11.26. a.
t(.975; 10) = 2.228, bw1 = 3.4211, s{bw1 } = .3703,
3.4211 ± 2.228(.3703), 2.5961 ≤ β1 ≤ 4.2461
11.27. a.
t(.95; 10) = 1.8125, bw1 = .18911, s{bw1 } = .05056,
.18911 ± 1.8125(.05056), .0975 ≤ β1 ≤ .2808
11-5
11.28. a.
b.
Ŷ = 38.64062 + .33143x − .09107x2 , R2 = .9474
X̄ = 47.5, b1 = .331429, b11 = −.091071,
X̂max = 47.5 − [.5(.331429)]/(−.091071) = 49.3196
Ŷh = 38.640625 + .331429(X̂max − 47.5) − .091071(X̂max − 47.5)2 = 38.942
11.29 a.
First split point at X = 57, SSE = 5108.14
b.
Second split point at X = 66, SSE = 4148.78
c.
Third split point at X = 47, SSE = 3511.66
11.30 a.
First split point at X1 = 37, SSE = 6753.62
b.
Second split point at X1 = 47, SSE = 5276.25
c.
Third split point at X1 = 30, SSE = 3948.85
d.
Fourth split point at X2 = 49, for the region defined by X1 < 30. SSE = 3563.79
11-6
Chapter 12
AUTOCORRELATION IN TIME
SERIES DATA
12.1.
a.
t: 1
2
3
4
εt : 3.5 2.8 3.1 3.1
εt−1 : 3.0 3.5 2.8 3.1
5
6
7
.8 −1.1 −.9
3.1
.8 −1.1
8
−1.2
−.9
9
−1.0
−1.2
10
−1.1
−1.0
b.
t:
ut :
εt−1 :
1
2
3
.5 −.7 .3
3.0 3.5 2.8
4
5
6
7
0 −2.3 −1.9
.2
3.1
3.1
.8 −1.1
8
−.3
−.9
9
.2
−1.2
10
−.1
−1.0
12.2. Yes.
12.5.
(1) H0 : ρ = 0, Ha : ρ =
6 0. dL = 1.12, dU = 1.45. If D > 1.45 and 4 − D > 1.45,
conclude H0 , if D < 1.12 or 4 − D < 1.12 conclude Ha , otherwise the test is
inconclusive.
(2) H0 : ρ = 0, Ha : ρ < 0. dL = 1.32, dU = 1.66. If 4 − D > 1.66 conclude H0 , if
4 − D < 1.32 conclude Ha , otherwise the test is inconclusive.
(3) H0 : ρ = 0, Ha : ρ > 0. dL = 1.12, dU = 1.45. If D > 1.45 conclude H0 , if
D < 1.12 conclude Ha , otherwise the test is inconclusive.
12.6. H0 : ρ = 0, Ha : ρ > 0. D = 2.4015, dL = 1.29, dU = 1.38. If D > 1.38 conclude H0 , if
D < 1.29 conclude Ha , otherwise the test is inconclusive. Conclude H0 .
12.7
H0 : ρ = 0, Ha : ρ > 0. D = 2.2984, dL = 1.51, dU = 1.59. If D > 1.59 conclude H0 , if
D < 1.51 conclude Ha , otherwise the test is inconclusive. Conclude H0 .
12.8. H0 : ρ = 0, Ha : ρ > 0. D = 2.652, dL = .83, dU = 1.52. If D > 1.52 conclude H0 , if
D < .83 conclude Ha , otherwise the test is inconclusive. Conclude H0 .
12.9.
a. Ŷ = −7.7385 + 53.9533X, s{b0 } = 7.1746, s{b1 } = 3.5197
t:
et :
1
−.0737
2
3
−.0709 .5240
4
.5835
12-1
5
.2612
6
−.5714
7
−1.9127
8
−.8276
t:
et :
9
−.6714
10
11
.9352 1.803
12
13
.4947 .9435
14
.3156
15
16
−.6714 −1.0611
c. H0 : ρ = 0, Ha : ρ > 0. D = .857, dL = 1.10, dU = 1.37. If D > 1.37 conclude H0 ,
if D < 1.10 conclude Ha , otherwise the test is inconclusive. Conclude Ha .
12.10.
a. r = .5784, 2(1 − .5784) = .8432, D = .857
b. b00 = −.69434, b01 = 50.93322
Ŷ 0 = −.69434 + 50.93322X 0
s{b00 } = 3.75590, s{b01 } = 4.34890
c. H0 : ρ = 0, Ha : ρ > 0. D = 1.476, dL = 1.08, dU = 1.36. If D > 1.36 conclude
H0 , if D < 1.08 conclude Ha , otherwise the test is inconclusive. Conclude H0 .
d. Ŷ = −1.64692 + 50.93322X
s{b0 } = 8.90868, s{b1 } = 4.34890
f. F17 = −1.64692 + 50.93322(2.210) + .5784(−.6595) = 110.534, s{pred} = .9508,
t(.975; 13) = 2.160, 110.534 ± 2.160(.9508), 108.48 ≤ Y17(new) ≤ 112.59
g. t(.975; 13) = 2.160, 50.93322 ± 2.160(4.349), 41.539 ≤ β1 ≤ 60.327.
12.11.
a.
ρ:
.1
SSE: 11.5073
.2
10.4819
.3
.4
9.6665 9.0616
.5
8.6710
ρ:
.6
.7
.8
.9
1.0
SSE: 8.5032 8.5718 8.8932 9.4811 10.3408
ρ = .6
b. Ŷ 0 = −.5574 + 50.8065X 0 , s{b00 } = 3.5967, s{b01 } = 4.3871
c. H0 : ρ = 0, Ha : ρ > 0. D = 1.499, dL = 1.08, dU = 1.36. If D > 1.36 conclude
H0 , if D < 1.08 conclude Ha , otherwise test is inconclusive. Conclude H0 .
d. Ŷ = −1.3935 + 50.8065X, s{b0 } = 8.9918, s{b1 } = 4.3871
f. F17 = −1.3935+50.8065(2.210)+.6(−.6405) = 110.505, s{pred} = .9467, t(.975; 13) =
2.160, 110.505 ± 2.160(.9467), 108.46 ≤ Y17(new) ≤ 112.55
12.12.
a. b1 = 49.80564, s{b1 } = 4.77891
b. H0 : ρ = 0, Ha : ρ 6= 0. D = 1.75 (based on regression with intercept term),
dL = 1.08, dU = 1.36. If D > 1.36 and 4 − D > 1.36 conclude H0 , if D < 1.08 or
4 − D < 1.08 conclude Ha , otherwise the test is inconclusive. Conclude H0 .
c. Ŷ = .71172 + 49.80564X, s{b1 } = 4.77891
e. F17 = .71172 + 49.80564(2.210) − .5938 = 110.188, s{pred} = .9078, t(.975; 14) =
2.145, 110.188 ± 2.145(.9078), 108.24 ≤ Y17(new) ≤ 112.14
f. t(.975; 14) = 2.145, 49.80564 ± 2.145(4.77891), 39.555 ≤ β1 ≤ 60.056
12.13.
a. Ŷ = 93.6865 + 50.8801X, s{b0 } = .8229, s{b1 } = .2634
t:
et :
1
−1.5552
2
3
−.2471 −.1526
4
5
−.2078 .3349
12-2
6
.6431
7
.2557
t:
et :
8
9
.5610 −.4949
10
−.6824
t:
et :
15
−.0299
17
.8066
16
.5671
11
.0747
12
13
−.0817 −.2336
18
19
.1203 .5750
14
−1.0425
20
.7294
c. H0 : ρ = 0, Ha : ρ > 0. D = .974, dL = .95, dU = 1.15. If D > 1.15 conclude H0 , if
D < .95 conclude Ha , otherwise the test is inconclusive. The test is inconclusive.
12.14.
a. r = .3319, 2(1 − .3319) = 1.3362, D = .974
b. b00 = 63.3840, b01 = 50.5470
Ŷ 0 = 63.3840 + 50.5470X 0
s{b00 } = .5592, s{b01 } = .2622
c. H0 : ρ = 0, Ha : ρ > 0. D = 1.76, dL = .93, dU = 1.13. If D > 1.13 conclude H0 ,
if D < .93 conclude Ha , otherwise the test is inconclusive. Conclude H0 .
d. Ŷ = 94.8720 + 50.5470X
s{b0 } = .8370, s{b1 } = .2622
f. F21 = 94.8720 + 50.5470(3.625) + .3319(.7490) = 278.3535, s{pred} = .4743,
t(.995; 17) = 2.898, 278.3535 ± 2.898(.4743), 276.98 ≤ Y21(new) ≤ 279.73
g. t(.995; 17) = 2.898, 50.5470 ± 2.898(.2622), 49.787 ≤ β1 ≤ 51.307
12.15.
a.
ρ:
.1
.2
.3
.4
.5
SSE: 4.0450 3.7414 3.5511 3.4685 3.4889
ρ:
.6
.7
.8
.9
1.0
SSE: 3.6126 3.8511 4.2292 4.7772 5.5140
ρ = .4
b. Ŷ 0 = 57.04056 + 50.49249X 0 , s{b00 } = .53287, s{b01 } = .27697
c. H0 : ρ = 0, Ha : ρ > 0. D = 1.905, dL = .93, dU = 1.13. If D > 1.13 conclude H0 ,
if D < .93 conclude Ha , otherwise test is inconclusive. Conclude H0 .
d. Ŷ = 95.0676 + 50.49249X, s{b0 } = .88812, s{b1 } = .27697
f. F21 = 95.0676+50.49249(3.625)+.4(.7506) = 278.403, s{pred} = .4703, t(.995; 17) =
2.898, 278.403 ± 2.898(.4703), 277.04 ≤ Y21(new) ≤ 279.77
g. t(.995; 17) = 2.898, 50.49249 ± 2.898(.27697), 49.690 ≤ β1 ≤ 51.295
12.16.
a. b01 = 50.16414, s{b01 } = .42496, Ŷ 0 = 50.16414X 0
b. H0 : ρ = 0, Ha : ρ 6= 0. D = 2.425 (based on regression with intercept term),
dL = .93, dU = 1.13. If D > 1.13 and 4 − D > 1.13 conclude H0 , if D < .93 or
4 − D < .93 conclude Ha , otherwise test is inconclusive. Conclude H0 .
c. Ŷ = 95.88984 + 50.16414X, s{b1 } = .42496
e. F21 = 95.88984+50.16414(3.625)+1.116 = 278.851, s{pred} = .5787, t(.995; 18) =
2.878, 278.851 ± 2.878(.5787), 277.19 ≤ Y21(new) ≤ 280.52
f. t(.995; 18) = 2.878, 50.16416 ± 2.878(.42496), 48.941 ≤ β1 ≤ 51.387
12-3
12.17.
a. Positive
b. Ŷ = −1.43484 + .17616X, s{b0 } = .24196, s{b1 } = .0016322
c.
t:
et :
1
−.0307
2
3
−.0664 .0180
t:
et :
8
−.0613
9
10
−.0969 −.1517
t:
et :
15
16
.1844 .1054
17
.0289
4
.1593
5
.0428
11
−.1501
18
.0422
6
.0429
12
−.0754
19
−.0439
7
.0582
13
−.0249
14
.1043
20
−.0852
d. H0 : ρ = 0, Ha : ρ > 0. D = .663, dL = .95, dU = 1.15. If D > 1.15 conclude H0 ,
if D < .95 conclude Ha , otherwise the test is inconclusive. Conclude Ha .
12.18.
a. r = .67296, 2(1 − .67296) = .65408, D = .663
b. Ŷ = −.29235 + .17261X 0 , s{b00 } = .17709, s{b01 } = .00351.
c. H0 : ρ = 0, Ha : ρ > 0. D = 1.364, dL = .93, dU = 1.13. If D > 1.13 conclude H0 ,
if D < .93 conclude Ha , otherwise test is inconclusive. Conclude H0 .
d. Ŷ = −.89390 + .17261X, s{b0 } = .54149, s{b1 } = .00351
f. F21 = −.89390 + .17261(181.0) + .67296(−.015405) = 30.338, s{pred} = .09155,
t(.95; 17) = 1.740, 30.338 ± 1.740(.09155), 30.179 ≤ Y21(new) ≤ 30.497.
g. t(.95; 17) = 1.740, .17261 ± 1.740(.00351), .1665 ≤ β1 ≤ .1787.
12.19.
a.
ρ:
.1
SSE: .1492
ρ:
SSE:
.6
.09275
.2
.3
.1318 .1176
.7
.08978
.4
.1064
.5
.09817
.8
.9
1.0
.08857 .08855 .09433
ρ = .9
b. Ŷ 0 = .04644 + .16484X 0 , s{b00 } = .11230, s{b01 } = .006538
c. H0 : ρ = 0, Ha : ρ > 0. D = 1.453, dL = .93, dU = 1.13. If D > 1.13 conclude H0 ,
if D < .93 conclude Ha , otherwise test is inconclusive. Conclude H0 .
d. Ŷ = .4644 + .16484X, s{b0 } = 1.1230, s{b1 } = .006538.
f. F21 = .4644+.16484(181.0)+.9(−.03688) = 30.267, s{pred} = .09545, t(.95; 17) =
1.740, 30.267 ± 1.740(.09545), 30.101 ≤ Y21(new) ≤ 30.433
g. t(.95; 17) = 1.740, .16484 ± 1.740(.006538), .1535 ≤ β1 ≤ .1762.
12.20.
a. b01 = .16883, s{b01 } = .0055426, Ŷ 0 = .16883X 0
b. H0 : ρ = 0, Ha : ρ > 0. D = 1.480 (based on regression with intercept term),
dL = .93, dU = 1.13. If D > 1.13 conclude H0 , if D < .93 conclude Ha , otherwise
test is inconclusive. Conclude H0 .
c. Ŷ = −.35222 + .16883X, s{b1 } = .0055426
e. F21 = −.35222 + .16883(181.0) + .0942 = 30.300, s{pred} = .0907, t(.95; 18) =
1.734, 30.300 ± 1.734(.0907), 30.143 ≤ Y21(new) ≤ 30.457
12-4
f. t(.95; 18) = 1.734, .16883 ± 1.734(.0055426), .1592 ≤ β1 ≤ .1784
12.22.
σ{εt , εt−2 } = E{εt εt−2 }
= E{[ut + ρut−1 + ρ2 ut−2 + ρ3 ut−3 + · · ·]
×[ut−2 + ρut−3 + ρ2 ut−4 + · · ·]}
= E{[(ut + ρut−1 ) + ρ2 (ut−2 + ρut−3 + · · ·)]
×[ut−2 + ρut−3 + ρ2 ut−4 + · · ·]}
= E{(ut + ρut−1 )(ut−2 + ρut−3 + ρ2 ut−4 + · · ·)}
+E{ρ2 (ut−2 + ρut−3 + ρ2 ut−4 + · · ·)2 }
= ρ2 E{ut−2 + ρut−3 + ρ2 ut−4 + · · ·}2 = ρ2 E{ε2t−2 }
!
Ã
σ2
2 2
2 2
2
= ρ σ {εt−2 } = ρ σ {εt } = ρ
1 − ρ2
12.23.
a. E{Y } = 100 − .35X
t:
1
2
3
4
5
6
Yt : 67.2058 61.5825 58.8570 67.2065 68.9889 73.4943
t:
7
Yt : 74.8076
8
9
10
66.7686 62.9622 61.3573
Ŷ = 96.08317 − .30839X
b.
t:
1
Yt : 65.7640
2
3
4
5
6
60.2590 57.7580 66.6920 69.7650 74.2510
t:
7
Yt : 74.9610
8
9
10
67.1840 62.9510 61.5300
Ŷ = 98.94338 − .34023X
c.
t:
1
Yt : 64.0819
2
3
4
5
6
60.9017 56.9518 67.4257 70.5170 74.0641
t:
7
Yt : 74.7411
8
9
10
67.7152 62.2754 62.2122
Ŷ = 99.45434 − .34576X
εt − εt−1 :
t:
1
ρ = .6: −.1972
ρ = 0: −1.6390
ρ = −.7: −3.3211
t:
6
ρ = .6: 1.0054
ρ = 0: .9860
ρ = −.7: .0471
2
−.3733
−.2550
2.0698
3
−.9750
−.7510
−2.1999
7
8
−.4367 .7110
−1.0400 .9730
−1.0730 1.7241
12-5
4
5
−2.1510 .0324
−1.5660 1.3230
−.0261 1.3413
9
−.3064
−.7330
−1.9398
10
.1451
.3290
1.6868
d.
ρ
.6
0
-.7
Σ(εt − εt−1 )2
7.579
11.164
32.687
12.24. Yt0 = Yt − ρYt−1
= β0 + β1 Xt1 + β2 Xt2 + εt − ρ(β0 + β1 Xt−1,1 + β2 Xt−1,2 + εt−1 )
= β0 (1 − ρ) + β1 (Xt1 − ρXt−1,1 ) + β2 (Xt2 − ρXt−1,2 ) + (εt − ρεt−1 )
Since εt − ρεt−1 = ut , we have:
0
0
+ ut
+ β20 Xt2
Yt0 = β00 + β10 Xt1
0
where β00 = β0 (1 − ρ), β10 = β1 , β20 = β2 , Xt1
= Xt1 − ρXt−1,1 , and
0
Xt2
= Xt2 − ρXt−1,2
12.25.
a. Yt0 = Yt − ρ1 Yt−1 − ρ2 Yt−2
Xt0 = Xt − ρ1 Xt−1 − ρ2 Xt−2
b. By regressing the residuals et against the two independent variables et−1 and et−2
with no intercept term in the regression model and obtaining the two regression
coefficients. The answer to Exercise 6.23a provides the explicit formulas, with Y ,
X1 , and X2 replaced by et , et−1 , and et−2 , respectively.
c. By minimizing SSE =
P
(Yt0 − b00 − b01 Xt0 )2 with respect to ρ1 and ρ2 .
12.26. Yn+1 = β0 + β1 Xn+1 + ρ1 εn + ρ2 εn−1 + un+1
since εn+1 = ρ1 εn + ρ2 εn−1 + un+1 . Therefore:
Fn+1 = Ŷn+1 + r1 en + r2 en−1
where r1 and r2 are point estimates of ρ1 and ρ2 , respectively, obtained by either the
Cochrane-Orcutt procedure or the Hildreth-Lu procedure.
12.27. c. E{b1 } = 24 even in presence of positive autocorrelation.
12-6
Chapter 13
INTRODUCTION TO NONLINEAR
REGRESSION AND NEURAL
NETWORKS
13.1.
a. Intrinsically linear
loge f (X, γ) = γ0 + γ1 X
b. Nonlinear
c. Nonlinear
13.2.
a. Intrinsically linear
loge f (X, γ) = γ0 + γ1 loge X
b. Intrinsically linear
loge f (X, γ) = loge γ0 + γ1 loge X1 + γ2 loge X2
c. Nonlinear
13.3. b. 300, 3.7323
13.4. b. 49, 2.2774
13.5.
(0)
(0)
(0)
a. b0 = −.5072512, b1 = −0.0006934571, g0 = 0, g1 = .0006934571, g2 = .6021485
b. g0 = .04823, g1 = .00112, g2 = .71341
13.6.
a. Ŷ = .04823 + .71341exp(−.00112X)
i:
1
Ŷi : .61877
ei : .03123
Exp. value: .04125
City A
2
3
.50451
.34006
−.04451 −.00006
−.04125 −.00180
i:
6
Ŷi : .12458
ei : .02542
Exp. value: .02989
7
8
.07320
.05640
−.01320 −.01640
−.01777 −.02304
13-1
4
.23488
.02512
.02304
5
.16760
.00240
.00180
City B
10
11
.50451
.34006
−.00451 −.04006
−.00545 −.02989
i:
9
Ŷi : .61877
ei : .01123
Exp. value: .01327
i:
14
Ŷi :
.12458
ei : −.00458
Exp. value: −.00923
12
.23488
.00512
.00545
13
.16760
.02240
.01777
15
16
.07320
.05640
.00680 −.00640
.00923 −.01327
13.7. H0 : E{Y } = γ0 + γ2 exp(−γ1 X), Ha : E{Y } 6= γ0 + γ2 exp(−γ1 X).
SSP E = .00290, SSE = .00707, M SP E = .00290/8 = .0003625,
M SLF = (.00707 − .00290)/5 = .000834, F ∗ = .000834/.0003625 = 2.30069,
F (.99; 5, 8) = 6.6318. If F ∗ ≤ 6.6318 conclude H0 , otherwise Ha . Conclude H0 .
13.8. s{g0 } = .01456, s{g1 } = .000092, s{g2 } = .02277, z(.9833) = 2.128
13.9.
.04823 ± 2.128(.01456)
.01725 ≤ γ0 ≤ .07921
.00112 ± 2.128(.000092)
.00092 ≤ γ1 ≤ .00132
.71341 ± 2.128(.02277)
.66496 ≤ γ2 ≤ .76186
a. g0 = .04948, g1 = .00112, g2 = .71341, g3 = −.00250
b. z(.975) = 1.96, s{g3 } = .01211, −.00250 ± 1.96(.01211), −.02624 ≤ γ3 ≤ .02124,
yes, no.
13.10.
(0)
(0)
a. b0 = .03376, b1 = .454, g0 = 29.6209, g1 = 13.4479
b. g0 = 28.13705, g1 = 12.57445
13.11.
a. Ŷ = 28.13705X/(12.57445 + X)
b.
i:
Ŷi :
ei :
Exp. value:
1
2.0728
.0272
−.1076
2
2.9987
−.4987
−.5513
3
3.8611
1.0389
.9447
i:
Ŷi :
ei :
Exp. value:
7
9.0890
.5110
.4390
8
10.5123
−.3123
−.3442
9
11.3486
.0514
.0356
i:
Ŷi :
ei :
Exp. value:
13
16.3726
.6274
.6983
14
17.2755
−.4755
−.4390
15
18.7209
−.1209
−.1817
c. No
13-2
4
5.4198
.0802
.1076
5
6.7905
.2095
.2597
10
12.4641
.0359
−.0356
16
19.8267
−.1267
−.2597
6
8.0051
.3949
.3442
11
14.0268
−.9268
−.9447
17
20.7001
.5999
.5513
12
15.3060
−.7060
−.6983
18
21.4074
.1926
.1817
13.12. s{g0 } = .72798, s{g1 } = .76305, z(.975) = 1.960
(1) 28.13705 ± 1.960(.72798), 26.7102 ≤ γ0 ≤ 29.5639
(2) H0 : γ1 = 20, Ha : γ1 6= 20. z ∗ = (12.57445 − 20)/.76305 = −9.731.
If |z ∗ | ≤ 1.960 conclude H0 , otherwise Ha . Conclude Ha .
13.13. g0 = 100.3401, g1 = 6.4802, g2 = 4.8155
13.14.
a. Ŷ = 100.3401 − 100.3401/[1 + (X/4.8155)6.4802 ]
b.
i:
Ŷi :
ei :
Expected Val.:
1
.0038
.4962
.3928
2
.3366
1.9634
1.6354
3
4.4654
−1.0654
−1.0519
4
11.2653
.2347
−.1947
i:
Ŷi :
ei :
Expected Val.:
8
39.3272
.2728
.1947
9
39.3272
−1.4272
−1.3183
10
56.2506
−1.5506
−1.6354
i:
Ŷi :
ei :
Expected Val.:
15
80.8876
−.2876
−.3928
16
87.7742
1.4258
1.3183
17
92.1765
2.6235
2.7520
5
11.2653
−.3653
−.5981
11
56.2506
.5494
.5981
18
96.7340
−.5340
−.8155
6
23.1829
.8171
.8155
12
70.5308
.2692
.0000
7
23.1829
2.1171
2.0516
13
70.5308
−2.1308
−2.0516
14
80.8876
1.2124
1.0519
19
98.6263
−2.2263
−2.7520
13.15. H0 : E{Y } = γ0 − γ0 /[1 + (X/γ2 )γ1 ], Ha : E{Y } 6= γ0 − γ0 /[1 + (X/γ2 )γ1 ].
SSP E = 8.67999, SSE = 35.71488, M SP E = 8.67999/6 = 1.4467, M SLF =
(35.71488 − 8.67999)/10 = 2.7035, F ∗ = 2.7035/1.4467 = 1.869,F (.99; 10, 6) = 7.87. If
F ∗ ≤ 7.87 conclude H0 , otherwise Ha . Conclude H0 .
13.16. s{g0 } = 1.1741, s{g1 } = .1943, s{g2 } = .02802, z(.985) = 2.17
100.3401 ± 2.17(1.1741)
13.17.
97.7923 ≤ γ0 ≤ 102.8879
6.4802 ± 2.17(.1943)
6.0586 ≤ γ1 ≤
6.9018
4.8155 ± 2.17(.02802)
4.7547 ≤ γ2 ≤
4.8763
(0)
(0)
(0)
a. b0 = .98187, b1 = .51485, b2 = .29845, g0 = 9.5911, g1 = .51485, g2 = .29845
b. g0 = 10.0797, g1 = .49871, g2 = .30199
13.18.
a. Ŷ = 10.0797X1.49871 X2.30199
b.
i:
Ŷi :
ei :
Exp.val:
1
10.0797
1.9203
1.4685
2
31.7801
.2199
.2880
3
100.1987
2.8013
2.7817
13-3
4
20.2039
−.2039
−.2880
5
63.7005
−2.7005
−2.7817
6
200.8399
−2.8399
−3.5476
i:
Ŷi :
ei :
Exp.val:
7
40.4970
−2.4970
−2.0992
8
127.6823
5.3177
5.6437
9
402.5668
3.4332
3.5476
10
10.0797
−2.0797
−.8696
i:
Ŷi :
ei :
Exp.val:
13
20.2039
−6.2039
−5.6437
14
63.7005
−7.7005
−7.6346
15
200.8399
4.1601
4.4559
16
40.4970
2.5030
2.0992
11
31.7801
6.2199
7.6346
12
100.1987
−2.1987
−1.4685
17
127.6823
.3177
.8696
18
402.5668
−4.5668
−4.4559
13.19. H0 : E{Y } = γ0 X1γ1 X2γ2 , Ha : E{Y } 6= γ0 X1γ1 X2γ2 . F (.95; 6, 9) = 3.37, SSP E = 150.5,
SSE = 263.443, SSLF = 112.943, F ∗ = [112.943/(15 − 9)] ÷ (150.5/9) = 1.126. If
F ∗ ≤ 3.37 conclude H0 , otherwise Ha . Conclude H0 .
13.20.
a. H0 : γ1 = γ2 , Ha : γ1 6= γ2 . F (.95; 1, 15) = 4.54, SSP E = 263.443, SSE =
9, 331.62, M SP E = 263.443/15 = 17.563, M SLF = (9, 331.62 − 263.443)/1 =
9, 068.177, F ∗ = 9, 068.177/17.563 = 516.327. If F ∗ ≤ 4.54 conclude H0 , otherwise
Ha . Conclude Ha .
b. s{g1 } = .00781, s{g2 } = .00485, z(.9875) = 2.24
.49871 ± 2.24(.00781)
.4812 ≤ γ1 ≤ .5162
.30199 ± 2.24(.00485)
.2911 ≤ γ2 ≤ .3129
c. γ1 6= γ2
13.21.
13.22.
P
a. Q = {Yi − [γ0 + γ2 exp(−γ1 Xi )]}2
X
∂Q
= −2 [Yi − γ0 − γ2 exp(−γ1 Xi )]
∂γ0
X
∂Q
= 2 [Yi − γ0 − γ2 exp(−γ1 Xi )][γ2 Xi exp(−γ1 Xi )]
∂γ1
X
∂Q
= −2 [Yi − γ0 − γ2 exp(−γ1 Xi )][exp(−γ1 Xi )]
∂γ2
Setting each derivative equal to zero, simplifying, and substituting the least squares
estimators g0 , g1 , and g2 yields:
P
P
Yi − ng0 − g2 exp(−g1 Xi ) = 0
P
P
P
g2 Yi Xi exp(−g1 Xi ) − g0 g2 Xi exp(−g1 Xi ) − g22 Xi exp(−2g1 Xi ) = 0
P
P
P
Yi exp(−g1 Xi ) − g0 exp(−g1 Xi ) − g2 exp(−2g1 Xi ) = 0
¾
½
1
1 X
2
exp
−
[Y
−
γ
−
γ
exp(−γ
X
)]
b. L(γ, σ 2 ) =
i
0
2
1 i
(2πσ 2 )n/2
2σ 2
a. Q =
P
Ã
γ0 Xi
Yi −
γ1 + Xi
!2
Ã
X
γ0 X i
∂Q
Yi −
= −2
∂γ0
γ1 + Xi
Ã
X
γ0 Xi
∂Q
Yi −
=2
∂γ1
γ1 + X i
!Ã
!"
Xi
γ1 + Xi
!
γ0 Xi
(γ1 + Xi )2
13-4
#
Setting the derivatives equal to zero, simplifying, and substituting the least squares
estimators g0 and g1 yields:
P
g0
Ã
X
Xi
Yi Xi
− g0
g1 + Xi
g1 + Xi
P
!2
=0
"
#
X
Xi2
Yi Xi
2
−
g
=0
0
(g1 + Xi )2
(g1 + Xi )3
Ã
1
1 X
γ0 Xi
−
b. L(γ, σ 2 ) =
exp
Yi −
2
n/2
2
(2πσ )
2σ
γ1 + X i
13.23.
!2
P
γ1 γ2 2
a. Q = (Yi − γ0 Xi1
Xi2 )
X
∂Q
γ1 γ2
γ1 γ2
= −2
(Yi − γ0 Xi1
Xi2 ) (Xi1
Xi2 )
∂γ0
X
∂Q
γ1 γ2
γ1 γ2
= −2
(Yi − γ0 Xi1
Xi2 ) (γ0 Xi1
Xi2 loge Xi1 )
∂γ1
X
∂Q
γ1 γ2
γ1 γ2
= −2
(Yi − γ0 Xi1
Xi2 ) (γ0 Xi1
Xi2 loge Xi2 )
∂γ2
Setting the derivatives equal to zero, simplifying, and substituting the least squares
estimators g0 , g1 , and g2 yields:
P
g0
g1 g2
Yi Xi1
Xi2 − g0
P
P
P
2g1 2g2
Xi1
Xi2 = 0
g1 g2
Yi Xi1
Xi2 loge Xi1 − g02
P
P
2g1 2g2
Xi1
Xi2 loge Xi1 = 0
g1 g2
2g1 2g2
Yi Xi1
Xi2 loge Xi2 − g02 Xi1
Xi2 loge Xi2 = 0
·
¸
1
1 X
γ1 γ2 2
exp
−
(Y
−
γ
X
X
)
b. L(γ, σ 2 ) =
i
0 i1
i2
(2πσ 2 )n/2
2σ 2
g0
(
13.24.
)
γ0
a. E{Y } = E γ0 −
+ε
1 + (X/γ2 )γ1
"
#
µ
γ0
(X/γ2 )γ1
A
= γ0 −
=
γ
= γ0
0
γ
γ
1
1
1 + (X/γ2 )
1 + (X/γ2 )
1+A
¶
since (X/γ2 )γ1 = exp[γ1 (loge X − loge γ2 )].
b. E{Y 0 } = (1/γ0 )E{Y 0 } = A/(1 + A); hence:
A
E{Y 0 }
= 1 + A = A = exp(β0 + β1 X 0 )
A
1 − E{Y 0 }
1−
1+A
!
Ã
!
Ã
0
Y
Y
= loge
, X 0 = loge X
c. loge
1−Y0
γ0 − Y
d. Since β0 = −γ1 loge γ2 or
0 /γ1 ) and γ1 = β1 , starting values are
³ γ2 = exp(−β
´
(0)
(0)
(0)
g1 = b1 and g2 = exp −b0 /g1 .
13.25.
13-5
(5, 5)
(5, 15)
(5, 25)
(5, 35)
(5, 45)
(5, 55)
(5, 65)
(15, 5)
(15, 15)
(15, 25)
(15, 35)
(15, 45)
(15, 55)
(15, 65)
(25, 5)
(25, 15)
(25, 25)
(25, 35)
(25, 45)
(25, 55)
(25, 65)
(35, 5)
(35, 15)
(35, 25)
(35, 35)
(35,
(35,
(35,
(45,
(45,
(45,
(45,
(45,
(45,
(45,
(55,
(55,
(55,
(55,
(55,
(55,
(55,
(65,
(65,
(65,
(65,
(65,
(65,
(65,
1,908.388
2,285.707
2,489.092
2,620.201
2,712.754
2,781.925
2,835.726
303.526
838.411
1,241.451
1,531.436
1,748.814
1,917.745
2,052.838
209.013
105.367
431.908
742.980
1,004.812
1,222.057
1,403.365
1,624.851
86.575
60.464
254.834
45)
55)
65)
5)
15)
25)
35)
45)
55)
65)
5)
15)
25)
35)
45)
55)
65)
5)
15)
25)
35)
45)
55)
65)
480.747
694.863
887.306
4,551.038
782.035
127.119
66.999
176.620
336.160
504.661
8,987.574
2,191.748
631.873
179.473
92.431
145.951
255.430
14,934.461
4,315.713
1,574.725
592.257
228.178
124.234
139.613
13.26.
(1, .2, .1)
(1, .2, .4)
(1, .2, .7)
(1, .5, .1)
(1, .5, .4)
(1, .5, .7)
(1, .8, .1)
(1, .8, .4)
(1, .8, .7)
(11, .2, .1)
(11, .2, .4)
(11, .2, .7)
(11, .5, .1)
(11, .5, .4)
459,935
433,916
345,157
429,284
342,964
119,656
322,547
98,262.9
728,313
348,524
153,117
494,720
124,813
201,515
(11,
(11,
(11,
(11,
(21,
(21,
(21,
(21,
(21,
(21,
(21,
(21,
(21,
13-6
.5,
.8,
.8,
.8,
.2,
.2,
.2,
.5,
.5,
.5,
.8,
.8,
.8,
.7)
.1)
.4)
.7)
.1)
.4)
.7)
.1)
.4)
.7)
.1)
.4)
.7)
12,640,200
649,132
13,211,900
238,296,000
257,136
57,435.2
3,225,660
46,639.7
2,152,210
54,335,000
4,290,060
56,967,000
903,149,000
Chapter 14
LOGISTIC REGRESSION,
POISSON REGRESSION,AND
GENERALIZED LINEAR MODELS
14.3. No
14.4.
a. E{Y } = [1 + exp(25 − .2X)]−1
b. 125
c. X = 150 :
π = .993307149, π/(1 − π) = 148.41316
X = 151 :
π = .994513701, π/(1 − π) = 181.27224
181.27224/148.41316 = 1.2214 = exp(.2)
14.5.
a. E{Y } = [1 + exp(−20 + .2X)]−1
b. 100
c. X = 125 :
π = .006692851, π/(1 − π) = .006737947
X = 126 :
π = .005486299, π/(1 − π) = .005516565
005516565/.006737947 = .81873 = exp(−.2)
14.6.
a. E{Y } = Φ(−25 + .2X)
b. 125
14.7.
a. b0 = −4.80751, b1 = .12508, π̂ = [1 + exp(4.80751 − .12508X)]−1
c. 1.133
d. .5487
e. 47.22
14.8.
a. b0 = −2.94964, b1 = .07666,
π̂ = Φ(−2.94964 + .07666X)
b. b0 = −3.56532, b1 = .08227,
π̂ = 1 − exp(− exp(−3.56532 + .08227X))
14-1
14.9.
a. b0 = −10.3089, b1 = .01892,
π̂ = [1 + exp(10.3089 − .01892X)]−1
c. 1.019
d. .5243
e. 589.65
14.10.
a. b0 = −6.37366, b1 = .01169,
π̂ = Φ(−6.37366 + .01169X)
b. b0 = −7.78587, b1 = .01344,
π̂ = 1 − exp(− exp(−7.78587 + .01344X))
14.11.
a.
j:
1
pj : .144
2
3
.206 .340
4
.592
5
6
.812 .898
b. b0 = −2.07656, b1 = .13585
π̂ = [1 + exp(2.07656 − .13585X)]−1
d. 1.1455
e. .4903
f. 23.3726
14.12. a&b.
j:
1
pj : .112
2
3
.212 .372
4
.504
5
6
.688 .788
b0 = −2.6437, b1 = .67399
π̂ = [1 + exp(2.6437 − .67399X)]−1
d. 1.962
e. .4293
f. 3.922
14.13.
a. b0 = −4.73931, b1 = .067733, b2 = .598632,
π̂ = [1 + exp(4.73931 − .067733X1 − .598632X2 )]−1
b. 1.070, 1.820
c. .6090
14.14.
a. b0 = −1.17717, b1 = .07279, b2 = −.09899, b3 = .43397
π̂ = [1 + exp(1.17717 − .07279X1 + .09899X2 − .43397X3 )]−1
b. exp(b1 ) = 1.0755, exp(b2 ) = .9058, exp(b3 ) = 1.5434
c. .0642
14.15.
a. z(.95) = 1.645, s{b1 } = .06676, exp[.12508 ± 1.645(.06676)],
14-2
1.015 ≤ exp(β1 ) ≤ 1.265
b. H0 : β1 = 0, Ha : β1 6= 0. b1 = .12508, s{b1 } = .06676, z ∗ = .12508/.06676 =
1.8736. z(.95) = 1.645, |z ∗ | ≤ 1.645, conclude H0 , otherwise conclude Ha . Conclude Ha . P -value=.0609.
c. H0 : β1 = 0, Ha : β1 6= 0. G2 = 3.99, χ2 (.90; 1) = 2.7055. If G2 ≤ 2.7055,
conclude H0 , otherwise conclude Ha . Conclude Ha . P -value=.046
14.16.
a. z(.975) = 1.960, s{b1 } = .007877, exp[.01892±1.960(.007877)], 1.0035 ≤ exp(β1 ) ≤
1.0350
b. H0 : β1 = 0, Ha : β1 6= 0. b1 = .01892, s{b1 } = .007877, z ∗ = .01892/.007877 =
2.402. z(.975) = 1.960, |z ∗ | ≤ 1.960, conclude H0 , otherwise conclude Ha . Conclude Ha . P -value= .0163.
c. H0 : β1 = 0, Ha : β1 6= 0. G2 = 8.151, χ2 (.95; 1) = 3.8415. If G2 ≤ 3.8415,
conclude H0 , otherwise conclude Ha . Conclude Ha . P -value=.004.
14.17.
a. z(.975) = 1.960, s{b1 } = .004772, .13585 ± 1.960(.004772),
.1265 ≤ β1 ≤ .1452,
1.1348 ≤ exp(β1 ) ≤ 1.1563.
b. H0 : β1 = 0, Ha : β1 6= 0. b1 = .13585, s{b1 } = .004772, z ∗ = .13585/.004772 =
28.468. z(.975) = 1.960, |z ∗ | ≤ 1.960, conclude H0 , otherwise conclude Ha .
Conclude Ha . P -value= 0+.
c. H0 : β1 = 0, Ha : β1 6= 0. G2 = 1095.99, χ2 (.95; 1) = 3.8415. If G2 ≤ 3.8415,
conclude H0 , otherwise conclude Ha . Conclude Ha . P -value= 0+.
14.18.
a. z(.995) = 2.576, s{b1 } = .03911, .67399 ± 2.576(.03911),
.5732 ≤ β1 ≤ .7747,
1.774 ≤ exp(β1 ) ≤ 2.170.
b. H0 : β1 = 0, Ha : β1 6= 0. b1 = .67399, s{b1 } = .03911, z ∗ = .67399/.03911 =
17.23. z(.995) = 2.576, |z ∗ | ≤ 2.576, conclude H0 , otherwise conclude Ha . Conclude Ha . P -value= 0+.
c. H0 : β1 = 0, Ha : β1 6= 0. G2 = 381.62, χ2 (.99; 1) = 6.6349. If G2 ≤ 6.6349,
conclude H0 , otherwise conclude Ha . Conclude Ha . P -value= 0+.
14.19.
a. z(1−.1/[2(2)]) = z(.975) = 1.960, s{b1 } = .02806, s{b2 } = .3901, exp{20[.067733±
1.960(.02806)]}, 1.29 ≤ exp(20β1 ) ≤ 11.64, exp{2[.5986 ± 1.960(.3901)]},.72 ≤
exp(2β2 ) ≤ 15.28.
b. H0 : β2 = 0, Ha : β2 6= 0. b2 = .5986, s{b2 } = .3901, z ∗ = .5986/.3901 = 1.53.
z(.975) = 1.96, |z ∗ | ≤ 1.96, conclude H0 , otherwise conclude Ha . Conclude H0 .
P -value= .125.
c. H0 : β2 = 0, Ha : β2 6= 0. G2 = 2.614, χ2 (.95; 1) = 3.8415. If G2 ≤ 3.8415,
conclude H0 , otherwise conclude Ha . Conclude H0 . P -value= .1059.
d. H0 : β3 = β4 = β5 = 0, Ha : not all βk = 0, for k = 3, 4, 5. G2 = 2.438,
χ2 (.95; 3) = 7.81. If G2 ≤ 7.81, conclude H0 , otherwise conclude Ha . Conclude
H0 . P -value= .4866.
14-3
14.20.
a. z(1−.1/[2(2)]) = z(.975) = 1.960, s{b1 } = .03036, s{b2 } = .03343, exp{30[.07279±
1.960(.03036)]}, 1.49 ≤ exp(30β1 ) ≤ 52.92, exp{25[−.09899±1.960(.03343)]},.016 ≤
exp(2β2 ) ≤ .433.
b. H0 : β3 = 0, Ha : β3 6= 0. b3 = .43397, s{b3 } = .52132, z ∗ = .43397/.52132 =
.8324. z(.975) = 1.96, |z ∗ | ≤ 1.96, conclude H0 , otherwise conclude Ha . Conclude
H0 . P -value= .405.
c. H0 : β3 = 0, Ha : β3 6= 0. G2 = .702, χ2 (.95; 1) = 3.8415. If G2 ≤ 3.8415,
conclude H0 , otherwise conclude Ha . Conclude H0 .
d. H0 : β3 = β4 = β5 = 0, Ha : not all βk = 0, for k = 3, 4, 5. G2 = 1.534,
χ2 (.95; 3) = 7.81. If G2 ≤ 7.81, conclude H0 , otherwise conclude Ha . Conclude
H0 .
14.21.
a. X1 enters in step 1;
no variables satisfy criterion for entry in step 2.
b. X22 is deleted in step 1; X11 is deleted in step 2; X12 is deleted in step 3; X2 is
deleted in step 4; X1 is retained in the model.
c. The best model according to the AICp criterion is based on X1 and X2 . AIC3 =
42.6896.
d. The best model according to the SBCp criterion is based on X1 . SBC2 = 46.2976.
14.22.
a. X1 enters in step 1; X2 enters in step 2;
no variables satisfy criterion for entry in step 3.
b. X11 is deleted in step 1; X12 is deleted in step 2; X3 is deleted in step 3; X22 is
deleted in step 4; X1 and X2 are retained in the model.
c. The best model according to the AICp criterion is based on X1 and X2 . AIC3 =
111.795.
d. The best model according to the SBCp criterion is based on X1 and X2 . SBC3 =
121.002.
14.23.
j:
Oj1 :
Ej1 :
Oj0 :
Ej0 :
1
2
3
4
5
72
103
170
296
406
71.0 99.5 164.1 327.2 394.2
428
397
330
204
94
429.0 400.5 335.9 172.9 105.8
6
449
440.0
51
60.0
H0 : E{Y } = [1 + exp(−β0 − β1 X)]−1 ,
Ha : E{Y } 6= [1 + exp(−β0 − β1 X)]−1 .
X 2 = 12.284, χ2 (.99; 4) = 13.28. If X 2 ≤ 13.28 conclude H0 , otherwise Ha . Conclude
H0 .
14.24.
14-4
j:
Oj1 :
Ej1 :
Oj0 :
Ej0 :
1
2
3
4
28
53
93
126
30.7 53.8 87.4 128.3
222
197
157
124
219.3 196.2 162.6 121.7
5
172
168.5
78
81.6
6
197
200.5
53
49.5
H0 : E{Y } = [1 + exp(−β0 − β1 X)]−1 ,
Ha : E{Y } 6= [1 + exp(−β0 − β1 X)]−1 .
X 2 = 1.452, χ2 (.99; 4) = 13.28. If X 2 ≤ 13.28 conclude H0 , otherwise Ha . Conclude
H0 .
14.25.
a.
Class j
1
2
3
0
π̂
−1.1
−.4
.6
Interval
Midpoint nj
- under −.4
−.75 10
- under .6
.10 10
- under 1.5
1.05 10
pj
.3
.6
.7
b.
i:
1
2
rSPi : −.6233 1.7905
14.26.
3
−.6233
···
···
28
.6099
29
.5754
30
−2.0347
a.
Class j
1
2
3
0
π̂
−2.80
−.70
.80
Interval
Midpoint nj
- under −.70
−1.75 9
- under .80
.05 9
- under 2.00
1.40 9
pj
.222
.556
.778
b.
i:
devi :
14.27.
1
−.6817
2
3
−.4727 −.5692
···
···
25
26
1.0433 −.8849
27
.7770
a.
Class j
1
2
3
0
π̂ Interval
Midpoint nj
−3.00 - under −1.10
−2.050 11
−1.10 - under .35
−.375 11
.35 - under 3.00
1.675 11
pj
.273
.182
.818
b.
i:
1
rSPi : −.7584
14.28.
2
−1.0080
3
.7622
···
31
32
33
· · · −.6014 1.3700 −.5532
a.
j:
1
2
Oj1 :
0
1
Ej1 : .2
.5
Oj0 : 19
19
Ej0 : 18.8 19.5
3
4
5
0
2
1
1.0 1.5 2.4
20
18
19
19.0 18.5 17.6
14-5
6
7
8
8
2
10
3.4 4.7 10.3
12
18
10
16.6 15.3 9.7
b. H0 : E{Y } = [1 + exp(−β0 − β1 X1 − β2 X2 − β3 X3 )]−1 ,
Ha : E{Y } 6= [1 + exp(−β0 − β1 X1 − β2 X2 − β3 X3 )]−1 .
X 2 = 12.116, χ2 (.95; 6) = 12.59. If X 2 ≤ 12.59, conclude H0 , otherwise conclude
Ha . Conclude H0 . P -value = .0594.
c.
i:
devi :
14.29
1
−.5460
2
3
···
−.5137 1.1526 · · ·
157
.4248
158
159
.8679 1.6745
a.
i:
1
hii : .1040
2
.1040
3
.1040
···
···
28
.0946
29
.1017
30
.1017
b.
i:
1
2
.3885 3.2058
∆devi : .6379 3.0411
Di : .0225 .1860
∆Xi2 :
14.30
3
.3885
.6379
.0225
···
···
···
···
28
4.1399
3.5071
.2162
29
.2621
.4495
.0148
30
.2621
.4495
.0148
a.
i:
1
hii : .0968
2
.1048
3
.1044
···
···
25
.0511
26
.0744
27
.0662
b.
i:
1
∆Xi2 : .2896
∆devi : .4928
Di : .0155
14.31
2
.1320
.2372
.0077
3
.1963
.3445
.0114
···
···
···
···
25
.7622
1.1274
.0205
26
.5178
.8216
.0208
27
.3774
.6287
.0134
a.
i:
1
hii : .0375
2
.0420
3
.0780
···
···
31
.0507
32
.0375
33
.0570
b.
i:
1
2
∆Xi : .5751 1.0161
∆devi2 : .9027 1.4022
Di : .0112 .0223
14.32
3
.5809
.9031
.0246
···
···
···
···
31
32
.3617 1.8769
.6087 2.1343
.0097 .0366
33
.3061
.5246
.0093
a.
i:
1
hii : .0197
2
.0186
3
.0992
···
···
157
.0760
158
.1364
159
.0273
b.
i:
1
.1340
∆devi : .2495
Di : .0007
∆Xi2 :
14.33.
2
3
···
.1775 1.4352 · · ·
.3245 1.8020 · · ·
.0008 .0395 · · ·
157
.0795
.1478
.0016
158
159
.6324 2.7200
.9578 2.6614
.0250 .0191
a. z(.95) = 1.645, π̂ 0h = .19561, s2 {b0 } = 7.05306, s2 {b1 } = .004457, s{b0 , b1 } =
−.175353, s{π̂ 0h } = .39428, .389 ≤ πh ≤ .699
14-6
b.
Cutoff Renewers Nonrenewers Total
.40
18.8
50.0
33.3
.45
25.0
50.0
36.7
.50
25.0
35.7
30.0
.55
43.8
28.6
36.7
.60
43.8
21.4
33.3
c. Cutoff = .50. Area = .70089.
14.34.
a. z(.975) = 1.960, s2 {b0 } = 19.1581, s2 {b1 } = .00006205, s{b0 , b1 } = −.034293
0
Xh
π̂ h
550 .0971
625 1.5161
0
s{π̂ h }
.4538
.7281
.312 ≤ πh ≤ .728
.522 ≤ πh ≤ .950
b.
Cutoff Able
.325
14.3
.425
14.3
.525
21.4
.625
42.9
Unable
46.2
38.5
30.8
30.8
Total
29.6
25.9
25.9
37.0
c. Cutoff = .525. Area = .79670.
14.35.
a. z(.975) = 1.960, π̂ 0h = −.04281, s2 {b0 } = .021824, s2 {b1 } = .000072174, s{b0 , b1 } =
−.0010644, s{π̂ 0h } = .0783, .451 ≤ πh ≤ .528
b.
Cutoff Purchasers Nonpurchasers Total
.15
4.81
71.54 76.36
.30
11.70
45.15 56.84
.45
23.06
23.30 46.27
.60
23.06
23.30 46.27
.75
48.85
9.64 52.49
c. Cutoff = .45 (or .60). Area = .82445.
14.36.
a. π̂ 0h = −1.3953, s2 {π̂ 0h } = .1613, s{ˆπh0 } = .4016, z(.95) = 1.645. L = −1.3953 −
1.645(.4016) = −2.05597, U = −1.3953 + 1.645(.4016) = −.73463.
L∗ = [1 + exp(2.05597)]−1 = .11345, U ∗ = [1 + exp(.73463)]−1 = .32418.
b.
Cutoff Received Not receive Total
.05
4.35
62.20 66.55
.10
13.04
39.37 52.41
.15
17.39
26.77 44.16
.20
39.13
15.75 54.88
c. Cutoff = .15. Area = .82222.
14.38.
a. b0 = 2.3529, b1 = .2638, s{b0 } = .1317, s{b1 } = .0792, µ̂ = exp(2.3529 + .2638X).
14-7
b.
i:
devi :
1
.6074
2
−.4796
3
···
−.1971 · · ·
8
.3482
9
10
.2752 .1480
c.
Xh :
0
1
2
3
Poisson: 10.5 13.7 17.8 23.2
Linear: 10.2 14.2 18.2 22.2
e. µ̂h = exp(2.3529) = 10.516
P (Y ≤ 10 | Xh = 0) =
10 (10.516)Y exp(−10.516)
P
Y =0
−5
= 2.7 × 10
Y!
+ · · · + .1235 = .5187
f. z(.975) = 1.96, .2638 ± 1.96(.0792), .1086 ≤ β1 ≤ .4190
14.39.
a. b0 = .4895, b1 = −1.0694, b2 = −.0466, b3 = .0095, b4 = .0086, s{b0 } = .3369,
s{b1 } = .1332, s{b2 } = .1200, s{b3 } = .0030, s{b4 } = .0043,
µ̂ = exp(.4895 − 1.0694X1 − .0466X2 + .0095X3 + .0086X4 )
b.
i:
devi :
1
−.4816
2
3
···
98
99
−.6328 .4857 · · · −.3452 .0488
100
−.9889
c. H0 : β2 = 0, Ha : β2 6= 0. G2 = .151, χ2 (.95; 1) = 3.84. If G2 ≤ 3.84 conclude
H0 , otherwise Ha . Conclude H0 .
d. b1 = −1.0778, s{b1 } = .1314, z(.975) = 1.96, −1.0778 ± 1.96(.1314), −1.335 ≤
β1 ≤ −.820.
"
#
exp(β0 + β1 X)
exp(−β0 − β1 X)
1
14.40. E{Y } =
=
1 + exp(β0 + β1 X) exp(−β0 − β1 X)
1 + exp(−β0 − β1 X)
= [1 + exp(−β0 − β1 X)]−1
14.41. Formula (14.26) holds for given observations Y1 , Y2 ,... Yn . Assembling all terms with
a given X value, Xj , we obtain:
y.j (β0 + β1 Xj ) − nj loge [1 + exp(β0 + β1 Xj )]
since
³ ´ there are nj cases with X value Xj , of which y.j have value Yi = 1. There are
nj
y.j ways of obtaining these y.j 1s out of nj , all of which are equally likely. Hence,
in the log-likelihood function of the y.j , we must add loge
given Xj :
loge
³
nj
y.j
´
³
nj
y.j
+ y.j (β0 + β1 Xj ) − nj loge [1 + exp(β0 + β1 Xj )]
Assembling the terms for all Xj , we obtain (14.34).
14.42. From (14.16) and (14.18), we have:
14-8
´
to the above term for
exp(π 0 )
1 + exp(π 0 )
Then:
1 + exp(π 0 ) − exp(π 0 )
1−π =
= [1 + exp(π 0 )]−1
0
1 + exp(π )
πi =
π
exp(π 0 )
=
× [1 + exp(π 0 )] = exp(π 0 )
1−π
1 + exp(π 0 )
Solving for π 0 = FL−1 (π) by taking logarithms of both sides yields the result.
14.43. From (14.26), we obtain:
n
X
∂ 2 loge L
exp(β0 + β1 Xi )
=
−
2
2
∂β0
i=1 [(1 + exp(β0 + β1 Xi )]
n
X
∂ 2 loge L
Xi2 exp(β0 + β1 Xi )
=
−
2
∂β12
i=1 [1 + exp(β0 + β1 Xi )]
n
X
∂ 2 loge L
Xi exp(β0 + β1 Xi )
=−
2
∂β0 ∂β1
i=1 [1 + exp(β0 + β1 Xi )]
Since these partial derivatives only involve the constants Xi , β0 , and β1 , the expectations of the partial derivatives are the partial derivatives themselves. Hence:
(
∂ 2 loge L
−E
∂β02
(
∂ 2 loge L
−E
∂β12
)
(
∂ 2 loge L
−E
∂β0 ∂β1
= −g00
)
)
= −g01 = −g10
= −g11
and the stated matrix reduces to (14.51).
"
14.44.
4.1762385 74.574657
74.574657 1, 568.4817
14.45. E{Y } =
"
#−1
=
1.58597 −.075406
−.075406 .0042228
#
γ0
1 + γ1 exp(γ2 X)
Consider γ2 < 0 and γ1 > 0; as X → ∞, E{Y } = π → 1 so that
"
1 = limX→∞
#
γ0
= γ0
1 + γ1 exp(γ2 X)
Therefore, letting γ2 = −β1 and γ1 = exp(−β0 ) we have:
E{Y } =
exp(β0 + β1 X)
1
=
1 + exp(−β0 − β1 X)
1 + exp(β0 + β1 X)
14.46. E{Y } = [1 + exp(−β0 − β1 X1 − β2 X2 − β3 X1 X2 )]−1
π 0 (X1 + 1) = β0 + β1 (X1 + 1) + β2 X2 + β3 (X1 + 1)X2
14-9
π 0 (X1 ) = β0 + β1 X1 + β2 X2 + β3 X1 X2
π 0 (X1 + 1) − π 0 (X1 ) = loge (odds ratio) = β1 + β3 X2
Hence the odds ratio for X1 is exp(β1 + β3 X2 ). No.
"
Ã
!#
Ã
!
Xi − γ0
14.47. 1 − πi = exp − exp
γ1
Xi − γ0
−loge (1 − πi ) = exp
γ1
Hence: loge [− loge (1 − πi )] =
where β0 = −
14.48.
γ0
1
Xi − γ0
= − + Xi = β0 + β1 Xi
γ1
γ1 γ1
1
γ0
and β1 = .
γ1
γ1
a. X1 = age
Socioeconomic
status
Upper
Middle
Lower
X2
0
1
0
Sector
1
2
X3
0
0
1
X4
0
1
b0 = .1932, b1 = .03476, b2 = −1.9092, b3 = −2.0940, b4 = .9508,
b5 = .02633, b6 = .007144, b7 = −.01721, b8 = .004107, b9 = .4145,
where X0 = (1 X1 X2 X3 X4 X1 X2 X1 X3 X1 X4 X2 X4 X3 X4 )
b. H0 : β5 = β6 = β7 = β8 = β9 = 0, Ha : not all equalities hold.
G2 = .858, χ2 (.99; 5) = 15.09. If G2 ≤ 15.09 conclude H0 , otherwise conclude
Ha . Conclude H0 . P -value = .973.
c. Retain socioeconomic status and age.
14.49.
a.
j:
1
2
3
4
5
Oj1 :
6
4
13
15
16
Ej1 : 4.6 7.3 11.7 14.8 15.7
Oj0 : 14
16
7
5
2
Ej0 : 15.4 12.7 8.3 5.2 2.3
nj :
20
20
20
20
18
H0 : E{Y } = [1 + exp(−β0 − β1 X1 − β2 X2 − β3 X3 )]−1 ,
Ha : E{Y } 6= [1 + exp(−β0 − β1 X1 − β2 X2 − β3 X3 )]−1 .
X 2 = 3.28, χ2 (.95; 3) = 7.81. If X 2 ≤ 7.81 conclude H0 , otherwise Ha . Conclude
H0 . P -value = .35.
b.
i:
devi :
1
.6107
2
3
.5905 −1.4368
···
96
· · · −.8493
14-10
97
−.7487
98
−1.0750
d&e.
i:
hii
∆Xi2 :
∆devi :
Di :
1
2
3
···
.0265 .0265 .0509 · · ·
.2106 .1956 1.9041 · · ·
.3785 .3538 2.1613 · · ·
.0014 .0013 .0255 · · ·
96
.0305
.4479
.7350
.0035
97
98
.0316 .0410
.3341 .8156
.5712 1.1891
.0027 .0087
f.
Cutoff Savings Account No Savings Account Total
.45
18.5
31.8
24.5
.50
22.2
31.8
26.5
.55
22.2
22.7
22.4
.60
29.6
22.7
26.5
Cutoff = .55. Area = .766.
14.50.
a.
Cutoff Savings Account No Savings Account Total
.55
24.5
28.9
26.5
b.
b0 :
s{b0 }:
b1 :
s{b1 }:
b2 :
s{b2 }:
b3 :
s{b3 }:
Model Building Combined
Data Set
Data Set
.3711
.3896
.5174
.3493
.03678
.03575
.01393
.00961
−1.2555
−1.1572
.5892
.4095
−1.9040
−2.0897
.5552
.3967
c. z(.9833) = 2.128, exp[.03575 ± 2.128(.00961)], 1.015 ≤ exp(β1 ) ≤ 1.058,
exp[−1.1572 ± 2.128(.4095)], .132 ≤ exp(β2 ) ≤ .751, exp [−2.0897 ± 2.128(.3967)],
.053 ≤ exp(β3 ) ≤ .288
14.51.
a. X1 = age, X2 = routine chest X-ray ratio, X3 = average daily census, X4 =
number of nurses
b0 = −8.8416, b1 = .02238, b2 = .005645, b3 = .14721, b4 = −.10475, b5 =
.0002529, b6 = −.001995, b7 = .0014375, b8 = −.000335, b9 = .0003912, b10 =
−.00000519,
where
X0 = (1 X1 X2 X3 X4 X1 X2 X1 X3 X1 X4 X2 X3 X2 X4 X3 X4 )
b. H0 : β5 = β6 = β7 = β8 = β9 = β10 = 0,
Ha : not all equalities hold. G2 = 7.45, χ2 (.95; 6) = 12.59. If G2 ≤ 12.59 conclude
H0 , otherwise Ha . Conclude H0 . P -value = .28
c. Retain age and average daily census.
14-11
d. The best subset: X3 , X6 , X10 , AIC4 = 59.6852;
The best subset: X10 , SBC2 = 66.963.
14.52.
a.
j:
1
2
Oj1 :
0
0
Ej1 :
.1
.4
Oj0 : 22
23
Ej0 : 21.9 22.6
nj :
22
23
3
4
5
1
3
13
.9
2.3 13.3
21
20
10
21.1 20.7 9.7
22
23
23
H0 : E{Y } = [1 + exp(−β0 − β1 X1 − β3 X3 )]−1 ,
Ha : E{Y } 6= [1 + exp(−β0 − β1 X1 − β3 X3 )]−1 ,
X 2 = .872, χ2 (.95; 3) = 7.81. If X 2 ≤ 7.81 conclude H0 , otherwise Ha . Conclude
H0 . P -value = .832
b
i:
devi :
1
−.3166
2
3
−.1039 −.1377
···
111
· · · −.1402
112
.1895
113
−.0784
112
.0279
.0186
.0364
.00018
113
.0041
.0031
.0062
.000004
d & e.
i:
1
hii .0168
∆Xi2 : .0523
∆devi : .1011
Di : .00030
2
.0056
.0054
.0108
.00001
3
.0074
.0096
.0190
.00002
···
···
···
···
···
111
.0076
.0100
.0197
.00003
f.
Cutoff Affiliation
.30
29.4
.40
29.4
.50
41.2
.60
52.9
No Affiliation Total
9.4
12.4
6.3
9.7
4.2
9.7
2.1
9.7
Cutoff = .40. Area = .923.
g. z(.95) = 1.645, π̂ 0h = .6622, s2 {b0 } = 17.1276, s2 {b1 } = .006744, s2 {b3 } =
.000006687, s{b0 , b1 } = −.33241, s{b0 , b3 } = .0003495, s{b1 , b3 } = −.00004731,
s{π̂ 0h } = .6193, −.35655 ≤ πh0 ≤ 1.68095, .70 ≤ πh /(1 − πh ) ≤ 5.37
14.57.
a. b1 =
33.249
−1.905
−.046
−.039
.039
−4.513
−.088
.039
−.085
, b2 =
12.387
−.838
−.016
−.028
.016
.590
.00008
−.009
−.097
, b3 =
14-12
13.505
−.562
−.095
−.010
.020
−.595
−.011
−.008
−.044
b. H0 : b13 = b23 = b33 = 0;
Ha : not all bk3 = 0, for k = 1, 2, 3. G2 = 2.34, conclude H0 .
P -value=.5049.
c. G2 = 10.3, conclude H0 . P -value=.1126.
−12.840
.585
.108
.007
−.017
d. N E = 1, N C = 0 :
b1 =
.231
.008
.009
.023
−14.087
.754
.016
.026
−.025
N E = 1, S = 0 :
b1 =
.567
.010
.010
.113
−48.020
3.014
.060
.012
N E = 1, W = 0 :
b1 = −.033
7.415
.079
−.038
.122
e&f. N E = 1, N C = 0 :
i:
1
2
3 ···
58
59
60
Devi : −1.137 .708 −1.200 · · · −.562 −1.406 .547
∆Xi2 :
1.061 .306
1.205 · · ·
.195
2.260 .347
∆devi :
1.445 .523
1.591 · · ·
.339
2.550 .485
Di :
.020 .003
.019 · · ·
.003
.085 .044
N E = 1, S = 0 :
i:
1
2
3 ···
63
64
65
Devi : −.327 .630 −1.153 · · · −.528 .696 −1.080
.058 .237
1.028 · · ·
.164 1.189
1.030
∆Xi2 :
∆devi :
.110 .415
1.413 · · ·
.293 1.400
1.404
Di : .0003 .0021
.0103 · · ·
.002 .441
.035
N E = 1, W = 0 :
14-13
i:
1
2
Devi : −.3762 −.3152
∆Xi2 :
.0936
.0852
∆devi :
.1618
.1336
Di :
.0029
.0064
14.58.
a. b1 =
−20.8100
−.0016
−.5738
−.2150
142.1400
.3998
.2751
.4516
.2236
−.0005
, b2 =
3
.0177
.0002
.0003
.0000
28.7900
−.0013
−.3878
−.1253
147.73
−.2426
.3778
.1510
−.6755
−.0004
···
···
···
···
···
42
.0000
.0000
.0000
.0000
, b3 =
b.
Row
T erm
1 X5 /X4
2
X6
3
X7
4 X10 /X5
5
X11
6
X12
7
X13
8
X14
9
X15
log L(b)
−189.129
−178.009
−166.716
−192.499
−197.042
−186.324
−168.769
−183.663
−172.189
G2 P -value
51.074
.0000
28.834
.0000
6.248
.1001
57.814
.0000
66.900
.0000
45.464
.0000
10.354
.0158
40.142
.0000
17.194
.0006
c. N E = 1, N C = 0 :
b1 =
7.8100
.0009
.1208
.9224
−107.4200
−.3536
.4683
−.6225
1.0985
−.0002
14-14
43
−.8576
1.1225
1.4135
.1903
−18.4800
−.0008
−.0354
−.1897
93.3700
.2884
−.2055
.2979
−.4803
.00008
44
.0000
.0000
.0000
.0000
−25.3800
.0015
.2399
−.0852
−172.7000
.2126
−.4522
−.4086
1.7355
.0006
b1 =
N E = 1, S = 0 :
b1 =
N E = 1, W = 0 :
−48.7700
.0054
1.9580
1.3413
−457.9000
.0917
−.6156
−.7196
−.3703
.0005
d&e. N E = 1, N C = 0 :
i:
1
2
Devi : −1.1205
.6339
∆Xi2 :
1.1715 6.5919
∆devi :
1.5536 6.7712
Di :
.0400 18.8671
3
−.0909
.0042
.0083
.000
···
101
· · · −.6718
···
.3024
···
.5006
···
.0059
102
.0464
.0011
.0022
.0000
103
−.4253
.1067
.1929
.0014
N E = 1, S = 0 :
i:
1
Devi :
.6801
2
∆Xi : 5.5413
∆devi : 5.7437
Di : 11.2465
2
−.0030
.0000
.0000
.0000
3
.0542
.0015
.0029
.0000
···
122
123
· · · −.5644 −.4215
···
.2338
.1275
···
.3797
.2123
···
.0083
.0047
124
−.4334
.1091
.1985
.0012
N E = 1, W = 0 :
i:
1
2
Devi : −.2713 .0000
.0506 .0000
∆Xi2 :
∆devi :
.0867 .0000
Di :
.0018 .0000
3
−.0011
.0000
.0000
.0000
···
···
···
···
···
14-15
87
.0713
.0027
.0052
.0000
88
−.2523
.0795
.1108
.0116
89
.0004
.0000
.0000
.0000
14.59.
a. b =
4.6970
7.5020
−.0509
−.0359
.0061
−.0710
−.0051
.3531
−.1699
b. X3 , orX4 , or X6 , or X7 , or X8 can be dropped.
c. Drop X6 , then X7 , then X3 , and then X4 , then stop.
d. The result is as follows:
Variable
Y(1)
Value
1
0
Total
Count
33
64
97
(Event)
Logistic Regression Table
Predictor
Coef
SE Coef
Constant
3.767
2.208
PSA
-0.03499
0.02208
age
-0.05548
0.03507
Capspen
-0.2668
0.1498
Response Information
Variable Value
Y(2)
1
0
Total
Count
76
21
97
Z
1.71
-1.59
-1.58
-1.78
P
0.088
0.113
0.114
0.075
(Event)
Logistic Regression Table
Predictor
Coef
SE Coef
Constant
8.704
3.595
PSA
-0.06045
0.01944
age
-0.08484
0.05253
Capspen
-0.14496
0.08098
Z
2.42
-3.11
-1.61
-1.79
P
0.015
0.002
0.106
0.073
Log-Likelihood = -32.633
Test that all slopes are zero: G = 36.086, DF = 3, P-Value = 0.000
e&f. Y (1)
i:
Devi :
∆Xi2 :
∆devi :
Di :
1
.8018
.4036
.6673
.0065
2
−.3651
.1476
.1431
.0026
···
95
96
· · · −.0233 −.0113
···
.0003
.0001
···
.0005
.0001
···
.0000
.0000
14-16
97
−.0012
.0000
.0000
.0000
Y (2)
i:
1
Devi : .1545
∆Xi2 : .0122
∆devi : .0240
Di : .00004
14.60.
a. b =
−133.0400
−123.4400
.00002
.0014
−.5250
.9014
1.1787
.5412
−.3977
.0585
.0000
.4336
2
.3077
.0492
.0960
.00033
···
95
· · · −.0297
···
.0004
···
.0009
· · · .00000
96
−.0031
.0000
.0000
.00000
97
−.0005
.0000
.0000
.00000
b. X13 , or X12 , or X8 , or X7 can be dropped.
c. Drop X12 , then X13 , then X8 , then X7 , then stop.
e&f. Y (1)
i:
1
2
Devi : −.4904 −.1791
∆Xi2 :
.1300
.0160
∆devi :
.2423
.0322
Di :
.0003
.0000
···
520
· · · −.0000
···
.0000
···
.0000
···
.0000
521
−.0143
.0000
.0002
.0000
522
−.0000
.0000
.0000
.0000
Y (2)
i:
1
Devi : .0162
∆Xi2 : .0000
∆devi : .0003
Di : .00000
14.61.
2
.2472
.0320
.0616
.00007
···
520
· · · −.2214
···
.0250
···
.0492
· · · .00003
521
−.4205
.0940
.1782
.00020
522
−.0850
.0040
.0072
.00000
a. The estimated regression coefficents and their estimated standard deviations are
as follows,
Poisson Regression
Coefficient Estimates
Label
Estimate
Constant
0.499446
Cost
0.0000149508
Age
0.00672387
Gender
0.181920
Interventions
0.0100748
Drugs
0.193237
14-17
Std. Error
0.176041
2.854645E-6
0.00296715
0.0439932
0.00380812
0.0126846
Complications
Comorbids
Duration
0.0612547
-0.000899912
0.000352919
0.0599478
0.00368517
0.000189870
b.
i:
1
2
Devi : .2813 1.7836
3
−1.0373
···
···
786
787
.6562 −1.2158
c. X3 , or X8 , or X9 or X10 can be dropped.
d. G2 = 5.262, conclude X0 , the P -value=.1536.
e. We drop X9 , then drop X8 , then stop.
14-18
788
−.0544
Chapter 15
INTRODUCTION TO THE DESIGN
OF EXPERIMENTAL AND
OBSERVATIONAL STUDIES
15.7. Panel.
15.8. a.
Mixed. Type of instruction is an experimental factor, and school is an observational
factor.
b.
Factor 1: type of instruction, two levels (standard curriculum, computer-based
curriculum).
Factor 2: school, three levels.
Randomized complete blocked design.
d.
15.9. a.
b.
Section.
Observational.
Factor: expenditures for research and development in the past three years.
Factor levels: low, moderate, and high.
c.
Cross-sectional study.
d.
Firm.
15.10. a.
b.
Mixed. Color of paper is experimental factor, and parking lot is an observational
factor.
Factor 1: color of paper, three levels (blue, green, orange).
Factor 2: supermarket parking lot, four levels.
c.
Randomized complete block design.
d.
Car.
15.11. a.
Observational.
b.
Fitness status, three levels (below average, average, above average).
c.
Cross-sectional study.
15-1
d.
15.12. a.
b.
Person
Mixed. Applicant’s eye contact is an experimental factor, and personnel officer’s
gender is an observational factor.
Factor 1: applicant’s eye contact, two levels (yes, no).
Factor 2: personnel officer’s gender (male, female).
c.
Randomized complete blocked design.
d.
Personnel officer.
15.13. a.
Mixed.
b.
Wheel.
c.
Four rubber compounds.
d.
Randomized complete blocked design.
e.
Balanced incomplete blocked design.
15.14. a.
b.
Experimental.
Factor 1: ingredient 1, with three levels (low, medium, high).
Factor 2: ingredient 2, with three levels (low, medium, high).
There are 9 factor-level combinations.
d.
Completely randomized design.
e.
Volunteer.
15.15. a.
b.
Observational.
Factor 1: treatment duration, with 2 levels (short, long).
Factor 2: weight gain, with 3 levels (slight, moderate, substantial)
c.
Cross-sectional study.
d.
Patient.
15.16. a.
Mixed.
b.
Factor: questionnaire, with 3 levels (A, B, C).
c.
Repeated measure design.
d.
Subject-time combination (i.e., the different occasions when a treatment is applied
to a subject).
15.17. a.
b.
Observational.
Factor 1: batch, with 5 levels.
Factor 2: barrel, with 4 levels (nested within batch).
c.
Nested design.
d.
Barrel.
15-2
15.18. a.
b.
Experimental.
Factor 1: poly-film thickness, with 2 levels (low, high).
Factor 2: old mixture ratio, with 2 levels (low, high).
Factor 3: operator glove type, with 2 levels (cotton, nylon).
Factor 4: underside oil coating, with 2 levels (no coating, coating).
c.
Fractional factorial design.
d.
1000 moldings in a batch.
15.19. a.
c.
15.20. a.
c.
15.23. a.
Randomized complete block design with four blocks and three treatments.
Assembler.
23 factorial design with two replicates.
Rod.
H0 : W̄ = 0, Ha : W̄ 6= 0. t∗ = −.1915/.0112 = −17.10, t(.975, 19) = 2.093. If
|t∗ | > 2.093 conclude H0 , otherwise conclude Ha . Conclude Ha . P -value = 0+.
Agree with results on page 670. They should agree.
b.
H0 : β2 = · · · = β20 = 0, Ha : not all βk (k = 2, 3, . . . , 20) equal zero. F ∗ =
[(.23586 − .023828)/(38 − 19)] ÷ [.023828/19] = 8.90, F (.95; 19, 19) = 2.17. If
F ∗ > 2.17 conclude H0 , otherwise conclude Ha . Conclude Ha . P -value = 0+.
Not of primary interest because blocking factor was used here to increase the
precision.
15.24.
Since X̄ =
P
n/2
n
= 1/2, it follows from the definition of Xi that:
(Xi − X̄) = n/2(1 − 1/2)2 + n/2(0 − 1/2)2 = n/4.
Then from (15.5a): σ 2 {b1 } = 4σ 2 /n.
15-3
15-4
Chapter 16
SINGLE-FACTOR STUDIES
16.4. b.
E{M ST R} = 9 +
25(450)
= 5, 634
2
E{M SE} = 9
16.5. b.
E{M ST R} = (2.8)2 +
100(11)
= 374.507
3
E{M SE} = 7.84
c.
16.7. b.
c.
E{M ST R} = (2.8)2 +
100(15.46)
= 523.173
3
Ŷ1j = Ȳ1. = 6.87778, Ŷ2j = Ȳ2. = 8.13333, Ŷ3j = Ȳ3. = 9.20000
eij :
i
1
2
3
j=1 j=2 j=3
.772 1.322 −.078
−1.433 −.033 1.267
−.700
.500
.900
i
1
2
j=7
−.578
.767
j=4
−1.078
.467
−1.400
j = 8 j = 9 j = 10
.822 −.878
−.233
.167
.567
j=5 j=6
.022 −.278
−.333 −.433
.400
.300
j = 11
j = 12
−1.033
.267
Yes
d.
Source
SS
df
MS
Between levels 20.125 2 10.0625
Error
15.362 24
.6401
Total
35.487 26
e.
H0 : all µi are equal (i = 1, 2, 3), Ha : not all µi are equal. F ∗ = 10.0625/.6401 =
15.720, F (.95; 2, 24) = 3.40. If F ∗ ≤ 3.40 conclude H0 , otherwise Ha . Conclude
Ha .
f.
P -value = 0+
16.8. b.
Ŷ1j = Ȳ1. = 29.4, Ŷ2j = Ȳ2. = 29.6, Ŷ3j = Ȳ3. = 28.0
16-1
c.
eij :
i j=1 j=2 j=3 j=4 j=5
1 −1.4 −3.4
1.6 −2.4
5.6
2
4.4
−.6 −4.6
1.4
−.6
3
3.0 −3.0 −1.0
1.0
0.0
d.
Source
SS
df M S
Between colors
7.60 2 3.80
Error
116.40 12 9.70
Total
124.00 14
e.
16.9. b.
c.
H0 : all µi are equal (i = 1, 2, 3), Ha : not all µi are equal. F ∗ = 3.80/9.70 = .392,
F (.90; 2, 12) = 2.81. If F ∗ ≤ 2.81 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .684
Ŷ1j = Ȳ1. = 38.0, Ŷ2j = Ȳ2. = 32.0, Ŷ3j = Ȳ3. = 24.0
eij :
i
1
2
3
j=1
−9.0
−2.0
2.0
i j=6
1
2.0
2 −1.0
3 −2.0
Yes
j=2
4.0
3.0
8.0
j=7
−8.0
−3.0
j=3 j=4 j=5
0.0
2.0
5.0
7.0 −4.0 −1.0
−3.0 −4.0 −1.0
j=8
4.0
3.0
j=9
j = 10
−3.0
1.0
d.
Source
SS
df
MS
Between treatments
672.0 2 336.00
Error
416.0 21 19.81
Total
1, 088.0 23
e.
16.10. b.
c.
H0 : all µi are equal (i = 1, 2, 3), Ha : not all µi are equal. F ∗ = 336.00/19.81 =
16.96, F (.99; 2, 21) = 5.78. If F ∗ ≤ 5.78 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
Ŷ1j = Ȳ1. = 21.500, Ŷ2j = Ȳ2. = 27.750, Ŷ3j = Ȳ3. = 21.417
eij :
i
1
2
3
i
1
2
3
j=1
1.500
.250
1.583
j=2 j=3 j=4
3.500 −.500
.500
−.750 −.750 1.250
−1.417 3.583 −.417
j=7
j=8
j=9
−1.500
1.500 −2.500
−.750
2.250
.250
−.417 −1.417 −2.417
j=5
−.500
−1.750
.583
j=6
.500
1.250
1.583
j = 10 j = 11
.500 −2.500
−.750 −1.750
−1.417
.583
16-2
j = 12
−.500
1.250
−.417
d.
Source
SS
df
Between ages 316.722 2
Error
82.167 33
Total
398.889 35
e.
16.11. b.
MS
158.361
2.490
H0 : all µi are equal (i = 1, 2, 3), Ha : not all µi are equal. F ∗ = 158.361/2.490 =
63.599, F (.99; 2, 33) = 5.31. If F ∗ ≤ 5.31 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
Ŷ1j = Ȳ1. = .0735, Ŷ2j = Ȳ2. = .1905, Ŷ3j = Ȳ3. = .4600, Ŷ4j = Ȳ4. = .3655,
Ŷ5j = Ȳ5. = .1250, Ŷ6j = Ȳ6. = .1515
c.
eij :
i
1
2
3
4
5
6
j=1
−.2135
.2695
−.2500
.1245
−.3150
−.1015
j=2
.1265
−.0805
.3200
.2145
.1450
−.2015
j=3
−.0035
−.0705
−.1400
.1545
−.0650
.1285
i
1
2
3
4
5
6
j=6
.0265
−.1305
−.1100
.1845
.0250
.1185
j=7
j=8
−.1135 −.3435
−.3105
.1395
.0800 −.2200
.0345 −.2255
−.1150
.0950
−.0715
.0185
j=9
.1965
−.1305
.0100
.1145
.1650
.2785
i
1
2
3
4
5
6
j = 11
.3165
−.1405
.0100
−.3555
.0750
.0485
j = 12
−.1435
.3395
.0900
−.0355
.1750
−.1415
j = 14 j = 15
.2065
.0165
.0995
.1695
.2500 −.0100
−.2355
.1145
.1450 −.3250
.0085 −.2115
i
1
2
3
4
5
6
j = 16
.0565
−.1505
.0200
.1745
.1150
−.0215
j = 17 j = 18
.1865 −.0035
−.0205 −.1705
−.0200
.0400
.1445
.0545
.0750
.0150
.2785
.1985
j = 13
−.0935
.2295
.1300
−.1855
−.2350
−.0515
j=4
j=5
.1065
.3065
.2795
.0495
−.0100 −.2400
−.0755 −.0955
−.0150
.1050
.3185 −.0315
j = 19
−.0835
−.0805
−.2600
.0845
.2250
−.2415
Yes
d.
16-3
j = 10
−.2835
−.2205
.1600
−.0255
.0150
−.2215
j = 20
−.2635
−.0705
.1500
−.1655
−.3050
−.1015
Source
SS
df
Between machines 2.28935
5
Error
3.53060 114
Total
5.81995 119
MS
.45787
.03097
e.
H0 : all µi are equal (i = 1, ..., 6), Ha : not all µi are equal. F ∗ = .45787/.03097 =
14.78, F (.95; 5, 114) = 2.29. If F ∗ ≤ 2.29 conclude H0 , otherwise Ha . Conclude
Ha .
f.
P -value = 0+
16.12. b.
Ŷ1j = Ȳ1. = 24.55, Ŷ2j = Ȳ2. = 22.55, Ŷ3j = Ȳ3. = 11.75, Ŷ4j = Ȳ4. = 14.80,
Ŷ5j = Ȳ5. = 30.10
c.
eij :
i
1
2
3
4
5
j=1 j=2 j=3 j=4
−.55 −.55
4.45 −4.55
−4.55 −2.55 −2.55
1.45
−1.75 −.75 −3.75
.25
.20 −1.80
3.20
1.20
2.90 −8.10 −2.10
4.90
j=5
−3.55
−.55
.25
−2.80
−1.10
i
1
2
3
4
5
j=6 j=7 j=8 j=9
.45
3.45
2.45 −1.55
6.45
.45
1.45
5.45
−1.75
2.25 −2.75 −3.75
4.20 −4.80
3.20 −3.80
−2.10 −.10
.90 −1.10
j = 10
−3.55
−3.55
−.75
2.20
−2.10
i
1
2
3
4
5
j = 11
−.55
1.45
4.25
.20
2.90
j = 12
1.45
2.45
.25
−2.80
−.10
j = 13
−1.55
−1.55
6.25
−1.80
1.90
j = 14
−.55
−2.55
2.25
−1.80
2.90
j = 15
3.45
1.45
1.25
−.80
−1.10
i
1
2
3
4
5
j = 16
−1.55
−.55
−.75
2.20
4.90
j = 17
−1.55
−3.55
2.25
1.20
1.90
j = 18
2.45
3.45
−2.75
2.20
−4.10
j = 19
1.45
−.55
−.75
−.80
−.10
j = 20
.45
−1.55
.25
1.20
−1.10
Yes
d.
Source
SS
df
Between agents 4, 430.10 4
Error
714.65 95
Total
5, 144.75 99
MS
1, 107.525
7.523
16-4
e.
H0 : all µi are equal (i = 1, ..., 5), Ha : not all µi are equal. F ∗ = 1, 107.525/7.523 =
147.22, F (.90; 4, 95) = 2.00. If F ∗ ≤ 2.00 conclude H0 , otherwise Ha . Conclude
Ha .
f.
P -value = 0+
16.15.
µ. = 80, τ1 = −15, τ2 = 0, τ3 = 15
16.16.
µ. = 7.2, τ1 = −2.1, τ2 = −.9, τ3 = .7, τ4 = 2.3
16.17. a.
µ̂. = 20.4725
b.
H0 : all τi equal zero (i = 1, ..., 5), Ha : not all τi equal zero.
No
16.18. a.
Y=
7.6
8.2
6.8
5.8
6.9
6.6
6.3
7.7
6.0
6.7
8.1
9.4
8.6
7.8
7.7
8.9
7.9
8.3
8.7
7.1
8.4
8.5
9.7
10.1
7.8
9.6
9.5
X=
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
−1
−1
−1
−1
−1
−1
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
1
−1
−1
−1
−1
−1
−1
b.
16-5
µ.
β = τ1
τ2
Xβ =
c.
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
=
µ1
µ1
µ1
µ1
µ1
µ1
µ1
µ1
µ1
µ2
µ2
µ2
µ2
µ2
µ2
µ2
µ2
µ2
µ2
µ2
µ2
µ3
µ3
µ3
µ3
µ3
µ3
Ŷ = 8.07037 − 1.19259X1 + .06296X2 , µ. defined in (16.63)
d.
Source
SS
df
MS
Regression 20.125 2 10.0625
Error
15.362 24
.6401
Total
35.487 26
e.
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero. F ∗ = 10.0625/.6401 = 15.720,
F (.95; 2, 24) = 3.40. If F ∗ ≤ 3.40 conclude H0 , otherwise Ha . Conclude Ha .
16.19. a.
16-6
Y=
28
26
31
27
35
34
29
25
31
29
31
25
27
29
28
X=
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
0
0
−1
−1
−1
−1
−1
0
0
0
0
0
1
1
1
1
1
−1
−1
−1
−1
−1
µ
.
β =
τ1
τ2
b.
Xβ =
c.
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ1
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. + τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
µ. − τ1 − τ2
=
µ1
µ1
µ1
µ1
µ1
µ2
µ2
µ2
µ2
µ2
µ3
µ3
µ3
µ3
µ3
Ŷ = 29.0 + .4X1 + .6X2 , µ. defined in (16.63)
d.
Source
SS
df M S
Regression
7.60 2 3.80
Error
116.40 12 9.70
Total
124.00 14
e.
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = 3.80/9.70 = .392, F (.90; 2, 12) = 2.81. If F ∗ ≤ 2.81 conclude H0 , otherwise
Ha . Conclude H0 .
16.20. a.
16-7
29
.
..
Y=
42
30
..
.
33
26
..
.
1
..
.
0
..
.
1
1
..
.
1
0
..
.
0
1
..
.
X=
22
b.
1
.
..
µ. + τ1
.
..
µ.
µ.
..
.
µ.
µ.
..
.
µ1
.
..
µ1
+ τ1
µ
+ τ2
2
= ..
.
µ2
+ τ2
µ
− 86 τ1 − 10
τ
3
2
6
.
.
.
µ. − 86 τ1 −
c.
µ
.
β =
τ1
τ2
1
0
1
10
8
1 −6 − 6
..
..
..
.
.
.
1 − 86 − 10
6
Xβ =
10
τ
6 2
µ3
Ŷ = 32.0 + 6.0X1 + 0.0X2 , µ. defined in (16.80a)
d.
Source
SS df
MS
Regression 672 2 336.00
Error
416 21 19.81
Total
1088 23
e.
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = 336.00/19.81 = 16.96, F (.99; 2, 21) = 5.78. If F ∗ ≤ 5.78 conclude H0 ,
otherwise Ha . Conclude H0 .
16.21. a.
Ŷ = 23.55556 − 2.05556X1 + 4.19444X2 , µ. defined in (16.63)
b.
Source
SS
df
MS
Regression 316.722 2 158.361
Error
82.167 33
2.490
Total
398.889 35
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero. F ∗ = 158.361/2.490 = 63.599,
F (.99; 2, 33) = 5.31. If F ∗ ≤ 5.31 conclude H0 , otherwise Ha . Conclude Ha .
16.22. a.
b.
Ŷ = 38X1 + 32X2 + 24X3
Ŷ = 32
16-8
c.
H0 : all µi are equal (i = 1, 2, 3), Ha : not all µi are equal. SSE(F ) = 416,
SSE(R) = 1, 088, F ∗ = (672/2) ÷ (416/21) = 16.96, F (.99; 2, 21) = 5.78. If
F ∗ ≤ 5.78 conclude H0 , otherwise Ha . Conclude Ha .
1−β ∼
= .878
16.23.
µ. = 15.5, φ = 1.58, 1 − β ∼
= .47
b. 1 − β ∼
= .18
16.24. a.
16.25.
µ. = 7.889, φ = 2.457, 1 − β ∼
= .95
16.26.
µ. = 33.917, φ = 2.214, 1 − β ∼
= .70
16.27.
µ. = 24, φ = 6.12, 1 − β > .99
16.29. a.
4 : 10 15 20 30
n : 51 23 14 7
b.
4 : 10 15 20 30
n : 39 18 11 6
16.30. a.
σ : 50 25 20
n : 34 10 7
b.
σ : 50
n : 30
25 20
9 6
16.31. a.
λ : 20 10 5
n : 10 38 150
b.
λ : 20 10 5
n : 22 85 337
16.32. a.
∆/σ = 4.5/3.0 = 1.5, n = 13
b.
∆/σ = 6.0/3.0 = 2.0, 1 − β ≥ .95
c.
n = [3.6173(3.0)/1.5]2 = 53
16.33. a.
∆/σ = 5.63/4.5 = 1.25, n = 20
·
b.
c.
16.34. a.
¸
1/2
1 20
φ=
(40.6667)
= 3.659, 1 − β ≥ .99
4.5 3
√
(2.0 n)/4.5 = 2.2302, n = 26
∆/σ = .15/.15 = 1.0, n = 22
16-9
·
b.
c.
¸
1/2
1 22
φ=
(.02968)
= 2.199, 1 − β ≥ .97
.15 6
√
(.10 n)/.15 = 3.1591, n = 23
16.35. a.
∆/σ = 3.75/3.0 = 1.25, n = 22
√
b. (1.0 n)/3.0 = 2.5997, n = 61
16.36.
L=
3 Y
2 µ
Y
i=1 j=1
1
2πσ 2
¶1/2
·
exp −
1
(Yij − µi )2
2σ 2
¸
3 X
2
1
1 X
−
=
exp
(Yij − µi )2
2σ 2 i=1 j=1
(2πσ 2 )3
loge L = −3 loge 2π − 3 loge σ 2 −
1 PP
(Yij − µi )2
2σ 2
2 P
∂(loge L)
= − 2 (Yij − µi )(−1)
∂µi
2σ j
Setting the partial derivatives equal to zero, simplifying, and substituting the
maximum likelihood estimators yields:
P
j
(Yij − µ̂i ) = 0
or:
µ̂i = Ȳi.
Yes
16.37.
t∗ =
Ȳ1. − Ȳ2.
Ȳ1. − Ȳ2.
s
=s
s{Ȳ1. − Ȳ2. }
nT
Σ(Y1j − Ȳ1. )2 + Σ(Y2j − Ȳ2. )2
n1 n2
nT − 2
F∗ = Ã
n1 (Ȳ1. − Ȳ.. )2 + n2 (Ȳ2. − Ȳ.. )2
!
Σ(Y1j − Ȳ1. )2 + Σ(Y2j − Ȳ2. )2
nT − 2
Therefore to show (t∗ )2 = F ∗ , it suffices to show:
n1 n2
(Ȳ1. − Ȳ2. )2 = n1 (Ȳ1. − Ȳ.. )2 + n2 (Ȳ2. − Ȳ.. )2
nT
Now, the right-hand side equals:
"
n1
Ã
= n1
·
= n1
=
Ã
n1 Ȳ1. + n2 Ȳ2.
Ȳ1. −
nT
"
!#2
nT Ȳ1. − n1 Ȳ1. − n2 Ȳ2.
nT
+ n2
n1 Ȳ1. + n2 Ȳ2.
Ȳ2. −
nT
Ã
!2
+ n2
Ã
!#2
nT Ȳ2. − n1 Ȳ1. − n2 Ȳ2.
nT
·
´¸2
´¸ 2
n1 ³
n2 ³
+ n2
Ȳ1. − Ȳ2.
Ȳ2. − Ȳ1.
nT
nT
´2
n1 n2 ³
Ȳ1. − Ȳ2.
nT
16-10
!2
P
16.38.
wi τi =
P
wi (µi − µ. ) =
= µ. − µ. = 0
since
16.39. a.
P
P
P
wi = 1 and µ. =
wi µi − µ.
P
wi
wi µi
Using (6.25) and substituting µ̂ for b:
µ̂ = (X0 X)−1 X0 Y
X0 X =
n1
n2
0
·
·
0
·
(X0 X)−1 =
n−1
1
n−1
2
·
·
0
X0 Y =
Y1·
Y2·
·
·
·
Yr·
16.40. a.
b.
16.41.
16.42.
µ̂ =
PP
Ȳ1·
Ȳ2·
·
·
·
Ȳr·
(Yij − Ȳi. )2 = SSE
1
PP
X0 X = nT , (X0 X)−1 =
, X0 Y =
Yij
nT
1 PP
µ̂c =
Yij = Ȳ..
nT
PP
SSE(R) =
(Yij − Ȳ.. )2 = SST O
SSE(F ) =
b.
·
n−1
r
nr
0
90, 15
Smallest P -value = .067
µ
¶
µ
¶
µ
¶
2µ2 − 1 2
2 − µ2 2
µ2 + 1 2
+
+
(µi − µ. ) = −
3
3
3
Differentiating with respect to µ2 yields:
12
6
µ2 −
9
9
Setting this derivative equal to zero and solving yields µ2 = .5.
P
2
H0 : all µi are equal (i = 1, ..., 4), Ha : not all µi are equal.
Source
SS
Between regions 13.9969
Error
187.3829
Total
201.3798
df
MS
3 4.6656
109 1.7191
112
F ∗ = 4.6656/1.7191 = 2.714, F (.95; 3, 109) = 2.688. If F ∗ ≤ 2.688 conclude H0 ,
otherwise Ha . Conclude Ha .
16.43.
H0 : all µi are equal (i = 1, ..., 4), Ha : not all µi are equal.
16-11
Source
SS
df
MS
Between ages
3.0677
3 1.02257
Error
198.3121 109 1.81938
Total
201.3798 112
F ∗ = 1.02257/1.81938 = .562, F (.90; 3, 109) = 2.135. If F ∗ ≤ 2.135 conclude H0 ,
otherwise Ha . Conclude H0 .
16.44.
H0 : all µi are equal (i = 1, ..., 4), Ha : not all µi are equal.
Source
Between regions
Error
Total
SS
df
.059181
3
.268666 436
.327847 439
MS
.019727
.000616
F ∗ = .019727/.000616 = 32.01, F (.95; 3, 436) = 2.6254. If F ∗ ≤ 2.6254 conclude
H0 , otherwise Ha . Conclude Ha .
16.45.
H0 : all µi are equal (i = 1, ..., 4), Ha : not all µi are equal.
Source
SS
df
Between treatments 1.6613 3
Error
.7850 32
Total
2.4463 35
MS
.5538
.0245
F ∗ = .5538/.0245 = 22.57, F (.95; 3, 32) = 2.9011. If F ∗ ≤ 2.9011 conclude H0 ,
otherwise Ha . Conclude Ha .
ν2
16.46. c. E{F ∗ } =
= 1.2
ν2 − 2
d. Expected proportion is .95.
e.
E{F ∗ } = 117.9; E{M ST R} = 14, 144, E{M SE} = 144
f.
φ = 8.05, expected proportion is 1 − β > .99.
16.47. a.
20, 6
b.
F∗ :
.29
.59
.97 1.06
P (F ∗ ) : 4/20 4/20 2/20 4/20
P -value = .10
c.
1.64
4/20
2.74
2/20
P {F (1, 4) ≥ 2.74} = .17
16.48. a.
F∗ :
P (F ∗ ) :
0
.1
.4
.98
2.0 3.85
8/70 18/70 12/70 14/70 8/70 6/70
7.71 19.60
2/70 2/70
H0 : µ1 = µ2 , Ha : µ1 6= µ2 . F ∗ = 7.71. P -value = P (F ∗ ≥ 7.71) = .0571. If
P -value ≥ .10 conclude H0 , otherwise Ha . Conclude Ha .
b&c.
F (.90; 1, 6) = 3.78
10/70 = .143
F (.95; 1, 6) = 5.99 F (.99; 1, 6) = 13.7
4/70 = .0571
2/70 = .0286
16-12
Chapter 17
ANALYSIS OF FACTOR LEVEL
MEANS
17.3. a.
b.
(i) and (iii) are contrasts.
(i) L̂ = Ȳ1. + 3Ȳ2. − 4Ȳ3. , s2 {L̂} =
26M SE
n
.36M SE
n
(Ȳ1. + Ȳ2. + Ȳ3. )
4M SE
(iii) L̂ =
− Ȳ4. , s2 {L̂} =
3
3n
(ii) L̂ = .3Ȳ1. + .5Ȳ2. + .1Ȳ3. + .1Ȳ4. , s2 {L̂} =
17.4. a.
q(.90; 6, 54) = 3.765, F (.90; 5, 54) = 1.96
g
T
S
B
2 2.66 3.13
t(.975; 54) = 2.00
5 2.66 3.13
t(.99; 54) = 2.40
15 2.66 3.13 t(.99667; 54) = 2.82
b.
Refer to part (a) for S and B multiples.
17.5. a.
q(.95; 5, 20) = 4.23, F (.95; 4, 20) = 2.87
g
T
2 2.99
5 2.99
10 2.99
b.
S
B
3.39 t(.9875; 20) = 2.42
3.39 t(.995; 20) = 2.845
3.39 t(.9975; 20) = 3.15
q(.95; 5, 95) = 3.94, F (.95; 4, 95) = 2.46
g
T
2 2.79
5 2.79
10 2.79
S
B
3.14 t(.9875; 95) = 2.28
3.14 t(.995; 95) = 2.63
3.14 t(.9975; 95) = 2.87
17.7. q(.99; 2, 18) = 4.07, F (.99; 1, 18) = 8.29, T = S = B = t(.995; 18) = 2.88
17.8. a.
b.
Ȳ1. = 6.878, Ȳ2. = 8.133, Ȳ3. = 9.200
s{Ȳ3. } = .327, t(.975; 24) = 2.064, 9.200 ± 2.064(.327), 8.525 ≤ µ3 ≤ 9.875
17-1
c.
D̂ = Ȳ2. − Ȳ1. = 1.255, s{D̂} = .353, t(.975; 24) = 2.064, 1.255 ± 2.064(.353),
.526 ≤ D ≤ 1.984
d.
D̂1 = Ȳ3. − Ȳ2. = 1.067, D̂2 = Ȳ3. − Ȳ1. = 2.322, D̂3 = Ȳ2. − Ȳ1. = 1.255, s{D̂1 } =
.400, s{D̂2 } = .422, s{D̂3 } = .353, q(.90; 3, 24) = 3.05, T = 2.157
1.067 ± 2.157(.400)
2.322 ± 2.157(.422)
1.255 ± 2.157(.353)
e.
.204 ≤ D1 ≤ 1.930
1.412 ≤ D2 ≤ 3.232
.494 ≤ D3 ≤ 2.016
F (.90; 2, 24) = 2.54, S = 2.25
B = t(.9833; 24) = 2.257
Yes
17.9. a.
Ȳ1. = 29.4, Ȳ2. = 29.6, Ȳ3. = 28.0, s{Ȳ1. } = s{Ȳ2. } = s{Ȳ3. } =
t(.975; 12) = 2.179
q
9.7
5
= 1.3928,
b.
s{Ȳ1. } = 1.393, t(.95; 12) = 1.782, 29.40 ± 1.782(1.393), 26.92 ≤ µ1 ≤ 31.88
c.
H0 : D = µ3 − µ2 = 0, Ha : D 6= 0. D̂ = −1.6, s{D̂} = 1.970,
t∗ = −1.6/1.970 = −.81, t(.95; 12) = 1.782.
If | t∗ | ≤ 1.782 conclude H0 , otherwise Ha . Conclude H0 . No
17.10. a.
Ȳ1. = 38.00, Ȳ2. = 32.00, Ȳ3. = 24.00
b.
M SE = 19.81, s{Ȳ2. } = 1.4075, t(.995; 21) = 2.831, 32.00±2.831(1.4075), 28.02 ≤
µ2 ≤ 35.98
c.
D̂1 = Ȳ2. − Ȳ3. = 8.00, D̂2 = Ȳ1. − Ȳ2. = 6.00, s{D̂1 } = 2.298, s{D̂2 } = 2.111,
B = t(.9875; 21) = 2.414
8.00 ± 2.414(2.298)
6.00 ± 2.414(2.111)
2.45 ≤ D1 ≤ 13.55
.90 ≤ D2 ≤ 11.10
d.
q(.95; 3, 21) = 3.57, T = 2.524, no
e.
Yes, no
f.
q(.95; 3, 21) = 3.57
Test Comparison
i
Di
1
µ1 − µ2
2
µ1 − µ3
3
µ2 − µ3
Group 1:
Group 2:
Group 3:
17.11. a.
b.
D̂i
6.00
14.00
8.00
s{D̂i }
2.111
2.404
2.298
qi∗ Conclusion
4.02
Ha
8.24
Ha
4.92
Ha
Below Average
Average
Above Average
Ȳ1. = 21.500, Ȳ2. = 27.750, Ȳ3. = 21.417
M SE = 2.490, s{Ȳ1. } = .456, t(.995; 33) = 2.733, 21.500 ± 2.733(.456), 20.254 ≤
µ1 ≤ 22.746
17-2
c.
D̂ = Ȳ3. − Ȳ1. = −.083, s{D̂} = .644, t(.995; 33) = 2.733, −.083 ± 2.733(.644),
−1.843 ≤ D ≤ 1.677
d.
H0 : 2µ2 − µ1 − µ3 = 0, Ha : 2µ2 − µ1 − µ3 6= 0. F ∗ = (12.583)2 /1.245 = 127.17,
F (.99; 1, 33) = 7.47. If F ∗ ≤ 7.47 conclude H0 , otherwise Ha . Conclude Ha .
e.
D̂1 = Ȳ3. − Ȳ1. = −.083, D̂2 = Ȳ3. − Ȳ2. = −6.333, D̂3 = Ȳ2. − Ȳ1. = 6.250,
s{D̂i } = .644 (i = 1, 2, 3), q(.90; 3, 33) = 3.01, T = 2.128
−.083 ± 2.128(.644)
−6.333 ± 2.128(.644)
6.250 ± 2.128(.644)
f.
17.12. a.
−1.453 ≤ D1 ≤ 1.287
−7.703 ≤ D2 ≤ −4.963
4.880 ≤ D3 ≤ 7.620
B = t(.9833; 33) = 2.220, no
Ȳ1. = .0735, Ȳ2. = .1905, Ȳ3. = .4600, Ȳ4. = .3655, Ȳ5. = .1250, Ȳ6. = .1515
b.
M SE = .03097, s{Ȳ1. } = .0394, t(.975; 114) = 1.981, .0735 ± 1.981(.0394),
−.005 ≤ µ1 ≤ .152
c.
D̂ = Ȳ2. − Ȳ1. = .1170, s{D̂} = .0557, t(.975; 114) = 1.981, .1170 ± 1.981(.0557),
.007 ≤ D ≤ .227
e.
D̂1 = Ȳ1. − Ȳ4. = −.2920, D̂2 = Ȳ1. − Ȳ5. = −.0515, D̂3 = Ȳ4. − Ȳ5. = .2405,
s{D̂i } = .0557 (i = 1, 2, 3), B = t(.9833; 114) = 2.178
Test Comparison
i
i
1
D1
2
D2
3
D3
f.
17.13. a.
t∗i
−5.242
−.925
4.318
Conclusion
Ha
H0
Ha
q(.90; 6, 114) = 3.71, T = 2.623, no
Ȳ1. = 24.55, Ȳ2. = 22.55, Ȳ3. = 11.75, Ȳ4. = 14.80, Ȳ5. = 30.10,
s{Ȳi. } =
q
7.52
20
= .6132, (i = 1, 2, 3, 4, 5), t(.975; 95) = 1.985
b.
Test 1: H0 :
Ha :
Test 2: H0 :
Ha :
Test 3: H0 :
Ha :
Test 4: H0 :
Ha :
Test 5: H0 :
Ha :
µ1 − µ2
µ1 − µ2
µ1 − µ3
µ1 − µ3
µ1 − µ4
µ1 − µ4
µ1 − µ5
µ1 − µ5
µ2 − µ3
µ2 − µ3
= 0 Test 6:
6= 0
= 0 Test 7:
6= 0
= 0 Test 8:
6= 0
= 0 Test 9:
6= 0
= 0 Test 10:
6= 0
H0 :
Ha :
H0 :
Ha :
H0 :
Ha :
H0 :
Ha :
H0 :
Ha :
µ2 − µ4
µ2 − µ4
µ2 − µ5
µ2 − µ5
µ3 − µ4
µ3 − µ4
µ3 − µ5
µ3 − µ5
µ4 − µ5
µ4 − µ5
=0
6= 0
=0
6= 0
=0
6= 0
=0
6= 0
=0
6= 0
D̂1 = 24.55 − 22.55 = 2.00, D̂2 = 24.55 − 11.75 = 12.80,
D̂3 = 24.55 − 14.80 = 9.75, D̂4 = 24.55 − 30.10 = −5.55,
D̂5 = 22.55 − 11.75 = 10.80, D̂6 = 22.55 − 14.80 = 7.75,
17-3
D̂7 = 22.55 − 30.10 = −7.55, D̂8 = 11.75 − 14.80 = −3.05,
D̂9 = 11.75 − 30.10 = −18.35, D̂10 = 14.80 − 30.10 = −15.30,
s{D̂i } = .8673 (i = 1, ..., 10), q(.90; 5, 95) = 3.54.
If |qi∗ | ≤ 3.54 conclude H0 , otherwise Ha .
Test
i
1
2
3
4
5
6
7
8
9
10
qi∗
Conclusion
3.26
Ho
20.87
Ha
15.90
Ha
−9.05
Ha
17.61
Ha
12.64
Ha
−12.31
Ha
−4.97
Ha
−29.92
Ha
−24.95
Ha
Group 1
Agent 3 Ȳ3. = 11.75
Group 2
Agent 4 Ȳ4. = 14.80
Group 3
Agent 1 Ȳ1. = 24.55
Agent 2 Ȳ2. = 22.55
Group 4
Agent 5 Ȳ5. = 30.10
c.
M SE = 7.523, s{Ȳ1. } = .6133, t(.95; 95) = 1.661, 24.550 ± 1.661(.6133), 23.531 ≤
µ1 ≤ 25.569
d.
D̂ = Ȳ2. − Ȳ1. = −2.000, s{D̂} = .8673, t(.95; 95) = 1.661, −2.000 ± 1.661(.8673),
−3.441 ≤ D ≤ −.559
e.
D̂1 = Ȳ1. − Ȳ3. = 12.800, D̂2 = Ȳ1. − Ȳ5. = −5.550, D̂3 = Ȳ3. − Ȳ5. = −18.350,
s{D̂i } = .8673 (i = 1, 2, 3), B = t(.9833; 95) = 2.158
12.800 ± 2.158(.8673)
−5.550 ± 2.158(.8673)
−18.350 ± 2.158(.8673)
f.
17.14. a.
10.928 ≤ D1 ≤ 14.672
−7.422 ≤ D2 ≤ −3.678
− 20.222 ≤ D3 ≤ −16.478
q(.90; 5, 95) = 3.54, T = 2.503, no
L̂ = (Ȳ1. + Ȳ2. )/2 − Ȳ3. = (6.8778 + 8.1333)/2 − 9.200 = −1.6945,
s{L̂} = .3712, t(.975; 24) = 2.064, −1.6945 ± 2.064(.3712), −2.461 ≤ L ≤ −.928
b.
L̂ = (3/9)Ȳ1. + (4/9)Ȳ2. + (2/9)Ȳ3. = 7.9518, s{L̂} = .1540, t(.975; 24) = 2.064,
7.9518 ± 2.064(.1540), 7.634 ≤ L ≤ 8.270
c.
F (.90; 2, 24) = 2.54, S = 2.254; see also part (a) and Problem 17.8.
1.067 ± 2.254(.400)
2.322 ± 2.254(.422)
.165 ≤ D1 ≤ 1.969
1.371 ≤ D2 ≤ 3.273
17-4
1.255 ± 2.254(.353)
−1.6945 ± 2.254(.3712)
17.15. a.
.459 ≤ D3 ≤ 2.051
−2.531 ≤ L1 ≤ −.858
L̂ = (Ȳ1. − Ȳ2. ) − (Ȳ2. − Ȳ3. ) = Ȳ1. − 2Ȳ2. + Ȳ3. = 38.000 − 2(32.000)
+24.000 = −2.000, s{L̂} = 3.7016, t(.995; 21) = 2.831, −2.000 ± 2.831(3.7016),
−12.479 ≤ L ≤ 8.479
b.
D̂1 = Ȳ1. − Ȳ2. = 6.000, D̂2 = Ȳ1. − Ȳ3. = 14.000, D̂3 = Ȳ2. − Ȳ3. = 8.000,
L̂1 = D̂1 − D̂3 = −2.000, s{D̂1 } = 2.1112, s{D̂2 } = 2.4037, s{D̂3 } = 2.2984,
s{L̂1 } = 3.7016, B = t(.99375; 21) = 2.732
6.000 ± 2.732(2.1112)
14.000 ± 2.732(2.4037)
8.000 ± 2.732(2.2984)
−2.000 ± 2.732(3.7016)
c.
17.16. a.
.232 ≤ D1 ≤ 11.768
7.433 ≤ D2 ≤ 20.567
1.721 ≤ D3 ≤ 14.279
−12.113 ≤ L1 ≤ 8.113
F (.95; 2, 21) = 3.47, S = 2.634, yes
L̂ = (Ȳ3. − Ȳ2. ) − (Ȳ2. − Ȳ1. ) = Ȳ3. − 2Ȳ2. + Ȳ1. = 21.4167−
2(27.7500) + 21.500 = −12.5833, s{L̂} = 1.1158, t(.995; 33) = 2.733,
−12.5833 ± 2.733(1.1158), −15.632 ≤ L ≤ −9.534
b.
D̂1 = Ȳ2. − Ȳ1. = 6.2500, D̂2 = Ȳ3. − Ȳ2. = −6.3333, D̂3 = Ȳ3. − Ȳ1. = −.0833,
L̂1 = D̂2 − D̂1 = −12.5833, s{D̂i } = .6442 (i = 1, 2, 3), s{L̂1 } = 1.1158,
F (.90; 2, 33) = 2.47, S = 2.223
6.2500 ± 2.223(.6442)
−6.3333 ± 2.223(.6442)
−.0833 ± 2.223(.6442)
−12.5833 ± 2.223(1.1158)
17.17. a.
4.818 ≤ D1 ≤ 7.682
−7.765 ≤ D2 ≤ −4.901
−1.515 ≤ D3 ≤ 1.349
−15.064 ≤ L1 ≤ −10.103
L̂ = (Ȳ1. + Ȳ2. )/2 − (Ȳ3. + Ȳ4. )/2 = (.0735 + .1905)/2 − (.4600 + .3655)/2
= −.28075, s{L̂} = .03935, t(.975; 114) = 1.981, −.28075±1.981(.03935), −.3587 ≤
L ≤ −.2028
b.
D̂1 = −.1170, D̂2 = .0945, D̂3 = −.0265, L̂1 = −.28075, L̂2 = −.00625, L̂3 =
−.2776, L̂4 = .1341, s{D̂i } = .0557 (i = 1, 2, 3), s{L̂1 } = s{L̂2 } = .03935,
s{L̂3 } = s{L̂4 } = .03408, B = t(.99286; 114) = 2.488
−.1170 ± 2.488(.0557)
.0945 ± 2.488(.0557)
−.0265 ± 2.488(.0557)
−.28075 ± 2.488(.03935)
−.00625 ± 2.488(.03935)
−.2776 ± 2.488(.03408)
.1341 ± 2.488(.03408)
17.18. a.
−.2556 ≤ D1 ≤ .0216
−.0441 ≤ D2 ≤ .2331
−.1651 ≤ D3 ≤ .1121
−.3787 ≤ L1 ≤ −.1828
−.1042 ≤ L2 ≤ .0917
−.3624 ≤ L3 ≤ −.1928
.0493 ≤ L4 ≤ .2189
L̂ = (Ȳ1. + Ȳ2. )/2 − (Ȳ3. + Ȳ4. )/2 = (24.55 + 22.55)/2 − (11.75 + 14.80)/2
17-5
= 10.275, s{L̂} = .6133, t(.95; 95) = 1.661, 10.275 ± 1.661(.6133), 9.256 ≤ L ≤
11.294
b.
D̂1 = 2.00, D̂2 = −3.05, L̂1 = −6.55, L̂2 = −16.825, L̂3 = 10.275, s{D̂i } = .8673
(i = 1, 2), s{L̂i } = .7511 (i = 1, 2), s{L̂3 } = .6133, F (.90; 4, 95) = 1.997, S =
2.826
2.00 ± 2.826(.8673)
−3.05 ± 2.826(.8673)
−6.55 ± 2.826(.7511)
−16.825 ± 2.826(.7511)
10.275 ± 2.826(.6133)
c.
17.19. a.
−.451 ≤ D1 ≤ 4.451
−5.501 ≤ D2 ≤ −.599
−8.673 ≤ L1 ≤ −4.427
−18.948 ≤ L2 ≤ −14.702
8.542 ≤ L3 ≤ 12.008
L̂ = .25Ȳ1. + .20Ȳ2. + .20Ȳ3. + .20Ȳ4. + .15Ȳ5. = 20.4725, s{L̂} = .2777, t(.95; 95) =
1.661, 20.4725 ± 1.661(.2777), 20.011 ≤ L ≤ 20.934
L1 = µ1 − µ.
L2 = µ2 − µ.
L3 = µ3 − µ.
L4 = µ4 − µ.
L5 = µ5 − µ.
L6 = µ6 − µ.
L̂1 = .0735 − .2277 = −.1542, L̂2 = .1905 − .2277 = −.0372
L̂3 = .4600 − .2277 = .2323, L̂4 = .3655 − .2277 = .1378
L̂5 = .1250 − .2277 = −.1027, L̂6 = .1515 − .2277 = −.0762
s
µ
¶
µ
¶
.03097 25
.03097 5
s{L̂i } =
+
= .0359
20
36
36
20
B = t(.99583; 114) = 2.685
−.1542 ± 2.685(.0359)
−.0372 ± 2.685(.0359)
.2323 ± 2.685(.0359)
.1378 ± 2.685(.0359)
−.1027 ± 2.685(.0359)
−.0762 ± 2.685(.0359)
.2506 ≤ L1 ≤ −.0578
−.1336 ≤ L2 ≤ .0592
.1359 ≤ L3 ≤ .3287
.0414 ≤ L4 ≤ .2342
−.1991 ≤ L5 ≤ −.0063
−.1726 ≤ L6 ≤ .0202
Conclude not all µi are equal.
17.20. a.
L1 = µ1 − µ.
L 2 = µ2 − µ.
L3 = µ3 − µ.
L4 = µ4 − µ.
L5 = µ5 − µ.
L̂1 = 24.55 − 20.75 = 3.80, L̂2 = 22.55 − 20.75 = 1.80
L̂3 = 11.75 − 20.75 = −9.00, L̂4 = 14.80 − 20.75 = −5.95
L̂5 = 30.10 − 20.75 = 9.35
s
µ
¶
µ
¶
7.5226 4
7.5226 16
s{L̂i } =
+
= .5485
20
25
25
20
B = t(.99; 95) = 2.366
17-6
3.80 ± 2.366(.5485)
1.80 ± 2.366(.5485)
−9.00 ± 2.366(.5485)
−5.95 ± 2.366(.5485)
9.35 ± 2.366(.5485)
2.502 ≤ L1 ≤ 5.098
.502 ≤ L2 ≤ 3.098
−10.298 ≤ L3 ≤ −7.702
−7.248 ≤ L4 ≤ −4.652
8.052 ≤ L5 ≤ 10.648
Conclude not all µi are equal.
17.21. a.
Yij = µi + ²ij
b.
i:
1
Ȳi. : .0800
2
3
4
5
.1800 .5333 1.1467 2.8367
c.
Source
SS
df
MS
Treatments 15.3644 4 3.8411
Error
.1574 10 .01574
Total
15.5218 14
d.
H0 : all µi are equal (i = 1, ..., 5), Ha : not all µi are equal. F ∗ = 3.8411/.01574 =
244.034, F (.975; 4, 10) = 4.47. If F ∗ ≤ 4.47 conclude H0 , otherwise Ha . Conclude
Ha .
e.
D̂1 = Ȳ1. − Ȳ2. = −.1000, D̂2 = Ȳ2. − Ȳ3. = −.3533, D̂3 = Ȳ3. − Ȳ4. = −.6134.
D4 = Ȳ4. − Ȳ5. = −1.6900, s{D̂i } = .1024 (i = 1, ..., 4), B = t(.99375; 10) = 3.038
−.1000 ± 3.038(.1024)
−.3533 ± 3.038(.1024)
−.6134 ± 3.038(.1024)
−1.6900 ± 3.038(.1024)
17.23.
n = 13
17.24.
Bonferroni, n = 24
17.25.
Bonferroni, n = 45
17.26.
Bonferroni, n = 92
17.27. a.
n = 20, 2n = 40, n = 20
b.
−.411 ≤ D1 ≤ .211
−.664 ≤ D2 ≤ −.042
−.924 ≤ D3 ≤ −.302
−2.001 ≤ D4 ≤ −1.379
(1) n = 26, n = 26, n = 26
(2) n = 18, 3n = 54, n = 18
c.
17.28. a.
b.
b(1)
Ŷ = 68.66655 − .36820X
eij :
17-7
i j=1
1 −9.106
2 −1.847
3
1.886
i j=6
1
1.894
2 −.847
3 −2.114
j=4
j=5
1.894
4.894
−3.847 −.847
−4.114 −1.114
j = 9 j = 10
−2.847
1.153
c.
H0 : E{Y } = β0 + β1 X, Ha : E{Y } 6= β0 + β1 X. SSP E = 416.0000, SSLF =
.4037, F ∗ = (.4037/1) ÷ (416.0000/21) = .020, F (.95; 1, 21) = 4.32. If F ∗ ≤ 4.32
conclude H0 , otherwise Ha . Conclude H0 .
d.
No
17.29. a.
b.
Ŷ = .18472 + .06199x + .01016x2
eij :
i
1
2
3
4
5
6
i
1
2
3
4
5
6
i
1
2
3
4
5
6
17.30.
j=2 j=3
3.894 −.106
3.153
7.153
7.886 −3.114
j=7 j=8
−8.106 3.894
−2.847 3.153
j=1
−.2310
.2393
−.2440
.1268
−.2969
−.0802
j=8
−.3610
.1093
−.2140
−.2232
.1131
.0398
j = 15
−.0010
.1393
−.0040
.1168
−.3069
−.1902
j=2
.1090
−.1107
.3260
.2168
.1631
−.1802
j=9
.1790
−.1607
.0160
.1168
.1831
.2998
j = 16
.0390
−.1807
.0260
.1768
.1331
−.0002
j=3
−.0210
−.1007
−.1340
.1568
−.0469
.1498
j = 10
−.3010
−.2507
.1660
−.0232
.0331
−.2002
j = 17
.1690
−.0507
−.0140
.1468
.0931
.2998
j=4
.0890
.2493
−.0040
−.0732
.0031
.3398
j = 11
.2990
−.1707
.0160
−.3532
.0931
.0698
j = 18
−.0210
−.2007
.0460
.0568
.0331
.2198
j=5
.2890
.0193
−.2340
−.0932
.1231
−.0102
j = 12
−.1610
.3093
.0960
−.0332
.1931
−.1202
j = 19
−.1010
−.1107
−.2540
.0868
.2431
−.2202
j=6
.0090
−.1607
−.1040
.1868
.0431
.1398
j = 13
−.1110
.1993
.1360
−.1832
−.2169
−.0302
j = 20
−.2810
−.1007
.1560
−.1632
−.2869
−.0802
j=7
−.1310
−.3407
.0860
.0368
−.0969
−.0502
j = 14
.1890
.0693
.2560
−.2332
.1631
.0298
c.
H0 : E{Y } = β0 + β1 x + β11 x2 , Ha : E{Y } 6= β0 + β1 x + β11 x2 . SSP E = 3.5306,
SSLF = .0408, F ∗ = (.0408/3) ÷ (3.5306/114) = .439, F (.99; 3, 114) = 3.96. If
F ∗ ≤ 3.96 conclude H0 , otherwise Ha . Conclude H0 .
d.
H0 : β11 = 0, Ha : β11 6= 0. s{b11 } = .00525, t∗ = .01016/.00525 = 1.935, t(.995; 117) =
2.619. If |t∗ | ≤ 2.619 conclude H0 , otherwise Ha . Conclude H0 .
With r = 2 and ni ≡ n, M SE = s2 as defined in (A.63) and
max(Ȳi. − µi ) − min(Ȳi. − µi ) = (Ȳi. − µi ) − (Ȳi0 . − µi0 ) =
17-8
= (Ȳi. − Ȳi0 . ) − (µi − µi0 ), i 6= i0 .
Thus:
q∗ =
17.31.
(Ȳi. − Ȳi0 . ) − (µi − µi0 ) √ ∗
√
= 2|t |
s/ n
Working within the probability expression, we obtain:
¯
¯
¯
¯
¯ (Ȳi. − µi ) − (Ȳi0 . − µi0 ) ¯
¯
¯ ≤ q(1 − α; r, nT − r)
q
¯
¯
¯
¯
M SE/n
³q
´
|(Ȳi. − µi ) − (Ȳi0 . − µi0 )| ≤
or
M SE/n q(1 − α; r, nT − r)
|(Ȳi. − µi ) − (Ȳi0 . − µi0 )| ≤ s{D̂}T
1
since T = √ q(1 − α; r, nT − r) and
2
s
M SE
s{D̂}
= √
n
2
−s{D̂}T ≤ (Ȳi. − Ȳi0 . ) − (µi − µi0 ) ≤ s{D̂}T
or
or
or
(Ȳi. − Ȳi0 . ) − T s{D̂} ≤ µi − µi0 ≤ (Ȳi. − Ȳi0 . ) + T s{D̂}
17.32.
When r = 2, S 2 = F (1 − α; 1, nT − 2) which by (A.50b) equals
[t(1 − α/2; nT − 2)]2 .
17.33.
2
2
σ {L̂i } = σ {Ȳi. −
r
X
Ȳh. /r}
h=1
2
= σ 2 {Ȳi. } + σ {
X
Ȳh. /r} − 2σ{Ȳi. ,
2
X
Ȳh. /r}
2
σ
1 X 2
2σ
+ 2
(σ /nh ) −
ni
r
rni
r
2
1 X
σ
σ2
2σ 2
= 2 σ 2 (1/nh ) +
+ 2 −
r
ni
r ni
rni
h6=i
=
µ
=
r
1 2X
σ2
1
2
σ
(1/n
)
+
1+ 2 −
h
2
r
ni
r
r
h6=i
=
r
1 2X
σ2
σ
(1/n
)
+
h
r2 h6=i
ni
µ
r−1
r
¶
¶2
Replacing σ 2 by the estimator M SE leads to (17.49).
17.34.
Given n1 = n3 = n and n2 = kn. Let c = kn/nT . Then n1 = n3 =
(nT − kn)/2 = nT (1 − c)/2 and n2 = cnT . Hence:
"
2
1
σ 2 {Ȳ1. − Ȳ2. } = σ 2 {Ȳ3. − Ȳ2. } = σ 2
+
nT (1 − c) cnT
Differentiating with respect to c yields:
σ2
2σ 2
(1 − c)−2 +
(−c−2 )
nT
nT
17-9
#
Setting the derivative equal to zero and solving yields c = .4142.
Hence, n2 = (.4142)nT and n1 = n3 = (.2929)nT .
(Note: This derivation treats n as a continuous variable. Since n2 must be an
even integer, appropriate rounding of the calculated sample sizes is required. For
example, if nT = 100, the calculated sample sizes are n1 = 29.29, n2 = 41.42, and
n3 = 29.29. The smallest variance is obtained for n1 = 29, n2 = 42, and n3 = 29.)
17.35.
Ȳ1. = 4.86071, Ȳ2. = 4.39375, Ȳ3. = 3.92703, Ȳ4. = 4.38125, M SE = 1.7191,
n1 = 28, n2 = 32, n3 = 37, n4 = 16,
D̂1 = Ȳ1. − Ȳ2. = .46696, D̂2 = Ȳ1. − Ȳ3. = .93368, D̂3 = Ȳ1. − Ȳ4. = .47946,
D̂4 = Ȳ2. − Ȳ3. = .46667, D̂5 = Ȳ2. − Ȳ4. = .01250, D̂6 = Ȳ3. − Ȳ4. = −.45422,
s{D̂1 } = .3393, s{D̂2 } = .3284, s{D̂3 } = .4109, s{D̂4 } = .3165, s{D̂5 } = .4015,
s{D̂6 } = .3923, q(.90; 4, 109) = 3.28, T = 2.319
.46696 ± 2.319(.3393)
.93368 ± 2.319(.3284)
.47946 ± 2.319(.4109)
.46667 ± 2.319(.3165)
.01250 ± 2.319(.4015)
−.45422 ± 2.319(.3923)
17.36.
−.320 ≤ D1 ≤ 1.254
.172 ≤ D2 ≤ 1.695
−.473 ≤ D3 ≤ 1.432
−.267 ≤ D4 ≤ 1.201
−.919 ≤ D5 ≤ .944
− 1.364 ≤ D6 ≤ .456
Ȳ1. = .04123, Ȳ2. = .05111, Ȳ3. = .07074, Ȳ4. = .06088, M SE = .000616,
n1 = 103, n2 = 108, n3 = 152, n4 = 77,
D̂1 = Ȳ1. − Ȳ2. = −.0099, s{D̂1 } = .0034
D̂2 = Ȳ1. − Ȳ3. = −.0295, s{D̂2 } = .0032
D̂3 = Ȳ1. − Ȳ4. = −.0196, s{D̂3 } = .0037
D̂4 = Ȳ2. − Ȳ3. = −.0196, s{D̂4 } = .0031
D̂5 = Ȳ2. − Ȳ4. = −.0098, s{D̂5 } = .0037
D̂6 = Ȳ3. − Ȳ4. = .0099, s{D̂6 } = .0035, q(.90; 4, 137) = 3.24, T = 2.291
−.01771 ≤ D1 ≤ −.00204, −.03677 ≤ D2 ≤ −.02225
−.02822 ≤ D3 ≤ −.01108, −.02679 ≤ D4 ≤ −.01247
−.01825 ≤ D5 ≤ −.00129, .00191 ≤ D6 ≤ .01782
17.37.
Ȳ1. = 2.4125, Ȳ2. = 2.7375, Ȳ3. = 2.4286, Ȳ4. = 2.9000, M SE = .0245,
n1 = 8, n2 = 8, n3 = 7, n4 = 13,
D̂1 = Ȳ1. − Ȳ2. = −.3250, s{D̂1 } = .0783
D̂2 = Ȳ1. − Ȳ3. = −.0161, s{D̂2 } = .0810
D̂3 = Ȳ1. − Ȳ4. = −.4875, s{D̂3 } = .0703
D̂4 = Ȳ2. − Ȳ3. = .3089, s{D̂4 } = .0810
D̂5 = Ȳ2. − Ȳ4. = −.1625, s{D̂5 } = .0703
17-10
D̂6 = Ȳ3. − Ȳ4. = −.4714, s{D̂6 } = .0734
q(.95; 4, 32) = 3.83, T = 2.708
−.5371 ≤ D1 ≤ −.1129, −.2356 ≤ D2 ≤ .2035
−.6781 ≤ D3 ≤ −.2969, .0894 ≤ D4 ≤ .5285
−.3531 ≤ D5 ≤ .0281, −.6703 ≤ D6 ≤ −.2726
17.38.
b. Expected proportion is .95.
17-11
17-12
Chapter 18
ANOVA DIAGNOSTICS AND
REMEDIAL MEASURES
18.4. a.
See Problem 16.7c.
b.
r = .992
c.
tij :
i
1
2
3
j=1
j=2
.9557 1.8377
−1.9821 −.0426
−.9568
.6768
i
1
2
j=7
−.7592
1.0009
j=3
−.1010
1.7197
1.2464
j=8
j=9
1.0945 −1.1728
−.2988
.2132
j=4
−1.4623
.6011
−2.0391
j=5
.0288
−.4277
.5395
j=6
−.3615
−.5575
.4035
j = 10
j = 11
j = 12
.7326
−1.3737
.3417
H0 : no outliers, Ha : at least one outlier. t(.999815; 23) = 4.17.
If | tij |≤ 4.17 conclude H0 , otherwise Ha . Conclude H0 .
18.5. a.
See Problem 16.8c.
b.
r = .991
d.
tij :
i
1
2
3
j=1
−.4863
1.6992
1.0849
j=2
−1.2486
−.2066
−1.0849
j=3
.5576
−1.7985
−.3456
j=4
−.8516
.4863
.3456
j=5
2.3634
−.2066
.0000
H0 : no outliers, Ha : at least one outlier. t(.99917; 11) = 4.13.
If |tij | ≤ 4.13 conclude H0 , otherwise Ha . Conclude H0 .
18.6. a.
See Problem 16.9c.
b.
r = .990
d.
tij :
18-1
i
1
2
3
j=1
j=2
−2.3926 .9589
−.4647 .7019
.4832 2.1280
j=3
.0000
1.7354
−.7301
j=4
.4714
−.9449
−.9837
i
1
2
j=7
j=8
−2.0656 .9589
−.7019 .7019
j=9
j = 10
−.7019
.2314
j=5
j=6
1.2145
.4714
−.2314 −.2314
−.2405 −.4832
H0 : no outliers, Ha : at least one outlier. t(.99979; 20) = 4.22.
If |tij | ≤ 4.22 conclude H0 , otherwise Ha . Conclude H0 .
18.7. a.
See Problem 16.10c.
b.
r = .984
d.
tij :
i j=1
1 .9927
2 .1630
3 1.0497
j=2
2.4931
−.4907
−.9360
i
1
2
3
j=8
j=9
.9927 −1.7017
1.5185
.1630
−.9360 −1.6401
j=7
−.9927
−.4907
−.2719
j=3
j=4
−.3265
.3265
−.4907
.8234
2.5645 −.2719
j=5
j=6
−.3265 .3265
−1.1646 .8234
.3811 1.0497
j = 10 j = 11
.3265 −1.7017
−.4907 −1.1646
−.9360
.3811
j = 12
−.3265
.8234
−.2719
H0 : no outliers, Ha : at least one outlier. t(.99965; 32) = 3.75.
If |tij | ≤ 3.75 conclude H0 , otherwise Ha . Conclude H0 .
18.8. a.
See Problem 16.11c.
b.
r = .992
d.
tij :
i
1
2
3
4
5
6
j=1
j=2
−1.2477
.7360
1.5815 −.4677
−1.4648
1.8864
.7243
1.2537
−1.8560
.8443
−.5901 −1.1767
j=3
−.0203
−.4095
−.8150
.9000
−.3775
.7477
j=4
.6192
1.6415
−.0580
−.4386
−.0871
1.8773
j=5
1.8045
.2874
−1.4052
−.5551
.6105
−.1829
j=6
.1538
−.7594
−.6396
1.0764
.1451
.6893
i
1
2
3
4
5
6
j=8
j=9
−2.0298 1.1472
.8121 −.7594
−1.2863
.0580
−1.3189
.6659
.5522
.9616
.1074 1.6355
j = 10
−1.6656
−1.2892
.9323
−.1480
.0871
−1.2952
j = 11
1.8651
−.8179
.0580
−2.1035
.4357
.2816
j = 12
−.8355
2.0053
.5230
−.2061
1.0204
−.8238
j = 13
−.5434
1.3427
.7565
−1.0823
−1.3754
−.2990
18-2
j=7
−.6601
−1.8287
.4648
.2003
−.6688
−.4153
j = 14
1.2063
.5784
1.4648
−1.3784
.8443
.0493
i
1
2
3
4
5
6
j = 15 j = 16
.0958
.3281
.9881 −.8765
−.0580
.1161
.6659 1.0175
−1.9168
.6688
−1.2359 −.1248
j = 17
1.0882
−.1190
−.1161
.8414
.4357
1.6355
j = 18 j = 19
−.0203 −.4852
−.9940 −.4677
.2322 −1.5246
.3165
.4910
.0871
1.3160
1.1590 −1.4141
j = 20
−1.5455
−.4095
.8736
−.9646
−1.7954
−.5901
H0 : no outliers, Ha : at least one outlier. t(.9999417; 113) = 4.08.
If |tij | ≤ 4.08 conclude H0 , otherwise Ha . Conclude H0 .
18.9. a.
See Problem 16.12c.
b.
r = .995
d.
tij :
i
1
2
3
4
5
j=1
j=2
−.2047 −.2047
−1.7195 −.9534
−.6526 −.2792
.0744 −.6714
1.0858 −3.1711
j=3
j=4
1.6805 −1.7195
−.9534
.5404
−1.4100
.0930
1.1998
.4470
−.7840
1.8564
i
1
2
3
4
5
j=8
j=9
.9157 −.5778
.5404
2.0738
−1.0290 −1.4100
1.1998 −1.4293
.3351 −.4097
j = 10 j = 11
−1.3334 −.2047
−1.3334
.5404
−.2792 1.6029
.8216
.0744
−.7840 1.0858
i
1
2
3
4
5
j = 15
1.2951
.5404
.4657
−.2978
−.4097
j = 16 j = 17
−.5778 −.5778
−.2047 −1.3334
−.2792
.8404
.8216
.4470
1.8564
.7089
j = 18
.9157
1.2951
−1.0290
.8216
−1.5448
j=5
j=6
−1.3334
.1675
−.2047 2.4771
.0930 −.6526
−1.0479 1.5835
−.4097 −.7840
j = 12 j = 13
.5404 −.5778
.9157 −.5778
.0930 2.3955
−1.0479 −.6714
−.0372
.7089
j = 19
.5404
−.2047
−.2792
−.2978
−.0372
j=7
1.2951
.1675
.8404
−1.8172
−.0372
j = 14
−.2047
−.9534
.8404
−.6714
1.0858
j = 20
.1675
−.5778
.0930
.4470
−.4097
H0 : no outliers, Ha : at least one outlier. t(.999875; 94) = 3.81.
If |tij | ≤ 3.81 conclude H0 , otherwise Ha . Conclude H0 .
18.11.
H0 : all σi2 are equal (i = 1, 2, 3), Ha : not all σi2 are equal.
Ỹ1 = 6.80, Ỹ2 = 8.20, Ỹ3 = 9.55, M ST R = .0064815, M SE = .26465,
∗
∗
≤ 3.40 conclude H0 ,
= .0064815/.26465 = .02, F (.95; 2, 24) = 3.40. If FBF
FBF
otherwise Ha . Conclude H0 . P -value = .98
18.12.
H0 : all σi2 are equal (i = 1, 2, 3), Ha : not all σi2 are equal.
Ỹ1 = 40.0, Ỹ2 = 31.0, Ỹ3 = 22.5, M ST R = 2.96667, M SE = 11.30476,
∗
∗
≤ 2.575 conclude
= 2.96667/11.30476 = .26, F (.90; 2, 21) = 2.575. If FBF
FBF
H0 , otherwise Ha . Conclude H0 . P -value = .77
18-3
18.13. a.
H0 : all σi2 are equal (i = 1, 2, 3), Ha : not all σi2 are equal.
s1 = 1.7321, s2 = 1.2881, s3 = 1.6765, ni ≡ 12, H ∗ = (1.7321)2 /(1.2881)2 = 1.808,
H(.99; 3, 11) = 6.75. If H ∗ ≤ 6.75 conclude H0 , otherwise Ha . Conclude H0 .
P -value > .05
b.
Ỹ1 = 21.5, Ỹ2 = 27.5, Ỹ3 = 21.0, M ST R = .19444, M SE = .93434,
∗
∗
FBF
= .19444/.93434 = .21, F (.99; 2, 33) = 5.31. If FBF
≤ 5.31 conclude H0 ,
otherwise Ha . Conclude H0 . P -value = .81
18.14. a.
H0 : all σi2 are equal (i = 1, ..., 6), Ha : not all σi2 are equal.
s1 = .1925, s2 = .1854, s3 = .1646, s4 = .1654, s5 = .1727, s6 = .1735, ni ≡ 20,
H ∗ = (.1925)2 /(.1646)2 = 1.3677, H(.99; 6, 19) = 5.2. If H ∗ ≤ 5.2 conclude H0 ,
otherwise Ha . Conclude H0 . P -value > .05
b.
18.15. a.
Ỹ1 = .08, Ỹ2 = .12, Ỹ3 = .47, Ỹ4 = .41, Ỹ5 = .175, Ỹ6 = .125, M ST R = .002336,
∗
∗
M SE = .012336, FBF
= .002336/.012336 = .19, F (.99; 5, 114) = 3.18. If FBF
≤
3.18 conclude H0 , otherwise Ha . Conclude H0 . P -value = .97
Ȳ1. = 3.90, Ȳ2. = 1.15, Ȳ3. = 2.00, Ȳ4. = 3.40
eij :
c.
i
1
2
3
4
j=1 j=2 j=3 j=4 j=5 j=6
.10 −.90
1.10
.10
2.10 −.90
−1.15
.85 −1.15
1.85
.85 −.15
0.0
−1.00 −2.00
1.00
2.00 −1.00
1.60 −1.40
.60
.60
2.60
1.60
i
1
2
3
4
j=8
1.10
1.85
2.00
1.60
i
1
2
3
4
j = 15
.10
−.15
−2.00
−.40
j=9
3.10
−.15
0.0
3.60
j = 16
1.10
1.85
−1.00
.60
j = 10
−2.90
−1.15
−2.00
−.40
j = 17
−3.90
−.15
1.00
−2.40
j = 11
−1.90
−1.15
−1.00
−2.40
j = 12
1.10
−.15
1.00
−3.40
j = 18
.10
.85
−2.00
1.60
j = 19
−2.90
.85
0.0
−1.40
j=7
−1.90
−1.15
1.00
−.40
j = 13
.10
−.15
0.0
−1.40
j = 14
3.10
−1.15
2.00
−.40
j = 20
2.10
−1.15
2.00
−.40
H0 : all σi2 are equal (i = 1, 2, 3, 4), Ha : not all σi2 are equal.
Ỹ1 = 4, Ỹ2 = 1, Ỹ3 = 2, Ỹ4 = 3, M ST R = 1.64583, M SE = .96776,
∗
∗
≤ 2.157 conclude H0 ,
= 1.64583/.96776 = 1.70, F (.90; 3, 76) = 2.157. If FBF
FBF
otherwise Ha . Conclude H0 . P -value = .17
d.
i
Ȳi.
si
1 3.9000 1.9708
2 1.1500 1.0894
3 2.0000 1.4510
4 3.4000 1.7889
18-4
e.
λ
−1.0
−.8
−.6
−.4
−.2
−.1
0
λ
.1
.2
.4
.6
.8
1.0
SSE
434.22
355.23
297.21
254.90
224.59
213.09
203.67
SSE
196.14
190.35
183.48
182.41
186.91
197.15
Yes
18.16. a.
Ȳ1.0 = 1.8714, Ȳ2.0 = .8427, Ȳ3.0 = 1.2293, Ȳ40 . = 1.7471
e0ij :
i
1
2
3
4
j=1
.129
−.843
.185
.489
j=2 j=3
−.139
.365
.572 −.843
−.229 −1.229
−.333
.253
i
1
2
3
4
j=8
.365
.889
.771
.489
j=9
.774
.157
.185
.899
i
1
2
3
4
j = 15 j = 16 j = 17
.129
.365 −1.871
.157
.889
.157
−1.229 −.229
.503
−.015
.253 −.747
j = 10
−.871
−.843
−1.229
−.015
j=4
.129
.889
.503
.253
j = 11
−.457
−.843
−.229
−.747
j=5 j=6
.578 −.139
.572
.157
.771 −.229
.702
.489
j=7
−.457
−.843
.503
−.015
j = 12 j = 13 j = 14
.365
.129
.774
.157
.157 −.843
.503
.185
.771
−1.747 −.333 −.015
j = 18
.129
.572
−1.229
.489
j = 19
−.871
.572
.185
−.333
j = 20
.578
−.843
.771
−.015
b.
r = .964
c.
H0 : all σi2 are equal (i = 1, 2, 3, 4), Ha : not all σi2 are equal.
Ỹ1 = 2.000, Ỹ2 = 1.000, Ỹ3 = 1.414, Ỹ4 = 1.732, M ST R = .07895, M SE = .20441,
∗
∗
FBF
= .07895/.20441 = .39, F (.90; 3, 76) = 2.157. If FBF
≤ 2.157 conclude H0 ,
otherwise Ha . Conclude H0 .
18.17. a.
Ȳ1. = 3.5625, Ȳ2. = 5.8750, Ȳ3. = 10.6875, Ȳ4. = 15.5625
eij :
i j=1
1 .4375
2 1.1250
3 1.3125
4 .4375
j=2
−.5625
.1250
−4.6875
−1.5625
j=3
−1.5625
−1.8750
3.3125
−9.5625
j=4
−.5625
.1250
1.3125
3.4375
18-5
j=5
.4375
1.1250
−.6875
−3.5625
j=6
.4375
−3.8750
−1.6875
−5.5625
c.
i
1
2
3
4
j=7
−.5625
3.1250
1.3125
−.5625
j=8
j=9
2.4375
1.4375
−.8750 −.8750
6.3125 −3.6875
8.4375 −5.5625
i
1
2
3
4
j = 13
j = 14
.4375 −1.5625
.1250 −1.8750
−4.6875
2.3125
−.5625
2.4375
j = 10
.4375
3.1250
−4.6875
7.4375
j = 11
−1.5625
−2.8750
1.3125
1.4375
j = 12
.4375
2.1250
.3125
4.4375
j = 15 j = 16
−.5625
.4375
1.1250
.1250
−.6875 3.3125
−7.5625 6.4375
H0 : all σi2 are equal (i = 1, 2, 3, 4), Ha : not all σi2 are equal.
Ỹ1 = 4.0, Ỹ2 = 6.0, Ỹ3 = 11.5, Ỹ4 = 16.5, M ST R = 37.1823, M SE = 3.8969,
∗
∗
FBF
= 37.1823/3.8969 = 9.54, F (.95; 3, 60) = 2.76. If FBF
≤ 2.76 conclude H0 ,
otherwise Ha . Conclude Ha . P -value = 0+
d.
i
1
2
3
4
Ȳi.
3.5625
5.8750
10.6875
16.5625
si
1.0935
1.9958
3.2397
5.3786
e.
λ
.1
.2
.4
.6
.8
1.0
λ
SSE
−1.0 1, 038.26
−.8
790.43
−.6
624.41
−.4
516.16
−.2
450.16
−.1
429.84
0
416.84
SSE
410.65
410.92
430.49
476.68
553.64
669.06
Yes
18.18. a.
Ȳ1.0 = .5314, Ȳ2.0 = .7400, Ȳ3.0 = 1.0080, Ȳ4.0 = 1.1943
e0ij :
i
1
2
3
4
j=1
.071
.105
.071
.036
j=2
−.054
.038
−.230
−.018
i
1
2
3
4
j=9
.168
−.041
−.163
−.153
j = 10
.071
.214
−.230
.186
j=3 j=4 j=5
−.230 −.054
.071
−.138
.038
.105
.138
.071 −.008
−.349
.107 −.080
j = 11
−.230
−.263
.071
.061
j=6
.071
−.439
−.054
−.153
j=7 j=8
−.054
.247
.214 −.041
.071
.222
.010
.204
j = 12 j = 13 j = 14 j = 15 j = 16
.071
.071 −.230 −.054
.071
.163
.038 −.138
.105
.038
.033 −.230
.106 −.008
.138
.128
.010
.085 −.240
.167
18-6
b.
r = .971
c.
H0 : all σi2 are equal (i = 1, 2, 3, 4), Ha : not all σi2 are equal.
Ỹ1 = .6021, Ỹ2 = .7782, Ỹ3 = 1.0603, Ỹ4 = 1.2173, M ST R = .001214,
∗
M SE = .01241, FBF
= .001214/.01241 = .10, F (.95; 3, 60) = 2.76.
∗
If FBF
≤ 2.76 conclude H0 , otherwise Ha . Conclude H0 .
18.19.
i:
1
si : 1.97084
wi : .25745
2
3
4
1.08942 1.45095 1.78885
.84257 .47500 .31250
H0 : all µi are equal (i = 1, 2, 3, 4), Ha : not all µi are equal.
SSEw (F ) = 76, dfF = 76, SSEw (R) = 118.54385, dfR = 79,
Fw∗ = [(118.54385 − 76)/3] ÷ (76/76) = 14.18, F (.95; 3, 76) = 2.725.
If Fw∗ ≤ 2.725 conclude H0 , otherwise Ha . Conclude Ha .
18.20.
i:
1
si : 1.09354
wi : .83624
2
3
4
1.99583 3.23973 5.37858
.25105 .09528 .034567
H0 : all µi are equal (i = 1, 2, 3, 4), Ha : not all µi are equal.
SSEw (F ) = 60, dfF = 60, SSEw (R) = 213.9541, dfR = 63,
Fw∗ = [(213.9541 − 60)/3] ÷ (60/60) = 51.32, F (.99; 3, 60) = 4.13.
If Fw∗ ≤ 4.13 conclude H0 , otherwise Ha . Conclude Ha .
18.23. a.
H0 : all µi are equal (i = 1, 2, 3, 4), Ha : not all µi are equal.
M ST R = 470.8125, M SE = 28.9740, FR∗ = 470.8125/28.9740 = 16.25,
F (.95; 2, 24) = 3.40. If FR∗ ≤ 3.40 conclude H0 , otherwise Ha . Conclude Ha .
b.
P -value = 0+
e.
R̄1. = 6.50, R̄2. = 15.50, R̄3. = 22.25, B = z(.9833) = 2.13
Comparison
1 and 2
1 and 3
2 and 3
Testing
−9.00 ± 2.13(3.500)
15.75 ± 2.13(4.183)
−6.75 ± 2.13(3.969)
Limits
−16.455 and −1.545
−24.660 and −9.840
−15.204 and 1.704
Group 1
Low Level i = 1
18.24. a.
Group 2
Moderate level i = 2
High level i = 3
H0 : all µi are equal (i = 1, 2, 3), Ha : not all µi are equal.
M ST R = 1, 297.0000, M SE = 37.6667, FR∗ = 1, 297.0000/37.6667 = 34.43,
F (.99; 2, 33) = 5.31. If FR∗ ≤ 5.31 conclude H0 , otherwise Ha . Conclude Ha .
18-7
b.
P -value = 0+
e.
R̄1. = 12.792, R̄2. = 30.500, R̄3. = 12.208, B = z(.9833) = 2.128
Comparison
1 and 2
1 and 3
2 and 3
Testing Limits
−17.708 ± 2.128(4.301) −26.861 and −8.555
.584 ± 2.128(4.301)
−8.569 and 9.737
18.292 ± 2.128(4.301)
9.140 and 27.445
Group 1
Young i = 1
Elderly i = 3
18.25. b.
Group 2
Middle i = 2
H0 : all µi are equal (i = 1, 2, 3), Ha : not all µi are equal.
M ST R = 465.6000, M SE = 48.7519, FR∗ = 465.6000/48.7519 = 9.550,
F (.95; 2, 27) = 3.354. If FR∗ ≤ 3.354 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .0007
c.
R̄1. = 21.1, R̄2. = 7.9, R̄3. = 17.5, B = z(.99167) = 2.394
Comparison
1 and 2
1 and 3
2 and 3
Testing Limits
13.2 ± 2.394(3.937)
3.775 and 22.625
3.6 ± 2.394(3.937) −5.825 and 13.025
−9.6 ± 2.394(3.937) −19.025 and −.175
Group 1
Production i = 2
18.26.
18.27.
Group 2
Sales i = 1
Research i = 3
Yij = µi + βtj + ²ij , tj = 1, ..., 7
nT
X
nT
X
nT (nT + 1)
i=
2
i=1
SST O =
nT
X
à n !2
T
X
i2 −
i
i=1
i2 =
i=1
nT (nT + 1)(2nT + 1)
6
/nT
i=1
nT (nT + 1)(2nT + 1) n2T (nT + 1)2
nT (nT + 1)(nT − 1)
−
=
6
4nT
12
SST O/(nT − 1) = nT (nT + 1)/12
=
18.28.
SST O
nT − 1
SST R SST O − SST R
÷
FR∗ =
r−1
nT − r
2
= SST R ÷
XKW
µ
·
Ã
=
=
2
XKW
r−1
!µ
¶
SST O
÷
nT − 1
SST O −
2
XKW
SST O
nT − 1
¶¸
nT − r
2
2
)
] SST O SST O(nT − 1 − XKW
[(nT − r)XKW
÷
(r − 1)(nT − 1)
nT − 1
18-8
2
2
= [(nT − r)XKW
] ÷ [(r − 1)(nT − 1 − XKW
)]
18.29. b.
c.
r = .994
H0 : all σi2 are equal (i = 1, ..., 4), Ha : not all σi2 are equal.
Ỹ1 = 4.85, Ỹ2 = 4.40, Ỹ3 = 4.20, Ỹ4 = 4.45, M ST R = .97716, M SE = .70526,
∗
∗
FBF
= .97716/.70526 = 1.39, F (.95; 3, 109) = 2.688. If FBF
≤ 2.688 conclude H0 ,
otherwise Ha . Conclude H0 .
P -value = .25
18.30. b.
i
Ȳi.
si
1 11.08893 2.66962
2 9.68344 1.19294
3 9.19135 1.22499
4 8.11375 1.00312
c.
λ
−1.0
−.8
−.6
−.4
−.2
−.1
0
SSE
206.15
208.55
212.09
216.87
223.09
226.79
230.93
λ
.1
.2
.4
.6
.8
1.0
SSE
235.54
240.65
252.56
267.04
284.55
305.66
Yes
e.
r = .995
f.
H0 : all σi2 are equal (i = 1, ..., 4), Ha : not all σi2 are equal.
Ỹ1 = .09332, Ỹ2 = .10199, Ỹ3 = .11111, Ỹ4 = .12799, M ST R = .00008213,
∗
M SE = .00008472, FBF
= .00008213/.00008472 = .97, F (.99; 3, 109) = 3.97.
∗
If FBF
≤ 3.97 conclude H0 , otherwise Ha . Conclude H0 . P -value = .41
g.
Source
Between regions
Error
Total
SS
df
.0103495
3
.0254284 109
.0357779 112
MS
.0034498
.0002333
H0 : all µi are equal (i = 1, ..., 4), Ha : not all µi are equal.
F ∗ = .0034498/.0002333 = 14.787, F (.99; 3, 109) = 3.967.
If F ∗ ≤ 3.967 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
18.31. b.
c.
r = .9154
H0 : all σi2 are equal (i = 1, ..., 4), Ha : not all σi2 are equal.
18-9
Ỹ1 = .03489, Ỹ2 = .04781, Ỹ3 = .06948, Ỹ4 = .05966, M ST R = .001001, M SE =
∗
∗
.000326, FBF
= .001001/.000326 = 3.07, F (.99; 3, 436) = 3.83. If FBF
≤ 3.83
conclude H0 , otherwise Ha . Conclude H0 . P -value = .028
18.32. b.
c.
r = .9902
H0 : all σi2 are equal (i = 1, ..., 4), Ha : not all σi2 are equal.
Ỹ1 = 2.415, Ỹ2 = 2.705, Ỹ3 = 2.480, Ỹ4 = 2.880, M ST R = .0106, M SE = .0085,
∗
∗
FBF
= .0106/.0085 = 1.25, F (.95; 3, 32) = 2.90. If FBF
≤ 2.90 conclude H0 ,
otherwise Ha . Conclude H0 . P -value = .31
18.33. a.
H0 : all µi are equal (i = 1, ..., 4), Ha : not all µi are equal. M ST R = 2, 582.575,
M SE = 1, 031.966, FR∗ = 2, 582.575/1, 031.966 = 2.50, F (.95; 3, 109) = 2.69. If
FR∗ ≤ 2.69 conclude H0 , otherwise Ha . Conclude H0 . P -value = .063
c.
R̄1. = 69.196, R̄2. = 57.797, R̄3. = 47.189, R̄4. = 56.750, B = z(.99167) = 2.394
Comparison
Testing Limits
1 and 2
11.399 ± 2.394(8.479) −8.900 and 31.698
1 and 3
22.007 ± 2.394(8.207)
2.359 and 41.655
1 and 4
12.446 ± 2.394(10.268) −12.136 and 37.028
2 and 3
10.608 ± 2.394(7.910) −8.329 and 29.545
2 and 4
1.047 ± 2.394(10.032) −22.970 and 25.064
3 and 4
−9.561 ± 2.394(9.803) −33.029 and 13.907
Group 1
Region 3
Region 4
Region 2
18.34. a.
Group 2
Region 4
Region 2
Region 1
H0 : all µi are equal (i = 1, ..., 4), Ha : not all µi are equal.
M ST R = 651, 049, M SE = 11, 802, FR∗ = 651, 049/11, 802 = 55.17,
F (.95; 3, 436) = 2.6254. If FR∗ ≤ 2.6254 conclude H0 , otherwise Ha . Conclude Ha .
P -value = 0+
c.
R̄1. = 120.4, R̄2. = 192.9, R̄3. = 290.7, R̄4. = 254.6,
n1 = 103, n2 = 108, n3 = 152, n4 = 77, B = z(.99583) = 2.638
Comparison
1 and 2
1 and 3
1 and 4
2 and 3
2 and 4
3 and 4
Testing Limits
−72.5 ± 2.638(17.51) −118.7 and −26.3
−170.3 ± 2.638(16.23) −213.1 and −127.5
−134.2 ± 2.638(19.16) −184.7 and −83.7
−97.8 ± 2.638(16.00) −140.0 and −55.6
−61.7 ± 2.638(18.97) −111.7 and −11.7
36.1 ± 2.638(17.79)
−10.8 and 83.0
Group 1
Region 1
Group 2
Region 2
18-10
Group 3
Region 3
Region 4
18.35. a.
H0 : all µi are equal (i = 1, ..., 4), Ha : not all µi are equal.
M ST R = 955.5, M SE = 31.8, FR∗ = 955.5/31.8 = 30.1,
F (.95; 3, 32) = 2.90. If FR∗ ≤ 2.90 conclude H0 , otherwise Ha . Conclude Ha .
P -value = 0+
c.
R̄1. = 7.938, R̄2. = 22.375, R̄3. = 8.571, R̄4. = 27.962,
n1 = 8, n2 = 8, n3 = 7, n4 = 13, B = z(.99583) = 2.638
Comparison
1 and 2
1 and 3
1 and 4
2 and 3
2 and 4
3 and 4
Testing Limits
−14.437 ± 2.638(5.268) −28.334 and −.541
−.633 ± 2.638(5.453) −15.017 and 13.751
−20.024 ± 2.638(4.734) −32.513 and −7.535
13.804 ± 2.638(5.453)
−.580 and 28.188
−5.587 ± 2.638(4.734) −18.076 and 6.902
−19.391 ± 2.638(4.939) −32.421 and −6.361
Group 1
Region 1
Region 3
18.36.
Group 2
Region 2
Region 3
Group 3
Region 2
Region 4
Under H0 , each arrangement of the ranks 1, ..., 4 into groups of size 2 are equally
likely and occur with probability 2!2!/4! = 1/6. The values of FR∗ computed for the
six arrangements are 0, .5, and 8, each occurring twice. Therefore the probability
function f (x) is:
x
0
.5
8
18.37. c.
f (x) = P (FR∗ = x)
1/3
1/3
1/3
For the F distribution with ν1 = 2 degrees of freedom and ν2 = 27 degrees of
freedom, the mean is:
ν2
= 1.08
ν2 − 2
and the standard deviation is:
"
ν2
2(ν1 + ν2 − 2)
ν2 − 2
ν1 (ν2 − 4)
d.
#1/2
= 1.17.
Expect 90% less than 2.51 and 99% less than 5.49.
18-11
18-12
Chapter 19
TWO-FACTOR ANALYSIS OF
VARIANCE WITH EQUAL
SAMPLE SIZES
19.1. a.
b.
19.3.
8
Infection risk
(αβ)11 = µ11 − (µ.. + α1 + β1 ) = 9 − (12 + 1 − 3) = −1
(αβ)12 = µ12 − (µ.. + α1 + β2 ) = 12 − (12 + 1 − 1) = 0
(αβ)13 = µ13 − (µ.. + α1 + β3 ) = 18 − (12 + 1 + 4) = 1
(αβ)21 = µ21 − (µ.. + α2 + β1 ) = 9 − (12 − 1 − 3) = 1
(αβ)22 = µ22 − (µ.. + α2 + β2 ) = 10 − (12 − 1 − 1) = 0
(αβ)23 = µ23 − (µ.. + α2 + β3 ) = 14 − (12 − 1 + 4) = −1
19.4. a.
b.
19.5. a.
c.
µ1. = 31, µ2. = 37
α1 = µ1. − µ.. = 31 − 34 = −3, α2 = µ2. − µ.. = 37 − 34 = 3
µ.j = 269 (j = 1, ..., 4), βj = µ.j − µ.. , βj = 0 (j = 1, ..., 4)
loge µij :
Factor A
A1
A2
Factor B
B1
B2
B3
B4
5.5215 5.5797 5.5910 5.5947
5.6630 5.6095 5.5984 5.5947
19.7. a.
E{M SE} = 1.96, E{M SA} = 541.96
19.8. a.
E{M SE} = 16, E{M SAB} = 952
19.10. a.
Ȳ11. = 21.66667, Ȳ12. = 21.33333, Ȳ21. = 27.83333,
Ȳ22. = 27.66667, Ȳ31. = 22.33333, Ȳ32. = 20.50000
b.
eijk :
19-1
i
1
d.
j=1
j=2
−.66667 −.33333
1.33333
.66667
−2.66667 −1.33333
.33333 −.33333
.33333 −2.33333
1.33333
3.66667
i
2
j=1
2.16667
1.16667
−1.83333
.16667
−.83333
−.83333
j=2
−1.66667
1.33333
−.66667
.33333
−.66667
1.33333
i
3
j=1
2.66667
−.33333
.66667
−1.33333
−.33333
−1.33333
r = .986
19.11. b.
Source
SS
df
MS
Treatments
327.222 5 65.444
A (age)
316.722 2 158.361
B (gender)
5.444 1
5.444
AB interactions
5.056 2
2.528
Error
71.667 30
2.389
Total
398.889 35
Yes, factor A (age) accounts for most of the total variability.
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = 2.528/2.389 = 1.06, F (.95; 2, 30) = 3.32.
If F ∗ ≤ 3.32 conclude H0 , otherwise Ha . Conclude H0 . P -value = .36
d.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero.
F ∗ = 158.361/2.389 = 66.29, F (.95; 2, 30) = 3.32.
If F ∗ ≤ 3.32 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = 5.444/2.389 = 2.28, F (.95; 1, 30) = 4.17.
If F ∗ ≤ 4.17 conclude H0 , otherwise Ha . Conclude H0 . P -value = .14
e.
α ≤ .143
g.
SSA = SST R, SSB + SSAB + SSE = SSE, yes
19.12. a.
b.
Ȳ11. = 9.2, Ȳ12. = 13.6, Ȳ21. = 13.0, Ȳ22. = 16.4
eijk :
i
1
d.
j=1
1.8
−2.2
2.8
−3.2
.8
j=2
1.4
−1.6
.4
−2.6
2.4
i j=1
2 −1.0
3.0
−3.0
0
1.0
r = .976
19.13. b.
19-2
j=2
−2.4
.6
−3.4
3.6
1.6
j=2
2.50000
−1.50000
−.50000
.50000
−.50000
−.50000
Source
SS
Treatments
131.75
A (eye contact)
54.45
B (gender)
76.05
AB interactions
1.25
Error
97.20
Total
228.95
df
MS
3 43.917
1 54.45
1 76.05
1
1.25
16 6.075
19
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = 1.25/6.075 = .21,
F (.99; 1, 16) = 8.53. If F ∗ ≤ 8.53 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .66
d.
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero. F ∗ = 54.45/6.075 = 8.96,
F (.99; 1, 16) = 8.53. If F ∗ ≤ 8.53 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .009
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero. F ∗ = 76.05/6.075 = 12.52,
F (.99; 1, 16) = 8.53. If F ∗ ≤ 8.53 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .003.
e.
19.14. a.
α ≤ .030
Ȳ11. = 2.475, Ȳ12. = 4.600, Ȳ13. = 4.575, Ȳ21. = 5.450, Ȳ22. = 8.925,
Ȳ23. = 9.125, Ȳ31. = 5.975, Ȳ32. = 10.275, Ȳ33. = 13.250
b.
eijk :
i
1
j=1
−.075
.225
−.175
.025
j=2 j=3
0
.225
−.400 −.075
.300 −.175
.100
.025
j=1
.125
−.275
−.075
.225
r = .988
j=2 j=3
−.375
.250
.225 −.250
.325
.050
−.175 −.050
i
3
d.
i
2
j=1 j=2
.350 −.025
−.250
.175
.050 −.225
−.150
.075
j=3
−.025
.175
−.425
.275
19.15. b.
Source
Treatments
A (ingredient 1)
B (ingredient 2)
AB interactions
Error
Total
c.
SS
df
MS
373.105 8 46.638
220.020 2 110.010
123.660 2 61.830
29.425 4
7.356
1.625 27
.0602
374.730 35
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = 7.356/.0602 =
122.19, F (.95; 4, 27) = 2.73. If F ∗ ≤ 2.73 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
19-3
d.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero. F ∗ = 110.010/.0602 =
1, 827.41, F (.95; 2, 27) = 3.35. If F ∗ ≤ 3.35 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : all βj equal zero (j = 1, 2, 3), Ha : not all βj equal zero. F ∗ = 61.830/.0602 =
1, 027.08, F (.95; 2, 27) = 3.35. If F ∗ ≤ 3.35 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
e.
19.16. a.
α ≤ .143
Ȳ11. = 59.8, Ȳ12. = 47.8, Ȳ13. = 58.4, Ȳ21. = 48.4, Ȳ22. = 61.2,
Ȳ23. = 56.2, Ȳ31. = 60.2, Ȳ32. = 60.8, Ȳ33. = 49.6
b.
d.
eijk :
i
1
j=1 j=2 j=3
2.2
9.2
.6
−11.8 −2.8 −5.4
3.2 −8.8
8.6
−2.8
6.2
7.6
9.2 −3.8 −11.4
i
3
j=1
−1.2
4.8
−5.2
−8.2
9.8
j=2
−2.8
2.2
9.2
−7.8
−.8
i
2
j=1
2.6
8.6
−3.4
1.6
−9.4
j=2 j=3
−.2 −1.2
−3.2
1.8
8.8 −6.2
4.8
12.8
−10.2 −7.2
j=3
−2.6
6.4
1.4
−5.6
.4
r = .989
19.17. b.
Source
Treatments
A (technician)
B (make)
AB interactions
Error
Total
SS
df
MS
1, 268.17778 8 158.52222
24.57778 2 12.28889
28.31111 2 14.15556
1, 215.28889 4 303.82222
1, 872.40000 36 52.01111
3, 140.57778 44
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = 303.82222/52.01111 =
5.84, F (.99; 4, 36) = 3.89. If F ∗ ≤ 3.89 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .001
d.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero. F ∗ = 12.28889/52.01111 =
.24, F (.99; 2, 36) = 5.25. If F ∗ ≤ 5.25 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .79
H0 : all βj equal zero (j = 1, 2, 3), Ha : not all βj equal zero. F ∗ = 14.15556/52.01111 =
.27, F (.99; 2, 36) = 5.25. If F ∗ ≤ 5.25 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .76
e.
α ≤ .003
19-4
19.18. a.
0
0
0
Ȳ11.
= .44348, Ȳ12.
= .80997, Ȳ13.
= 1.10670,
0
0
0
Ȳ21.
= .39823, Ȳ22.
= .58096, Ȳ23.
= .86639
e0ijk :
i
1
j=1
−.44348
.03364
.03364 −.44348
−.14245
.33467
.15858
.40162
−.44348
.51076
j=2
−.33285 −.11100
−.11100 −.20791
.09312 −.50894
.30398 −.03182
.39415
.51225
j=3
.09742
.12375
−.06531 −.20361
−.15246
.38466
−.32855 −.50464
.30827
.34046
j=1
−.39823
.07889
−.09720
.50486
−.09720
.30074
−.39823 −.39823
.30074
.20383
r = .987
j=2
.19719 −.27993
.02110
.02110
−.10384
.26413
−.58096
.32213
−.27993
.41904
j=3
.17500
.33773
.08785 −.16742
.24755
.13361
−.26433 −.02130
.03670 −.56536
i
2
c.
19.19. b.
Source
Treatments
A (duration)
B (weight gain)
AB interactions
Error
Total
SS
3.76217
.44129
3.20098
.11989
5.46770
9.22987
df
MS
5 .75243
1 .44129
2 1.60049
2 .05995
54 .10125
59
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = .05995/.10125 = .59,
F (.95; 2, 54) = 3.17. If F ∗ ≤ 3.17 conclude H0 , otherwise Ha . Conclude H0 . P value = .56
d.
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero. F ∗ = .44129/.10125 = 4.36,
F (.95; 1, 54) = 4.02. If F ∗ ≤ 4.02 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .04
H0 : all βj equal zero (j = 1, 2, 3), Ha : not all βj equal zero. F ∗ = 1.60049/.10125 =
15.81, F (.95; 2, 54) = 3.17. If F ∗ ≤ 3.17 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
e.
19.20. a.
b.
α ≤ .143
Ȳ11. = 222.00, Ȳ12. = 106.50, Ȳ13. = 60.50, Ȳ21. = 62.25, Ȳ22. = 44.75, Ȳ23. = 38.75
eijk :
i
1
j=1
18
−16
−5
3
j=2 j=3
3.5 −4.5
11.5
−.5
−3.5
7.5
−11.5 −2.5
i
2
j=1
8.75
−9.25
5.75
−5.25
19-5
j=2 j=3
2.25 −1.75
7.25 −5.75
−13.75
1.25
4.25
6.25
d.
r = .994
19.21. b.
Source
Treatments
A (type)
B (years)
AB interactions
Error
Total
SS
df
96, 024.37500 5
39, 447.04167 1
36, 412.00000 2
20, 165.33333 2
1, 550.25000 18
97, 574.62500 23
MS
19, 204.87500
39, 447.04167
18, 206.00000
10, 082.66667
86.12500
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = 10, 082.66667/86.12500 =
117.07, F (.99; 2, 18) = 6.01. If F ∗ ≤ 6.01 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
d.
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero. F ∗ = 39, 447.04167/86.12500 =
458.02, F (.99; 1, 18) = 8.29. If F ∗ ≤ 8.29 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : all βj equal zero (j = 1, 2, 3), Ha : not all βj equal zero. F ∗ = 18, 206.00000/86.12500 =
211.39, F (.99; 2, 18) = 6.01. If F ∗ ≤ 6.01 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
e.
α ≤ .030
19.27. a.
B = t(.9975; 75) = 2.8925, q(.95; 5, 75) = 3.96, T = 2.800
b.
B = t(.99167; 27) = 2.552, q(.95; 3, 27) = 3.51, T = 2.482
19.28.
(1) B = t(.9972; 324) = 2.791,
(2) F (.975; 5, 324) = 2.606, S = 3.6097
(3) F (.95; 10, 324) = 1.86, S = 4.3128
19.30. a.
s{Ȳ11. } = .631, t(.975; 30) = 2.042, 21.66667 ± 2.042(.631), 20.378 ≤ µ11 ≤ 22.955
b.
Ȳ.1. = 23.94, Ȳ.2. = 23.17
c.
D̂ = .77, s{D̂} = .5152, t(.975; 30) = 2.042, .77±2.042(.5152), −.282 ≤ D ≤ 1.822
d.
Ȳ1.. = 21.50, Ȳ2.. = 27.75, Ȳ3.. = 21.42
e.
D̂1 = Ȳ1.. −Ȳ2.. = −6.25, D̂2 = Ȳ1.. −Ȳ3.. = .08, D̂3 = Ȳ2.. −Ȳ3.. = 6.33, s{D̂i } = .631
(i = 1, 2, 3), q(.90; 3, 30) = 3.02, T = 2.1355
−6.25 ± 2.1355(.631)
.08 ± 2.1355(.631)
6.33 ± 2.1355(.631)
−7.598 ≤ D1 ≤ −4.902
−1.268 ≤ D2 ≤ 1.428
4.982 ≤ D3 ≤ 7.678
f.
Yes
g.
L̂ = −6.29, s{L̂} = .5465, t(.976; 30) = 2.042, −6.29 ± 2.042(.5465), −7.406 ≤
L ≤ −5.174
h.
L = .3µ12 + .6µ22 + .1µ32 , L̂ = 25.05000, s{L̂} = .4280, t(.975; 30) = 2.042,
25.05000 ± 2.042(.4280), 24.176 ≤ L ≤ 25.924
19-6
19.31. a.
s{Ȳ21. } = 1.1023, t(.995; 16) = 2.921, 13.0 ± 2.921(1.1023), 9.780 ≤ µ21 ≤ 16.220
b.
s{Ȳ1.. } = .7794, t(.995; 16) = 2.921, 11.4 ± 2.921(.7794), 9.123 ≤ µ1. ≤ 13.677
c.
Ȳ.1. = 11.1, Ȳ.2. = 15.0
d.
s{Ȳ.1. } = s{Ȳ.2. } = .7794, t(.995; 16) = 2.921
11.1 ± 2.921(.7794)
15.0 ± 2.921(.7794)
98 percent
8.823 ≤ µ.1 ≤ 13.377
12.723 ≤ µ.2 ≤ 17.277
e.
Ȳ1.. = 11.4, Ȳ2.. = 14.7
f.
D̂1 = 3.3, D̂2 = 3.9, s{D̂i } = 1.1023 (i = 1, 2), B = t(.9875; 16) = 2.473
3.3 ± 2.473(1.1023)
3.9 ± 2.473(1.1023)
g.
19.32. a.
.574 ≤ D1 ≤ 6.026
1.174 ≤ D2 ≤ 6.626
Yes
s{Ȳ23. } = .1227, t(.975; 27) = 2.052, 9.125 ± 2.052(.1227), 8.873 ≤ µ23 ≤ 9.377
b.
D̂ = 2.125, s{D̂} = .1735, t(.975; 27) = 2.052, 2.125 ± 2.052(.1735), 1.769 ≤ D ≤
2.481
c.
L̂1 = 2.1125, L̂2 = 3.5750, L̂3 = 5.7875, L̂4 = 1.4625, L̂5 = 3.6750, L̂6 = 2.2125,
s{L̂i } = .1502 (i = 1, 2, 3), s{L̂i } = .2125 (i = 4, 5, 6), F (.90; 8, 27) = 1.90,
S = 3.899
2.1125 ± 3.899(.1502)
1.527 ≤ L1 ≤ 2.698
3.5750 ± 3.899(.1502)
2.989 ≤ L2 ≤ 4.161
5.7875 ± 3.899(.1502)
5.202 ≤ L3 ≤ 6.373
1.4625 ± 3.899(.2125)
.634 ≤ L4 ≤ 2.291
3.6750 ± 3.899(.2125)
2.846 ≤ L5 ≤ 4.504
2.2125 ± 3.899(.2125)
1.384 ≤ L6 ≤ 3.041
d.
s{D̂i } = .1735, q(.90; 9, 27) = 4.31, T = 3.048, T s{D̂i } = .529, Ȳ33. = 13.250
e.
q
i j
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3
19.33. a.
1/Ȳij.
.404
.217
.219
.183
.112
.110
.167
.097
.075
Ȳij.
1.573
2.145
2.139
2.335
2.987
3.021
2.444
3.205
3.640
s{Ȳ11. } = 3.2252, t(.995; 36) = 2.7195, 59.8 ± 2.7195(3.2252),
51.029 ≤ µ11 ≤ 68.571
b.
D̂ = 12.8, s{D̂} = 4.5612, t(.995; 36) = 2.7195, 12.8 ± 2.7195(4.5612), .396 ≤ D ≤
25.204
19-7
c.
D̂1 = Ȳ11. − Ȳ12. = 12.0, D̂2 = Ȳ11. − Ȳ13. = 1.4, D̂3 = Ȳ12. − Ȳ13. = −10.6,
D̂4 = Ȳ21. − Ȳ22. = −12.8, D̂5 = Ȳ21. − Ȳ23. = −7.8, D̂6 = Ȳ22. − Ȳ23. = 5.0,
D̂7 = Ȳ31. − Ȳ32. = −.6, D̂8 = Ȳ31. − Ȳ33. = 10.6, D̂9 = Ȳ32. − Ȳ33. = 11.2,
s{D̂i } = 4.5612 (i = 1, ..., 9), B = t(.99167; 36) = 2.511
12.0 ± 2.511(4.5612)
1.4 ± 2.511(4.5612)
−10.6 ± 2.511(4.5612)
−12.8 ± 2.511(4.5612)
−7.8 ± 2.511(4.5612)
5.0 ± 2.511(4.5612)
−.6 ± 2.511(4.5612)
10.6 ± 2.511(4.5612)
11.2 ± 2.511(4.5612)
d.
.547 ≤ D1 ≤ 23.453
−10.053 ≤ D2 ≤ 12.853
−22.053 ≤ D3 ≤ .853
−24.253 ≤ D4 ≤ −1.347
−19.253 ≤ D5 ≤ 3.653
−6.453 ≤ D6 ≤ 16.453
−12.053 ≤ D7 ≤ 10.853
−.853 ≤ D8 ≤ 22.053
−.253 ≤ D9 ≤ 22.653
Ȳ... = 55.8222, 90Ȳ... = 5, 024, s{90Ȳ... } = 96.7574, t(.995; 36) = 2.7195,
5, 024 ± 2.7195(96.7574), 4, 760.87 ≤ 90µ.. ≤ 5, 287.13
e.
L = 10µ11 + 10µ13 + 10µ22 + 10µ23 + 10µ31 + 10µ32 − 20µ12 − 20µ21 − 20µ33 ,
L̂ = 650, s{L̂} = 136.8357, t(.995; 36) = 2.7195,
650 ± 2.7195(136.8357), 277.875 ≤ L ≤ 1, 022.125
f.
i
1
1
1
2
2
2
3
3
3
19.34. a.
j
1
2
3
1
2
3
1
2
3
1/Ȳij.
.0167
.0209
.0171
.0207
.0163
.0178
.0166
.0164
.0202
log10 Ȳij.
1.777
1.679
1.766
1.685
1.787
1.750
1.780
1.784
1.695
0
s{Ȳ22.
} = .1006, t(.975; 54) = 2.005,
.58096 ± 2.005(.1006), .37926 ≤ µ22 ≤ .78266
b.
D̂ = .46816, s{D̂} = .1423, t(.975; 54) = 2.005,
.46816 ± 2.005(.1423), .18285 ≤ D ≤ .75347
c.
0
0
= .61519
= .78672, Ȳ2..
Ȳ1..
0
0
0
= .98655
= .69547, Ȳ.3.
= .42086, Ȳ.2.
Ȳ.1.
d.
B = t(.9875; 54) = 2.306, q(.95; 2, 54) = 2.84, T = 2.008, q(.95; 3, 54) = 3.41,
T = 2.411, F (.90; 3, 54) = 2.20, S = 2.569
e.
0
0
0
0
= −.27461,
− Ȳ.2.
= .17153, D̂2 = Ȳ.1.
− Ȳ2..
D̂1 = Ȳ1..
0
0
0
0
= −.29108,
− Ȳ.3.
= −.56569, D̂4 = Ȳ.2.
− Ȳ.3.
D̂3 = Ȳ.1.
s{D̂1 } = .0822, s{D̂i } = .1006 (i = 2, 3, 4), B = 2.306
19-8
.17153 ± 2.306(.0822)
−.27461 ± 2.306(.1006)
−.56569 ± 2.306(.1006)
−.29108 ± 2.306(.1006)
f.
−.0180 ≤ D1
−.5066 ≤ D2
−.7977 ≤ D3
−.5231 ≤ D4
≤ .3611
≤ −.0426
≤ −.3337
≤ −.0591
L = .3µ.1 + .4µ.2 + .3µ.3 , L̂ = .70041, s{L̂} = .04149, t(.975; 54) = 2.005,
.70041 ± 2.005(.04149), .6172 ≤ L ≤ .7836, (3.142, 5.076), yes
19.35. a.
s{Ȳ23. } = 4.6402, t(.995; 18) = 2.878,
38.75 ± 2.878(4.6402), 25.3955 ≤ µ23 ≤ 52.1045
b.
D̂ = 46.00, s{D̂} = 6.5622, t(.995; 18) = 2.878,
46.00 ± 2.878(6.5622), 27.114 ≤ D ≤ 64.886
c.
F (.95; 5, 18) = 2.77, S = 3.7216, B = t(.99583; 18) = 2.963
d.
D̂1 = 159.75, D̂2 = 61.75, D̂3 = 21.75, L̂1 = 98.00, L̂2 = 138.00, L̂3 = 40.00,
s{D̂i } = 6.5622 (i = 1, 2, 3), s{L̂i } = 9.2804 (i = 1, 2, 3), B = t(.99583; 18) =
2.963
159.75 ± 2.963(6.5622)
140.31 ≤ D1 ≤ 179.19
61.75 ± 2.963(6.5622)
42.31 ≤ D2 ≤ 81.19
21.75 ± 2.963(6.5622)
2.31 ≤ D3 ≤ 41.19
98.00 ± 2.963(9.2804)
70.50 ≤ L1 ≤ 125.50
138.00 ± 2.963(9.2804)
110.50 ≤ L2 ≤ 165.50
40.00 ± 2.963(9.2804)
12.50 ≤ L3 ≤ 67.50
e.
q(.95; 6, 18) = 4.49, T = 3.1749, s{D̂} = 6.5622, T s{D̂} = 20.834, Ȳ23. = 38.75,
Ȳ22. = 44.75
f.
B = t(.9875; 18) = 2.445, s{Ȳij. } = 4.6402
44.75 ± 2.445(4.6402)
38.75 ± 2.445(4.6402)
33.405 ≤ µ22 ≤ 56.095
27.405 ≤ µ23 ≤ 50.095
g.
i
1
1
1
2
2
2
19.36. a.
j
1
2
3
1
2
3
1/Ȳij.
.00450
.00939
.01653
.01606
.02235
.02581
log10 Ȳij.
2.346
2.027
1.782
1.794
1.651
1.588
Yijk = µ.. + αi + βj + (αβ)ij + ²ijk , i = 1, ..., 4; j = 1, 2; k = 1, 2
b.
Source
SS
df
MS
Treatments
1, 910.00 7 272.85714
A (moisture content) 1, 581.50 3 527.16667
B (sweetness)
306.25 1 306.25000
AB interactions
22.25 3
7.41667
Error
57.00 8
7.12500
Total
1, 967.00 15
19-9
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = 7.41667/7.125 =
1.04, F (.99; 3, 8) = 7.59. If F ∗ ≤ 7.59 conclude H0 , otherwise Ha . Conclude H0 .
d.
L̂ = −1.500, s{L̂} = 2.669, t(.975; 8) = 2.306,
−1.500 ± 2.306(2.669), −7.655 ≤ L ≤ 4.655
e.
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero. F ∗ = 306.25/7.125 = 42.98,
F (.99; 1, 8) = 11.3. If F ∗ ≤ 11.3 conclude H0 , otherwise Ha . Conclude Ha .
19.37.
n = 21
19.38.
∆/σ = 2, 2n = 8, n = 4
19.39.
n = 21
√
.5 n/.29 = 4.1999, n = 6
19.40.
19.41.
19.42.
n = 14
√
8 n/9.1 = 3.1591, n = 13
19.43.
Using (19.4) and (19.5), we have:
µij = µ.. + αi + βj
= µ.. + (µi. − µ.. ) + (µ.j − µ.. ) = µi. + µ.j − µ..
19.44.
P
P
j
j
(αβ)ij =
(µij − µi. − µ.j + µ.. )
= bµi. − bµi. − bµ.. + bµ.. = 0
19.45.
L=
·
2 Y
2 Y
2
Y
1
1
exp − 2 (Yijk − µij )2
2
1/2
2σ
i=1 j=1 k=1 (2πσ )
·
¸
¸
1
1 PPP
exp
−
(Yijk − µij )2
(2πσ 2 )4
2σ 2
1 PPP
loge L = −4 loge (2πσ 2 ) − 2
(Yijk − µij )2
2σ
∂(loge L)
2 P
= − 2 (Yijk − µij )(−1)
∂µij
2σ k
Setting the derivatives equal to zero, simplifying, and solving for the maximum
=
likelihood estimators µ̂ij yields:
2
X
µ̂ij =
Yijk
k=1
2
= Ȳij.
Yes
19.46.
PPP
Q=
(Yijk − µij )2
X
∂Q
= 2 (Yijk − µij )(−1)
∂µij
k
19-10
Setting the derivatives equal to zero, simplifying, and solving for the least squares
estimators yields:
P
Yijk
k
µ̂ij =
= Ȳij.
n
PPP
19.47.
(Ȳij. − Ȳ... )2 =
=
PPP
PPP
[(Ȳi.. − Ȳ... ) + (Ȳ.j. − Ȳ... ) + (Ȳij. − Ȳi.. − Ȳ.j. + Ȳ... )]2
[(Ȳi.. − Ȳ... )2 + (Ȳ.j. − Ȳ... )2 + (Ȳij. − Ȳi.. − Ȳ.j. + Ȳ... )2
+2(Ȳi.. − Ȳ... )(Ȳ.j. − Ȳ... ) + 2(Ȳi.. − Ȳ... )(Ȳij. − Ȳi.. − Ȳ.j. + Ȳ... )
+2(Ȳ.j. − Ȳ... )(Ȳij. − Ȳi.. − Ȳ.j. + Ȳ... )]
= nb
P
(Ȳi.. −Ȳ... )2 +na
P
(Ȳ.j. −Ȳ... )2 +n
PP
(Ȳij. −Ȳi.. −Ȳ.j. +Ȳ... )2
since all summations of cross-product terms equal zero.
19.48.
E{L̂} = E{
P
cj Ȳ.j. } =
P
σ 2 {L̂} = σ 2 {
=
P
cj Ȳ.j. } =
cj E{Ȳ.j. } =
P
P 2 2
cj σ {Ȳ.j. }
because of independence
σ2 P 2
P 2 σ2
cj
=
cj
an
an
PP
PP 2 2
σ2{
cij Ȳij. } =
cij σ {Ȳij. }
19.49.
cj µ.j = L
=
because of independence
σ2 P P 2
P P 2 σ2
cij
=
cij
n
n
By (19.9a), (αβ)11 + (αβ)21 = 0; hence (αβ)21 = −(αβ)11 .
19.50.
By (19.9b), (αβ)11 + (αβ)12 = 0; hence (αβ)12 = −(αβ)11 .
19.51. a.
Ȳ11. = 10.05875, Ȳ12. = 11.45500, Ȳ21. = 9.84000, Ȳ22. = 9.57250,
Ȳ31. = 9.68250, Ȳ32. = 9.52375, Ȳ41. = 8.21250, Ȳ42. = 8.01500
d.
r = .996
19.52. b.
Source
SS
df
MS
Treatments
65.08508 7 9.29787
A (region)
56.74396 3 18.91465
B (average age)
.59676 1
.59676
AB interactions
7.74436 3 2.58145
Error
76.03013 56 1.35768
Total
141.11521 63
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = 2.58145/1.35768 =
1.90, F (.95; 3, 56) = 2.77. If F ∗ ≤ 2.77 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .14
d.
H0 : all αi equal zero (i = 1, ..., 4), Ha : not all αi equal zero. F ∗ = 18.91465/1.35768 =
13.93, F (.95; 3, 56) = 2.77. If F ∗ ≤ 2.77 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
19-11
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero. F ∗ = .59676/1.35768 = .44,
F (.95; 1, 56) = 4.01. If F ∗ ≤ 4.01 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .51
e.
19.53. a.
α ≤ .143
Ȳ11. = .0359, Ȳ12. = .0454, Ȳ21. = .0516, Ȳ22. = .0515,
Ȳ31. = .0758, Ȳ32. = .1015, Ȳ41. = .0673, Ȳ42. = .0766
d.
19.54. a.
r = .993
Ȳ1.. = .0406, Ȳ2.. = .0515, Ȳ3.. = .0886, Ȳ4.. = .0719
Ȳ.1. = .0576, Ȳ.2. = .0687
b.
c.
Source
SS
df
MS
A
.019171 3 .006390
B
.001732 1 .001732
AB
.001205 3 .000402
Error
.011042 48 .000230
Total
.033151 55
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = .000402/.000230 = 1.75, F (.99; 3, 48) = 4.22.
If F ∗ ≤ 4.22 conclude H0 , otherwise Ha . Conclude H0 . P -value = .170
d.
(i) Test for factor A (region effect)
H0 : all αi equal zero (i = 1, ..., 4), Ha : not all αi equal zero.
F ∗ = .006390/.000230 = 27.79, F (.99; 3, 48) = 4.22.
If F ∗ ≤ 4.22 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
(ii) Test for factor B (% below poverty)
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = .001732/.000230 = 7.53, F (.99; 1, 48) = 7.19.
If F ∗ ≤ 7.19 conclude H0 , otherwise Ha . Conclude Ha . P -value = .009
e.
19.55. a.
d.
19.56. a.
α ≤ .030
Ȳ11. = 2.4386, Ȳ12. = 2.4286, Ȳ21. = 2.7286, Ȳ22. = 2.9786,
r = .994
Ȳ1.. = 2.4336, Ȳ2.. = 2.8536
Ȳ.1. = 2.5836, Ȳ.2. = 2.7036
b.
Source
SS df
MS
A
1.2348 1 1.2348
B
.1008 1 .1008
AB
.1183 1 .1183
Error
.5341 24 .0223
Total
1.9880 27
19-12
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = .1183/.0223 = 5.32, F (.95; 1, 24) = 4.26.
If F ∗ ≤ 4.26 conclude H0 , otherwise Ha . Conclude Ha . P -value = .030
d.
(i) Test for factor A (region effect)
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero.
F ∗ = 1.2348/.0223 = 55.48, F (.95; 1, 24) = 4.26.
If F ∗ ≤ 4.26 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
(ii) Test for factor B (% below poverty)
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = .1008/.0223 = 4.53, F (.95; 1, 24) = 4.26.
If F ∗ ≤ 4.26 conclude H0 , otherwise Ha . Conclude Ha . P -value = .044
e.
19.57 b.
α ≤ .143
D̂1 = Ȳ1.. − Ȳ2.. = 1.0506, D̂2 = Ȳ1.. − Ȳ3.. = 1.1538,
D̂3 = Ȳ1.. − Ȳ4.. = 2.6431, D̂4 = Ȳ2.. − Ȳ3.. = .1032,
D̂5 = Ȳ2.. − Ȳ4.. = 1.5925, D̂6 = Ȳ3.. − Ȳ4.. = 1.4893,
s{D̂i } = .41196 (i = 1, ..., 6), q(.90; 4, 56) = 3.31, T = 2.341
1.0506 ± 2.341(.41196)
1.1538 ± 2.341(.41196)
2.6431 ± 2.341(.41196)
.1032 ± 2.341(.41196)
1.5925 ± 2.341(.41196)
1.4893 ± 2.341(.41196)
19.58. b.
.0862 ≤ D1
.1894 ≤ D2
1.6787 ≤ D3
−.8612 ≤ D4
.6281 ≤ D5
.5249 ≤ D6
≤ 2.0150
≤ 2.1182
≤ 3.6075
≤ 1.0676
≤ 2.5569
≤ 2.4537
D̂1 = Ȳ1.. − Ȳ2.. = −.0109, D̂2 = Ȳ1.. − Ȳ3.. = −.0480,
D̂3 = Ȳ1.. − Ȳ4.. = −.0313, D̂4 = Ȳ2.. − Ȳ3.. = −.0371,
D̂5 = Ȳ2.. − Ȳ4.. = −.0204, D̂6 = Ȳ3.. − Ȳ4.. = .0167,
s{D̂i } = .005732 (i = 1, ..., 6), q(.95; 4, 48) = 3.79, T = 2.680
−.0109 ± 2.680(.0057)
−.0480 ± 2.680(.0057)
−.0313 ± 2.680(.0057)
−.0371 ± 2.680(.0057)
−.0204 ± 2.680(.0057)
.0167 ± 2.680(.0057)
−.0262 ≤ D1 ≤ .0043
−.0633 ≤ D2 ≤ −.0328
−.0466 ≤ D3 ≤ −.0161
−.0524 ≤ D4 ≤ −.0219
−.0356 ≤ D5 ≤ −.0052
.0015 ≤ D6 ≤ .0320
19-13
19-14
Chapter 20
TWO-FACTOR STUDIES – ONE
CASE PER TREATMENT
20.1.
0
20.2. b.
Source
SS
df
Location 37.0050 3
Week
47.0450 1
Error
.3450 3
Total
84.3950 7
MS
12.3350
47.0450
.1150
H0 : all αi equal zero (i = 1, ..., 4), Ha : not all αi equal zero.
F ∗ = 12.3350/.1150 = 107.26, F (.95; 3, 3) = 9.28. If F ∗ ≤ 9.28 conclude H0 ,
otherwise Ha . Conclude Ha . P -value = .0015
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = 47.0450/.1150 = 409.09, F (.95; 1, 3) = 10.1. If F ∗ ≤ 10.1 conclude H0 ,
otherwise Ha . Conclude Ha . P -value = .0003. α ≤ .0975
c.
D̂1 = Ȳ1. − Ȳ2. = 18.95 − 14.55 = 4.40, D̂2 = Ȳ1. − Ȳ3. = 18.95 − 14.60 = 4.35,
D̂3 = Ȳ1. −Ȳ4. = 18.95−18.80 = .15, D̂4 = Ȳ2. −Ȳ3. = −.05, D̂5 = Ȳ2. −Ȳ4. = −4.25,
D̂6 = Ȳ3. − Ȳ4. = −4.20, D̂7 = Ȳ.1 − Ȳ.2 = 14.30 − 19.15 = −4.85, s{D̂i } = .3391
(i = 1, ..., 6), s{D̂7 } = .2398, B = t(.99286; 3) = 5.139
4.40 ± 5.139(.3391)
4.35 ± 5.139(.3391)
.15 ± 5.139(.3391)
−.05 ± 5.139(.3391)
−4.25 ± 5.139(.3391)
−4.20 ± 5.139(.3391)
−4.85 ± 5.139(.2398)
20.3. a.
2.66 ≤ D1 ≤ 6.14
2.61 ≤ D2 ≤ 6.09
−1.59 ≤ D3 ≤ 1.89
−1.79 ≤ D4 ≤ 1.69
−5.99 ≤ D5 ≤ −2.51
−5.94 ≤ D6 ≤ −2.46
−6.08 ≤ D7 ≤ −3.62
µ̂32 = Ȳ3. + Ȳ.2 − Ȳ.. = 14.600 + 19.150 − 16.725 = 17.025
b.
s2 {µ̂32 } = .071875
c.
s{µ̂32 } = .2681, t(.975; 3) = 3.182, 17.025 ± 3.182(.2681), 16.172 ≤ µ32 ≤ 17.878
20-1
D̂ = (−4.13473)/(18.5025)(11.76125) = −.019, SSAB ∗ = .0786, SSRem∗ =
.2664.
20.4.
H0 : D = 0, Ha : D 6= 0. F ∗ = (.0786/1) ÷ (.2664/2) = .59, F (.975; 1, 2) = 38.5.
If F ∗ ≤ 38.5 conclude H0 , otherwise Ha . Conclude H0 .
20.5. b.
Source
SS
df
Type of group
1.125 1
Size of group
318.375 3
Error
6.375 3
Total
325.875 7
MS
1.125
106.125
2.125
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero. F ∗ = 1.125/2.125 = .53,
F (.99; 1, 3) = 34.1. If F ∗ ≤ 34.1 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .52
H0 : all βj equal zero (j = 1, ..., 4), Ha : not all βj equal zero. F ∗ = 106.125/2.125 =
49.94, F (.99; 3, 3) = 29.5. If F ∗ ≤ 29.5 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .005. α ≤ .0199
c.
D̂1 = Ȳ.2 − Ȳ.1 = 22.5 − 16.5 = 6.0, D̂2 = Ȳ.3 − Ȳ.2 = 30.0 − 22.5 = 7.5, D̂3 =
Ȳ.4 − Ȳ.3 = 32.5−30.0 = 2.5, s{D̂i } = 1.4577 (i = 1, 2, 3), B = t(.99167; 3) = 4.857
6.0 ± 4.857(1.4577)
7.5 ± 4.857(1.4577)
2.5 ± 4.857(1.4577)
d.
20.6. a.
−1.08 ≤ D1 ≤ 13.08
.42 ≤ D2 ≤ 14.58
−4.58 ≤ D3 ≤ 9.58
No, q(.95; 4, 3) = 6.82, T = 4.822
µ̂14 = Ȳ1. + Ȳ.4 − Ȳ.. = 25.750 + 32.500 − 25.375 = 32.875
b.
s2 {µ̂14 } = 1.3281
c.
s{µ̂14 } = 1.1524, t(.995; 3) = 5.841, 32.875 ± 5.841(1.1524), 26.144 ≤ µ14 ≤ 39.606
20.7.
D̂ = (−8.109375)/(.28125)(159.1875) = −.1811, SSAB ∗ = 1.4688,
SSRem∗ = 4.9062. H0 : D = 0, Ha : D 6= 0. F ∗ = (1.4688/1) ÷ (4.9062/2) = .60,
F (.99; 1, 2) = 98.5. If F ∗ ≤ 98.5 conclude H0 , otherwise Ha . Conclude H0 .
20.8. b.
Source
SS
df
Humidity
2.12167 2
Temperature 202.20000 3
Error
6.58500 6
Total
210.90667 11
MS
1.06083
67.40000
1.09750
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero.
F ∗ = 1.06083/1.09750 = .97, F (.975; 2, 6) = 7.26. If F ∗ ≤ 7.26 conclude H0 ,
otherwise Ha . Conclude H0 . P -value = .43
H0 : all βj equal zero (j = 1, ..., 4), Ha : not all βj equal zero.
20-2
F ∗ = 67.40000/1.09750 = 61.41, F (.975; 3, 6) = 6.60. If F ∗ ≤ 6.60 conclude H0 ,
otherwise Ha . Conclude Ha . P -value = 0+
c.
D̂1 = Ȳ.2 − Ȳ.1 = 15.30 − 14.90 = .40, D̂2 = Ȳ.3 − Ȳ.2 = 20.70 − 15.30 = 5.40,
D̂3 = Ȳ.4 − Ȳ.3 = 24.83 − 20.70 = 4.13, s{D̂i } = .8554 (i = 1, 2, 3), B =
t(.99167; 6) = 3.2875
.40 ± 3.2875(.8554)
5.40 ± 3.2875(.8554)
4.13 ± 3.2875(.8554)
d.
20.9. a.
20.10.
−2.41 ≤ D1 ≤ 3.21
2.59 ≤ D2 ≤ 8.21
1.32 ≤ D3 ≤ 6.94
Yes
µ̂23 = Ȳ2. + Ȳ.3 − Ȳ.. = 19.325 + 20.700 − 18.933 = 21.092
b.
s2 {µ̂23 } = .54875
c.
s{µ̂23 } = .7408, t(.99; 6) = 3.143, 21.092 ± 3.143(.7408), 18.764 ≤ µ23 ≤ 23.420,
(2.66%, 4.12%)
D̂ = (−8.27113)/(.5304)(67.4000) = −.2314, SSAB ∗ = 1.9137, SSRem∗ =
4.6713.
H0 : D = 0, Ha : D 6= 0. F ∗ = (1.9137/1) ÷ (4.6713/5) = 2.05,
F (.995; 1, 5) = 22.8. If F ∗ ≤ 22.8 conclude H0 , otherwise Ha . Conclude H0 .
20.11.
SSA = b
P
(Ȳi. − Ȳ.. )2 , SSB = a
P
(Ȳ.j − Ȳ.. )2
20.12.
(αβ)ij = A + Bαi + Cβj + Dαi βj + Eαi2 + F βj2
Averaging (1) over i yields:
P
(αβ).j = A + Cβj + E αi2 /a + F βj2 = 0
P
P
because αi = 0 and (αβ)ij = 0. Similarly:
i
P
(αβ)i. = A + Bαi + Eαi2 + F βj2 /b
P
P
because βj = 0 and (αβ)ij = 0. From (2) and (3) we obtain:
j
P
Cβj + F βj2 = −A − E αi2 /a
P
Bαi + Eαi2 = −A − F βj2 /b
Substituting (4) and (5) in (1) yields:
P
P
(αβ)ij = −A − E αi2 /a − F βj2 /b + Dαi βj
Averaging (6) over j yields:
P
P
(αβ)i. = −A − E αi2 /a − F βj2 /b = 0
Using (7) in (6) yields:
(αβ)ij = Dαi βj
20-3
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
20-4
Chapter 21
RANDOMIZED COMPLETE
BLOCK DESIGNS
21.5. b.
eij :
i
j=1
1 −2.50000
2
1.50000
3
2.16667
4
.16667
5
4.16667
6
1.50000
7 −1.50000
8 −2.83333
9 −1.50000
10 −1.16667
r = .984
d.
j=2
1.50000
−.50000
−.83333
−.83333
−4.83333
−.50000
−1.50000
3.16667
2.50000
1.83333
j=3
1.00000
−1.00000
−1.33333
.66667
.66667
−1.00000
3.00000
−.33333
−1.00000
−.66667
H0 : D = 0, Ha : D 6= 0. SSBL.T R∗ = .13, SSRem∗ = 112.20,
F ∗ = (.13/1) ÷ (112.20/17) = .02, F (.99; 1, 17) = 8.40. If F ∗ ≤ 8.40 conclude H0 ,
otherwise Ha . Conclude H0 . P -value = .89
21.6. a.
Source
SS
df
MS
Blocks
433.36667 9 48.15185
Training methods 1, 295.00000 2 647.50000
Error
112.33333 18
6.24074
Total
1, 840.70000 29
b.
Ȳ.1 = 70.6, Ȳ.2 = 74.6, Ȳ.3 = 86.1
c.
H0 : all τj equal zero (j = 1, 2, 3), Ha : not all τj equal zero.
F ∗ = 647.50000/6.24074 = 103.754, F (.95; 2, 18) = 3.55. If F ∗ ≤ 3.55 conclude
H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
D̂1 = Ȳ.1 − Ȳ.2 = −4.0, D̂2 = Ȳ.1 − Ȳ.3 = −15.5, D̂3 = Ȳ.2 − Ȳ.3 = −11.5,
s{D̂i } = 1.1172 (i = 1, 2, 3), q(.90; 3, 18) = 3.10, T = 2.192
21-1
−4.0 ± 2.192(1.1172)
−15.5 ± 2.192(1.1172)
−11.5 ± 2.192(1.1172)
e.
−6.45 ≤ D1 ≤ −1.55
−17.95 ≤ D2 ≤ −13.05
−13.95 ≤ D3 ≤ −9.05
H0 : all ρi equal zero (i = 1, ..., 10), Ha : not all ρi equal zero.
F ∗ = 48.15185/6.24074 = 7.716, F (.95; 9, 18) = 2.46.
If F ∗ ≤ 2.46 conclude H0 , otherwise Ha . Conclude Ha . P -value = .0001
21.7. b.
eij :
i
1
2
3
4
5
j=1
−.05267
−.01267
.00400
−.02267
.08400
j=2
.00533
−.00467
−.00800
.01533
−.00800
j=3
.04733
.01733
.00400
.00733
−.07600
r = .956
d.
H0 : D = 0, Ha : D 6= 0. SSBL.T R∗ = .0093, SSRem∗ = .01002,
F ∗ = (.0093/1) ÷ (.01002/7) = 6.50, F (.99; 1, 7) = 12.2.
If F ∗ ≤ 12.2 conclude H0 , otherwise Ha . Conclude H0 . P -value = .038
21.8. a.
Source
SS
df
Blocks
1.41896 4
Fat content 1.32028 2
Error
.01932 8
Total
2.75856 14
MS
.35474
.66014
.002415
b.
Ȳ.1 = 1.110, Ȳ.2 = .992, Ȳ.3 = .430
c.
H0 : all τj equal zero (j = 1, 2, 3), Ha : not all τj equal zero.
F ∗ = .66014/.002415 = 273.35, F (.95; 2, 8) = 4.46.
If F ∗ ≤ 4.46 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
D̂1 = .118, D̂2 = .562, s{D̂i } = .03108 (i = 1, 2), B = t(.9875; 8) = 2.7515
.118 ± 2.7515(.03108)
.562 ± 2.7515(.03108)
e.
.032 ≤ D1 ≤ .204
.476 ≤ D2 ≤ .648
H0 : all ρi equal zero (i = 1, ..., 5), Ha : not all ρi equal zero.
F ∗ = .35474/.002415 = 146.89, F (.95; 4, 8) = 3.84.
If F ∗ ≤ 3.84 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
21.9. c.
ejik :
21-2
i
1
2
3
4
5
6
7
8
k=1
j=1
j=2
−.01875
.01875
.13125 −.03125
.05625 −.00625
.08125 −.08125
−.09375 −.15625
−.09375 −.05625
−.09375
.24375
.03125
.06875
i
1
2
3
4
5
6
7
8
k=2
j=1
j=2
.00625 −.00625
−.04375 −.05625
−.11875
.06875
.00625 −.00625
.23125
.01875
.03125
.11875
.03125 −.18125
−.14375
.04375
r = .984
e.
H0 : D = 0, Ha : D 6= 0. SSBL.T R∗ = .00503, SSRem∗ = .29872,
F ∗ = (.00503/1) ÷ (.29872/20) = .337, F (.99; 1, 20) = 8.10.
If F ∗ ≤ 8.10 conclude H0 , otherwise Ha . Conclude H0 . P -value = .57
21.10. a.
Source
SS
df
MS
Blocks
5.59875 7
.79982
A
2.31125 1 2.31125
B
3.38000 1 3.38000
AB interactions
.04500 1
.04500
Error
.30375 21 .01446
Total
11.63875 31
b.
H0 : all (αβ)jk equal zero, Ha : not all (αβ)jk equal zero.
F ∗ = .04500/.01446 = 3.112, F (.99; 1, 21) = 8.017.
If F ∗ ≤ 8.017 conclude H0 , otherwise Ha . Conclude H0 . P -value = .092
c.
Ȳ.1. = .88750, Ȳ.2. = 1.42500, Ȳ..1 = .83125, Ȳ..2 = 1.42500
d.
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero.
F ∗ = 2.31125/.01446 = 159.84, F (.99; 1, 21) = 8.017.
If F ∗ ≤ 8.017 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = 3.38000/.01446 = 233.75, F (.99; 1, 21) = 8.017.
If F ∗ ≤ 8.017 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
L̂1 = Ȳ.1. − Ȳ.2. = −.53750, L̂2 = Ȳ..1 − Ȳ..2 = −.65000, s{L̂1 } = s{L̂2 } = .0425,
B = t(.9875; 21) = 2.414
−.53750 ± 2.414(.0425)
−.65000 ± 2.414(.0425)
f.
−.640 ≤ L1 ≤ −.435
−.753 ≤ L2 ≤ −.547
H0 : all ρi equal zero (i = 1, ..., 8), Ha : not all ρi equal zero.
F ∗ = .79982/.01446 = 55.31, F (.99; 7, 21) = 3.64.
If F ∗ ≤ 3.64 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
21-3
21.12. b.
Ȳ.1.. = 7.25, Ȳ.2.. = 12.75, L̂ = Ȳ.1.. − Ȳ.2.. = −5.50, s{L̂} = 1.25,
t(.995; 8) = 3.355, −5.50 ± 3.355(1.25), −9.69 ≤ L ≤ −1.31
21.13. a.
Yijk = µ.. + ρi + τj + (ρτ )ij + ²ijk
b.
Source
SS
df
MS
Blocks
523.20000 4 130.80000
Treatments
1, 796.46667 2 898.23333
BL.T R interactions
87.20000 8 10.90000
Error
207.00000 15 13.80000
Total
2, 613.86667 29
c.
H0 : all τj equal zero (j = 1, 2, 3), Ha : not all τj equal zero.
F ∗ = 898.23333/13.80000 = 65.089, F (.99; 2, 15) = 6.36.
If F ∗ ≤ 6.36 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
e.
Ȳ.1. = 68.9, Ȳ.2. = 77.1, Ȳ.3. = 87.8, L̂1 = Ȳ.1. − Ȳ.2. = −8.2, L̂2 = Ȳ.1. − Ȳ.3. = −18.9,
L̂3 = Ȳ.2. − Ȳ.3. = −10.7, s{L̂i } = 1.6613 (i = 1, 2, 3), q(.95; 3, 15) = 3.67,
T = 2.595
−8.2 ± 2.595(1.6613)
−12.51 ≤ L1 ≤ −3.89
−18.9 ± 2.595(1.6613)
−23.21 ≤ L2 ≤ −14.59
−10.7 ± 2.595(1.6613)
−15.01 ≤ L3 ≤ −6.39
eijk :
i
1
2
3
4
5
j=1
k=1
k=2
1.5
−1.5
2.0
−2.0
4.0
−4.0
−2.5
2.5
1.5
−1.5
j=2
k=1
k=2
3.0
−3.0
−4.0
4.0
3.5
−3.5
−2.5
2.5
−1.5
1.5
j=3
k=1
k=2
−2.0
2.0
−2.5
2.5
1.5
−1.5
−1.5
1.5
3.5
−3.5
r = .956
f.
H0 : all (ρτ )ij equal zero, Ha : not all (ρτ )ij equal zero.
F ∗ = 10.90000/13.80000 = .7899, F (.99; 8, 15) = 4.00.
If F ∗ ≤ 4.00 conclude H0 , otherwise Ha . Conclude H0 . P -value = .62
s
21.14.
1
φ=
2.5
21.15.
1
φ=
.04
21.16.
nb = 49 blocks
21.17. a.
nb = 21 blocks
b.
s
10(18)
= 3.098, ν1 = 2, ν2 = 27, 1 − β > .99
3
5(.02)
= 4.564, ν1 = 2, ν2 = 12, 1 − β > .99
3
nb = 7 blocks
21-4
21.18.
Ê = 3.084
21.19.
Ê 0 = 40.295
21.20.
Ê = 13.264
21.21.
L=
3 Y
2
Y
·
√
i=1 j=1
1
1
exp − 2 (Yij − µ.. − ρi − τj )2
2σ
2πσ 2
loge L = −3 loge 2π − 3 loge σ 2 −
∂(loge L)
∂µ..
∂(loge L)
∂ρi
∂(loge L)
∂τj
Setting each
P
ρi =
P
¸
1 PP
(Yij − µ.. − ρi − τj )2
2σ 2
2 PP
(Yij − µ.. − ρi − τj )(−1)
2σ 2
2 P
= − 2 (Yij − µ.. − ρi − τj )(−1)
2σ j
2 P
= − 2 (Yij − µ.. − ρi − τj )(−1)
2σ i
partial derivative equal to zero, utilizing the constraints
=−
τj = 0, simplifying, and substituting the maximum likelihood
estimators yields:
PP
P
j
P
i
Yij =
Yij =
Yij =
P
j
P
i
PP
Ȳ.. = µ̂..
or Ȳi. − µ̂.. = ρ̂i
(µ̂.. + τ̂ j )
or
E{M ST R} = E
=
or
(µ̂.. + ρ̂i )
(
21.22.
µ̂..
nb
P
Ȳ.j − µ̂.. = τ̂ j
(Ȳ.j − Ȳ.. )2
r−1
)
nb
P
E{ (Ȳ.j − Ȳ.. )2 }
r−1
Since:
(Ȳ.j − Ȳ.. ) = (µ.. + τj + ²̄.j ) − (µ.. + ²̄.. ) = τj + (²̄.j − ²̄.. )
and:
P
(Ȳ.j − Ȳ.. )2 =
P 2 P
P
τj + (²̄.j − ²̄.. )2 + 2 τj (²̄.j − ²̄.. )
we find:
E{
P
P 2
τj } = τj2
Ã
σ2
E { (²̄.j − ²̄.. ) } = (r − 1)
nb
P
E {2
P
!
2
τj (²̄.j − ²̄.. )} = 0
Hence:
¸
·
nb P 2
nb P 2 r − 1 2
E{M ST R} =
σ =
τj +
τj + σ 2
r−1
nb
r−1
21.23.
From (A.69):
21-5
"
W̄
(t ) =
s{W̄ }
From (27.6b):
#2
∗ 2
=
nb (nb − 1)(Ȳ.1 − Ȳ.2 )2
Σ[(Yi1 − Yi2 ) − (Ȳ.1 − Ȳ.2 )]2
"
Ã
Ȳ.1 + Ȳ.2
M ST R = nb Ȳ.1 −
2
nb
= (Ȳ.1 − Ȳ.2 )2
2
From (27.6c):
·
P ³
M SBL.T R =
i
!#2
"
Ã
Ȳ.1 + Ȳ.2
+ Ȳ.2 −
2
Yi1 − Ȳi. − Ȳ.1 + Ȳ..
´2
!#2
¸
2
+ (Yi2 − Ȳi. − Ȳ.2 + Ȳ.. )
(nb − 1)(2 − 1)
Using:
Yi1 + Yi2
2
we obtain:
Ȳi. =
Ȳ.. =
Ȳ.1 + Ȳ.2
2
i
1 P1h
(Yi1 − Yi2 − Ȳ.1 + Ȳ.2 )2 + (Yi2 − Yi1 − Ȳ.2 + Ȳ.1 )2
nb − 1 i 4
i2
1
Ph
=
(Yi1 − Yi2 ) − (Ȳ.1 − Ȳ.2 )
2(nb − 1) i
M SBL.T R =
Therefore:
F∗ =
21.24.
nb (nb − 1)(Ȳ.1 − Ȳ.2 )2
= (t∗ )2
Σ[(Yi1 − Yi2 ) − (Ȳ.1 − Ȳ.2 )]2
When there are no ties:
R..2
P
2
= [nb r(r + 1)/2]
PP
"
Ri.2
= nb
r(r + 1)
2
#2
2
Rij
= nb [r(r + 1)(2r + 1)/6]
Then:
nb (r − 1)SST R
SST R + SSBL.T R Ã
!
R..2
2
(r − 1) ΣR.j −
r
Ã
!
Ã
!
=
2
2
2
2
2
ΣR.j
ΣR
R..
ΣR
R
.j
i.
2
−
+ ΣΣRij
−
−
+ ..
nb
rnb
r
nb
rnb
Ã
P
!
R2
− ..
(r − 1)
r
=
2
ΣRi.
PP 2
Rij −
r
"
# "
#
n2b (r + 1)2 r
nb r(r + 1)(2r + 1) nb r(r + 1)2
P 2
= (r − 1)
÷
−
R.j −
4
6
4
P 2
12 R.j
− 3nb (r + 1)
=
nb r(r + 1)
R.j2
21-6
Chapter 22
ANALYSIS OF COVARIANCE
22.5. a.
B = t(.9917; 11) = 2.820
22.6.
Yij = µ. + τi + γ1 (Xij1 − X̄..1 ) + γ2 (Xij2 − X̄..2 ) + γ3 (Xij1 − X̄..1 )2
+γ4 (Xij2 − X̄..2 )2 + ²ij , i = 1, ..., 4
22.7. a.
eij :
i
1
2
3
j=1
−.5281
−.2635
−.1615
j=2
j=3
.4061
.0089
−.2005
.3196
.2586 −.0099
j=4
j=5
.4573 −.1140
.2995 −.1662
−.3044
.0472
j=6
−.1911
.0680
.1700
i
1
2
j=7
.0660
−.0690
j=8
j=9
−.0939 −.0112
−.1776 −.0005
j = 10
j = 11
j = 12
.0653
.0251
.0995
b.
r = .988
c.
Yij = µ. + τ1 Iij1 + τ2 Iij2 + γxij + β1 Iij1 xij + β2 Iij2 xij + εij
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
SSE(F ) = .9572, SSE(R) = 1.3175,
F ∗ = (.3603/2) ÷ (.9572/21) = 3.95, F (.99; 2, 21) = 5.78.
If F ∗ ≤ 5.78 conclude H0 , otherwise Ha . Conclude H0 . P -value = .035
d.
22.8. b.
Yes, 5
Full model: Yij = µ. + τ1 Iij1 + τ2 Iij2 + γxij + εij , (X̄.. = 9.4).
Reduced model: Yij = µ. + γxij + εij .
c.
Full model: Ŷ = 7.80627 + 1.65885I1 − .17431I2 + 1.11417x, SSE(F ) = 1.3175
Reduced model: Ŷ = 7.95185 + .54124x, SSE(R) = 5.5134
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (4.1959/2) ÷ (1.3175/23) = 36.625, F (.95; 2, 23) = 3.42.
If F ∗ ≤ 3.42 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
M SE(F ) = .0573, M SE = .6401
22-1
e.
Ŷ = µ̂. + τ̂ 2 − .4γ̂ = 7.18629, s2 {µ̂. } = .00258, s2 {τ̂ 2 } = .00412, s2 {γ̂} = .00506,
s{µ̂. , τ̂ 2 } = −.00045, s{τ̂ 2 , γ̂} = −.00108, s{µ̂. , γ̂} = −.00120, s{Ŷ } = .09183,
t(.975; 23) = 2.069, 7.18629 + 2.069(.09183), 6.996 ≤ µ. + τ2 − .4γ ≤ 7.376
f.
D̂1 = τ̂ 1 − τ̂ 2 = 1.83316, D̂2 = τ̂ 1 − τ̂ 3 = 2τ̂ 1 + τ̂ 2 = 3.14339, D̂3 = τ̂ 2 − τ̂ 3 =
2τ̂ 2 + τ̂ 1 = 1.31023, s2 {τ̂ 1 } = .03759, s{τ̂ 1 , τ̂ 2 } = −.00418, s{D̂1 } = .22376,
s{D̂2 } = .37116, s{D̂3 } = .19326, F (.90; 2, 23) = 2.55, S = 2.258
1.83316 ± 2.258(.22376)
3.14339 ± 2.258(.37116)
1.31023 ± 2.258(.19326)
22.9. a.
1.328 ≤ D1 ≤ 2.338
2.305 ≤ D2 ≤ 3.981
.874 ≤ D3 ≤ 1.747
eij :
i
1
2
3
j=1
−.5474
−.2747
−.4225
j=2 j=3 j=4
−.1325 .2465 −.0567
.1215 .2655
.0346
−.0128 .1090
.5290
j=5
.4901
−.1468
−.2027
b.
r = .994
c.
Yij = µ. + τ1 Iij1 + τ2 Iij2 + γxij + β1 Iij1 xij + β2 Iij2 xij + εij .
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
SSE(F ) = .7682, SSE(R) = 1.3162,
F ∗ = (.5480/2) ÷ (.7682/9) = 3.21, F (.995; 2, 9) = 10.1.
If F ∗ ≤ 10.1 conclude H0 , otherwise Ha . Conclude H0 . P -value = .089
d.
22.10. b.
No
Full model: Yij = µ. + τ1 Iij1 + τ2 Iij2 + γxij + εij , (X̄.. = 280).
Reduced model: Yij = µ. + γxij + εij .
c.
Full model: Ŷ = 29.00000 + .14361I1 + 1.48842I2 − .02981x, SSE(F ) = 1.3162
Reduced model: Ŷ = 29.00000 − .02697x, SSE(R) = 24.7081
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (23.3919/2) ÷ (1.3162/11) = 97.748, F (.90; 2, 11) = 2.86.
If F ∗ ≤ 2.86 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
M SE(F ) = .1197, M SE = 9.70
e.
Ŷ = µ̂. + τ̂ 1 = 29.14361, s2 {µ̂. } = .00798, s2 {τ̂ 1 } = .01602, s{µ̂. , τ̂ 1 } = 0, s{Ŷ } =
.15492, t(.95; 11) = 1.796, 29.14361 ± 1.796(.15492), 28.865 ≤ µ. + τ1 ≤ 29.422
f.
D̂1 = τ̂ 1 − τ̂ 2 = −1.34481, D̂2 = τ̂ 1 − τ̂ 3 = 2τ̂ 1 + τ̂ 2 = 1.77564, D̂3 = τ̂ 2 − τ̂ 3 =
2τ̂ 2 + τ̂ 1 = 3.12045, s2 {τ̂ 2 } = .01678, s{τ̂ 1 , τ̂ 2 } = −.00822, s{D̂1 } = .2219,
s{D̂2 } = .2190, s{D̂3 } = .2242, F (.90; 2, 11) = 2.86, S = 2.392
−1.34481 ± 2.392(.2219)
1.77564 ± 2.392(.2190)
3.12045 ± 2.392(.2242)
22.11. a.
−1.876 ≤ D1 ≤ −.814
1.252 ≤ D2 ≤ 2.299
2.584 ≤ D3 ≤ 3.657
eij :
22-2
i
1
2
3
j=1
.2070
.1361
.7938
i
1
2
3
j=6
.1178
−.3813
−.0545
j=2
−.4503
.0001
−.2099
j=3
j=4
−.3648 −.2324
−.6691 −.9300
.2295
.2801
j=7 j=8 j=9
−.5440 .3668
.4201 .5837 −.1635
j=5
.8999
.3190
−1.0389
j = 10
.6848
b.
r = .995
c.
Yij = µ. + τ1 Iij1 + τ2 Iij2 + γxij + β1 Iij1 xij + β2 Iij2 xij + εij
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
SSE(F ) = 5.94391, SSE(R) = 6.16575,
F ∗ = (.221834/2) ÷ (5.94391/18) = .336, F (.95; 2, 18) = 3.55.
If F ∗ ≤ 3.55 conclude H0 , otherwise Ha . Conclude H0 . P -value = .72
d.
22.12. b.
No
Full model: Yij = µ. + τ1 Iij1 + τ2 Iij2 + γxij + εij , (X̄.. = 23.575).
Reduced model: Yij = µ. + γxij + εij .
c.
Full model: Ŷ = 31.42704 + 3.52342I1 + 1.67605I2 + 1.16729x, SSE(F ) = 6.16575
Reduced model: Ŷ = 32.00000 + 1.47113x, SSE(R) = 252.24945
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (246.08370/2) ÷ (6.16575/20) = 399.114, F (.99; 2, 20) = 5.85.
If F ∗ ≤ 5.85 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
M SE(F ) = .30829, M SE = 19.8095
e.
Ŷ = µ̂. + τ̂ 2 + .425γ̂ = 33.59919, s2 {µ̂. } = .013423, s2 {τ̂ 2 } = .024459, s2 {γ̂} =
.001025, s{µ̂. , τ̂ 2 } = −.003069, s{µ̂. , γ̂} = .000082, s{τ̂ 2 , γ̂} = .000886, s{Ŷ } =
.180975, t(.995; 20) = 2.845, 33.59919±2.845(.180975), 33.0843 ≤ µ. +τ2 +.425γ ≤
34.1141
f.
D̂1 = τ̂ 1 − τ̂ 2 = 1.84738, D̂2 = τ̂ 1 − τ̂ 3 = 2τ̂ 1 + τ̂ 2 = 8.72289, D̂3 = τ̂ 2 − τ̂ 3 =
2τ̂ 2 + τ̂ 1 = 6.87551, s2 {τ̂ 1 } = .0336934, s{τ̂ 1 , τ̂ 2 } = −.0120919, s{D̂1 } = .28705,
s{D̂2 } = .33296, s{D̂3 } = .28838, B = t(.99167; 20) = 2.613
1.84738 ± 2.613(.28705)
8.72289 ± 2.613(.33296)
6.87551 ± 2.613(.28838)
22.13. a.
1.097 ≤ D1 ≤ 2.597
7.853 ≤ D2 ≤ 9.529
6.122 ≤ D3 ≤ 7.629
eij :
i
1
2
3
j=1
j=2
−1.7973 −6.7636
−3.4017 −4.9059
−3.0314 −2.5019
j=3 j=4 j=5
2.2280 .5922 5.7406
1.4415 3.9373 2.9288
.7781 2.6297 2.1255
22-3
b.
r = .983
c.
Yij = µ. + τ1 Iij1 + τ2 Iij2 + γxij + β1 Iij1 xij + β2 Iij2 xij + εij
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
SSE(F ) = 145.2007, SSE(R) = 176.5300,
F ∗ = (31.3293/2) ÷ (145.2007/9) = .971, F (.95; 2, 9) = 4.26.
If F ∗ ≤ 4.26 conclude H0 , otherwise Ha . Conclude H0 . P -value = .415
d.
22.14. b.
No
Full model: Yij = µ. + τ1 Iij1 + τ2 Iij2 + γxij + εij , (X̄.. = 70.46667).
Reduced model: Yij = µ. + γxij + εij .
c.
Full model: Ŷ = 66.40000−13.57740I1 +5.54806I2 +.83474x, SSE(F ) = 176.5300
Reduced model: Ŷ = 66.40000 + .81587x, SSE(R) = 1, 573.8109
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (1, 397.2809/2) ÷ (176.5300/11) = 43.53, F (.95; 2, 11) = 3.98.
If F ∗ ≤ 3.98 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
M SE(F ) = 16.0482, M SE = 113.9333
e.
Ŷ = µ̂. + τ̂ 2 + 4.5333γ̂ = 75.7322, s2 {µ̂. } = 1.06988, s2 {τ̂ 2 } = 2.40689, s2 {γ̂} =
.00939, s{µ̂. , τ̂ 2 } = s{µ̂. , γ̂} = 0, s{τ̂ 2 , γ̂} = −.05009, s{Ŷ } = 1.7932, t(.975; 11) =
2.201, 75.7322 ± 2.201(1.7932), 71.785 ≤ µ. + τ2 + 4.5333γ ≤ 79.679
f.
D̂1 = τ̂ 1 − τ̂ 2 = −19.12546, D̂2 = τ̂ 1 − τ̂ 3 = 2τ̂ 1 + τ̂ 2 = −21.60674, D̂3 = τ̂ 2 − τ̂ 3 =
2τ̂ 2 + τ̂ 1 = −2.48128, s2 {τ̂ 1 } = 2.14043, s{τ̂ 1 , τ̂ 2 } = −1.08324, s{D̂1 } = 2.5911,
s{D̂2 } = 2.5760, s{D̂3 } = 2.7267, F (.90; 2, 11) = 2.86, S = 2.392
−19.12546 ± 2.392(2.5911)
−21.60674 ± 2.392(2.5760)
−2.48128 ± 2.392(2.7267)
22.15. a.
−25.323 ≤ D1 ≤ −12.928
−27.769 ≤ D2 ≤ −15.445
−9.004 ≤ D3 ≤ 4.041
eijk :
i
1
j=1
−.1184
−.3469
.0041
−.6041
1.2000
−.1347
j=2
−.3510
−.0939
.0286
.0735
−.0163
.3592
i j=1
j=2
2 −.6809
.2082
.8660 −.1877
−.1177
.2531
.2905 −.2327
−.3912 −.2776
.0333
.2367
i
3
b.
r = .974
c.
Yijk = µ.. + α1 Iijk1 + α2 Iijk2 + β1 Iijk3 + (αβ)11 Iijk1 Iijk3
j=1
.9687
−.0150
.8789
−1.1211
.0912
−.8027
+(αβ)21 Iijk2 Iijk3 + γxijk + δ1 Iijk1 xijk + δ2 Iijk2 xijk
+δ3 Iijk3 xijk + δ4 Iijk1 Iijk3 xijk + δ5 Iijk2 Iijk3 xijk + ²ijk
H0 : all δi equal zero (i = 1, ..., 5), Ha : not all δi equal zero.
22-4
j=2
.6606
.0565
−.1109
−.0660
−.4293
−.1109
SSE(R) = 8.2941, SSE(F ) = 6.1765,
F ∗ = (2.1176/5) ÷ (6.1765/24) = 1.646, F (.99; 5, 24) = 3.90.
If F ∗ ≤ 3.90 conclude H0 , otherwise Ha . Conclude H0 . P -value = .19
22.16. a.
Yijk = µ.. + α1 Iijk1 + α2 Iijk2 + β1 Iijk3 + (αβ)11 Iijk1 Iijk3
+(αβ)21 Iijk2 Iijk3 + γxijk + ²ijk
1 if case from level 1 for factor A
Iijk1 = −1 if case from level 3 for factor A
0 otherwise
Iijk2
1 if case from level 2 for factor A
−1 if case from level 3 for factor A
=
0 otherwise
(
Iijk3 =
1 if case from level 1 for factor B
−1 if case from level 2 for factor B
xijk = Xijk − X̄...
(X̄... = 3.4083)
Ŷ = 23.55556−2.15283I1 +3.68152I2 +.20907I3 −.06009I1 I3 −.04615I2 I3 +1.06122x
SSE(F ) = 8.2941
b.
Interactions:
Yijk = µ.. + α1 Iijk1 + α2 Iijk2 + β1 Iijk3 + γxijk + ²ijk
Ŷ = 23.55556 − 2.15400I1 + 3.67538I2 + .20692I3 + 1.07393x
SSE(R) = 8.4889
Factor A:
Yijk = µ.. + β1 Iijk3 + (αβ)11 Iijk1 Iijk3 + (αβ)21 Iijk2 Iijk3 + γxijk + ²ijk
Ŷ = 23.55556 + .12982I3 + .01136I1 I3 + .06818I2 I3 + 1.52893x
SSE(R) = 240.7835
Factor B:
Yijk = µ.. + α1 Iijk1 + α2 Iijk2 + (αβ)11 Iijk1 Iijk3
+(αβ)21 Iijk2 Iijk3 + γxijk + ²ijk
Ŷ = 23.55556 − 2.15487I1 + 3.67076I2 − .05669I1 I3 − .04071I2 I3 + 1.08348x
SSE(R) = 9.8393
c.
H0 : (αβ)11 = (αβ)21 = 0, Ha : not both (αβ)11 and (αβ)21 equal zero.
F ∗ = (.1948/2) ÷ (8.2941/29) = .341, F (.95; 2, 29) = 3.33.
If F ∗ ≤ 3.33 conclude H0 , otherwise Ha . Conclude H0 . P -value = .714
d.
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero.
F ∗ = (232.4894/2) ÷ (8.2941/29) = 406.445, F (.95; 2, 29) = 3.33.
If F ∗ ≤ 3.33 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
22-5
e.
H0 : β1 = 0, Ha : β1 6= 0.
F ∗ = (1.5452/1) ÷ (8.2941/29) = 5.403, F (.95; 1, 29) = 4.18.
If F ∗ ≤ 4.18 conclude H0 , otherwise Ha . Conclude Ha . P -value = .027
f.
D̂1 = α̂1 − α̂2 = −5.83435, D̂2 = α̂1 − α̂3 = 2α̂1 + α̂2 = −.62414,
D̂3 = α̂2 − α̂3 = 2α̂2 + α̂1 = 5.21021, D̂4 = β̂ 1 − β̂ 2 = 2β̂ 1 = .41814,
s2 {α̂1 } = .01593, s2 {α̂2 } = .01708, s{α̂1 , α̂2 } = −.00772, s2 {β̂ 1 } = .00809,
s{D̂1 } = .22011, s{D̂2 } = .22343, s{D̂3 } = .23102, s{D̂4 } = .17989,
B = t(.9875; 29) = 2.364
−5.83435 ± 2.364(.22011)
−.62414 ± 2.364(.22343)
5.21021 ± 2.364(.23102)
.41814 ± 2.364(.17989)
22.17. a.
−6.355 ≤ D1 ≤ −5.314
−1.152 ≤ D2 ≤ −.096
4.664 ≤ D3 ≤ 5.756
−.007 ≤ D4 ≤ .843
eijk :
i
1
j=1
j=2
.1707 −.8159
.0810 1.3979
−.4586 −.8383
−1.2448 −.5796
1.4517
.8359
i
2
j=1
.3035
1.0448
−.7190
−.9776
.3483
j=2
.2069
1.5776
−2.0965
.6672
−.3552
b.
r = .988
c.
Yijk = µ.. + α1 Iijk1 + β1 Iijk2 + (αβ)11 Iijk1 Iijk2 + γxijk
+δ1 Iijk1 xijk + δ2 Iijk2 xijk + δ3 Iijk1 Iijk2 xijk + ²ijk
H0 : all δi equal zero (i = 1, 2, 3), Ha : not all δi equal zero.
SSE(F ) = 16.8817, SSE(R) = 18.5364,
F ∗ = (1.6547/3) ÷ (16.8817/12) = .392, F (.995; 3, 12) = 7.23.
If F ∗ ≤ 7.23 conclude H0 , otherwise Ha . Conclude H0 . P -value = .76
22.18. a.
Yijk = µ. + α1 Iijk1 + β1 Iijk2 + (αβ)11 Iijk1 Iijk2 + γxijk + εijk , (X̄... = 44.55).
Ŷ = 13.05000 − .36284I1 − 1.11905I2 + .09216I1 I2 + .32586x
SSE(F ) = 18.5364
b.
Interactions:
Yijk = µ.. + α1 Iijk1 + β1 Iijk2 + γxijk + ²ijk
Ŷ = 13.05000 − .37286I1 − 1.12552I2 + .32333x
SSE(R) = 18.7014
Factor A:
Yijk = µ.. + β1 Iijk2 + (αβ)11 Iijk1 Iijk2 + γxijk + ²ijk
Ŷ = 13.05000 − 1.04962I2 + .12074I1 I2 + .35309x
22-6
SSE(R) = 20.3891
Factor B:
Yijk = µ.. + α1 Iijk1 + (αβ)11 Iijk1 Iijk2 + γxijk + ²ijk
Ŷ = 13.05000 − .10397I1 + .16097I1 I2 + .39140x
SSE(R) = 39.8416
c.
H0 : (αβ)11 = 0, Ha : (αβ)11 6= 0.
F ∗ = (.1650/1) ÷ (18.5364/15) = .1335, F (.99; 1, 15) = 8.68.
If F ∗ ≤ 8.68 conclude H0 , otherwise Ha . Conclude H0 . P -value = .72
d.
H0 : α1 = 0, Ha : α1 6= 0.
F ∗ = (1.8527/1) ÷ (18.5364/15) = 1.499, F (.99; 1, 15) = 8.68.
If F ∗ ≤ 8.68 conclude H0 , otherwise Ha . Conclude H0 . P -value = .24
e.
H0 : β1 = 0, Ha : β1 6= 0.
F ∗ = (21.3052/1) ÷ (18.5364/15) = 17.241, F (.99; 1, 15) = 8.68.
If F ∗ ≤ 8.68 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
f.
D̂ = β̂ 1 − β̂ 2 = 2β̂ 1 = −2.2381, s{D̂} = .539, t(.995; 15) = 2.947,
−2.2381 ± 2.947(.539), −3.8265 ≤ D ≤ −.6497
g.
2
2
d
Ŷ = µ̂.. + α̂1 − β̂ 1 − (αβ)
11 − 4.55γ̂ = 12.2314, s {µ̂.. } = .06179, s {α̂1 } = .08782,
d } = .06363, s{α̂ , β̂ } = .01680,
s2 {β̂ 1 } = .07264, s2 {γ̂} = .00167, s2 {(αβ)
11
1
1
d
d
s{α̂1 , (αβ)11 } = .00692, s{β̂ 1 , (αβ)11 } = .00447, s{α̂1 , γ̂} = .00659, s{β̂ 1 , γ̂} =
d } = .00175, s{µ̂ , α̂ } = s{µ̂ , β̂ } = s{µ̂ , (αβ)
d } = s{µ̂ , γ̂} =
.00425, s{γ̂, (αβ)
11
1
11
..
..
1
..
..
0, s{Ŷ } = .5259, t(.995; 15) = 2.947,
12.2314 ± 2.947(.5259), 10.682 ≤ µ.. + α1 − β1 − (αβ)11 − 4.55γ ≤ 13.781
22.19. b.
Yij = µ.. + ρ1 Iij1 + ρ2 Iij2 + ρ3 Iij3 + ρ4 Iij4 + ρ5 Iij5 + ρ6 Iij6
+ρ7 Iij7 + ρ8 Iij8 + ρ9 Iij9 + τ1 Iij10 + τ2 Iij11 + γxij + ²ij
Iij1
1
−1
=
0
if experimental unit from block 1
if experimental unit from block 10
otherwise
Iij2 , . . . , Iij9 are defined similarly
Iij10
Iij11
1 if experimental unit received treatment 1
−1 if experimental unit received treatment 3
=
0 otherwise
1 if experimental unit received treatment 2
= −1 if experimental unit received treatment 3
0 otherwise
xij = Xij − X̄..
c.
(X̄.. = 80.033333)
Ŷ = 77.10000 + 4.87199I1 + 3.87266I2 + 2.21201I3 + 3.22003I4
+1.23474I5 + .90876I6 − 1.09124I7 − 3.74253I8 − 4.08322I9
22-7
−6.50033I10 − 2.49993I11 + .00201x
SSE(F ) = 112.3327
d.
Yij = µ.. + ρ1 Iij1 + ρ2 Iij2 + ρ3 Iij3 + ρ4 Iij4 + ρ5 Iij5 + ρ6 Iij6
+ρ7 Iij7 + ρ8 Iij8 + ρ9 Iij9 + γxij + ²ij
Ŷ = 77.10000 + 6.71567I1 + 5.67233I2 + 3.61567I3 + 4.09567I4
+1.14233I5 + .33233I6 − 1.66767I7 − 5.33100I8 − 5.18767I9 − .13000x
SSE(R) = 1, 404.5167
e.
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (1, 292.18/2) ÷ (112.3327/17) = 97.777, F (.95; 2, 17) = 3.59.
If F ∗ ≤ 3.59 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
f.
τ̂ 1 = −6.50033, τ̂ 2 = −2.49993, L̂ = −4.0004, L2 {τ̂ 1 } = .44162, s2 {τ̂ 2 } = .44056,
s{τ̂ 1 , τ̂ 2 } = −.22048, s{L̂} = 1.1503, t(.975; 17) = 2.11,
−4.0004 ± 2.11(1.1503), −6.43 ≤ L ≤ −1.57
22.20. a.
Yij = µ.. + ρ1 Iij1 + ρ2 Iij2 + ρ3 Iij3 + ρ4 Iij4 + τ1 Iij5 + τ2 Iij6 + γxij + ²ij
Iij1
1
= −1
0
if experimental unit from block 1
if experimental unit from block 5
otherwise
Iij2 , . . . , Iij4 are defined similarly
1
= −1
0
if experimental unit received treatment 1
if experimental unit received treatment 3
otherwise
1
Iij6 = −1
0
if experimental unit received treatment 2
if experimental unit received treatment 3
otherwise
Iij5
xij = Xij − X̄..
b.
(X̄.. = 104.46667)
Ŷ = .84400 − .25726I1 − .18916I2 − .16649I3 + .27012I4 + .26663I5
+.15238I6 + .009385x
SSE(F ) = .007389
c.
Yij = µ.. + ρ1 Iij1 + ρ2 Iij2 + ρ3 Iij3 + ρ4 Iij4 + γxij + ²ij
Ŷ = .84400 − .34176I1 − .24725I2 − .17555I3 + .32143I4 − .00193x
SSE(R) = 1.339085
d.
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (1.331696/2) ÷ (.007389/7) = 630.79, F (.95; 2, 7) = 4.737.
If F ∗ ≤ 4.737 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
τ̂ 1 = .26663, τ̂ 2 = .15238, L̂1 = τ̂ 1 − τ̂ 2 = .11425, L̂2 = τ̂ 1 + 2τ̂ 2 = .57139,
s2 {τ̂ 1 } = .0001408, s2 {τ̂ 2 } = .0001424, s{τ̂ 1 , τ̂ 2 } = −.0000701, s{L̂1 } = .02058,
s{L̂2 } = .02074, B = t(.9875; 7) = 2.841
22-8
.11425 ± 2.841(.02058)
.57139 ± 2.841(.02074)
.0558 ≤ L1 ≤ .1727
.5125 ≤ L2 ≤ .6303
22.21. a.
Source
SS
df
Between treatments 25.5824 2
Error
1.4650 24
Total
27.0474 26
b.
MS
12.7912
.0610
Covariance: M SE = .0573, γ̂ = 1.11417
22.22. a.
Source
SS
df
MS
Between treatments 1, 417.7333 2 708.8667
Error
223.2000 12 18.6000
Total
1, 640.9333 14
b.
Covariance: M SE = 16.048, γ̂ = .83474
Yij = µ. + τi + γ(Xij − X̄.. ) + ²ij = ∆i + γ(Xij − X̄.. ) + ²ij
22.23.
Q=
PP
[Yij − ∆i − γ(Xij − X̄.. )]2
∂Q
P
= 2 [Yij − ∆i − γ(Xij − X̄.. )](−1)
∂∆i
j
∂Q
PP
=2
[Yij − ∆i − γ(Xij − X̄.. )][−(Xij − X̄.. )]
∂γ
Setting the partial derivatives equal to zero, simplifying, and substituting the least
squares estimators yields:
P
P
ˆi
Yij − γ̂ (Xij − X̄.. ) = ni ∆
j
j
or:
ˆ i = Ȳi. − γ̂(X̄i. − X̄.. )
∆
and:
PP
ˆ i − γ̂(Xij − X̄.. )](Xij − X̄.. ) = 0
[Yij − ∆
or:
PP
or:
Yij (Xij − X̄.. ) −
PP
[Ȳi. − γ̂(X̄i. − X̄.. )](Xij − X̄.. ) = γ̂
PP
(Xij − X̄.. )2
PP
(Yij − Ȳi. )(Xij − X̄i. )
(Xij − X̄i. )2
It needs to be recognized in the development that:
γ̂ =
PP
PP
22.24. b.
c.
PP
(Yij − Ȳi. )(Xij − X̄.. ) =
PP
(Xij − X̄.. )(Xij − X̄i. ) =
(Yij − Ȳi. )(Xij − X̄i. )
PP
(Xij − X̄i. )2
r = .907
Yij = µ. + τ1 Iij1 + τ2 Iij2 + τ3 Iij3 + γxij + β1 Iij1 xij + β2 Iij2 xij + β3 Iij3 xij + ²ij
22-9
H0 : β1 = β2 = β3 = 0, Ha : not all βi equal zero.
SSE(F ) = 147.8129, SSE(R) = 151.3719,
F ∗ = (3.5590/3) ÷ (147.8129/56) = .449, F (.995; 3, 56) = 4.76.
If F ∗ ≤ 4.76 conclude H0 , otherwise Ha . Conclude H0 . P -value = .72
22.25. b.
Full model: Yij = µ. + τ1 Iij1 + τ2 Iij2 + τ3 Iij3 + γxij + ²ij
Iij1 =
Iij2
Iij3
1
−1
0
1
= −1
0
1
−1
=
0
if case from region NE
if case from region W
otherwise
if case from region NC
if case from region W
otherwise
if case from region S
if case from region W
otherwise
xij = Xij − X̄..
(X̄.. = 42.75625)
Reduced model: Yij = µ. + γxij + ²ij
c.
Full model: Ŷ = 9.58406 + 1.60061I1 + .05250I2 − .26776I3 + .02579x,
SSE(F ) = 151.3719
Reduced model: Ŷ = 9.58406 + .04013x,
SSE(R) = 221.2543
H0 : all τi equal zero (i = 1, 2, 3), Ha : not all τi equal zero.
F ∗ = (69.8824/3) ÷ (151.3719/59) = 9.079, F (.95; 3, 59) = 2.76.
If F ∗ ≤ 2.76 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
D̂1 = τ̂ 1 − τ̂ 2 = 1.54811, D̂2 = τ̂ 1 − τ̂ 3 = 1.86837, D̂3 = τ̂ 1 − τ̂ 4 = 2τ̂ 1 + τ̂ 2 + τ̂ 3 =
2.98596, D̂4 = τ̂ 2 − τ̂ 3 = .32026, D̂5 = τ̂ 2 − τ̂ 4 = 2τ̂ 2 + τ̂ 1 + τ̂ 3 = 1.43785,
D̂6 = τ̂ 3 − τ̂ 4 = 2τ̂ 3 + τ̂ 1 + τ̂ 2 = 1.11759, s2 {τ̂ 1 } = .12412, s2 {τ̂ 2 } = .12188,
s2 {τ̂ 3 } = .12355, s{τ̂ 1 , τ̂ 2 } = −.03759, s{τ̂ 1 , τ̂ 3 } = −.04365, s{τ̂ 2 , τ̂ 3 } = −.04240,
s{D̂1 } = .56673, s{D̂2 } = .57877, s{D̂3 } = .57632, s{D̂4 } = .57466, s{D̂5 } =
.57265, s{D̂6 } = .56641, B = t(.99167; 59) = 2.464
1.54811 ± 2.464(.56673)
1.86837 ± 2.464(.57877)
2.98596 ± 2.464(.57632)
.32026 ± 2.464(.57466)
1.43785 ± 2.464(.57265)
1.11759 ± 2.464(.56641)
22.26. b.
c.
.1517 ≤ D1
.4423 ≤ D2
1.5659 ≤ D3
−1.0957 ≤ D4
.0268 ≤ D5
−.2780 ≤ D6
≤ 2.9445
≤ 3.2945
≤ 4.4060
≤ 1.7362
≤ 2.8489
≤ 2.5132
r = .9914
Yij = µ. + τ1 Iij1 + τ2 Iij2 + τ3 Iij3 + γxij + β1 Iij1 xij + β2 Iij2 xij + β3 Iij3 xij + ²ij
H0 : β1 = β2 = β3 = 0, Ha : not all βi equal zero.
22-10
SSE(F ) = .6521, SSE(R) = .6778,
F ∗ = (.0257/3) ÷ (.6521/28) = .37, F (.95; 3, 28) = 2.95.
If F ∗ ≤ 2.95 conclude H0 , otherwise Ha . Conclude H0 . P -value = .78
22.27. b.
Full model: Yij = µ. + τ1 Iij1 + τ2 Iij2 + τ3 Iij3 + γxij + ²ij
1
Iij1 = −1
0
Iij2
Iij3
1
= −1
0
1
−1
=
0
if case from (Var5 Var6)=(0,0)
if case from (Var5 Var6)=(1,1)
otherwise
if case from (Var5 Var6)=(1,0)
if case from (Var5 Var6)=(1,1)
otherwise
if case from (Var5 Var6)=(0,1)
if case from (Var5 Var6)=(1,1)
otherwise
xij = Xij − X̄..
(X̄.. = 2.3244)
Reduced model: Yij = µ. + γxij + ²ij
c.
Full model: Ŷ = 2.619 − .217I1 + .109I2 − .178I3 − .344x, SSE(F ) = .6778
Reduced model: Ŷ = 2.664 − .306x, SSE(R) = 2.3593
H0 : τ1 = τ2 = τ3 = 0, Ha : not all τi equal zero.
F ∗ = (1.6815/3) ÷ (.6778/31) = 25.64, F (.99; 3, 31) = 4.51.
If F ∗ ≤ 4.51 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
In Project 16.45, SSE = .7850.
No, almost none.
e.
(2.5005, 2.8353)
f.
D̂1 = τ̂ 1 − τ̂ 2 = −.326, D̂2 = τ̂ 1 − τ̂ 3 = −.039,
D̂3 = τ̂ 1 − τ̂ 4 = 2τ̂ 1 + τ̂ 2 + τ̂ 3 = −.503, D̂4 = τ̂ 2 − τ̂ 3 = .287,
D̂5 = τ̂ 2 − τ̂ 4 = 2τ̂ 2 + τ̂ 1 + τ̂ 3 = −.177, D̂6 = τ̂ 3 − τ̂ 4 = 2τ̂ 3 + τ̂ 1 + τ̂ 2 = −.464,
s2 {τ̂ 1 } = .002028, s2 {τ̂ 2 } = .002024, s2 {τ̂ 3 } = .002239,
s{τ̂ 1 , τ̂ 2 } = −.00071, s{τ̂ 1 , τ̂ 3 } = −.00085, s{τ̂ 2 , τ̂ 3 } = −.00085,
s{D̂1 } = .073927, s{D̂2 } = .077232, s{D̂3 } = .066790,
s{D̂4 } = .077165, s{D̂5 } = .066743, s{D̂6 } = .069400,
B = t(.99583; 31) = 2.818 (S 2 = 3F (.95; 3, 31) = 3(2.9113), S = 2.955)
−.326 ± 2.818(.073927)
−.039 ± 2.818(.077232)
−.503 ± 2.818(.066790)
.287 ± 2.818(.077165)
−.177 ± 2.818(.066743)
−.464 ± 2.818(.069400)
−.534 ≤ D1 ≤ −.118
−.257 ≤ D2 ≤ .179
−.691 ≤ D3 ≤ −.315
.070 ≤ D4 ≤ .504
−.365 ≤ D5 ≤ .011
−.660 ≤ D6 ≤ −.268
22-11
22.28. b.
c.
r = .991
Yijk = µ.. + α1 Iijk1 + α2 Iijk2 + α3 Iijk3 + β1 Iijk4
+(αβ)11 Iijk1 Iijk4 + (αβ)21 Iijk2 Iijk4 + (αβ)31 Iijk3 Iijk4
+γxijk + δ1 Iijk1 xijk + δ2 Iijk2 xijk + δ3 Iijk3 xijk
+δ4 Iijk4 xijk + δ5 Iijk1 Iijk4 xijk + δ6 Iijk2 Iijk4 xijk + δ7 Iijk3 Iijk4 xijk + ²ijk
H0 : all δi equal zero (i = 1, ..., 7), Ha : not all δi equal zero.
SSE(F ) = .0093126, SSE(R) = .0108089,
F ∗ = (.0014963/7) ÷ (.0093126/40) = .92, F (.999; 7, 40) = 4.436.
If F ∗ ≤ 4.436 conclude H0 , otherwise Ha . Conclude H0 . P -value = .46
22.29. a.
Yijk = µ.. + α1 Iijk1 + α2 Iijk2 + α3 Iijk3 + β1 Iijk4 + (αβ)11 Iijk1 Iijk4
+(αβ)21 Iijk2 Iijk4 + (αβ)31 Iijk3 Iijk4 + γxijk + ²ijk
Iijk1
1 if case from region NE
−1 if case from region W
=
0 otherwise
Iijk2 =
1 if case from region NC
−1 if case from region W
0 otherwise
1 if case from region S
Iijk3 = −1 if case from region W
0 otherwise
(
1
−1
Iijk4 =
if percent of poverty less than 8.0 percent
if percent of poverty 8.0 percent or more
xijk = Xijk − X̄...
(X̄... = 12.521)
Ŷ = .0632 − .0239I1 − .0115I2 + .0254I3 − .00548I4
+.00149I1 I4 + .00643I2 I4 − .00904I3 I4 + .000627x
SSE(F ) = .0108089
b.
Interactions:
Yijk = µ.. + α1 Iijk1 + α2 Iijk2 + α3 Iijk3 + β1 Iijk4 + γxijk + ²ijk
Ŷ = .0632 − .0224I1 − .0117I2 + .0255I3 − .00557I4 − .000061x
SSE(R) = .0122362
Factor A:
Yijk = µ.. + β1 Iijk4 + (αβ)11 Iijk1 Iijk4 + (αβ)21 Iijk2 Iijk4
+(αβ)31 Iijk3 Iijk4 + γxijk + ²ijk
Ŷ = .0632 − .00565I4 + .00001I1 I4 + .00463I2 I4 − .00532I3 I4 − .000719x
SSE(R) = .0298356
Factor B:
22-12
Yijk = µ.. + α1 Iijk1 + α2 Iijk2 + α3 Iijk3 + (αβ)11 Iijk1 Iijk4
+(αβ)21 Iijk2 Iijk4 + (αβ)31 Iijk3 Iijk4 + γxijk + ²ijk
Ŷ = .0632 − .0241I1 − .0115I2 + .0254I3 + .00157I1 I4
+.00652I2 I4 − .00923I3 I4 + .000695x
SSE(R) = .012488
c.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = (.0014273/3) ÷ (.0108089/47) = 2.069, F (.99; 3, 47) = 4.23.
If F ∗ ≤ 4.23 conclude H0 , otherwise Ha . Conclude H0 . P -value = .12
d.
H0 : α1 = α2 = α3 = 0, Ha : not all αi equal zero.
F ∗ = (.0190267/3) ÷ (.0108089/47) = 27.57, F (.99; 3, 47) = 4.23.
If F ∗ ≤ 4.23 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
H0 : β1 = 0, Ha : β1 6= 0.
F ∗ = (.0016791/1) ÷ (.0108089/47) = 7.30, F (.99; 1, 47) = 7.21.
If F ∗ ≤ 7.21 conclude H0 , otherwise Ha . Conclude Ha . P -value = .0096
22.30. b.
c.
r = .983
Yijk = µ.. + α1 Iijk1 + β1 Iijk2 + (αβ)11 Iijk1 Iijk4 + γxijk
+δ1 Iijk1 xijk + δ2 Iijk2 xijk + δ3 Iijk1 Iijk2 xijk + ²ijk
H0 : δ1 = δ2 = δ3 = 0, Ha : not all δi equal zero.
SSE(F ) = .48044, SSE(R) = .51032,
F ∗ = (.02988/3) ÷ (.48044/20) = .41, F (.95; 3, 20) = 3.10.
If F ∗ ≤ 3.10 conclude H0 , otherwise Ha . Conclude H0 . P -value = .75
22.31. a.
Yijk = µ.. + α1 Iijk1 + β1 Iijk2 + (αβ)11 Iijk1 Iijk2 + ²ijk
(
Iijk1 =
(
Iijk2 =
1
−1
no discount price
discount price
1
−1
no package promotion
package promotion
xijk = Xijk − X̄...
(X̄... = 2.2716)
Ŷ = 2.644 − .197I1 − .0605I2 + .0533I1 I2 − .276x
SSE(F ) = .51032
b.
Interactions:
Yijk = µ.. + α1 Iijk1 + β1 Iijk2 + γxijk + ²ijk
Ŷ = 2.644 − .189I1 − .0608I2 − .451x
SSE(R) = .57864
Factor A:
22-13
Yijk = µ.. + β1 Iijk2 + (αβ)11 Iijk1 Iijk2 + γxijk + ²ijk
Ŷ = 2.644 − .0616I2 + .0241I1 I2 − .962x
SSE(R) = 1.42545
Factor B:
Yijk = µ.. + α1 Iijk1 + (αβ)11 Iijk1 Iijk2 + γxijk + ²ijk
Ŷ = 2.644 − .198I1 + .0536I1 I2 − .267x
SSE(R) = .61269
c.
H0 : (αβ)11 = 0, Ha : (αβ)11 6= 0
F ∗ = (.06832/1) ÷ (.51032/23) = 3.08, F (.99; 1, 23) = 7.88.
If F ∗ ≤ 7.88 conclude H0 , otherwise Ha . Conclude H0 . P -value = .09
d.
H0 : α1 = 0, Ha : α1 6= 0
F ∗ = (.91513/1) ÷ (.51032/23) = 41.24, F (.99; 1, 23) = 7.88.
If F ∗ ≤ 7.88 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
H0 : β1 = 0, Ha : β1 6= 0.
F ∗ = (.10237/1) ÷ (.51032/23) = 4.61, F (.99; 1, 23) = 7.88.
If F ∗ ≤ 7.88 conclude H0 , otherwise Ha . Conclude H0 . P -value = .04
22-14
Chapter 23
TWO-FACTOR STUDIES WITH
UNEQUAL SAMPLE SIZES
23.3. a.
Yijk = µ.. + α1 Xijk1 + β1 Xijk2 + (αβ)11 Xijk1 Xijk2 + ²ijk
(
Xijk1 =
(
1
−1
if case from level 1 for factor A
if case from level 2 for factor A
1 if case from level 1 for factor B
−1 if case from level 2 for factor B
Y entries: in order Y111 , ..., Y115 , Y121 , ..., Y125 , Y211 , ...
β entries: µ.. , α1 , β1 , (αβ)11
X entries:
A B Freq.
X1 X2 X1 X2
1 1
5
1
1
1
1
1 2
5
1
1 −1
−1
2 1
5
1 −1
1
−1
2 2
5
1 −1 −1
1
Xβ entries:
A B
1 1 µ.. + α1 + β1 + (αβ)11
1 2 µ.. + α1 − β1 − (αβ)11 = µ.. + α1 + β2 + (αβ)12
2 1 µ.. − α1 + β1 − (αβ)11 = µ.. + α2 + β1 + (αβ)21
2 2 µ.. − α1 − β1 + (αβ)11 = µ.. + α2 + β2 + (αβ)22
Xijk2 =
b.
c.
d.
Ŷ = 13.05 − 1.65X1 − 1.95X2 − .25X1 X2
µ..
e.
Source
Regression
X1
X2 |X1
X1 X2 |X1 , X2
Error
Total
SS
131.75
54.45
76.05
1.25
97.20
228.95
df
3
1] A
1] B
1] AB
16
19
23-1
Yes.
f.
23.4. a.
See Problem 19.13c and d.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + β1 Xijk3 + β2 Xijk4 + (αβ)11 Xijk1 Xijk3
+(αβ)12 Xijk1 Xijk4 + (αβ)21 Xijk2 Xijk3 + (αβ)22 Xijk2 Xijk4 + ²ijk
1 if case from level 1 for factor A
Xijk1 = −1 if case from level 3 for factor A
0 otherwise
Xijk2
1 if case from level 2 for factor A
−1 if case from level 3 for factor A
=
0 otherwise
Xijk3 =
Xijk4
b.
1 if case from level 1 for factor B
−1 if case from level 3 for factor B
0 otherwise
1 if case from level 2 for factor B
−1 if case from level 3 for factor B
=
0 otherwise
Y entries: in order Y111 , ..., Y114 , Y121 , ..., Y124 , Y131 , ..., Y134, Y211 , ...
β entries: µ.. , α1 , α2 , β1 , β2 , (αβ)11 , (αβ)12 , (αβ)21 , (αβ)22
X entries:
A B Freq.
1 1
4
1 2
4
1 3
4
2 1
4
2 2
4
2 3
4
3 1
4
3 2
4
3 3
4
c.
1
1
1
1
1
1
1
1
1
X1 X2 X3
1
0
1
1
0
0
1
0 −1
0
1
1
0
1
0
0
1 −1
−1 −1
1
−1 −1
0
−1 −1 −1
X4
0
1
−1
0
1
−1
0
1
−1
X1 X3
1
0
−1
0
0
0
−1
0
1
X1 X4 X2 X3 X2 X4
0
0
0
1
0
0
−1
0
0
0
1
0
0
0
1
0
−1
−1
0
−1
0
−1
0
−1
1
1
1
Xβ entries:
A
1
1
1
2
2
2
3
3
3
B
1
2
3
1
2
3
1
2
3
µ.. + α1 + β1 + (αβ)11
µ.. + α1 + β2 + (αβ)12
µ.. + α1 − β1 − β2 − (αβ)11 − (αβ)12 = µ.. + α1 + β3 + (αβ)13
µ.. + α2 + β1 + (αβ)21
µ.. + α2 + β2 + (αβ)22
µ.. + α2 − β1 − β2 − (αβ)21 − (αβ)22 = µ.. + α2 + β3 + (αβ)23
µ.. − α1 − α2 + β1 − (αβ)11 − (αβ)21 = µ.. + α3 + β1 + (αβ)31
µ.. − α1 − α2 + β2 − (αβ)12 − (αβ)22 = µ.. + α3 + β2 + (αβ)32
µ.. − α1 − α2 − β1 − β2 + (αβ)11 + (αβ)12 + (αβ)21 + (αβ)22
= µ.. + α3 + β3 + (αβ)33
23-2
d.
Ŷ = 7.18333 − 3.30000X1 + .65000X2 − 2.55000X3 + .75000X4
+1.14167X1 X3 − .03333X1 X4 + .16667X2 X3 + .34167X2 X4
α1 = µ1. − µ..
e.
Source
Regression
X1
X2 | X1
X3 | X1 , X2
X4 | X1 , X2 , X3
X1 X3 | X1 , X2 , X3 , X4
X1 X4 | X1 , X2 , X3 , X4 , X1 X3
X2 X3 | X1 , X2 , X3 , X4 , X1 X3 , X1 X4
X2 X4 | X1 , X2 , X3 , X4 , X1 X3 , X1 X4 , X2 X3
Error
Total
SS
373.125
212.415
7.605
113.535
10.125
26.7806
.2269
1.3669
1.0506
1.625
374.730
df
8
1] A
1] A
1] B
1] B
1] AB
1] AB
1] AB
1] AB
27
35
Yes.
f.
23.5. a.
b.
See Problem 19.15c and d.
See Problem 23.4a.
Ŷ = 55.82222 − .48889X1 − .55556X2 + .31111X3 + .77778X4
+4.15556X1 X3 − 8.31111X1 X4 − 7.17778X2 X3 + 5.15556X2 X4
β1 = µ.1 − µ..
c.
Source
Regression
X1
X2 | X1
X3 | X1 , X2
X4 | X1 , X2 , X3
X1 X3 | X1 , X2 , X3 ,
X1 X4 | X1 , X2 , X3 ,
X2 X3 | X1 , X2 , X3 ,
X2 X4 | X1 , X2 , X3 ,
Error
Total
X4
X4 , X1 X3
X4 , X1 X3 , X1 X4
X4 , X1 X3 , X1 X4 , X2 X3
SS
1, 268.17778
17.63333
6.94445
14.70000
13.61111
105.80000
493.06667
317.40000
299.02222
1, 872.40000
3, 140.57778
Yes.
d.
23.6. a.
See Problem 19.17c and d.
Yijk = µ.. + αi + βj + (αβ)ij + ²ijk
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + β1 Xijk3 + (αβ)11 Xijk1 Xijk3
+(αβ)21 Xijk2 Xijk3 + ²ijk
23-3
df
8
1] A
1] A
1] B
1] B
1] AB
1] AB
1] AB
1] AB
36
44
Xijk1 =
Xijk2
1 if case from level 1 for factor A
−1 if case from level 3 for factor A
0 otherwise
1 if case from level 2 for factor A
−1 if case from level 3 for factor A
=
0 otherwise
(
Xijk3 =
b.
1
−1
β entries: µ.. , α1 , α2 , β1 , (αβ)11 , (αβ)21
X entries:
A B Freq.
1 1
6
1 2
6
2 1
5
2 2
6
3 1
6
3 2
5
c.
if case from level 1 for factor B
if case from level 2 for factor B
1
1
1
1
1
1
X1 X2 X3 X1 X3 X2 X3
1
0
1
1
0
1
0 −1
−1
0
0
1
1
0
1
0
1 −1
0
−1
−1 −1
1
−1
−1
−1 −1 −1
1
1
Xβ entries:
A
1
1
2
2
3
3
B
1
2
1
2
1
2
µ.. + α1 + β1 + (αβ)11
µ.. + α1 − β1 − (αβ)11 = µ.. + α1 + β2 + (αβ)12
µ.. + α2 + β1 + (αβ)21
µ.. + α2 − β1 − (αβ)21 = µ.. + α2 + β2 + (αβ)22
µ.. − α1 − α2 + β1 − (αβ)11 − (αβ)21 = µ.. + α3 + β1 + (αβ)31
µ.. − α1 − α2 − β1 + (αβ)11 + (αβ)21 = µ.. + α3 + β2 + (αβ)32
d.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + β1 Xijk3 + ²ijk
e.
Full model:
Ŷ = 23.56667 − 2.06667X1 + 4.16667X2 + .36667X3 − .20000X1 X3 − .30000X2 X3 ,
SSE(F ) = 71.3333
Reduced model:
Ŷ = 23.59091 − 2.09091X1 + 4.16911X2 + .36022X3 ,
SSE(R) = 75.5210
H0 : (αβ)11 = (αβ)21 = 0, Ha : not both (αβ)11 and (αβ)21 equal zero.
F ∗ = (4.1877/2) ÷ (71.3333/28) = .82, F (.95; 2, 28) = 3.34.
If F ∗ ≤ 3.34 conclude H0 , otherwise Ha . Conclude H0 . P -value = .45
f.
A effects:
Ŷ = 23.50000 + .17677X3 − .01010X1 X3 − .49495X2 X3 ,
SSE(R) = 359.9394
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero.
23-4
F ∗ = (288.6061/2) ÷ (71.3333/28) = 56.64, F (.95; 2, 28) = 3.34.
If F ∗ ≤ 3.34 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
B effects:
Ŷ = 23.56667 − 2.06667X1 + 4.13229X2 − .17708X1 X3 − .31146X2 X3 ,
SSE(R) = 75.8708
H0 : β1 = 0, Ha : β1 6= 0.
F ∗ = (4.5375/1) ÷ (71.3333/28) = 1.78, F (.95; 1, 28) = 4.20.
If F ∗ ≤ 4.20 conclude H0 , otherwise Ha . Conclude H0 . P -value = .19
g.
D̂1 = α̂1 − α̂2 = −6.23334, D̂2 = α̂1 − α̂3 = 2α̂1 + α̂2 = .03333, D̂3 = α̂2 − α̂3 =
2α̂2 + α̂1 = 6.26667, s2 {α̂1 } = .14625, s2 {α̂2 } = .15333, s{α̂1 , α̂2 } = −.07313,
s{D̂1 } = .6677, s{D̂2 } = .6677, s{D̂3 } = .6834, q(.90; 3, 28) = 3.026, T = 2.140
−6.23334 ± 2.140(.6677)
.03333 ± 2.140(.6677)
6.26667 ± 2.140(.6834)
h.
23.7. a.
−7.662 ≤ D1 ≤ −4.804
−1.396 ≤ D2 ≤ 1.462
4.804 ≤ D3 ≤ 7.729
L̂ = .3Ȳ12. +.6Ȳ22. +.1Ȳ32. = .3(21.33333)+.6(27.66667)+.1(20.60000) = 25.06000,
s{L̂} = .4429, t(.975; 28) = 2.048, 25.06000 ± 2.048(.4429), 24.153 ≤ L ≤ 25.967
Yijk = µ.. + αi + βj + (αβ)ij + ²ijk
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + β1 Xijk3 + β2 Xijk4
+ (αβ)11 Xijk1 Xijk3 + (αβ)12 Xijk1 Xijk4 + (αβ)21 Xijk2 Xijk3
+ (αβ)22 Xijk2 Xijk4 + ²ijk
Xijk1 =
1 if case from level 1 for factor A
−1 if case from level 3 for factor A
0 otherwise
1 if case from level 2 for factor A
Xijk2 = −1 if case from level 3 for factor A
0 otherwise
Xijk3
Xijk4
b.
1 if case from level 1 for factor B
−1
if case from level 3 for factor B
=
0 otherwise
1 if case from level 2 for factor B
−1 if case from level 3 for factor B
=
0 otherwise
β entries: µ.. , α1 , α2 , β1 , β2 , (αβ)11 , (αβ)12 , (αβ)21 , (αβ)22
X entries:
23-5
A
1
1
1
2
2
2
3
3
3
c.
B
1
2
3
1
2
3
1
2
3
Freq.
3
4
4
4
2
4
4
4
4
1
1
1
1
1
1
1
1
1
X1 X2 X3 X4 X1 X3
1
0
1
0
1
1
0
0
1
0
1
0 −1 −1
−1
0
1
1
0
0
0
1
0
1
0
0
1 −1 −1
0
−1 −1
1
0
−1
−1 −1
0
1
0
−1 −1
1 −1
1
X1 X4
0
1
−1
0
0
0
0
−1
1
X2 X3
0
0
0
1
0
−1
−1
0
1
X2 X4
0
0
0
0
1
−1
0
−1
1
Xβ entries:
A
1
1
1
2
2
2
3
3
3
B
1
2
3
1
2
3
1
2
3
µ.. + α1 + β1 + (αβ)11
µ.. + α1 + β2 + (αβ)12
µ.. + α1 − β1 − β2 − (αβ)11 − (αβ)12 = µ.. + α1 + β3 + (αβ)13
µ.. + α2 + β1 + (αβ)21
µ.. + α2 + β2 + (αβ)22
µ.. + α2 − β1 − β2 − (αβ)21 − (αβ)22 = µ.. + α2 + β3 + (αβ)23
µ.. − α1 − α2 + β1 − (αβ)11 − (αβ)21 = µ.. + α3 + β1 + (αβ)31
µ.. − α1 − α2 + β2 − (αβ)12 − (αβ)22 = µ.. + α3 + β2 + (αβ)32
µ.. − α1 − α2 − β1 − β2 + (αβ)11 + (αβ)12 + (αβ)21 + (αβ)22
= µ.. + α3 + β3 + (αβ)33
d.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + β1 Xijk3 + β2 Xijk4 + ²ijk
e.
Full model:
Ŷ = 7.18704 − 3.28426X1 + .63796X2 − 2.53426X3 + .73796X4
+1.16481X1 X3 − .04074X1 X4 + .15926X2 X3 + .33704X2 X4 ,
SSE(F ) = 1.5767
Reduced model:
Ŷ = 7.12711 − 3.33483X1 + .62861X2 − 2.58483X3 + .72861X4 ,
SSE(R) = 29.6474
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = (28.0707/4) ÷ (1.5767/24) = 106.82, F (.95; 4, 24) = 2.78.
If F ∗ ≤ 2.78 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
f.
Ȳ11. = 2.5333, Ȳ12. = 4.6000, Ȳ13. = 4.57500, Ȳ21. = 5.45000, Ȳ22. = 8.90000,
Ȳ23. = 9.12500, Ȳ31. = 5.97500, Ȳ32. = 10.27500, Ȳ33. = 13.25000, L̂1 = 2.0542,
L̂2 = 3.5625, L̂3 = 5.7875, L̂4 = 1.5083, L̂5 = 3.7333, L̂6 = 2.2250, s{L̂1 } = .1613,
s{L̂2 } = .1695, s{L̂3 } = .1570, s{L̂4 } = .2340, s{L̂5 } = .2251, s{L̂6 } = .2310,
F (.90; 8, 24) = 1.94, S = 3.9395
23-6
2.0542 ± 3.9395(.1613)
3.5625 ± 3.9395(.1695)
5.7875 ± 3.9395(.1570)
1.5083 ± 3.9395(.2340)
3.7333 ± 3.9395(.2251)
2.2250 ± 3.9395(.2310)
23.8. a.
1.419 ≤ L1
2.895 ≤ L2
5.169 ≤ L3
.586 ≤ L4
2.846 ≤ L5
1.315 ≤ L6
≤ 2.690
≤ 4.230
≤ 6.406
≤ 2.430
≤ 4.620
≤ 3.135
Yijk = µ.. + αi + βj + (αβ)ij + ²ijk
Regression model: see (22.4).
b.
β entries: µ.. , α1 , β1 , β2 , (αβ)11 , (αβ)12
X entries:
A B Freq.
1 1
10
1 2
9
1 3
10
2 1
9
2 2
10
2 3
9
c.
1
1
1
1
1
1
X1 X2 X3 X1 X2 X1 X3
1
1
0
1
0
1
0
1
0
1
1 −1 −1
−1
−1
−1
1
0
−1
0
−1
0
1
0
−1
−1 −1 −1
1
1
Xβ entries:
A
1
1
1
2
2
2
B
1
2
3
1
2
3
µ.. + α1 + β1 + (αβ)11
µ.. + α1 + β2 + (αβ)12
µ.. + α1 − β1 − β2 − (αβ)11 − (αβ)12 = µ.. + α1 + β3 + (αβ)13
µ.. − α1 + β1 − (αβ)11 = µ.. + α2 + β1 + (αβ)21
µ.. − α1 + β2 − (αβ)12 = µ.. + α2 + β2 + (αβ)22
µ.. − α1 − β1 − β2 + (αβ)11 + (αβ)12 = µ.. + α2 + β3 + (αβ)23
d.
Yijk = µ.. + α1 Xijk1 + β1 Xijk2 + β2 Xijk3 + ²ijk
e.
Full model:
Ŷ = .69139 + .08407X1 − .27492X2 − .01281X3 − .05706X1 X2 + .01355X1 X3 ,
SSE(F ) = 5.3383
Reduced model:
Ŷ = .69092 + .08407X1 − .27745X2 − .01305X3 ,
SSE(R) = 5.4393
H0 : (αβ)11 = (αβ)12 = 0, Ha : not both (αβ)11 and (αβ)12 equal zero.
F ∗ = (.1010/2) ÷ (5.3383/51) = .48, F (.95; 2, 51) = 3.179.
If F ∗ ≤ 3.179 conclude H0 , otherwise Ha . Conclude H0 . P -value = .62
f.
Duration:
Yijk = µ.. + β1 Xijk2 + β2 Xijk3 + (αβ)11 Xijk1 Xijk2 + (αβ)12 Xijk1 Xijk3 + ²ijk
Ŷ = .69287 − .27197X2 − .01871X3 − .05706X1 X2 + .01355X1 X3 ,
SSE(R) = 5.7400
23-7
H0 : α1 = 0, Ha : α1 6= 0.
F ∗ = (.4017/1) ÷ (5.3383/51) = 3.84, F (.95; 1, 51) = 4.03.
If F ∗ ≤ 4.03 conclude H0 , otherwise Ha . Conclude H0 . P -value = .06
Weight gain:
Yijk = µ.. + α1 Xijk1 + (αβ)11 Xijk1 Xijk2 + (αβ)12 Xijk1 Xijk3 + ²ijk
Ŷ = .69139 + .08452X1 − .07198X1 X2 + .01377X1 X3 ,
SSE(R) = 8.3421
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = (3.0038/2) ÷ (5.3383/51) = 14.35, F (.95; 2, 51) = 3.179.
If F ∗ ≤ 3.179 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
g.
H0 : µ.1 ≤ .5, Ha : µ.1 > .5. µ̂.1 = (Ȳ11. + Ȳ21. )/2 = (.44348 + .38946)/2 = .41647,
s{µ̂.1 } = .0743, t∗ = −.08353/.0743 = −1.12, t(.95; 51) = 1.675. If t∗ ≤ 1.675
conclude H0 , otherwise Ha . Conclude H0 . P -value = .87
h.
Ȳ11. = .44348, Ȳ12. = .77619, Ȳ13. = 1.10670, Ȳ21. = .38946, Ȳ22. = .58096, Ȳ23. =
.85155, D̂1 = .16813, D̂2 = .26211, D̂3 = .56266, D̂4 = .30055, s{D̂1 } = .08582,
s{D̂i } = .10511 (i = 2, 3, 4), B = t(.9875; 51) = 2.3096
.16813 ± 2.3096(.08582)
.26211 ± 2.3096(.10511)
.56266 ± 2.3096(.10511)
.30055 ± 2.3096(.10511)
23.9. a.
−.0301 ≤ D1
.0193 ≤ D2
.3199 ≤ D3
0578 ≤ D4
≤ .3663
≤ .5049
≤ .8054
≤ .5433
Yijk = µ.. + αi + βj + (αβ)ij + ²ijk
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + β1 Xijk4 + β2 Xijk5
+(αβ)11 Xijk1 Xijk4 + (αβ)12 Xijk1 Xijk5 + (αβ)21 Xijk2 Xijk4
+(αβ)22 Xijk2 Xijk5 + (αβ)31 Xijk3 Xijk4 + (αβ)32 Xijk3 Xijk5 + ²ijk
a.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + β1 Xijk3 + β2 Xijk4 + (αβ)11 Xijk1 Xijk3
+(αβ)12 Xijk1 Xijk4 + (αβ)21 Xijk2 Xijk3 + (αβ)22 Xijk2 Xijk4 + ²ijk
1 if case from level 1 for factor A
Xijk1 = −1 if case from level 4 for factor A
0 otherwise
Xijk2 and Xijk3 are defined similarly
Xijk4
1 if case from level 1 for factor B
= −1 if case from level 3 for factor B
0 otherwise
1 if case from level 2 for factor B
Xijk5 = −1 if case from level 3 for factor B
0 otherwise
b.
β entries: µ.. , α1 , α2 , α3 , β1 , β2 , (αβ)11 , (αβ)12 , (αβ)21 ,
(αβ)22 , (αβ)31 , (αβ)32
23-8
X entries:
A B Freq.
1 1
2
1 2
2
1 3
8
2 1
4
2 2
5
2 3
4
3 1
2
3 2
4
3 3
5
4 1
2
4 2
2
4 3
5
A
1
1
1
2
2
2
3
3
3
4
4
4
c.
d.
B
1
2
3
1
2
3
1
2
3
1
2
3
1
1
1
1
1
1
1
1
1
1
1
1
X1 X2 X3 X4
1
0
0
1
1
0
0
0
1
0
0 −1
0
1
0
1
0
1
0
0
0
1
0 −1
0
0
1
1
0
0
1
0
0
0
1 −1
−1 −1 −1
1
−1 −1 −1
0
−1 −1 −1 −1
X5
0
1
−1
0
1
−1
0
1
−1
0
1
−1
X1 X4
1
0
−1
0
0
0
0
0
0
−1
0
1
X1 X5 X2 X4 X2 X5 X3 X4 X3 X5
0
0
0
0
0
1
0
0
0
0
−1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
−1
−1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
−1
−1
0
−1
0
−1
0
−1
0
−1
0
−1
1
1
1
1
1
Xβ entries:
A
1
1
1
2
2
2
3
3
3
4
B
1
2
3
1
2
3
1
2
3
1
4
2
4
3
µ.. + α1 + β1 + (αβ)11
µ.. + α1 + β2 + (αβ)12
µ.. + α1 − β1 − β2 − (αβ)11 − (αβ)12 = µ.. + α1 + β3 + (αβ)13
µ.. + α2 + β1 + (αβ)21
µ.. + α2 + β2 + (αβ)22
µ.. + α2 − β1 − β2 − (αβ)21 − (αβ)22 = µ.. + α2 + β3 + (αβ)23
µ.. + α3 + β1 + (αβ)31
µ.. + α3 + β2 + (αβ)32
µ.. + α3 − β1 − β2 − (αβ)31 − (αβ)32 = µ.. + α3 + β3 + (αβ)33
µ.. − α1 − α2 − α3 + β1 − (αβ)11 − (αβ)21 − (αβ)31
= µ.. + α4 + β1 + (αβ)41
µ.. − α1 − α2 − α3 + β2 − (αβ)12 − (αβ)22 − (αβ)32
= µ.. + α4 + β2 + (αβ)42
µ.. − α1 − α2 − α3 − β1 − β2 + (αβ)11 + (αβ)12 + (αβ)21 + (αβ)22
+(αβ)31 + (αβ)32 = µ.. + α4 + β3 + (αβ)43
See Problem 23.10c for fitted model.
23-9
eijk :
e.
23.10. a.
k
1
2
3
4
5
6
7
8
i=1
j=1 j=2 j=3
−.10 −.15 −.20
.10
.15
.00
.20
−.20
−.10
.10
.00
.20
k
1
2
3
4
5
i=3
j=1 j=2 j=3
−.05
.05 −.04
.05
.15 −.14
−.05 −.04
−.15
.06
.16
k
1
2
3
4
5
k
1
2
3
4
5
i=2
j=1 j=2
.05
.18
−.15 −.12
.15
.08
.05 −.12
−.02
j=1
−.05
.05
i=4
j=2
−.25
.25
j=3
.05
−.15
.15
−.05
j=3
−.12
−.02
−.12
.08
.18
r = .986
Ȳ11. = 1.80, Ȳ12. = 1.95, Ȳ13. = 2.70, Ȳ21. = 2.45, Ȳ22. = 2.52,
Ȳ23. = 3.45, Ȳ31. = 2.75, Ȳ32. = 2.85, Ȳ33. = 3.74, Ȳ41. = 2.55,
Ȳ42. = 2.55, Ȳ43. = 3.42
b.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + β1 Xijk4 + β2 Xijk5 + ²ijk
c.
Full model:
Ŷ = 2.72750 − .57750X1 + .07917X2 − .38583X3 − .34000X4 − .26000X5
−.01000X1 X4 + .06000X1 X5 − .01667X2 X4 − .02667X2 X5
−.02333X3 X4 − .00333X3 X5 ,
SSE(F ) = .7180
Reduced Model:
Ŷ = 2.72074 − .59611X1 + .08412X2 + .39964X3 − .33756X4 − .26317X5 ,
SSE(R) = .7624
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = (.0444/6) ÷ (.7180/33) = .34, F (.99; 6, 33) = 3.41.
If F ∗ ≤ 3.41 conclude H0 , otherwise Ha . Conclude H0 . P -value = .91
d.
Subject matter:
Yijk = µ.. + β1 Xijk4 + β2 Xijk5 + (αβ)11 Xijk1 Xijk4 + (αβ)12 Xijk1 Xijk5
+(αβ)21 Xijk2 Xijk4 + (αβ)22 Xijk2 Xijk5 + (αβ)31 Xijk3 Xijk4
+(αβ)32 Xijk3 Xijk5 + ²ijk
23-10
Ŷ = 2.75121 − .34885X4 − .19441X5 + .19925X1 X4 + .19481X1 X5
−.01178X2 X4 − .08433X2 X5 − .22413X3 X4 + .00731X3 X5 ,
SSE(R) = 4.9506
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero.
F ∗ = (4.2326/3) ÷ (.7180/33) = 64.845, F (.99; 3, 33) = 4.437.
If F ∗ ≤ 4.437 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
Degree:
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + (αβ)11 Xijk1 Xijk4
+(αβ)12 Xijk1 Xijk5 + (αβ)21 Xijk2 Xijk4 + (αβ)22 Xijk2 Xijk5
+(αβ)31 Xijk3 Xijk4 + (αβ)32 Xijk3 Xijk5 + ²ijk
Ŷ = 2.88451 − .44871X1 − .09702X2 + .36160X3 − .06779X1 X4 + .08939X1 X5
−.05349X2 X4 − .03742X2 X5 + .07190X3 X4 − .10851X3 X5 ,
SSE(R) = 8.9467
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = (8.2287/2) ÷ (.7180/33) = 189.10, F (.99; 2, 33) = 5.321.
If F ∗ ≤ 5.321 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
D̂1 = µ̂1. − µ̂2. = 2.1500 − 2.8067 = −.6567, D̂2 = µ̂1. − µ̂3. = 2.1500 − 3.1133 =
−.9633, D̂3 = µ̂1. − µ̂4. = 2.1500 − 2.8400 = −.6900, D̂4 = µ̂2. − µ̂3. = −.3066,
D̂5 = µ̂2. − µ̂4. = −.0333, D̂6 = µ̂3. − µ̂4. = .2733, s{D̂1 } = .06642, s{D̂2 } =
.07083, s{D̂3 } = .07497, s{D̂4 } = .06316, s{D̂5 } = .06777, s{D̂6 } = .07209,
q(.95; 4, 33) = 3.825, T = 2.705
−.6567 ± 2.705(.06642)
−.9633 ± 2.705(.07083)
−.6900 ± 2.705(.07497)
−.3066 ± 2.705(.06316)
−.0333 ± 2.705(.06777)
.2733 ± 2.705(.07209)
f.
D̂1 = µ̂.1 − µ̂.2 = 2.3875 − 2.4675 = −.0800, D̂2 = µ̂.1 − µ̂.3 = 2.3875 − 3.3350 =
−.9475, D̂3 = µ̂.2 − µ̂.3 = −.8675, s{D̂1 } = .06597, s{D̂2 } = .05860, s{D̂3 } =
.05501, q(.95; 3, 33) = 3.470, T = 2.4537
−.0800 ± 2.4537(.06597)
−.9475 ± 2.4537(.05860)
−.8675 ± 2.4537(.05501)
23.11. a.
−.836 ≤ D1 ≤ −.477
−1.155 ≤ D2 ≤ −.772
−.893 ≤ D3 ≤ −.487
−.477 ≤ D4 ≤ −.136
−.217 ≤ D5 ≤ .150
.078 ≤ D6 ≤ .468
−.242 ≤ D1 ≤ .082
−1.091 ≤ D2 ≤ −.804
−1.002 ≤ D3 ≤ −.733
Full model:
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + β1 Xijk4 + β2 Xijk5 + ²ijk
Xijk1 , Xijk2 Xijk3 , Xijk4 , Xijk5 defined same as in Problem 23.9a
Reduced models:
Factor A: Yijk = µ.. + β1 Xijk4 + β2 Xijk5 + ²ijk
23-11
Factor B: Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + ²ijk
b.
Full model:
Ŷ = 2.72074 − .59611X1 + .08412X2 + .33964X3 − .33756X4 − .26317X5 ,
SSE(F ) = .762425, dfF = 39
Reduced models:
Factor A:
Ŷ = 2.72494 − .32494X4 − .18648X5 , SSE(R) = 6.741678, dfR = 42
H0 : α1 = α2 = α3 = 0, Ha : not all αi equal zero.
F ∗ = (5.979253/3) ÷ (.762425/39) = 101.95, F (.95; 3, 39) = 2.845.
If F ∗ ≤ 2.845 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
Factor B:
Ŷ = 2.86983 − .44483X1 − .08521X2 + .36654X3 , SSE(R) = 9.144878, dfR = 41
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = (8.382453/2) ÷ (.762425/39) = 214.39, F (.95; 2, 39) = 3.238.
If F ∗ ≤ 3.238 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
23.12. a.
See Problem 19.14a. D̂1 = Ȳ13. − Ȳ11. = 2.100, D̂2 = Ȳ23. − Ȳ21. = 3.675,
D̂3 = Ȳ33. − Ȳ31. = 7.275, L̂1 = D̂1 − D̂2 = −1.575, L̂2 = D̂1 − D̂3 = −5.175,
M SE = .06406, s{D̂i } = .1790 (i = 1, 2, 3), s{L̂i } = .2531 (i = 1, 2), B =
t(.99; 24) = 2.492
2.100 ± 2.492(.1790)
3.675 ± 2.492(.1790)
7.275 ± 2.492(.1790)
−1.575 ± 2.492(.2531)
−5.175 ± 2.492(.2531)
b.
1.654 ≤ D1 ≤ 2.546
3.229 ≤ D2 ≤ 4.121
6.829 ≤ D3 ≤ 7.721
−2.206 ≤ L1 ≤ −.944
−5.806 ≤ L2 ≤ −4.544
H0 : µ12 − µ13 = 0, Ha : µ12 − µ13 6= 0. D̂ = Ȳ12. − Ȳ13. = .025, s{D̂} = .1790,
t∗ = .025/.1790 = .14, t(.99; 24) = 2.492. If |t∗ | ≤ 2.492 conclude H0 , otherwise
Ha . Conclude H0 .
H0 : µ32 − µ33 = 0, Ha : µ32 − µ33 6= 0. D̂ = Ȳ32. − Ȳ33. = −2.975, s{D̂} = .1790,
t∗ = −2.975/.1790 = −16.62, t(.99; 24) = 2.492. If |t∗ | ≤ 2.492 conclude H0 ,
otherwise Ha . Conclude Ha . α ≤ .04
23.13. a.
Yijk = µ.. + α1 Xijk1 + β1 Xijk2 + β2 Xijk3 + ²ijk
(
Xijk1 =
Xijk2 =
1
−1
if case from level 1 for factor A
if case from level 2 for factor A
1 if case from level 1 for factor B
−1 if case from level 3 for factor B
0 otherwise
1 if case from level 2 for factor B
Xijk3 = −1 if case from level 3 for factor B
0 otherwise
23-12
Ŷ = .66939 + .11733X1 − .34323X2 + .02608X3 , SSE(F ) = 4.4898
Factor A:
Yijk = µ.. + β1 Xijk2 + β2 Xijk3 + ²ijk
Ŷ = .70850 − .26502X2 − .01303X3 , SSE(R) = 5.0404
Factor B:
Yijk = µ.. + α1 Xijk1 + ²ijk
Ŷ = .77520 + .03152X1 , SSE(R) = 7.1043
b.
H0 : α1 = 0, Ha : α1 6= 0.
F ∗ = (.5506/1) ÷ (4.4898/46) = 5.641, F (.95; 1, 46) = 4.05.
If F ∗ ≤ 4.05 conclude H0 , otherwise Ha . Conclude Ha . P -value = .022
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = (2.6145/2) ÷ (4.4898/46) = 13.393, F (.95; 2, 46) = 3.20.
If F ∗ ≤ 3.20 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
23.14. a.
See Problem 19.20a. D̂1 = Ȳ12. − Ȳ13. = 46.0, D̂2 = Ȳ22. − Ȳ23. = 6.0,
L̂1 = D̂1 − D̂2 = 40.0, M SE = 88.50, s{D̂1 } = s{D̂2 } = 6.652, s{L̂1 } = 9.407,
B = t(.99167; 15) = 2.694
46.0 ± 2.694(6.652)
28.080 ≤ D1 ≤ 63.920
6.0 ± 2.694(6.652) −11.920 ≤ D2 ≤ 23.920
40.0 ± 2.694(9.407)
14.658 ≤ L1 ≤ 65.342
b.
23.15. a.
H0 : µ22 − µ23 ≤ 0, Ha : µ22 − µ23 > 0. D̂ = Ȳ22. − Ȳ23. = 6.0, s{D̂} = 6.652,
t∗ = 6.0/6.652 = .90, t(.95; 15) = 1.753. If t∗ ≤ 1.753 conclude H0 otherwise Ha .
Conclude H0 . P -value = .19
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + β1 Xijk4 + β2 Xijk5 + ²ijk
Xijk1
1 if case from level 1 for factor A
−1 if case from level 4 for factor A
=
0 otherwise
Xijk2 and Xijk3 are defined similarly
Xijk4
Xijk5
1 if case from level 1 for factor B
= −1 if case from level 3 for factor B
0 otherwise
1 if case from level 2 for factor B
= −1 if case from level 3 for factor B
0 otherwise
Ŷ = 2.71932 − .59897X1 + .08546X2 + .40036X3 − .34043X4 − .26218X5 ,
SSE(F ) = .7419
Factor A:
23-13
Yijk = µ.. + β1 Xijk4 + β2 Xijk5 + ²ijk
Ŷ = 2.77494 − .22494X4 − .23648X5 , SSE(R) = 5.8217
Factor B:
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + ²ijk
Ŷ = 2.90108 − .35108X1 − .11646X2 + .33529X3 , SSE(R) = 8.1874
b.
H0 : α1 = α2 = α3 = 0, Ha : not all αi equal zero.
F ∗ = (5.0798/3) ÷ (.7419/37) = 84.45, F (.99; 3, 37) = 4.360.
If F ∗ ≤ 4.360 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = (7.4455/2) ÷ (.7419/37) = 185.66, F (.99; 2, 37) = 5.229.
If F ∗ ≤ 5.229 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
23.16. a.
Yij = µ.. + ρ1 Xij1 + ρ2 Xij2 + ρ3 Xij3 + ρ4 Xij4 + ρ5 Xij5 + ρ6 Xij6
+ρ7 Xij7 + ρ8 Xij8 + ρ9 Xij9 + τ1 Xij10 + τ2 Xij11 + ²ij
Iij1
1
−1
=
0
if experimental unit from block 1
if experimental unit from block 10
otherwise
Iij2 , . . . , Iij9 are defined similarly
Iij10 =
Iij11
b.
1 if experimental unit received treatment 1
−1 if experimental unit received treatment 3
0 otherwise
1 if experimental unit received treatment 2
= −1 if experimental unit received treatment 3
0 otherwise
Ŷ = 77.10000 + 4.90000X1 + 3.90000X2 + 2.23333X3 + 3.23333X4 + 1.23333X5
+.90000X6 − 1.10000X7 − 3.76667X8 − 4.10000X9 − 6.50000X10 − 2.50000X11
c.
Source
Regression
X1 , X2 , X3 , X4 , X5 , X6, X7 , X8, X9
X10 , X11 |X1 , X2 , X3 , X4 , X5 , X6, X7 , X8, X9
Error
Total
d.
SS
1, 728.3667
433.3667
1, 295.0000
112.3333
1, 840.7000
df
MS
1 157.1242
9 48.1519
2 647.5000
18
6.2407
29
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (1, 295.0000/2 ÷ (112.3333/18) = 103.754, F (.95; 2, 18) = 3.55.
If F ∗ ≤ 3.55 conclude H0 , otherwise Ha . Conclude Ha .
23.17. a.
Yij = µ.. + ρ1 Xij1 + ρ2 Xij2 + ρ3 Xij3 + ρ4 Xij4 + τ1 Xij5 + τ2 Xij6 + ²ij
23-14
Iij1 =
1
−1
0
if experimental unit from block 1
if experimental unit from block 5
otherwise
Iij2 , . . . , Iij4 are defined similarly
Iij5
Iij6
b.
1
= −1
0
1
−1
=
0
if experimental unit received treatment 1
if experimental unit received treatment 3
otherwise
if experimental unit received treatment 2
if experimental unit received treatment 3
otherwise
Ŷ = .84400 − .32733X1 − .23733X2 − .17400X3 + .31267X4 + .26600X5 + .14800X6
c.
Source
Regression
X 1 , X2 , X 3 , X4
X5 , X6 , | X1 , X2 , X3 , X4
Error
Total
d.
SS
2.7392
1.4190
1.3203
.0193
2.7585
df
6
4
2
8
14
MS
.4565
.35475
.6602
.0024
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (1.3203/2) ÷ (.0193/8) = 273.637, F (.95; 2, 8) = 4.46.
If F ∗ ≤ 4.46 conclude H0 , otherwise Ha . Conclude Ha .
23.18. a.
Yij = µ.. + ρi + τj + ²ij
Yij = µ.. + ρ1 Xij1 + ρ2 Xij2 + ρ3 Xij3 + ρ4 Xij4 + ρ5 Xij5 + ρ6 Xij6
+ρ7 Xij7 + ρ8 Xij8 + ρ9 Xij9 + τ1 Xij10 + τ2 Xij11 + ²ij
Iij1
1
−1
=
0
if experimental unit from block 1
if experimental unit from block 10
otherwise
Iij2 , . . . , Iij9 are defined similarly
Iij10 =
Iij11
b.
1 if experimental unit received treatment 1
−1 if experimental unit received treatment 3
0 otherwise
1 if experimental unit received treatment 2
−1 if experimental unit received treatment 3
=
0 otherwise
Yij = µ.. + ρ1 Xij1 + ρ2 Xij2 + +ρ3 Xij3 + ρ4 Xij4 + ρ5 Xij5 + ρ6 Xij6
+ρ7 Xij7 + ρ8 Xij8 + ρ9 Xij9 + ²ij
c.
Full model: Ŷ = 77.15556 + 4.84444X1 + 4.40000X2 + 2.17778X3
+3.17778X4 + 1.17778X5 + .84444X6 − 1.15556X7
−3.82222X8 − 4.15556X9 − 6.55556X10 − 2.55556X11
23-15
SSE(F ) = 110.6667
Reduced model: Ŷ = 76.70000 + 5.30000X1 + .30000X2 + 2.63333X3
+3.63333X4 + 1.63333X5 + 1.30000X6 − .70000X7
−3.36667X8 − 3.70000X9
SSE(R) = 1, 311.3333
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (1, 200.6666/2) ÷ (110.6667/17) = 92.22, F (.95; 2, 17) = 3.59.
If F ∗ ≤ 3.59 conclude H0 , otherwise Ha . Conclude Ha .
d.
L̂ = τ̂ 2 − τ̂ 3 = 2τ̂ 2 + τ̂ 1 = −11.66667, s2 {τ̂ i } = .44604 (i = 1, 2), s{τ̂ 1 , τ̂ 2 } =
−.20494, s{L̂} = 1.1876, t(.975; 17) = 2.11,
−11.66667 ± 2.11(1.1876), −14.17 ≤ L ≤ −9.16
23.19. a.
Yij = µ.. + ρi + τj + ²ij
Yij = µ.. + ρ1 Xij1 + ρ2 Xij2 + ρ3 Xij3 + ρ4 Xij4 + τ1 Xij5 + τ2 Xij6 + ²ij
Iij1
1
= −1
0
if experimental unit from block 1
if experimental unit from block 5
otherwise
Iij2 , . . . , Iij4 are defined similarly
Iij5
1
−1
=
0
Iij6 =
1
−1
0
if experimental unit received treatment 1
if experimental unit received treatment 3
otherwise
if experimental unit received treatment 2
if experimental unit received treatment 3
otherwise
b.
Yij = µ.. + ρ1 Xij1 + ρ2 Xij2 + ρ3 Xij3 + ρ4 Xij4 + ²ij
c.
Full model: Ŷ = .82941 − .33613X1 − .22274X2 − .15941X3 + .32726X4
+.25085X5 + .16259X6
SSE(F ) = .0035
Reduced model: Ŷ = .84567 − 14567X1 − .23900X2 − .17567X3 + .31100X4
SSE(R) = .9542
H0 : τ1 = τ2 = 0, Ha : not both τ1 and τ2 equal zero.
F ∗ = (.9507/2) ÷ (.0035/6) = 814.89, F (.95; 2, 6) = 5.14.
If F ∗ ≤ 5.14 conclude H0 , otherwise Ha . Conclude Ha .
d.
L̂ = τ̂ 1 − τ̂ 3 = 2τ̂ 1 + τ̂ 2 = .66429, s2 {τ̂ 1 } = .000105, s2 {τ̂ 2 } = .000087, s{τ̂ 1 , τ̂ 2 } =
−.000043, s{L̂} = .0183, t(.99; 6) = 3.143,
.66429 ± 3.143(.0183), .607 ≤ L ≤ .722
23.20.
See Problem 19.10a. L1 = .3µ11 + .6µ21 + .1µ31 , L2 = .3µ12 +.6µ22 + .1µ32 .
23-16
H0 : L1 = L2 , Ha : L1 6= L2 .
L̂1 − L̂2 = 25.43332 − 25.05001 = .38331, M SE = 2.3889, s{L̂1 − L̂2 } = .6052,
t∗ = .38331/.6052 = .63, t(.975; 30) = 2.042.
If |t∗ | ≤ 2.042 conclude H0 , otherwise Ha . Conclude H0 . P -value = .53
23.21. a.
3µ11 + µ21 3µ12 + µ22
−
=0
4
4
H0 :
3µ11 + µ21 3µ13 + µ23
L2 =
−
=0
4
4
Ha : not both L1 and L2 equal zero
L1 =
b.
Regression model equivalent to (19.15) using 1, 0 indicator variables:
Full model:
Yijk = µ11 Xijk1 + µ12 Xijk2 + µ13 Xijk3 + µ21 Xijk4 + µ22 Xijk5
+µ23 Xijk6 + ²ijk
(
Xijk1 =
(
Xijk2 =
(
Xijk3 =
(
Xijk4 =
(
Xijk5 =
(
Xijk6 =
c.
1 if case from level 1 for factor A and level 1 for factor B
0 otherwise
1 if case from level 1 for factor A and level 2 for factor B
0 otherwise
1 if case from level 1 for factor A and level 3 for factor B
0 otherwise
1 if case from level 2 for factor A and level 1 for factor B
0 otherwise
1 if case from level 2 for factor A and level 2 for factor B
0 otherwise
1 if case from level 2 for factor A and level 3 for factor B
0 otherwise
Reduced model:
µ11 = µ12 + µ22 /3 − µ21 /3
µ13 = µ12 + µ22 /3 − µ23 /3
Yijk = (µ12 + µ22 /3 − µ21 /3)Xijk1 + µ12 Xijk2 + (µ12 + µ22 /3 − µ23 /3)Xijk3
+µ21 Xijk4 + µ22 Xijk5 + µ23 Xijk6 + ²ijk
or
Yijk = µ12 Zijk1 + µ21 Zijk2 + µ22 Zijk3 + µ23 Zijk4 + ²ijk
where:
Zijk1
Zijk2
Zijk3
Zijk4
=
=
=
=
Xijk1 + Xijk2 + Xijk3
−Xijk1 /3 + Xijk4
(Xijk1 + Xijk3 )/3 + Xijk5
(−Xijk3 /3) + Xijk6
23-17
d.
SSE(F ) = 5.468, dfF = 54, SSE(R) = 8.490, dfR = 56,
F ∗ = [(8.490 − 5.468)/2] ÷ (5.468/54) = 14.92, F (.95; 2, 54) = 3.17.
If F ∗ ≤ 3.17 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
See Problem 19.18a. L̂2 = (.75Ȳ11. + .25Ȳ21. ) − (.75Ȳ13. + .25Ȳ23. ) = −.61446,
s{L̂2 } = .1125, t(.975; 54) = 2.005, −.61446 ± 2.005(.1125), −.840 ≤ L2 ≤ −.389
23.22. a.
µ11 + 2µ12 + 7µ13 µ21 + 2µ22 + 7µ23
−
=0
10
10
µ11 + 2µ12 + 7µ13 µ31 + 2µ32 + 7µ33
H0 : L 2 =
−
=0
10
10
µ11 + 2µ12 + 7µ13 µ41 + 2µ42 + 7µ43
L3 =
−
=0
10
10
Ha : not all Li equal zero (i = 1, 2, 3)
L1 =
b.
β entries: µ11 , µ12 , µ13 , µ21 , µ22 , µ23 , µ31 , µ32 , µ33 , µ41 , µ42 , µ43
X entries:
A B Freq.
1 1
2
1 2
2
1 3
8
2 1
4
2 2
5
2 3
4
3 1
2
3 2
4
3 3
5
4 1
2
4 2
2
4 3
5
c.
.1
C = .1
.1
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
1
.2 .7 −.1 −.2 −.7
0
0
0
0
0
0
0
.2 .7
0
0
0 −.1 −.2 −.7
0
0
0
h= 0
.2 .7
0
0
0
0
0
0 −.1 −.2 −.7
0
d.
SSE(R) − SSE(F ) = 5.6821
e.
SSE(F ) = .7180, F ∗ = (5.6821/3) ÷ (.7180/33) = 87.05, F (.99; 3, 33) = 4.437.
If F ∗ ≤ 4.437 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
f.
See Problem 23.10a. L̂2 = (.1Ȳ11. + .2Ȳ12. + .7Ȳ13. ) − (.1Ȳ31. + .2Ȳ32. + .7Ȳ33. ) =
−1.003, s{L̂2 } = .0658, t(.995; 33) = 2.733,
−1.003 ± 2.733(.0658), −1.183 ≤ L2 ≤ −.823
23.23.
H0 :
4µ11 +4µ12 +2µ13
10
=
4µ21 +4µ22 +3µ23
,
11
Ha : equality does not hold.
Ȳ... = 93.714, Ȳ1.. = 143, Ȳ2.. = 48.91
SSA = 10(143 − 93.714)2 + 11(48.91 − 93.714)2 = 46, 372
23-18
F ∗ = (46, 372/1) ÷ (1, 423.1667/15) = 488.8, F (.99; 1, 15) = 8.68.
If F ∗ ≤ 8.68 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
2
2
H0 : 12
µ11 + 12
µ12 +
2
2
µ + 9 µ42 + 95 µ43 ,
9 41
23.24.
8
µ
12 13
=
4
µ
13 21
+
5
µ
13 22
+
4
µ
13 23
=
2
µ
11 31
+
4
µ
11 32
+
5
µ
11 33
=
Ha : not all equalities hold.
Ȳ... = 2.849, Ȳ1.. = 2.425, Ȳ2.. = 2.785, Ȳ3.. = 3.236, Ȳ4.. = 3.033
SSA = 12(2.425 − 2.849)2 + 13(2.785 − 2.849)2 + 11(3.236 − 2.849)2 + 9(3.033 −
2.849)2 = 4.163
F ∗ = (4.163/3) ÷ (.718/33) = 63.78, F (.95; 3, 33) = 2.89.
If F ∗ ≤ 2.89 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
P
σ 2 {L̂} = σ 2 {
23.25.
ci µ̂i. } =
P 2 2
P
ci σ {µ̂i. } = c2i σ 2
i
P
Ȳij.
j
b
1 P 2 P σ2
σ2 P 2 P 1
=
c
c
b2 i i j nij
b2 i i j nij
=
because of independence of Ȳij. .
n
(
o
2
2
P P cij
E s {L̂} = E M SE
23.26.
23.27. a.
X=
b.
1
1
1
1
1
1
1
1
1
Xβ =
1
1
1
0
0
0
0
0
0
0
0
0
1
1
1
0
0
0
1
0
0
1
0
0
1
0
0
0
1
0
0
1
0
0
1
0
β0 + β1 + β3
β0 + β1 + β4
β0 + β1
β0 + β2 + β3
β0 + β2 + β4
β0 + β2
β0
+ β3
β0
+ β4
β0
β0 = µ.. + ρ3 + τ3
β4 = τ 2 − τ 3
β3 = τ 1 − τ 3
nij
)
=
2
P P cij
nij
=
µ.. + ρ1 + τ1
µ.. + ρ1 + τ2
µ.. + ρ1 + τ3
µ.. + ρ2 + τ1
µ.. + ρ2 + τ2
µ.. + ρ2 + τ3
µ.. + ρ3 + τ1
µ.. + ρ3 + τ2
µ.. + ρ3 + τ3
β2 = ρ2 − ρ3
β1 = ρ1 − ρ3
23-19
E{M SE} = σ
2
2
P P cij
nij
= σ 2 {L̂}
23.28.
Y111
Y112
Y121
Y122
Y211
Y212
Y221
Y=
Cβ =
h
X=
1
1
0
0
0
0
0
0
0
1
1
0
0
0
i
1
1 −1 −1
0
0
0
0
1
1
0
µ11
µ12
µ21
µ22
0
0
0
0
0
0
1
bR =
Ȳ11.
Ȳ12.
Ȳ21.
Ȳ22.
=
−
Ȳ11.
Ȳ12.
Ȳ21.
Ȳ22.
β=
µ11
µ12
µ21
µ22
bF =
Ȳ11.
Ȳ12.
Ȳ21.
Ȳ22.
= µ11 − µ12 − µ21 + µ22 = 0
From (23.46):
1/2 0
0 0
0 1/2 0 0
0
0 1/2 0
0
0
0 1
1
−1
−1
1
µ 5 ¶−1 ³
´
Ȳ11. − Ȳ12. − Ȳ21. + Ȳ22. − 0
2
´
2³
Ȳ11. − Ȳ12. − Ȳ21. + Ȳ22.
−
5
1/2
−1/2
−1/2
1
2
µ̂22 = Ȳ22. − (Ȳ11. − Ȳ12. − Ȳ21. + Ȳ22. )
5
23.29. a.
β entries: µ11 , µ12 , µ13 , µ22
X entries:
A B Freq.
1 1
10
1 2
10
1 3
10
2 2
10
2 3
10
b.
23.30. a.
C = [0
1
1
0
0
0
0
0
1
0
0
−1
0 − 1]
0
0
1
0
1
0
0
0
1
1
h=0
Yijk = µ.. + αi + βj + (αβ)ij + ²ijk
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + β1 Xijk4 + β2 Xijk5
+(αβ)11 Xijk1 Xijk4 + (αβ)12 Xijk1 Xijk5 + (αβ)21 Xijk2 Xijk4
+(αβ)22 Xijk2 Xijk5 + (αβ)31 Xijk3 Xijk4 + (αβ)32 Xijk3 Xijk4 + ²ijk
1 if case from NE
Iijk1 = −1 if case from W
0 otherwise
1 if case from NC
Iijk2 = −1 if case from W
0 otherwise
23-20
Iijk3 =
Iijk4 =
Iijk5
b.
1 if case from S
−1 if case from W
0 otherwise
1 if average age under 52.0
−1 if average age 55.0 or more
0 otherwise
1 if average age 52.0 - under 55.0
−1 if average age 55.0 or more
=
0 otherwise
Ŷ = 9.40661 + 1.45009X1 + .23601X2 − .24406X3 − .38373X4
+.19446X5 − .76296X1 X4 − .57198X1 X5 + .44674X2 X4
+.17515X2 X5 + .35707X3 X4 − .38986X3 X5 , SSE(F ) = 261.2341
c.
23.31. a.
r = .959
Ȳ11. = 9.71000, Ȳ12. = 10.47917, Ȳ13. = 12.38091, Ȳ21. = 9.70563,
Ȳ22. = 10.01222, Ȳ23. = 9.21000, Ȳ31. = 9.13588, Ȳ32. = 8.96714,
Ȳ33. = 9.38462, Ȳ41. = 7.54000, Ȳ42. = 8.94571, Ȳ43. = 7.40800
b.
Yijk = µ.. + α1 Xijk1 + α1 Xijk2 + α3 Xijk3 + β1 Xijk4 + β2 Xijk5 + ²ijk
c.
Ŷ = 9.52688 + 1.50561X1 + .23119X2 − .30781X3 − .26423X4 − .00518X5 ,
SSE(R) = 300.4100
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = (39.1759/6) ÷ (261.2341/101) = 2.524, F (.99; 6, 101) = 2.99.
If F ∗ ≤ 2.99 conclude H0 , otherwise Ha . Conclude H0 . P -value = .026
d.
Yijk = µ.. + β1 Xijk4 + β2 Xijk5 + (αβ)11 Xijk1 Xijk4
+ (αβ)12 Xijk1 Xijk5 + (αβ)21 Xijk2 Xijk4 + (αβ)22 Xijk2 Xijk5
+ (αβ)31 Xijk3 Xijk4 + (αβ)32 Xijk3 Xijk5 + ²ijk
Ŷ = 9.52473 − .50240X4 + .24409X5 − 1.44778X1 X4 − .17942X1 X5
+.61345X2 X4 + .11924X2 X5 + .34296X3 X4 − .24453X3 X5 ,
SSE(R) = 345.4833
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero.
F ∗ = (84.2492/3) ÷ (261.2341/101) = 10.858, F (.99; 3, 101) = 3.98.
If F ∗ ≤ 3.98 conclude H0 , otherwise Ha . Conclude Ha . P -value 0+
e.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + (αβ)11 Xijk1 Xijk4
+ (αβ)12 Xijk1 Xijk5 + (αβ)21 Xijk2 Xijk4 + (αβ)22 Xijk2 Xijk5
+ (αβ)31 Xijk3 Xijk4 + (αβ)32 Xijk3 Xijk5 + ²ijk
Ŷ = 9.42456 + 1.53811X1 + .15829X2 − .31129X3 − .69021X1 X4
−.60603X1 X5 + .29855X2 X4 + .26600X2 X5 + .18805X3 X4 − .35618X3 X5 ,
23-21
SSE(R) = 267.7103
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = (6.4762/2) ÷ (261.2341/101) = 1.252, F (.99; 2, 101) = 4.82.
If F ∗ ≤ 4.82 conclude H0 , otherwise Ha . Conclude H0 . P -value = .29
f.
n11 = 5, n12 = 12, n13 = 11, n21 = 16, n22 = 9, n23 = 7,
n31 = 17, n32 = 7, n33 = 13, n41 = 4, n42 = 7, n43 = 5,
D̂1 = µ̂1. − µ̂2. = 10.85669 − 9.64262 = 1.21407,
D̂2 = µ̂1. − µ̂3. = 10.85669 − 9.16255 = 1.69414,
D̂3 = µ̂1. − µ̂4. = 10.85669 − 7.96457 = 2.89212,
D̂4 = µ̂2. − µ̂3. = .48007, D̂5 = µ̂2. − µ̂4. = 1.67805,
D̂6 = µ̂3. − µ̂4. = 1.19798, M SE = 2.5865, s{D̂1 } = .4455, s{D̂2 } = .4332,
s{D̂3 } = .5272, s{D̂4 } = .4135, s{D̂5 } = .5112, s{D̂6 } = .5004,
q(.95; 4, 101) = 3.694, T = 2.612
1.21407 ± 2.612(.4455)
1.69414 ± 2.612(.4332)
2.89212 ± 2.612(.5272)
.48007 ± 2.612(.4135)
1.67805 ± 2.612(.5112)
1.19798 ± 2.612(.5004)
23.32. a.
.050 ≤ D1
.563 ≤ D2
1.515 ≤ D3
−.600 ≤ D4
.343 ≤ D5
−.109 ≤ D6
≤ 2.378
≤ 2.826
≤ 4.269
≤ 1.560
≤ 3.013
≤ 2.505
ANOVA model: Yijk = µ.. + αi + βj + (αβ)ij + ²ijk
Regression: Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + β1 Xijk4 + β2 Xijk5
+ (αβ)11 Xijk1 Xijk4 + (αβ)12 Xijk1 Xijk5 + (αβ)21 Xijk2 Xijk4
Iijk1
1 if case from NE
−1 if case from W
=
0 otherwise
Iijk2 =
Iijk3
Iijk4
+ (αβ)22 Xijk2 Xijk5 + (αβ)31 Xijk3 Xijk4 + (αβ)32 Xijk3 Xijk5 +²ijk
1 if case from NC
−1 if case from W
0 otherwise
1 if case from S
= −1 if case from W
0 otherwise
1 if poverty level below 6.0 percent
−1 if poverty level is 10 percent or more
=
0 otherwise
1 if poverty level between 6.0 and under 10.0 percent
Iijk5 = −1 if poverty level is 10 percent or more
0 otherwise
23-22
b.
Ŷ = .0568 − .00852X1 − .00475X2 + .00983X3 − .0114X4 − .00173X5 − .00206X1 X4
−.00629X1 X5 − .00106X2 X4 + .00069X2 X5 − .00102X3 X4 + .00133X3 X5 ,
SSE(F ) = .23111
c.
23.33. a.
r = .932
Ȳ11. = .0348, Ȳ12. = .0402, Ȳ13. = .0697, Ȳ21. = .0396, Ȳ22. = .0510, Ȳ23. = .0655,
Ȳ31. = .0542, Ȳ32. = .0662, Ȳ33. = .0794, Ȳ41. = .0530, Ȳ42. = .0627, Ȳ43. = .0649
b.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + β1 Xijk4 + β2 Xijk5 +²ijk
c.
Ŷ = .0563 − .0105X1 − .0043X2 + .0106X3 − .0111X4 − .0013X5 ,
SSE(R) = .23589
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = (.00478/6) ÷ (.23111/428) = 1.476. F (.995; 6, 428) = 3.14.
If F ∗ ≤ 3.14 conclude H0 , otherwise Ha . Conclude H0 . P -value = .18
d.
Yijk = µ.. + β1 Xijk4 + β2 Xijk5 + (αβ)11 Xijk1 Xijk4
+ (αβ)12 Xijk1 Xijk5 + (αβ)21 Xijk2 Xijk4 + (αβ)22 Xijk2 Xijk5
+ (αβ)31 Xijk3 Xijk4 + (αβ)32 Xijk3 Xijk5 + ²ijk
Ŷ = .0578 − .0141X4 − .0024X5 − .0043X1 X4 − .0090X1 X5 + .0020X2 X4
−.0005X2 X5 − .0040X3 X4 + .0039X3 X5 ,
SSE(R) = .25118
H0 : α1 = α2 = α3 , Ha : not all αi equal zero.
F ∗ = (.02007/3) ÷ (.23111/428) = 12.39, F (.995; 3, 428) = 4.34.
If F ∗ ≤ 4.34 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + (αβ)11 Xijk1 Xijk4
+ (αβ)12 Xijk1 Xijk5 + (αβ)21 Xijk2 Xijk4 + (αβ)22 Xijk2 Xijk5
+ (αβ)31 Xijk3 Xijk4 + (αβ)32 Xijk3 Xijk5 + ²ijk
Ŷ = .0558 − .0138X1 − .0047X2 + .0135X3 − .0015X1 X4 − .0009X1 X5
+3.15459X2 X4 − .23458X2 X5 − .0032X3 X4 − .0025X3 X5 ,
SSE(R) = .26209
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = (.03098/2) ÷ (.23111/428) = 28.69, F (.995; 2, 428) = 5.36.
If F ∗ ≤ 5.36 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
f.
n11 = 52, n12 = 38, n13 = 13, n21 = 32, n22 = 50, n23 = 26,
n31 = 25, n32 = 52, n33 = 75, n41 = 20, n42 = 33, n43 = 24,
D̂1 = µ̂1. − µ̂2. = −.0038, D̂2 = µ̂1. − µ̂3. = −.0184, D̂3 = µ̂1. − µ̂4. = −.0120,
D̂4 = µ̂2. − µ̂3. = −.0146, D̂5 = µ̂2. − µ̂4. = −.0082, D̂6 = µ̂3. − µ̂4. = .0064,
23-23
M SE = .00054, q(.95; 4, 428) = 3.63, T = 2.567
s{D̂1 } = .00357, s{D̂2 } = .00342, s{D̂3 } = .00383,
s{D̂4 } = .00312, s{D̂5 } = .00356, s{D̂6 } = .00342,
−.0038 ± 2.567(.00357)
−.0184 ± 2.567(.00342)
−.0120 ± 2.567(.00383)
−.0146 ± 2.567(.00312)
−.0082 ± 2.567(.00356)
.0064 ± 2.567(.00342)
23.34. a.
−.0130 ≤ D1 ≤ .0054
−.0271 ≤ D2 ≤ −.0096
−.0218 ≤ D3 ≤ −.0021
−.0226 ≤ D4 ≤ −.0066
−.0173 ≤ D5 ≤ .0010
−.0024 ≤ D6 ≤ .0152
ANOVA model: Yijk = µ.. + αi + βj + (αβ)ij + ²ijk
Regression: Yijk = µ.. + α1 Xijk1 + β1 Xijk2 + (αβ)11 Xijk1 Xijk2 + ²ijk
(
Xijk1 =
(
Xijk2 =
b.
1 if no discount price (level 0 of variable 5)
−1 if discount price (level 1 of variable 5)
1 if no package promotion (level 0 of variable 6)
−1 if package promotion (level 1 of variable 6)
Ŷ = 2.620 − .199X1 − .0446X2 + .0366X1 X2
SSE(F ) = .7850
c.
23.35. a.
r = .990
Ȳ00. = 2.4125, Ȳ01. = 2.4286, Ȳ10. = 2.7375, Ȳ11. = 2.9000,
b.
Yijk = µ.. + α1 Xijk1 + β1 Xijk2 + ²ijk
c.
Ŷ = 2.625 − .201X1 − .0498X2 , SSE(R) = .8306
H0 : (αβ)11 = 0, Ha : (αβ)11 6= 0.
F ∗ = (.0456/1) ÷ (.7850/32) = 1.86. F (.95; 1, 32) = 4.15.
If F ∗ ≤ 4.15 conclude H0 , otherwise Ha . Conclude H0 . P -value = .18
d.
Yijk = µ.. + β1 Xijk2 + (αβ)11 Xijk1 Xijk2 + ²ijk
Ŷ = 2.648 − .0726X2 + .0494X1 X2 , SSE(R) = 2.1352
H0 : α1 = 0, Ha : α1 6= 0.
F ∗ = (1.3502/1) ÷ (.7850/32) = 55.04, F (.95; 1, 32) = 4.15.
If F ∗ ≤ 4.15 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
Yijk = µ.. + α1 Xijk1 + (αβ)11 Xijk1 Xijk2 + ²ijk
Ŷ = 2.623 − .205X1 + .0429X1 X2 , SSE(R) = .8529
H0 : β1 = 0, Ha : β1 6= 0
F ∗ = (.0679/1) ÷ (.7850/32) = 2.77, F (.95; 1, 32) = 4.15.
If F ∗ ≤ 4.15 conclude H0 , otherwise Ha . Conclude H0 . P -value = .11
23.36. a.
H0 :
5µ11 +12µ12 +11µ13
28
=
16µ21 +9µ22 +7µ23
32
=
23-24
17µ31 +7µ32 +13µ33
37
=
4µ41 +7µ42 +5µ43
,
16
Ha : not all equalities hold.
F ∗ = (103.55418/3) ÷ (261.23406/101) = 13.346, F (.99; 3, 101) = 3.98.
If F ∗ ≤ 3.98 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
b.
H0 :
5µ11 +16µ21 +17µ31 +4µ41
42
=
12µ12 +9µ22 +7µ32 +7µ42
35
=
11µ13 +7µ23 +13µ33 +5µ43
,
36
Ha : not all equalities hold.
F ∗ = (10.63980/2) ÷ (261.23406/101) = 2.057, F (.99; 2, 101) = 4.82.
If F ∗ ≤ 4.82 conclude H0 , otherwise Ha . Conclude H0 . P -value = .13
23.37. a.
H0 :
52µ11 +38µ12 +13µ13
103
=
32µ21 +50µ22 +26µ23
108
=
25µ31 +52µ32 +75µ33
152
=
20µ41 +33µ42 +24µ43
,
77
Ha : not all equalities hold.
Ȳ... = .05729, Ȳ1.. = .04123, Ȳ2.. = .05111, Ȳ3.. = .07074, Ȳ4.. = .06088
F ∗ = (.0592/3) ÷ (.23111/428) = 36.54, F (.995; 3, 428) = 4.34.
If F ∗ ≤ 4.34 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
b.
H0 :
52µ11 +32µ21 +25µ31 +20µ41
129
=
38µ12 +50µ22 +52µ32 +33µ42
173
=
13µ13 +26µ23 +75µ33 +24µ43
,
138
Ha : not all equalities hold.
Ȳ... = .05729, Ȳ.1. = .04259, Ȳ.2. = .05544, Ȳ.3. = .07334,
F ∗ = (.0640/2) ÷ (.23111/428) = 59.26, F (.995; 2, 428) = 5.36.
If F ∗ ≤ 5.36 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
23-25
23-26
Chapter 24
MULTIFACTOR STUDIES
24.1. a.
β1 = µ.1. − µ... = −2, β2 = µ.2. − µ... = −.5, β3 = µ.3. − µ... = 2.5
b.
(βγ)12 = µ.12 − µ.1. − µ..2 + µ... = 1
c.
(αβγ)212 = µ212 − µ21. − µ.12 − µ2.2 + µ2.. + µ.1. + µ..2 − µ... = −.5
24.4. a.
α1 = µ1.. − µ... = 138 − 131.5 = 6.5, α2 = µ2.. − µ... = 131.5 − 131.5 = 0
α3 = µ3.. − µ... = 125 − 131.5 = −6.5
b.
β2 = µ.2. − µ... = 134 − 131.5 = 2.5, γ1 = µ..1 − µ... = 128.5 − 131.5 = −3
c.
(αβ)12 = µ12. − µ1.. − µ.2. + µ... = 141 − 138 − 134 + 131.5 = .5
(αγ)21 = µ2.1 − µ2.. − µ..1 + µ... = 128 − 131.5 − 128.5 + 131.5 = −.5
(βγ)12 = µ.12 − µ.1. − µ..2 + µ... = 132 − 129 − 134.5 + 131.5 = 0
d.
(αβγ)111 = µ111 − µ.11 − µ1.1 − µ11. + µ1.. + µ.1. + µ..1 − µ...
= 130 − 126 − 134 − 135 + 138 + 129 + 128.5 − 131.5 = −1
(αβγ)322 = µ322 − µ32. − µ3.2 − µ.22 + µ3.. + µ.2. + µ..2 − µ...
= 131 − 128 − 126.5 − 137 + 125 + 134 + 134.5 − 131.5 = 1.5
24.6. a.
eijkm :
i
1
2
b.
24.7. a.
k=1
j=1
j=2
3.7667
1.1667
−3.9333 −1.6333
.1667
.4667
−1.7000
1.1000
.6000
−.8333
1.3667
−.5333
i
1
2
k=2
j=1
j=2
−.5000 −1.0333
.4000
1.3667
.1000 −.3333
1.1333
−1.6667
.5333
−.5667
1.7333
−1.1667
r = .973
Ȳ111. = 36.1333, Ȳ112. = 56.5000, Ȳ121. = 52.3333, Ȳ122. = 71.9333,
Ȳ211. = 46.9000, Ȳ212. = 68.2667, Ȳ221. = 64.1333, Ȳ222. = 83.4667
24-1
b.
Source
Between treatments
A (chemical)
B (temperature)
C (time)
AB interactions
AC interactions
BC interactions
ABC interactions
Error
Total
SS
df
MS
4, 772.25835 7
681.75119
788.90667 1
788.90667
1, 539.20167 1 1, 539.20167
2, 440.16667 1 2, 440.16667
.24000 1
.24000
.20167 1
.20167
2.94000 1
2.94000
.60167 1
.60167
53.74000 16
3.35875
4, 825.99835 23
c.
H0 : all (αβγ)ijk equal zero, Ha : not all (αβγ)ijk equal zero. F ∗ = .60167/3.35875 =
.18, F (.975; 1, 16) = 6.12. If F ∗ ≤ 6.12 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .68
d.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = .24000/3.35875 =
.07, F (.975; 1, 16) = 6.12. If F ∗ ≤ 6.12 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .79
H0 : all (αγ)ik equal zero, Ha : not all (αγ)ik equal zero. F ∗ = .20167/3.35875 =
.06, F (.975; 1, 16) = 6.12. If F ∗ ≤ 6.12 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .81
H0 : all (βγ)jk equal zero, Ha : not all (βγ)jk equal zero. F ∗ = 2.94000/3.35875 =
.875, F (.975; 1, 16) = 6.12. If F ∗ ≤ 6.12 conclude H0 , otherwise Ha . Conclude
H0 . P -value = .36
e.
H0 : all αi equal zero (i = 1, 2), Ha : not all αi equal zero. F ∗ = 788.90667/3.35875 =
234.88, F (.975; 1, 16) = 6.12. If F ∗ ≤ 6.12 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : all βj equal zero (j = 1, 2), Ha : not all βj equal zero. F ∗ = 1, 539.20167/3.35875 =
458.27, F (.975; 1, 16) = 6.12. If F ∗ ≤ 6.12 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : all γk equal zero (k = 1, 2), Ha : not all γk equal zero. F ∗ = 2, 440.1667/3.35875 =
726.51, F (.975; 1, 16) = 6.12. If F ∗ ≤ 6.12 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
f.
24.8. a.
α ≤ .1624
D̂1 = 65.69167 − 54.22500 = 11.46667, D̂2 = 67.96667 − 51.95000 = 16.01667,
D̂3 = 70.04167 − 49.87500 = 20.16667, M SE = 3.35875,
s{D̂i } = .7482 (i = 1, 2, 3), B = t(.99167; 16) = 2.673
11.46667 ± 2.673(.7482) 9.467 ≤ D1 ≤ 13.467
16.01667 ± 2.673(.7482) 14.017 ≤ D2 ≤ 18.017
20.16667 ± 2.673(.7482) 18.167 ≤ D3 ≤ 22.167
b.
Ȳ222. = 83.46667, s{Ȳ222. } = 1.0581, t(.975; 16) = 2.120,
24-2
83.46667 ± 2.120(1.0581), 81.2235 ≤ µ222 ≤ 85.7098
24.9. a.
eijkm :
i
1
b.
24.10. a.
k=1
j=1
j=2
2.250 −1.825
−1.450
2.975
−1.350 −1.525
.550
.375
i
1
k=2
j=1
j=2
1.450 −4.475
−1.050
3.325
2.250
3.725
−2.650 −2.575
2
−1.925
.950
−2.325 −1.850
4.075 −2.850
.175
3.750
2
2.625
2.100
−1.875
.500
−2.075
.100
1.325 −2.700
3
−.850
4.375
3.550 −2.525
−2.950
.975
.250 −2.825
3 −1.350 −2.450
.650
2.450
3.550 −1.250
−2.850
1.250
r = .974
Ȳ111. = 122.050, Ȳ112. = 111.250, Ȳ121. = 116.925, Ȳ122. = 92.675,
Ȳ211. = 121.225, Ȳ212. = 110.975, Ȳ221. = 116.250, Ȳ222. = 90.600,
Ȳ311. = 91.750, Ȳ312. = 79.950, Ȳ321. = 85.525, Ȳ322. = 61.050
b.
Ȳ1... = 110.7250, Ȳ2... = 109.7625, Ȳ3... = 79.5688
c.
Source
SS
df
MS
Between treatments 16, 291.75564 11 1, 481.06870
A (fee)
10, 044.27125 2 5, 022.13563
B (scope)
1, 833.97688 1 1, 833.97688
C (supervision)
3, 832.40021 1 3, 832.40021
AB interactions
1.60125 2
.80062
AC interactions
.78792 2
.39396
BC interactions
574.77521 1
574.77521
ABC interactions
3.94292 2
1.97146
Error
266.13750 36
7.39271
Total
16, 557.89314 47
d.
H0 : all (αβγ)ijk equal zero, Ha : not all (αβγ)ijk equal zero. F ∗ = 1.97146/7.39271 =
.27, F (.99; 2, 36) = 5.25. If F ∗ ≤ 5.25 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .77
e.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = .80062/7.39271 =
.11, F (.99; 2, 36) = 5.25. If F ∗ ≤ 5.25 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .90
24-3
H0 : all (αγ)ik equal zero, Ha : not all (αγ)ik equal zero. F ∗ = .39396/7.39271 =
.05, F (.99; 2, 36) = 5.25. If F ∗ ≤ 5.25 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .95
H0 : all (βγ)jk equal zero, Ha : not all (βγ)jk equal zero. F ∗ = 574.77521/7.39271 =
77.75, F (.99; 1, 36) = 7.40. If F ∗ ≤ 7.40 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
f.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero. F ∗ = 5, 022.13563/7.39271 =
679.34, F (.99; 2, 36) = 5.25. If F ∗ ≤ 5.25 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
g.
α ≤ .049
24.11. a.
D̂1 = .9625, D̂2 = 30.1937, D̂3 = 31.1562, D̂4 = 111.6750 − 100.7250 = 10.9500
D̂5 = 106.2333 − 81.4417 = 24.7916, L̂1 = −13.8416,
s{D̂i } = .9613 (i = 1, 2, 3), s{D̂4 } = s{D̂5 } = 1.1100, s{L̂1 } = 1.5698,
B = t(.99167; 36) = 2.511
.9625 ± 2.511(.9613)
30.1937 ± 2.511(.9613)
31.1562 ± 2.511(.9613)
10.9500 ± 2.511(1.1100)
24.7916 ± 2.511(1.1100)
−13.8416 ± 2.511(1.5698)
b.
−1.451 ≤ D1 ≤ 3.376
27.780 ≤ D2 ≤ 32.608
28.742 ≤ D3 ≤ 33.570
8.163 ≤ D4 ≤ 13.737
22.004 ≤ D5 ≤ 27.579
−17.783 ≤ L1 ≤ −9.900
D̂ = 116.925 − 116.250 = .675, s{D̂} = 1.9226, t(.975; 36) = 2.028,
.675 ± 2.028(1.9226), −3.224 ≤ D ≤ 4.574
c.
s{D̂} = 1.9226, q(.90; 12, 36) = 4.52, T = 3.196, T s{D̂} = 6.14,
Ȳ111. = 122.050, Ȳ211. = 121.225, Ȳ121. = 116.925, Ȳ221. = 116.250
24.12. a.
eijkm :
i
1
i
2
b.
j=1
31.4
−43.6
17.4
20.4
−25.6
k=1
j=2 j=3
44.8 −1.2
−23.2 −28.2
−33.2 −17.2
20.8
13.8
−9.2
32.8
j=1
29.6
39.6
−32.4
−34.4
−2.4
k=1
j=2 j=3
27.6
.6
−34.4
−.4
−26.4
14.6
50.6 −20.4
−17.4
5.6
i j=1
1 −30.0
48.0
18.0
−55.0
19.0
i
2
j=1
−6.6
−22.6
10.4
21.4
−2.6
r = .992
24-4
k=2
j=2
−3.4
−12.4
.6
−25.4
40.6
j=3
−18.2
15.8
5.8
25.8
−29.2
k=2
j=2
−4.6
12.4
25.4
−34.6
1.4
j=3
−19.4
4.6
−43.4
50.6
7.6
24.13. a.
Ȳ111. = 1, 218.6, Ȳ112. = 1, 051.0, Ȳ121. = 1, 274.2, Ȳ122. = 1, 122.4,
Ȳ131. = 1, 218.2, Ȳ132. = 1, 051.2, Ȳ211. = 1, 036.4, Ȳ212. = 870.6,
Ȳ221. = 1, 077.4, Ȳ222. = 931.6, Ȳ231. = 1, 020.4, Ȳ232. = 860.4
b.
Source
SS
df
MS
Between treatments
973, 645.933 11 88, 513.267
A (gender)
540, 360.600 1 540, 360.600
B (sequence)
49, 319.633 2 24, 659.817
C (experience)
382, 401.667 1 382, 401.667
AB interactions
542.500 2
271.250
AC interactions
91.267 1
91.267
BC interactions
911.233 2
455.617
ABC interactions
19.033 2
9.517
Error
41, 186.000 48
858.042
Total
1, 014, 831.933 59
c.
H0 : all (αβγ)ijk equal zero, Ha : not all (αβγ)ijk equal zero. F ∗ = 9.517/858.042 =
.01, F (.95; 2, 48) = 3.19. If F ∗ ≤ 3.19 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .99
d.
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = 271.250/858.042 =
.32, F (.95; 2, 48) = 3.19. If F ∗ ≤ 3.19 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .73
H0 : all (αγ)ik equal zero, Ha : not all (αγ)ik equal zero. F ∗ = 91.267/858.042 =
.11, F (.95; 1, 48) = 4.04. If F ∗ ≤ 4.04 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .75
H0 : all (βγ)jk equal zero, Ha : not all (βγ)jk equal zero. F ∗ = 455.617/858.042 =
.53, F (.95; 2, 48) = 3.19. If F ∗ ≤ 3.19 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .59
e.
H0 : all αi equal zero (i = 1, 2), Ha : not all αi equal zero. F ∗ = 540, 360.600/858.042 =
629.76, F (.95; 1, 48) = 4.04. If F ∗ ≤ 4.04 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : all βj equal zero (j = 1, 2, 3), Ha : not all βj equal zero. F ∗ = 24, 659.817/858.042 =
28.74, F (.95; 2, 48) = 3.19. If F ∗ ≤ 3.19 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : all γk equal zero (k = 1, 2), Ha : not all γk equal zero. F ∗ = 382, 401.667/858.042 =
445.67, F (.95; 1, 48) = 4.04. If F ∗ ≤ 4.04 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
f.
24.14. a.
α ≤ .302
Ȳ1... = 1, 155.933, Ȳ2... = 966.133, Ȳ.1.. = 1, 044.150, Ȳ.2.. = 1, 101.400,
Ȳ.3.. = 1, 037.550, Ȳ..1. = 1, 140.867, Ȳ..2. = 981.200
D̂1 = 189.800, D̂2 = −57.250, D̂3 = 6.600, D̂4 = 63.850, D̂5 = 159.667,
M SE = 858.042, s{D̂1 } = 7.5633, s{D̂i } = 9.2631 (i = 2, 3, 4),
24-5
s{D̂5 } = 7.5633, B = t(.99; 48) = 2.406
189.800 ± 2.406(7.5633)
−57.250 ± 2.406(9.2631)
6.600 ± 2.406(9.2631)
63.850 ± 2.406(9.2631)
159.667 ± 2.406(7.5633)
b.
171.603 ≤ D1 ≤ 207.997
−79.537 ≤ D2 ≤ −34.963
−15.687 ≤ D3 ≤ 28.887
41.563 ≤ D4 ≤ 86.137
141.470 ≤ D5 ≤ 177.864
Ȳ231. = 1, 020.4, s{Ȳ231. } = 13.0999, t(.975; 48) = 2.011,
1, 020.4 ± 2.011(13.0999), 994.056 ≤ µ231 ≤ 1, 046.744
24.15. a.
Yijkm = µ... +α1 Xijkm1 +β1 Xijkm2 +γ1 Xijkm3 +(αβ)11 Xijkm1 Xijkm2 +(αγ)11 Xijkm1 Xijkm3
(
Xijk1 =
(
Xijk2 =
(
Xijk3 =
b.
+(βγ)11 Xijkm2 Xijkm3 + (αβγ)111 Xijkm1 Xijkm2 Xijkm3 + ²ijkm
1
−1
if case from level 1 for factor A
if case from level 2 for factor A
1
−1
if case from level 1 for factor B
if case from level 2 for factor B
1
−1
if case from level 1 for factor C
if case from level 2 for factor C
Yijkm = µ... + β1 Xijkm2 + γ1 Xijkm3 + (αβ)11 Xijkm1 Xijkm2 + (αγ)11 Xijkm1 Xijkm3
+(βγ)11 Xijkm2 Xijkm3 + (αβγ)111 Xijkm1 Xijkm2 Xijkm3 + ²ijkm
c.
Full model:
Ŷ = 60.01667 − 5.67500X1 − 8.06667X2 − 10.02500X3 + .04167X1 X2
+.15000X1 X3 − .40833X2 X3 + .10000X1 X2 X3 ,
SSE(F ) = 49.4933
Reduced model:
Ŷ = 61.15167 − 9.20167X2 − 8.89000X3 − 1.09333X1 X2 + 1.28500X1 X3
−1.54333X2 X3 − 1.03500X1 X2 X3 ,
SSE(R) = 667.8413
H0 : α1 = 0, Ha : α1 6= 0.
F ∗ = (618.348/1) ÷ (49.4933/14) = 174.91, F (.975; 1, 14) = 6.298.
If F ∗ ≤ 6.298 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
D̂ = µ̂2.. − µ̂1.. = α̂2 − α̂1 = −2α̂1 = 11.35000, s2 {α̂1 } = .18413, s{D̂} = .8582,
t(.975; 14) = 2.145,
11.35000 ± 2.145(.8582), 9.509 ≤ D ≤ 13.191
24.16. a.
Yijkm = µ... + α1 Xijkm1 + β1 Xijkm2 + β2 Xijkm3 + γ1 Xijkm4
+(αβ)11 Xijkm1 Xijkm2 + (αβ)12 Xijkm1 Xijkm3 + (αγ)11 Xijkm1 Xijkm4
+(βγ)11 Xijkm2 Xijkm4 + (βγ)21 Xijkm3 Xijkm4
+(αβγ)111 Xijkm1 Xijkm2 Xijkm4 + (αβγ)121 Xijkm1 Xijkm3 Xijkm4 + ²ijkm
24-6
(
Xijkm1 =
Xijkm2 =
Xijkm3
1 if case from level 1 for factor B
−1 if case from level 3 for factor B
0 otherwise
1 if case from level 2 for factor B
−1 if case from level 3 for factor B
=
0 otherwise
(
Xijkm4 =
b.
1 if case from level 1 for factor A
−1 if case from level 2 for factor A
1 if case from level 1 for factor C
−1 if case from level 2 for factor C
Yijkm = µ... + α1 Xijkm1 + β1 Xijkm2 + β2 Xijkm3 + (αβ)11 Xijkm1 Xijkm2
+(αβ)12 Xijkm1 Xijkm3 + (αγ)11 Xijkm1 Xijkm4 + (βγ)11 Xijkm2 Xijkm4
+(βγ)21 Xijkm3 Xijkm4 + (αβγ)111 Xijkm1 Xijkm2 Xijkm4
+(αβγ)121 Xijkm1 Xijkm3 Xijkm4 + ²ijkm
c.
Full model:
Ŷ = 1, 062.16667 + 94.82500X1 − 17.85417X2 + 42.47083X3 + 79.80000X4
−4.33750X1 X2 + 2.01250X1 X3 + .20833X1 X4 + 3.38750X2 X4
−5.33750X3 X4 + .40417X1 X2 X4 − 1.94583X1 X3 X4 ,
SSE(F ) = 39, 499.9000
Reduced model:
Ŷ = 1, 063.73137 + 96.38971X1 − 14.72475X2 + 40.90613X3 − 10.59632X1 X2
+9.83603X1 X3 + 1.77304X1 X4 + 3.38750X2 X4 − 10.03162X3 X4
+3.53358X1 X2 X4 − 3.51054X1 X3 X4 ,
SSE(R) = 399, 106.8647
H0 : γ1 = 0, Ha : γ1 6= 0.
F ∗ = (359, 606.9647/1) ÷ (39, 499.9000/45) = 409.68, F (.95; 1, 45) = 4.06.
If F ∗ ≤ 4.06 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
24.17.
24.18.
24.19.
D̂ = µ̂..1 − µ̂..2 = γ̂ 1 − γ̂ 2 = 2γ̂ 1 = 159.60000, s2 {γ̂ 1 } = 15.54394, s{D̂} = 7.8852,
t(.975; 45) = 2.014,
159.60000 ± 2.014(7.8852), 143.719 ≤ D ≤ 175.481
√
2 n
= 4.1475, n = 14
1.8
q
t[.99; 12(n − 1)] (29)2 /2n = ±20, n = 6
P
P
i
i
(αβγ)ijk =
(µijk − µij. − µi.k − µ.jk + µi.. + µ.j. + µ..k − µ... )
= aµ.jk − aµ.j. − aµ..k − aµ.jk + aµ... + aµ.j. + aµ..k − aµ... = 0
24-7
24.20.
Yijk = µ... + αi + βj + γk + (αβ)ij + (αγ)ik + (βγ)jk + ²ijk
Source
A
B
C
AB
AC
BC
Error
Total
SS
SSA
SSB
SSC
SSAB
SSAC
SSBC
SSE
SST O
PP
σ 2 {L̂} = σ 2 {
24.21.
=
df
a−1
b−1
c−1
(a − 1)(b − 1)
(a − 1)(c − 1)
(b − 1)(c − 1)
(a − 1)(b − 1)(c − 1)
abc − 1
cij Ȳij.. } =
2
2
cn
cn
MS
M SA
M SB
M SC
M SAB
M SAC
M SBC
M SE
PP 2 2
cij σ {Ȳij.. }
(because of independence)
σ PP 2
PP 2 σ
cij
=
cij
24.22. c.
r = .992
24.23. a.
Ȳ111. = 8.80000, Ȳ112. = 9.68667, Ȳ113. = 8.33000, Ȳ114. = 7.50333,
Ȳ121. = 10.07667, Ȳ122. = 9.56333, Ȳ123. = 10.02667, Ȳ124. = 8.16000,
Ȳ211. = 10.55333, Ȳ212. = 8.79000, Ȳ213. = 8.77333, Ȳ214. = 8.00667,
Ȳ221. = 12.48000, Ȳ222. = 10.01667, Ȳ223. = 10.20000, Ȳ224. = 8.33000
b.
c.
d.
Source
SS
df
MS
Between treatments 69.63346 15 4.64223
A (age)
4.69375 1 4.69375
B (facilities)
13.26152 1 13.26152
C (region)
37.43491 3 12.47830
AB interactions
.36575 1
.36575
AC interactions
9.03731 3 3.01244
BC interactions
3.38421 3 1.12807
ABC interactions
1.45601 3
.48534
Error
34.18440 32 1.06826
Total
103.81786 47
H0 : all (αβγ)ijk equal zero, Ha : not all (αβγ)ijk equal zero. F ∗ = .48534/1.06826 =
.45, F (.99; 3, 32) = 4.46. If F ∗ ≤ 4.46 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .72
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero. F ∗ = .36575/1.06826 =
.34, F (.99; 1, 32) = 7.50. If F ∗ ≤ 7.50 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .56
H0 : all (αγ)ik equal zero, Ha : not all (αγ)ik equal zero. F ∗ = 3.01244/1.06826 =
2.82, F (.99; 3, 32) = 4.46. If F ∗ ≤ 4.46 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .055
H0 : all (βγ)jk equal zero, Ha : not all (βγ)jk equal zero. F ∗ = 1.12807/1.06826 =
1.06, F (.99; 3, 32) = 4.46. If F ∗ ≤ 4.46 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .38
24-8
e.
H0 : all αi equal zero (i = 1, 2), Ha : not all αi equal zero. F ∗ = 4.69375/1.06826 =
4.39, F (.99; 1, 32) = 7.50. If F ∗ ≤ 7.50 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .044
H0 : all βj equal zero (j = 1, 2), Ha : not all βj equal zero. F ∗ = 13.26152/1.06826 =
12.41, F (.99; 1, 32) = 7.50. If F ∗ ≤ 7.50 conclude H0 , otherwise Ha . Conclude
Ha . P -value = .0013
H0 : all γk equal zero (k = 1, ..., 4), Ha : not all γk equal zero. F ∗ = 12.47830/1.06826 =
11.68, F (.99; 3, 32) = 4.46. If F ∗ ≤ 4.46 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
f.
Ȳ1... = 9.01833, Ȳ2... = 9.64375, Ȳ.1.. = 8.80542, Ȳ.2.. = 9.85667,
Ȳ..1. = 10.47750, Ȳ..2. = 9.51417, Ȳ..3. = 9.33250, Ȳ..4. = 8.00000
D̂1 = Ȳ.1.. − Ȳ.2.. = −1.05125, D̂2 = Ȳ..1. − Ȳ..2. = .96333, D̂3 = Ȳ..1. − Ȳ..3. = 1.14500,
D̂4 = Ȳ..1. − Ȳ..4. = 2.47750, D̂5 = Ȳ..2. − Ȳ..3. = .18167, D̂6 = Ȳ..2. − Ȳ..4. = 1.51417,
D̂7 = Ȳ..3. − Ȳ..4. = 1.33250, M SE = 1.06826,
s{D̂1 } = .29836, s{D̂i } = .42195 (i = 2, ..., 7), B = t(.99286; 32) = 2.5915
−1.05125 ± 2.5915(.29836)
.96333 ± 2.5915(.42195)
1.14500 ± 2.5915(.42195)
2.47750 ± 2.5915(.42195)
.18167 ± 2.5915(.42195)
1.51417 ± 2.5915(.42195)
1.33250 ± 2.5915(.42195)
−1.824 ≤ D1
−.130 ≤ D2
.052 ≤ D3
1.384 ≤ D4
−.912 ≤ D5
.421 ≤ D6
.239 ≤ D7
≤ −.278
≤ 2.057
≤ 2.238
≤ 3.571
≤ 1.275
≤ 2.608
≤ 2.426
24.24. c.
r = .920
24.25. a.
Ȳ111. = .03303, Ȳ112. = .03886, Ȳ121. = .03553, Ȳ122. = .05415,
Ȳ211. = .04076, Ȳ212. = .05128, Ȳ221. = .05516, Ȳ222. = .06056,
Ȳ311. = .05841, Ȳ312. = .05997, Ȳ321. = .07738, Ȳ322. = .07915,
Ȳ411. = .05655, Ȳ412. = .04688, Ȳ421. = .06755, Ȳ422. = .06442
b.
Yijkm = µ... + α1 Xijk1 + α2 Xijk2 + α3 Xijk3 + β1 Xijk4 + γ1 Xijk5
+(αβ)11 Xijk1 Xijk4 + (αβ)21 Xijk2 Xijk4 + (αβ)31 Xijk3 Xijk4
+(αγ)11 Xijk1 Xijk5 + (αγ)21 Xijk2 Xijk5 + (αγ)31 Xijk3 Xijk5
+(βγ)11 Xijk4 Xijk5 + (αβγ)111 Xijk1 Xijk4 Xijk5
Xijk1
+(αβγ)211 Xijk2 Xijk4 Xijk5 + (αβγ)311 Xijk3 Xijk4 Xijk5 + ²ijkm
1 if case from NE
= −1 if case from W
0 otherwise
1 if case from NC
Xijk2 = −1 if case from W
0 otherwise
24-9
Xijk3 =
1 if case from S
−1 if case from W
0 otherwise
(
Xijk4 =
(
Xijk5 =
1
−1
if poverty level below 8 percent
if poverty level 8 percent or higher
1
−1
if percent of population 65 or older < 12.0%
if percent of population 65 or older ≥ 12.0%
Ŷ = .05498 − .0146X1 − .00303X2 + .0137X3 − .00676X4
−.00193X5 + .00231X1 X4 + .00084X2 X4 − .00278X3 X4 − .00418X1 X5
−.00205X2 X5 + .00110X3 X5 + .00090X4 X5 + .00230X1 X4 X5
−.00218X2 X4 X5 − .00085X3 X4 X5 ,
SSE(F ) = .23779
c.
ABC interactions;
Ŷ = .0552 − .0133X1 − .00412X2 + .0135X3 − .00712X4
−.00157X5 + .00085X1 X4 + .00128X2 X4 − .00254X3 X4
−.00278X1 X5 − .00251X2 X5 + .00083X3 X5 + .00046X4 X5 ,
SSE(R) = .23849
H0 : all (αβγ)ijk equal zero, Ha : not all (αβγ)ijk equal zero.
F ∗ = (.0007/3) ÷ (.23779/424) = .42, F (.975; 3, 424) = 3.147.
If F ∗ ≤ 3.147 conclude H0 , otherwise Ha . Conclude H0 . P -value = .74
AB interactions:
Ŷ = 0.0556 − 0.0133X1 − 0.00355X2 + 0.0136X3 − 0.00743X4
−0.00135X5 − 0.00315X1 X5 − 0.00190X2 X5 +0.00040X3 X5 +0.00026X4 X5
+0.00095X1 X4 X5 − 0.00161X2 X4 X5 − 0.00081X3 X4 X5 ,
SSE(R) = .23897
H0 : all (αβ)ij equal zero, Ha : not all (αβ)ij equal zero.
F ∗ = (.00118/3) ÷ (.23779/424) = .70, F (.975; 3, 424) = 3.147.
If F ∗ ≤ 3.147 conclude H0 , otherwise Ha . Conclude H0 . P -value = .55
AC interactions:
Ŷ = 0.0562 − 0.0129X1 − 0.00419X2 + 0.0127X3 − 0.00727X4
−0.00161X5 + 0.00038X1 X4 + 0.00021X2 X4 − 0.00222X3 X4
−0.00023X4 X5 + 0.00052X1 X4 X5 − 0.00119X2 X4 X5 + 0.00011X3 X4 X5 ,
SSE(R) = .24070
H0 : all (αγ)ik equal zero, Ha : not all (αγ)ik equal zero.
F ∗ = (.00291/3) ÷ (.23779/424) = 1.73, F (.975; 3, 424) = 3.147.
If F ∗ ≤ 3.147 conclude H0 , otherwise Ha . Conclude H0 . P -value = .16
24-10
BC interactions:
Ŷ = 0.0553 − 0.0142X1 − 0.00303X2 + 0.0134X3 − 0.00687X4 − 0.00179X5
+0.00152X1 X4 +0.00092X2 X4 −0.00253X3 X4 −0.00344X1 X5 −0.00214X2 X5
+0.00085X3 X5 + 0.00183X1 X4 X5 − 0.00204X2 X4 X5 − 0.00042X3 X4 X5 ,
SSE(R) = .23801
H0 : all (βγ)jk equal zero, Ha : not all (βγ)jk equal zero.
F ∗ = (.00022/1) ÷ (.23779/424) = .39, F (.975; 1, 424) = 5.06.
If F ∗ ≤ 5.06 conclude H0 , otherwise Ha . Conclude H0 . P -value = .53
d.
A effects:
Ŷ = 0.0585 − 0.0104X4 + 0.00191X5 − 0.00575X1 X4 + 0.00409X2 X4
−0.00150X3 X4 + 0.00401X1 X5 − 0.00558X2 X5 + 0.00016X3 X5
−0.00218X4 X5 − 0.00445X1 X4 X5 − 0.00227X2 X4 X5 + 0.00280X3 X4 X5 ,
SSE(R) = .27011
H0 : all αi equal zero (i = 1, ..., 4), Ha : not all αi equal zero.
F ∗ = (.03232/3) ÷ (.23779/424) = 19.21, F (.975; 3, 424) = 3.147.
If F ∗ ≤ 3.147 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
B effects:
Ŷ = 0.0539 − 0.0202X1 − 0.00235X2 + 0.0156X3 − 0.00465X5
+0.00583X1 X4 − 0.00024X2 X4 − 0.00598X3 X4 − 0.00705X1 X5
−0.00208X2 X5 + 0.00362X3 X5 + 0.00174X4 X5 + 0.00828X1 X4 X5
−0.00275X2 X4 X5 − 0.00270X3 X4 X5 ,
SSE(R) = .25047
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = (.01268/1) ÷ (.23779/424) = 22.61, F (.975; 1, 424) = 5.06.
If F ∗ ≤ 5.06 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
C effects:
Ŷ = 0.0552 − 0.0129X1 − 0.00320X2 + 0.0132X3 − 0.00754X4
+0.00150X1 X4 + 0.00083X2 X4 − 0.00206X3 X4 − 0.00318X1 X5
−0.00236X2 X5 + 0.00018X3 X5 + 0.00061X4 X5 + 0.00069X1 X4 X5
−0.00198X2 X4 X5 − 0.00032X3 X4 X5 ,
SSE(R) = .23882
H0 : γ1 = γ2 = 0, Ha : not both γ1 and γ2 equal zero.
F ∗ = (.00103/1) ÷ (.23779/424) = 1.84, F (.975; 1, 424) = 5.06.
If F ∗ ≤ 5.06 conclude H0 , otherwise Ha . Conclude H0 . P -value = .175
e.
D̂1 = µ̂1.. − µ̂2.. = α̂1 − α̂2 = −.01155, D̂2 = µ̂1.. − µ̂3.. = α̂1 − α̂3 = −.02834,
24-11
D̂3 = µ̂1.. − µ̂4.. = −.01846, D̂4 = µ̂2.. − µ̂3.. = −.016784,
D̂5 = µ̂2.. − µ̂4.. = −.006907, D̂6 = µ̂3.. − µ̂4.. = .009877,
−.02276 ≤ D1
−.03878 ≤ D2
−.03030 ≤ D3
−.02531 ≤ D4
−.01711 ≤ D5
.00053 ≤ D6
≤
≤
≤
≤
≤
≤
−.00034
−.01790
−.00662
−.00826
.00329
.01922
24-12
Chapter 25
RANDOM AND MIXED EFFECTS
MODELS
25.3.
(1) I, (2) II, (3) I, (4) II
25.5. b.
H0 : σµ2 = 0, Ha : σµ2 > 0. F ∗ = .45787/.03097 = 14.78, F (.95; 5, 114) = 2.29.
If F ∗ ≤ 2.29 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
Ȳ.. = .22767, nT = 120, s{Ȳ.. } = .06177, t(.975; 5) = 2.571,
.22767 ± 2.571(.06177), .0689 ≤ µ. ≤ .3865
25.6. a.
F (.025; 5, 114) = .1646, F (.975; 5, 114) = 2.680, L = .22583, U = 4.44098
σ2
.1842 ≤ 2 µ 2 ≤ .8162
σµ + σ
b.
χ2 (.025; 114) = 90.351, χ2 (.975; 114) = 145.441, .02427 ≤ σ 2 ≤ .03908
c.
s2µ = .02135
d.
Satterthwaite:
df = (ns2µ )2 ÷ [(M ST R)2 /(r − 1) + (M SE)2 /r(n − 1)]
= [20(.02135)]2 ÷ [(.45787)2 /5 + (.03907)2 /6(19)] = 4.35,
χ2 (.025; 4) = .484, χ2 (.975; 4) = 11.143
4.35(.02135)
4.35(.02135)
.0083 =
≤ σµ2 ≤
= .192
11.143
.484
M LS: c1 = .05, c2 = −.05, M S1 = .45787, M S2 = .03907, df1 = 5, df2 = 114,
F1 = F (.975; 5, ∞) = 2.57, F2 = F (.975; 114, ∞) = 1.28, F3 = F (.975; ∞, 5) =
6.02, F4 = F (.975; ∞, 114) = 1.32, F5 = F (.975; 5, 114) = 2.68, F6 = F (.975; 114, 5) =
6.07, G1 = .6109, G2 = .2188, G3 = .0147, G4 = −.2076, HL = .014, HU = .115,
.02135 − .014, .02135 + .115, .0074 ≤ σµ2 ≤ .1364
25.7. a.
Source
SS
df
MS
Between brands 854.52917 5 170.90583
Error
30.07000 42
.71595
Total
884.59917 47
25-1
H0 : σµ2 = 0, Ha : σµ2 > 0. F ∗ = 170.90583/.71595 = 238.71, F (.99; 5, 42) = 3.49.
If F ∗ ≤ 3.49 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
b.
Ȳ.. = 17.62917, nT = 48, s{Ȳ.. } = 1.8869, t(.995; 5) = 4.032,
17.62917 ± 4.032(1.8869), 10.021 ≤ µ. ≤ 25.237
25.8. a.
F (.005; 5, 42) = .0795, F (.995; 5, 42) = 3.95, L = 7.4292, U = 375.20828
.8814 ≤
σµ2
≤ .9973
σµ2 + σ 2
b.
M SE = .71595, s2µ = 21.27374
c.
χ2 (.005; 42) = 22.138, χ2 (.995; 42) = 69.336, .4337 ≤ σ 2 ≤ 1.3583
d.
H0 : σµ2 ≤ 2σ 2 , Ha : σµ2 > 2σ 2 . F ∗ = [M ST R/(2n + 1)] ÷ M SE = 14.042,
F (.99; 5, 42) = 3.49. If F ∗ ≤ 3.49 conclude H0 , otherwise Ha . Conclude Ha .
e.
c1 = .125, c2 = −.125, df1 = 5, df2 = 42, F1 = F (.995; 5, ∞) = 3.35, F2 =
F (.995; 42, ∞) = 1.66, F3 = F (.995; ∞, 5) = 12.1, F4 = F (.995; ∞, 42) = 1.91,
F5 = F (.995; 5, 42) = 3.95, F6 = F (.995; 42, 5) = 12.51, G1 = .7015, G2 = .3976,
G3 = .0497, G4 = −1.2371, HL = 14.990, HU = 237.127, 21.2737 − 14.990,
21.2737 + 237.127, 6.284 ≤ σµ2 ≤ 258.401
25.9. a.
Source
SS
df
MS
Between machines 602.5000 3 200.8333
Error
257.4000 36
7.1500
Total
859.9000 39
H0 : σµ2 = 0, Ha : σµ2 > 0. F ∗ = 200.8333/7.1500 = 28.09, F (.90; 3, 36) = 2.25.
If F ∗ ≤ 2.25 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
b.
Ȳ.. = 205.05, nT = 40, s{Ȳ.. } = 2.2407, t(.95; 3) = 2.353,
205.05 ± 2.353(2.2407), 199.778 ≤ µ. ≤ 210.322
25.10. a.
F (.05; 3, 36) = .117, F (.95; 3, 36) = 2.87, L = .8787, U = 23.9073
.4677 ≤
b.
σµ2
≤ .9599
σµ2 + σ 2
H0 : σµ2 = σ 2 , Ha : σµ2 6= σ 2 .
F ∗ = [M ST R/(n + 1)] ÷ M SE = 2.554, F (.05; 3, 36) = .117, F (.95; 3, 36) = 2.87.
If .117 ≤ F ∗ ≤ 2.87 conclude H0 , otherwise Ha . Conclude H0 .
c.
χ2 (.05; 36) = 23.269, χ2 (.95; 36) = 50.998, 5.047 ≤ σ 2 ≤ 11.062
d.
s2µ = 19.3683
e.
Satterthwaite:
df = [10(19.3683)]2 ÷ [(200.8333)2 /3 + (7.1500)2 /36] = 2.79,
χ2 (.05; 3) = .352, χ2 (.95; 3) = 7.815,
25-2
2.79(19.3683)
2.79(19.3683)
≤ σµ2 ≤
= 153.516
7.815
.352
M LS: c1 = .10, c2 = −.10, df1 = 3, df2 = 36, F1 = F (.95; 3, ∞) = 2.60,
F2 = F (.95; 36, ∞) = 1.42, F3 = F (.95; ∞, 3) = 8.53, F4 = F (.95; ∞, 36) = 1.55,
F5 = F (.95; 3, 36) = 2.87, F6 = F (.95; 36, 3) = 8.60, G1 = .6154, G2 = .2958,
G3 = .0261, G4 = −.6286, HL = 12.381, HU = 151.198, 19.3683 − 12.381,
19.3683 + 151.198, 6.987 ≤ σµ2 ≤ 170.566
6.915 =
25.13. a.
b.
E{M SA} = 115, E{M SB} = 185, E{M SAB} = 35
E{M SA} = 85, E{M SB} = 155, E{M SAB} = 5
25.15. a.
Source
SS
df
MS
Factor A (driver) 280.28475 3 93.42825
Factor B (car)
94.71350 4 23.67838
AB interactions
2.44650 12
.20388
Error
3.51500 20
.17575
Total
380.95975 39
2
2
> 0. F ∗ = .20388/.17575 = 1.16, F (.95; 12, 20) = 2.28.
= 0, Ha : σαβ
H0 : σαβ
If F ∗ ≤ 2.28 conclude H0 , otherwise Ha . Conclude H0 . P -value = .37
b.
H0 : σα2 = 0, Ha : σα2 > 0. F ∗ = 93.42825/.20388 = 458.25, F (.95; 3, 12) = 3.49.
If F ∗ ≤ 3.49 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
H0 : σβ2 = 0, Ha : σβ2 > 0. F ∗ = 23.67838/.20388 = 116.14, F (.95; 4, 12) = 3.26.
If F ∗ ≤ 3.26 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
s2α = 9.3224, s2β = 2.9343
d.
c1 = .10, c2 = −.10, M S1 = 93.42825, M S2 = .203875, df1 = 3, df2 = 12, F1 =
F (.975; 3, ∞) = 3.12, F2 = F (.975; 12, ∞) = 1.94, F3 = F (.975; ∞, 3) = 13.9,
F4 = F (.975; ∞, 12) = 2.72, F5 = F (.975; 3, 12) = 4.47, F6 = F (.975; 12, 3) =
14.3, G1 = .6795, G2 = .4845, G3 = −.0320, G4 = −2.6241, HL = 6.348,
HU = 120.525, 9.3244 − 6.348, 9.3224 + 120.525, 2.974 ≤ σα2 ≤ 129.847
e.
df = [8(2.9343)]2 ÷ [(23.678375)2 /4 + (.203875)2 /12] = 3.93
χ2 (.025; 4) = .484, χ2 (.975; 4) = 11.143,
1.03 =
25.16. a.
3.93(2.9343)
3.93(2.9343)
≤ σβ2 ≤
= 23.83
11.143
.484
2
2
> 0. F ∗ = 303.822/52.011 = 5.84, F (.99; 4, 36) = 3.89.
= 0, Ha : σαβ
H0 : σαβ
If F ∗ ≤ 3.89 conclude H0 , otherwise Ha . Conclude Ha . P -value = .001
b.
s2αβ = 50.362
c.
H0 : σα2 = 0, Ha : σα2 > 0. F ∗ = 12.289/52.011 = .24, F (.99; 2, 36) = 5.25.
If F ∗ ≤ 5.25 conclude H0 , otherwise Ha . Conclude H0 .
d.
H0 : all βj equal zero (j = 1, 2, 3), Ha : not all βj equal zero.
25-3
F ∗ = 14.156/303.822 = .047, F (.99; 2, 4) = 18.0.
If F ∗ ≤ 18.0 conclude H0 , otherwise Ha . Conclude H0 .
e.
Ȳ.1. = 56.133, Ȳ.2. = 56.600, Ȳ.3. = 54.733, D̂1 = Ȳ.1. − Ȳ.2. = −.467, D̂2 =
Ȳ.1. − Ȳ.3. = −1.400, D̂3 = Ȳ.2. − Ȳ.3. = 1.867, s{D̂i } = 6.3647 (i = 1, 2, 3),
q(.95; 3, 4) = 5.04, T = 3.5638
−.467 ± 3.5638(6.3647)
−1.400 ± 3.5638(6.3647)
1.867 ± 3.5638(6.3647)
f.
−23.150 ≤ D1 ≤ 22.216
−24.083 ≤ D2 ≤ 21.283
−20.816 ≤ D3 ≤ 24.550
µ̂.1 = 56.1333, M SA = 12.28889, M SAB = 303.82222, s2 {µ̂.1 } = (2/45)(303.82222)+
(1/45)(12.28889) = 13.7763, s{û.1 } = 3.712, df = (13.7763)2 ÷{[(2/45)(303.82222)]2 /4+
[(1/45)(12.28889)]2 /2} = 4.16, t(.995; 4) = 4.60,
56.1333 ± 4.60(3.712), 39.06 ≤ µ.1 ≤ 73.21
g.
M SA = 12.28889, M SE = 52.01111, s2α = (M SA − M SE)/nb = −2.648,
c1 = 1/15, c2 = −1/15, df1 = 2, df2 = 36, F1 = F (.995; 2, ∞) = 5.30, F2 =
F (.995; 36, ∞) = 1.71, F3 = F (.995; ∞, 2) = 200, F4 = F (.995; ∞, 36) = 2.01,
F5 = F (.995; 2, 36) = 6.16, F6 = F (.995; 36, 2) = 199.5, G1 = .8113, G2 = .4152,
G3 = .1022, G4 = −35.3895, HL = 3.605, HU = 162.730,−2.648 − 3.605,
−2.648 + 162.730, −6.253 ≤ σα2 ≤ 160.082
25.17. a.
Source
Factor A (coats)
Factor B (batch)
AB interactions
Error
Total
SS
df
150.3879 2
152.8517 3
1.8521 6
173.6250 36
478.7167 47
MS
75.1940
50.9506
.3087
4.8229
2
2
H0 : σαβ
= 0, Ha : σαβ
> 0. F ∗ = .3087/4.8229 = .06, F (.95; 6, 36) = 2.36.
If F ∗ ≤ 2.36 conclude H0 , otherwise Ha . Conclude H0 . P -value = .999
b.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero. F ∗ = 75.1940/.3087 =
243.58, F (.95; 2, 6) = 5.14. If F ∗ ≤ 5.14 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : σβ2 = 0, Ha : σβ2 > 0. F ∗ = 50.9506/4.8229 = 10.56, F (.95; 3, 36) = 2.87. If
F ∗ ≤ 2.87 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
Ȳ1.. = 73.10625, Ȳ2.. = 76.79375, Ȳ3.. = 76.92500, D̂1 = Ȳ2.. − Ȳ1.. = 3.68750,
D̂2 = Ȳ3.. − Ȳ2.. = .13125, s{D̂i } = .1964 (i = 1, 2), B = t(.975; 6) = 2.447
3.68750 ± 2.447(.1964)
.13125 ± 2.447(.1964)
3.2069 ≤ D1 ≤ 4.1681
−.3493 ≤ D2 ≤ .6118
d.
µ̂2. = 76.79375, s2 {µ̂2. } = (2/48)(.30868) + (1/48)(50.95056) = 1.0743, s{µ̂2. } =
1.0365, df = (1.0743)2 ÷ {[(2/48)(.30868)]2 /6 + [(1/48)(50.95056)]2 /3} = 3.07,
t(.975; 3) = 3.182, 76.79375 ± 3.182(1.0365), 73.496 ≤ µ2. ≤ 80.092
e.
s2β = (M SB − M SE)/na = 3.844, c1 = 1/12, c2 = −1/12, df1 = 3, df2 = 36,
F1 = F (.95; 3, ∞) = 2.60, F2 = F (.95; 36, ∞) = 1.42, F3 = F (.95; ∞, 3) = 8.53,
25-4
F4 = F (.95; ∞, 36) = 1.55, F5 = F (.95; 3, 36) = 2.87, F6 = F (.95; 36, 3) = 8.60,
G1 = .6154, G2 = .2958, G3 = .0261, G4 = −.6286, HL = 2.631, HU = 31.989,
3.844 − 2.631, 3.844 + 31.989, 1.213 ≤ σβ2 ≤ 35.833
25.18. a.
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero.
F ∗ = 47.0450/.1150 = 409.09, F (.95; 1, 3) = 10.1.
If F ∗ ≤ 10.1 conclude H0 , otherwise Ha . Conclude Ha . P -value = .0003
25.19. a.
eij :
i j=1
1 −.175
2
.025
3 −.575
4
.025
5
.025
6
.025
7 −.175
8
.825
r = .985
c.
j=2
j=3
−1.300
.325
4.900 −3.475
−1.700
.925
1.900 −1.475
−1.100
.525
−1.100
2.525
−.300 −.675
−1.300
1.325
j=4
j=5
−2.050
3.200
1.150 −2.600
−1.450
2.800
2.150 −2.600
−.850
1.400
−.850 −.600
2.950 −1.800
−1.050
.200
H0 : D = 0, Ha : D 6= 0. SSBL.T R∗ = 27.729, SSRem∗ = 94.521,
F ∗ = (27.729/1) ÷ (94.521/27) = 7.921, F (.995; 1, 27) = 9.34.
If F ∗ ≤ 9.34 conclude H0 , otherwise Ha . Conclude H0 .
25.20. a.
Source
SS
Blocks
4, 826.375
Paint type
531.350
Error
122.250
Total
5, 479.975
b.
df
MS
7 689.48214
4 132.83750
28
4.36607
39
H0 : all τj equal zero (j = 1, ..., 5), Ha : not all τj equal zero.
F ∗ = 132.83750/4.36607 = 30.425, F (.95; 4, 28) = 2.71.
If F ∗ ≤ 2.71 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
d.
25.21. a.
Ȳ.1 = 20.500, Ȳ.2 = 23.625, Ȳ.3 = 19.000, Ȳ.4 = 29.375, Ȳ.5 = 21.125, L̂1 =
Ȳ.1 − Ȳ.2 = −3.125, L̂2 = Ȳ.1 − Ȳ.3 = 1.500, L̂3 = Ȳ.1 − Ȳ.4 = −8.875, L̂4 =
Ȳ.1 − Ȳ.5 = −.625, s{L̂i } = 1.0448 (i = 1, ..., 4), B = t(.9875; 28) = 2.369
−3.125 ± 2.369(1.0448)
−5.60 ≤ L1 ≤ −.65
1.500 ± 2.369(1.0448)
−.98 ≤ L2 ≤ 3.98
−8.875 ± 2.369(1.0448)
−11.35 ≤ L3 ≤ −6.40
−.625 ± 2.369(1.0448)
−3.10 ≤ L4 ≤ 1.85
1
1
L̂ = (Ȳ.1 + Ȳ.3 + Ȳ.5 ) − (Ȳ.2 + Ȳ.4 ) = −6.29167, s{L̂} = .6744, t(.975; 28) = 2.048,
3
2
−6.29167 ± 2.048(.6744), −7.67 ≤ L ≤ −4.91
eij :
25-5
i
j=1
j=2
1
−.1333
.4667
2
−.1333 −.5333
3
−.4667
1.1333
4
.8667 −.5333
5
−.4667
1.1333
6
1.2000 −1.2000
7
−.1333
1.4667
8
.8667 −1.5333
9 −2.1333 −.5333
10
.5333
.1333
r = .985
c.
j=3
−.3333
.6667
−.6667
−.3333
−.6667
.0000
−1.3333
.6667
2.6667
−.6667
H0 : D = 0, Ha : D 6= 0. SSBL.T R∗ = 4.5365, SSRem∗ = 24.6635,
F ∗ = (4.5365/1) ÷ (24.6635/17) = 3.127, F (.975; 1, 17) = 6.042.
If F ∗ ≤ 6.042 conclude H0 , otherwise Ha . Conclude H0 . P -value = .095
25.22. a.
Source
SS
df
Blocks
1, 195.5000 9
Reagents
123.4667 2
Error
29.2000 18
Total
1, 348.1667 29
b.
MS
132.8333
61.7333
1.6222
H0 : all τj equal zero (j = 1, 2, 3), Ha : not all τj equal zero.
F ∗ = 61.7333/1.6222 = 38.055, F (.975; 2, 18) = 4.56.
If F ∗ ≤ 4.56 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
Ȳ.1 = 15.3, Ȳ.2 = 19.7, Ȳ.3 = 19.5, B = t(.9875; 18) = 2.445, L̂1 = .2, L̂2 = 4.3,
s{L̂1 } = .5696, s{L̂2 } = .4933
.2 ± 2.445(.5696)
4.3 ± 2.445(.4933)
d.
25.23. a.
−1.193 ≤ L1 ≤ 1.593
3.094 ≤ L2 ≤ 5.506
H0 : σρ2 = 0, Ha : σρ2 > 0. F ∗ = 132.8333/1.6222 = 81.885, F (.975; 9, 18) = 2.929.
If F ∗ ≤ 2.929 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
2
2
H0 : σαβγ
= 0, Ha : σαβγ
> 0. F ∗ = M SABC/M SE = 1.49/2.30 = .648,
F (.975; 8, 60) = 2.41. If F ∗ ≤ 2.41 conclude H0 , otherwise Ha . Conclude H0 .
P -value=.27.
b.
2
2
> 0. F ∗ = M SAB/M SABC = 2.40/1.49 = 1.611,
= 0, Ha : σαβ
H0 : σαβ
F (.99; 2, 8) = 8.65. If F ∗ ≤ 8.65 conclude H0 , otherwise Ha . Conclude H0 .
c.
H0 : σβ2 = 0, Ha : σβ2 > 0. F ∗∗ = M SB/(M SAB +M SBC −M SABC) =
4.20/(2.40 + 3.13 − 1.49) = 1.04, df = 16.32161/5.6067 = 2.91, F (.99; 1, 3) = 34.1.
If F ∗∗ ≤ 34.1 conclude H0 , otherwise Ha . Conclude H0 .
d.
s2α = (M SA −M SAB − M SAC +M SABC)/nbc = .126,
df = [(8.650/30) − (2.40/30) − (3.96/30) + (1.49/30)]2
25-6
"
#
(8.65/30)2 (2.40/30)2 (3.96/30)2 (1.49/30)2
÷
+
+
+
= .336
2
2
8
8
χ2 (.025; 1) = .001, χ2 (.975; 1) = 5.02
.008 =
25.24. a.
.336(.126)
.336(.126)
≤ σα2 ≤
= 42.336
5.02
.001
F ∗ = M SAC/M SABC, F ∗ = M SB/M SE
b.
H0 : all (αγ)ik equal zero, Ha : not all (αγ)ik equal zero. F ∗ = 91.267/9.517 = 9.59,
F (.95; 1, 2) = 18.5. If F ∗ ≤ 18.5 conclude H0 , otherwise Ha . Conclude H0 .
c.
H0 : σβ2 = 0, Ha : σβ2 > 0. F ∗ = 24, 659.817/858.042 = 28.74, F (.95; 2, 48) = 3.19.
If F ∗ ≤ 3.19 conclude H0 , otherwise Ha . Conclude Ha .
d.
s2β = (M SB − M SE)/acn = (24, 659.817 − 858.042)/20 = 1, 190.09, c1 = .05, c2 =
−.05, df1 = 2, df2 = 48, F1 = F (.975; 2, ∞) = 3.69, F2 = F (.975; 48, ∞) = 1.44,
F3 = F (.975; ∞, 2) = 39.5, F4 = F (.975; ∞, 48) = 1.56, F5 = F (.975; 2, 48) =
3.99, F6 = F (.975; 48, 2) = 39.5, G1 = .7290, G2 = .3056, G3 = .0416, G4 =
−3.6890, HL = 900.39, HU = 47, 468.09, 1,190.09 − 900.39, 1, 190.09 + 47, 468.09,
289.70 ≤ σβ2 ≤ 48, 658.18
F ∗∗ = M SA/(M SAB +M SAC − M SABC)
25.25.
Ã
(M SAB)2 (M SAC)2 (M SABC)2
df = (M SAB +M SAC −M SABC) ÷
+
+
dfAB
dfAC
dfABC
!
2
25.26. a.
µ̂.. = 55.593, β̂ 1 = .641, β̂ 2 = .218, σ̂ 2α = 5.222, σ̂ 2αβ = 15.666, σ̂ 2 = 55.265, no
(Note: Unrestricted estimators are same except that variance component for
random effect A is zero.)
b.
Estimates remain the same.
c.
2
2
H0 : σαβ
= 0, Ha : σαβ
> 0. z(.99) = 2.326, s{σ̂ 2αβ } = 13.333, z ∗ = 15.666/13.333 =
1.175. If z ∗ ≤ 2.326 conclude H0 , otherwise Ha . Conclude H0 . P -value = .12.
d.
H0 : β1 = β2 = β3 = 0, Ha : not all βj = 0 (j = 1, 2, 3). −2loge L(R) = 295.385,
−2loge L(F ) = 295.253, X 2 = 295.385 − 295.253 = .132, χ2 (.99; 2) = 9.21. If
X 2 ≤ 9.21 conclude H0 , otherwise Ha . Conclude H0 . P -value = .94.
e.
2
z(.995) = 2.576, 15.666 ± 2.576(13.333), −18.680 ≤ ααβ
≤ 50.012
25.27. a.
µ̂.. = 75.817, α̂1 = −2.398, α̂2 = .977, σ̂ 2β = 2.994, σ̂ 2αβ = 0, σ̂ 2 = 3.103, yes
b.
Estimates remain the same.
c.
H0 : σβ2 = 0, Ha : σβ2 > 0. −2loge L(R) = 214.034, −2loge L(F ) = 192.599, X 2 =
214.034−192.599 = 21.435, χ2 (.95; 1) = 3.84. If X 2 ≤ 3.84 conclude H0 , otherwise
Ha . Conclude Ha . P -value = 0+
d.
H0 : α1 = α2 = α3 = 0, Ha : not all αi = 0 (i = 1, 2, 3). −2loge L(R) = 221.722,
−2loge L(F ) = 192.599, X 2 = 221.722 − 192.599 = 29.123, χ2 (.95; 2) = 5.99. If
X 2 ≤ 5.99 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
s{σ̂ 2β } = 2.309, z(.975) = 1.96, 2.994 ± 1.96(2.309), −1.532 ≤ σβ2 ≤ 7.520
25-7
(Note: Answers to parts (c) and (e) are not consistent; may be due to
large-sample approximation not being appropriate here.)
"
P
#
25.28.
1
n2
1
1
P
n0 =
( ni ) − P i =
(rn − rn2 /rn) =
(rn − n) = n
r−1
ni
r−1
r−1
25.29.
From (25.12), σ 2 {Ȳ.. } = (σµ2 /r) + (σ 2 /nT ). When nT is fixed, σ 2 {Ȳ.. }
is minimized by making r as large as possible, i.e., r = nT .
In that case, n = 1 since rn = nT .
25.30.
L≤
σµ2
1
σ2
1
1
σ2
1
≤
U
or
≥
≥
or
+
1
≥
+ 1 ≥ + 1 or
2
2
2
σ
L
σµ
U
L
σµ
U
σ 2 + σµ2
σµ2
1+L
1+U
U
L
∗
≥
≤
≥
or
L
=
≤
= U∗
L
σµ2
U
1+L
σ 2 + σµ2
1+U
σ 2 {Ȳi.. } = σ 2 {µ.. + αi + β̄ . + (αβ)i. + ²̄i.. }
25.31.
=
σα2
2
σβ2 σσβ
σ2
+
+
+
b
b
bn
because of independence.
σ 2 {Yij } = σ 2 {µ.. + ρi + τj + ²ij } = στ2 + σ 2
25.32.
P
2
2
σ {Ȳ.j } = σ {µ.. +
25.33. a.
b.
25.34.
ρi
nb
+ τj + ²̄.j } = στ2 +
σ2
nb
Yijk = µ... + ρi + αj + βk + (αβ)jk + ²ijk
F ∗ = M SAB/M SBL.T R, F ∗ = M SA/M SBL.T R, F ∗ = M SB/M SBL.T R
σ{Yij , Yij 0 } = E{(Yij − E{Yij })(Yij 0 − E{Yij 0 })}
= E{[(µ.. + ρi + τj + ²ij ) − (µ.. + τj )][(µ.. + ρi + τj 0 + ²ij 0 ) − (µ.. + τj 0 )]}
= E{(ρi + ²ij )(ρi + ²ij 0 )}
= E{ρ2i } + E{ρi ²ij } + E{ρi ²ij 0 } + E{²ij ²ij 0 } = σρ2
since ρi , ²ij , and ²ij 0 are pairwise independent and have expectations equal to zero.
25.35.
σ 2 {Ȳi... } = σ 2 {µ... + αi + β̄ . + γ̄ . + (αβ)i. + (αγ)i. + (βγ).. + (αβγ)i.. + ²̄i... }
= σα2 +
25.36. e.
2
σ2
σ2
σ2
σβ2 σγ2 σαβ
σ2
+
+
+ αγ + βγ + αβγ +
b
c
b
c
bc
bc
nbc
E{M SA} = 248.5, E{M SAB} = 8.5
25.37. a.
1
2
3
4
5
137.4286 145.3571 131.2857 156.6429
158.5536 135.4286 167.1607
128.5714 148.4286
179.4107
25.38. a.
25-8
117.2143
121.6250
113.8571
133.2321
102.9821
1
37.5667 42.9889 40.8333
2
51.1222 47.3889
3
47.3889
25.39. a.
µ̂.. = 30.051, σ̂ 2α = 7.439, σ̂ 2β = 2.757, σ̂ 2αβ = .011, σ̂ 2 = .183,
s{σ̂ 2α } = 5.570, s{σ̂ 2β } = 1.958, s{σ̂ 2αβ } = .053, s{σ̂ 2 } = .059.
25-9
25-10
Chapter 26
NESTED DESIGNS,
SUBSAMPLING, AND PARTIALLY
NESTED DESIGNS
26.4. a.
eijk :
k
1
2
3
4
5
k
1
2
3
4
5
j=1
3.2
−3.8
1.2
−4.8
4.2
i=1
j=2 j=3 j=4
.2 −6.6 −7.6
−5.8
2.4
3.4
7.2 −4.6
1.4
−3.8
7.4 −4.6
2.2
1.4
7.4
j=1
−7.8
6.2
−2.8
1.2
3.2
i=3
j=2 j=3 j=4
−6.6
6.6 −6.4
.4 −2.4
5.6
2.4 −1.4
−.4
−1.6
1.6
3.6
5.4 −4.4 −2.4
k
1
2
3
4
5
r = .986
26.5. a.
b.
No
Ȳij. :
i
1
2
3
j=1
61.8
75.8
76.8
j=2
67.8
75.2
69.6
j=3 j=4
62.6
52.6
55.8
77.0
74.4
73.4
c.
26-1
j=1
−1.8
5.2
.2
4.2
−7.8
i=2
j=2 j=3 j=4
−6.2 −3.8 −4.0
.8
.2
1.0
4.8
6.2
6.0
2.8
2.2 −2.0
−2.2 −4.8 −1.0
Source
SS
df
MS
Machines (A)
1, 695.63 2 847.817
Operators, within mahcines [B(A)] 2, 272.30 9 252.478
Error (E)
1, 132.80 48 23.600
Total
5, 100.73 59
d.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero.
F ∗ = 847.817/23.600 = 35.924, F (.99; 2, 48) = 5.075.
If F ∗ ≤ 5.075 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
e.
H0 : all βj(i) equal zero, Ha : not all βj(i) equal zero.
F ∗ = 252.478/23.600 = 10.698, F (.99; 9, 48) = 2.802.
If F ∗ ≤ 2.802 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
f.
i SSB(Ai )
1
599.20
2 1, 538.55
3
134.55
H0 : all βj(1) equal zero, Ha : not all βj(1) equal zero.
F ∗ = (599.20/3) ÷ 23.600 = 8.46, F (.99; 3, 48) = 4.22.
If F ∗ ≤ 4.22 conclude H0 , otherwise Ha . Conclude Ha .
H0 : all βj(2) equal zero, Ha : not all βj(2) equal zero.
F ∗ = (1, 538.55/3) ÷ 23.600 = 21.73, F (.99; 3, 48) = 4.22.
If F ∗ ≤ 4.22 conclude H0 , otherwise Ha . Conclude Ha .
H0 : all βj(3) equal zero, Ha : not all βj(3) equal zero.
F ∗ = (134.55/3) ÷ 23.600 = 1.90, F (.99; 3, 48) = 4.22.
If F ∗ ≤ 4.22 conclude H0 , otherwise Ha . Conclude H0 .
g.
26.6. a.
α ≤ .05
Ȳ1.. = 61.20, Ȳ2.. = 70.95, Ȳ3.. = 73.55, L̂1 = Ȳ1.. − Ȳ2.. = −9.75,
L̂2 = Ȳ1.. − Ȳ3.. = −12.35, L̂3 = Ȳ2.. − Ȳ3.. = −2.60, s{L̂i } = 1.536 (i = 1, 2, 3),
q(.95; 3, 48) = 3.42, T = 2.418
−9.75 ± 2.418(1.536)
−12.35 ± 2.418(1.536)
−2.60 ± 2.418(1.536)
b.
−13.46 ≤ L1 ≤ −6.04
−16.06 ≤ L2 ≤ −8.64
−6.31 ≤ L3 ≤ 1.11
Ȳ11. = 61.8, Ȳ12. = 67.8, Ȳ13. = 62.6, Ȳ14. = 52.6, L̂1 = Ȳ11. − Ȳ12. = −6.0,
L̂2 = Ȳ11. − Ȳ13. = −.8, L̂3 = Ȳ11. − Ȳ14. = 9.2, L̂4 = Ȳ12. − Ȳ13. = 5.2, L̂5 =
Ȳ12. − Ȳ14. = 15.2, L̂6 = Ȳ13. − Ȳ14. = 10.0, s{L̂i } = 3.0725 (i = 1, ..., 6), B =
t(.99583; 48) = 2.753
26-2
−6.0 ± 3.0725(2.753)
−.8 ± 3.0725(2.753)
9.2 ± 3.0725(2.753)
5.2 ± 3.0725(2.753)
15.2 ± 3.0725(2.753)
10.0 ± 3.0725(2.753)
c.
−14.46 ≤ L1 ≤ 2.46
−9.26 ≤ L2 ≤ 7.66
.74 ≤ L3 ≤ 17.66
−3.26 ≤ L4 ≤ 13.66
6.74 ≤ L5 ≤ 23.66
1.54 ≤ L6 ≤ 18.46
L̂ = 11.467, s{L̂} = 2.5087, t(.995; 48) = 2.682,
11.467 ± 2.682(2.5087), 4.74 ≤ L ≤ 18.20
26.7. a.
βj(i) are independent N (0, σβ2 ); βj(i) are independent of ²k(ij) .
b.
σ̂ 2β = 45.7756
c.
H0 : σβ2 = 0, Ha : σβ2 > 0. F ∗ = 252.478/23.600 = 10.698, F (.90; 9, 48) = 1.765.
If F ∗ ≤ 1.765 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
c1 = .2, c2 = −.2, M S1 = 252.478, M S2 = 23.600, df1 = 9, df2 = 48,
F1 = F (.95; 9, ∞) = 1.88, F2 = F (.95; 48, ∞) = 1.36, F3 = F (.95; ∞, 9) = 2.71,
F4 = F (.95; ∞, 48) = 1.45, F5 = F (.95; 9, 48) = 2.08, F6 = F (.95; 48, 9) = 2.81,
G1 = .4681, G2 = .2647, G3 = .00765, G4 = −.07162, HL = 23.771, HU = 86.258,
45.7756 − 23.771, 45.7756 + 86.258, 22.005 ≤ σβ2 ≤ 132.034
e.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero.
F ∗ = 847.817/252.478 = 3.358, F (.90; 2, 9) = 3.01.
If F ∗ ≤ 3.01 conclude H0 , otherwise Ha . Conclude Ha . P -value = .081
f.
See Problem 26.6a. s{L̂i } = 5.025 (i = 1, 2, 3), q(.90; 3, 9) = 3.32, T = 2.348
−9.75 ± 2.348(5.025)
−12.35 ± 2.348(5.025)
−2.60 ± 2.348(5.025)
g.
26.8. a.
−21.55 ≤ L1 ≤ 2.05
−21.15 ≤ L2 ≤ −.55
−14.40 ≤ L3 ≤ 9.20
H0 : all σ 2 {βj(i) } are equal (i = 1, 2, 3), Ha : not all σ 2 {βj(i) } are equal. Ye1 =
∗
62.2, Ye2 = 75.5, Ye3 = 73.9, M ST R = 11.6433, M SE = 38.0156, FBF
=
∗
11.6433/38.0156 = .31, F (.99; 2, 9) = 8.02. If FBF ≤ 8.02 conclude H0 , otherwise
Ha . Conclude H0 .
αi are independent N (0, σα2 ); βj(i) are independent N (0, σβ2 );
αi , βj(i) , and ²k(ij) are independent.
b.
σ̂ 2β = 45.7756, σ̂ 2α = 29.7669
c.
H0 : σα2 = 0, Ha : σα2 > 0. F ∗ = 847.817/252.478 = 3.358, F (.95; 2, 9) = 4.26.
If F ∗ ≤ 4.26 conclude H0 , otherwise Ha . Conclude H0 . P -value = .081
d.
c1 = .2, c2 = −.2, M S1 = 252.478, M S2 = 23.600, df1 = 9, df2 = 48,
F1 = F (.975; 9, ∞) = 2.11, F2 = F (.975; 48, ∞) = 1.44, F3 = (.975; ∞, 9) = 3.33,
F4 = F (.975; ∞, 48) = 1.56, F5 = F (.975; 9, 48) = 2.39, F6 = F (.975; 48, 9) =
3.48,
26-3
G1 = .5261, G2 = .3056, G3 = .01577, G4 = −.1176, HL = 26.766, HU = 117.544,
45.7756 − 26.766, 45.7756 + 117.544, 19.01 ≤ σβ2 ≤ 163.32
e.
26.9. a.
Ȳ... = 68.56667, s{Ȳ... } = 3.759, t(.975; 2) = 4.303, 68.56667 ± 4.303(3.759),
52.392 ≤ µ.. ≤ 84.742
eijk :
k
1
2
3
4
5
j=1
1.8
15.8
−5.2
−.2
−12.2
i=1
j=2 j=3
−12.8 −9.6
−.8
7.4
3.2
16.4
−3.8 −14.6
14.2
.4
k
1
2
3
4
5
j=1
−5.8
11.2
−.8
−12.8
8.2
i=3
j=2
−9.8
12.2
−.8
3.2
−4.8
k
1
2
3
4
5
j=1
−7.2
3.8
−15.2
7.8
10.8
i=2
j=2 j=3
−2.6
8.8
−15.6 −8.2
6.4 −10.2
11.4
11.8
.4 −2.2
j=3
−12.0
0.0
17.0
2.0
−7.0
r = .987
26.10. a.
Source
SS
df
States (A)
6, 976.84 2
Cities within states [B(A)]
167.60 6
Error (E)
3, 893.20 36
Total
11, 037.64 44
MS
3, 488.422
27.933
108.144
b.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero. F ∗ = 3, 488.422/108.144 =
32.257, F (.95; 2, 36) = 3.26. If F ∗ ≤ 3.26 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
c.
H0 : all βj(i) equal zero, Ha : not all βj(i) equal zero. F ∗ = 27.933/108.144 = .258,
F (.95; 6, 36) = 2.36. If F ∗ ≤ 2.36 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .95
d.
α ≤ .10
26.11. a.
Ȳ11. = 40.2, s{Ȳ11. } = 4.6507, t(.975; 36) = 2.0281,
40.2 ± 2.0281(4.6507), 30.77 ≤ µ11 ≤ 49.63
b.
Ȳ1.. = 40.8667, Ȳ2.. = 57.3333, Ȳ3.. = 26.8667, s{Ȳi.. } = 2.6851 (i = 1, 2, 3),
t(.995; 36) = 2.7195
40.8667 ± 2.7195(2.6851)
57.3333 ± 2.7195(2.6851)
26.8667 ± 2.7195(2.6851)
33.565 ≤ µ1. ≤ 48.169
50.031 ≤ µ2. ≤ 64.635
19.565 ≤ µ3. ≤ 34.169
26-4
c.
L̂1 = Ȳ1.. − Ȳ2.. = −16.4666, L̂2 = Ȳ1.. − Ȳ3.. = 14.0000, L̂3 = Ȳ2.. − Ȳ3.. = 30.4666,
s{L̂i } = 3.7973 (i = 1, 2, 3), q(.90; 3, 36) = 2.998, T = 2.120
−16.4666 ± 2.120(3.7973)
14.0000 ± 2.120(3.7973)
30.4666 ± 2.120(3.7973)
d.
26.12. a.
−24.52 ≤ L1 ≤ −8.42
5.95 ≤ L2 ≤ 22.05
22.42 ≤ L3 ≤ 38.52
L̂ = 12.4, s{L̂} = 6.5771, t(.975; 36) = 2.0281, 12.4 ± 2.0281(6.5771), −.94 ≤ L ≤
25.74
βj(i) are independent N (0, σβ2 ); βj(i) are independent of ²k(j) .
b.
σ̂ 2β = 0, yes.
c.
H0 : σβ2 = 0, Ha : σβ2 > 0. F ∗ = 27.933/108.144 = .258, F (.90; 6, 36) = 1.94.
If F ∗ ≤ 1.94 conclude H0 , otherwise Ha . Conclude H0 . P -value = .95
d.
H0 : all αi equal zero (i = 1, 2, 3), Ha : not all αi equal zero. F ∗ = 3, 488.422/27.933 =
124.885, F (.90; 2, 6) = 3.46. If F ∗ ≤ 3.46 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
e.
See Problem 26.11c. s{L̂i } = 1.9299 (i = 1, 2, 3), q(.90; 3, 6) = 3.56, T = 2.5173
−16.4666 ± 2.5173(1.9299)
14.0000 ± 2.5173(1.9299)
30.4666 ± 2.5173(1.9299)
f.
−21.32 ≤ L1 ≤ −11.61
9.14 ≤ L2 ≤ 18.86
25.61 ≤ L3 ≤ 35.32
H0 : all σ 2 {βj(i) } are equal (i = 1, 2, 3), Ha : not all σ 2 {βj(i) } are equal.
H ∗ = 37.27/16.07 = 2.32, H(.95; 3, 2) = 87.5.
If H ∗ ≤ 87.5 conclude H0 , otherwise Ha . Conclude H0 .
26.13. a.
αi are independent N (0, σα2 ); βj(i) are independent N (0, σβ2 );
αi , βj(i) , and ²k(ij) are independent.
b.
σ̂ 2β = 0, σ̂ 2α = 230.699
c.
H0 : σα2 = 0, Ha : σα2 > 0. F ∗ = 3, 488.422/27.933 = 124.885, F (.99; 2, 6) = 10.9.
If F ∗ ≤ 10.9 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
c1 = 1/15, c2 = −1/15, M S1 = 3488.422, M S2 = 27.933, df1 = 2, df2 = 6,
F1 = F (.995; 2, ∞) = 5.30, F2 = F (.995; 6, ∞) = 3.09, F3 = F (.995; ∞, 2) = 200,
F4 = F (.995; ∞, 6) = 8.88, F5 = F (.995; 2, 6) = 14.5, F6 = F (.995; 6, 2) = 199,
G1 = .8113, G2 = .6764, G3 = −1.2574, G4 = −93.0375, HL = 187.803, HU =
46, 279.30, 230.699 − 187.803, 230.699 + 46, 279.30, 42.90 ≤ σα2 ≤ 46, 510.00
e.
Ȳ... = 41.6889, s{Ȳ... } = 8.8046, t(.995; 2) = 9.925, 41.6889 ± 9.925(8.8046),
−45.70 ≤ µ.. ≤ 129.07
26.14. a.
Yijk = µ.. + α1 Xijk1 + α2 Xijk2 + β1(1) Xijk3 + β2(1) Xijk4 + β1(2) Xijk5 + β1(3) Xijk6 + ²ijk
1 if case from region 1
Xijk1 = −1 if case from region 3
0 otherwise
26-5
Xijk2 =
Xijk3
1 if case for team 1 from region 1
−1 if case for team 3 from region 1
=
0 otherwise
1 if case for team 2 from region 1
−1 if case for team 3 from region 1
0 otherwise
1 if case for team 1 from region 2
−1 if case for team 2 from region 2
=
0 otherwise
Xijk6 =
b.
1 if case from region 2
−1 if case from region 3
0 otherwise
Xijk4 =
Xijk5
1 if case for team 1 from region 3
−1 if case for team 2 from region 3
0 otherwise
Full model: Ŷ = 150.01667 − 9.21667X1 + 5.28333X2 + 6.60000X3
+.50000X4 + 3.70000X5 − 1.85000X6
eijk :
k
1
2
3
j=1
4.20
−6.20
2.00
i=1
j=2
1.90
−1.90
i=2
i=3
j=3 j=1 j=2 j=1 j=2
−2.30
4.80
0.00
4.90
4.20
2.30 −4.80
−4.90 −4.20
r = .962
26.15. a.
SSE(F ) = 207.2600
Reduced model: Ŷ = 147.60248+4.89938X3 +1.35031X4 +6.26584X5 −1.85000X6
SSE(R) = 838.7766
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero. F ∗ = (631.5166/2) ÷
(207.2600/7) = 10.664, F (.975; 2, 7) = 6.54. If F ∗ ≤ 6.54 conclude H0 , otherwise
Ha . Conclude Ha . P -value = .0075
b.
Reduced model: Ŷ = 150.74206 − 8.99921X1 + 5.79127X2
SSE(R) = 483.2338
H0 : β1(1) = β2(1) = β1(2) = β1(3) = 0, Ha : not all βj(i) equal zero. F ∗ =
(275.9738/4) ÷ (207.26/7) = 2.33, F (.975; 4, 7) = 5.52. If F ∗ ≤ 5.52 conclude
H0 , otherwise Ha . Conclude H0 .
c.
26.17. a.
L̂ = α̂1 − α̂2 = −14.5, s2 {α̂1 } = 4.0057, s2 {α̂2 } = 6.2446, s{α̂1 , α̂2 } = −2.6197,
s{L̂} = 3.9357, t(.99; 7) = 2.998, −14.5 ± 2.998(3.9357), −26.30 ≤ L ≤ −2.70
eijk :
26-6
k
1
2
i=1
j=1 j=2 j=3 j=4 j=5
−2.0
1.5
1.0
1.5 −1.0
2.0 −1.5 −1.0 −1.5
1.0
k
1
2
i=2
j=1 j=2 j=3 j=4
.5
1.0
1.5 −1.5
−.5 −1.0 −1.5
1.5
j=5
2.0
−2.0
i=3
k j=1 j=2 j=3 j=4 j=5
1 −2.0 −1.5
1.0
2.0 −1.5
2
2.0
1.5 −1.0 −2.0
1.5
r = .957
b.
H0 : all σ 2 {²j(i) } are equal (i = 1, 2, 3), Ha : not all σ 2 {²j(i) } are equal. Ye1 = 30,
∗
Ye2 = 28, Ye3 = 27, M ST R = 2.2167, M SE = 6.8750, FBF
= 2.2167/6.8750 = .32,
∗
F (.99; 2, 12) = 6.93. If FBF ≤ 6.93 conclude H0 , otherwise Ha . Conclude H0 .
26.18. a.
Source
SS
df
MS
Treatments (colors)
3.2667 2 1.63335
Experimental error 369.4000 12 30.78333
Observational error
67.5000 15 4.50000
Total
440.1667 29
b.
H0 : τ1 = τ2 = τ3 = 0, Ha : not all τi equal zero.
F ∗ = 1.63335/30.78333 = .053, F (.95; 2, 12) = 3.89.
If F ∗ ≤ 3.89 conclude H0 , otherwise Ha . Conclude H0 . P -value = .95
c.
H0 : σ 2 = 0, Ha : σ 2 > 0. F ∗ = 30.78333/4.50000 = 6.841, F (.95; 12, 15) = 2.48.
If F ∗ ≤ 2.48 conclude H0 , otherwise Ha . Conclude Ha . P -value = .0004
d.
Ȳ1.. = 29.2, s{Ȳ1.. } = 1.7545, t(.975; 12) = 2.179,
29.2 ± 2.179(1.7545), 25.38 ≤ µ1. ≤ 33.02
e.
σ̂ 2 = 13.1417, σ̂ 2η = 4.5
f.
For σ 2 : c1 = .5, c2 = −.5, M S1 = 30.7833, M S2 = 4.5000, df1 = 12, df2 = 15,
F1 = F (.975; 12, ∞) = 1.94, F2 = F (.975; 15, ∞) = 1.83, F3 = F (.975; ∞, 12) =
2.72, F4 = F (.975; ∞, 15) = 2.40, F5 = F (.975; 12, 15) = 2.96, F6 = F (.975; 15, 12) =
3.18, G1 = .4845, G2 = .4536, G3 = −.05916, G4 = −.0906, HL = 7.968,
HU = 26.434, 13.1417 − 7.968, 13.1417 + 26.434, 5.174 ≤ σ 2 ≤ 39.576
For ση2 : df = 15, χ2 (.025; 15) = 6.26, χ2 (.975; 15) = 27.49,
2.455 =
26.19.
15(4.5)
15(4.5)
≤ ση2 ≤
= 10.783
27.49
6.26
eijk :
k
1
2
3
j=1
−.4000
.0000
.4000
i=1
j=2
.0333
.3333
−.3667
j=3
−.3667
.0333
.3333
k
1
2
3
26-7
j=1
.0667
−.2333
.1667
i=2
j=2
.4333
.0667
−.3667
j=3
−.2000
.3000
−.1000
k
1
2
3
j=1
−.4333
.1667
.2667
i=3
j=2
−.1333
.4667
−.3333
j=3
−.3667
.3333
−.0667
k
1
2
3
j=1
−.0667
.4333
−.3667
i=4
j=2
−.3000
.2000
.1000
j=3
.4000
.0000
−.4000
r = .972
26.20. a.
Source
SS
df
Plants
343.1789 3
Leaves, within plants
187.4533 8
Observations, within leaves
3.0333 24
Total
533.6655 35
b.
MS
114.3930
23.4317
.1264
H0 : στ2 = 0, Ha : στ2 > 0. F ∗ = 114.3930/23.4317 = 4.88, F (.95; 3, 8) = 4.07.
If F ∗ ≤ 4.07 conclude H0 , otherwise Ha . Conclude Ha . P -value = .03
c.
H0 : σ 2 = 0, Ha : σ 2 > 0. F ∗ = 23.4317/.1264 = 185.38, F (.95; 8, 24) = 2.36.
If F ∗ ≤ 2.36 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
Ȳ... = 14.26111, s{Ȳ... } = 1.7826, t(.975; 3) = 3.182,
14.26111 ± 3.182(1.7826), 8.59 ≤ µ.. ≤ 19.93
e.
σ̂ 2τ = 10.1068, σ̂ 2 = 7.7684, σ̂ 2η = .1264
f.
c1 = 1/9 = .1111, c2 = −1/9 = −.1111, M S1 = 114.3930, M S2 = 23.4317,
df1 = 3, df2 = 8, F1 = F (.95; 3, ∞) = 2.60, F2 = F (.95; 8, ∞) = 1.94, F3 =
F (.95; ∞, 3) = 8.53, F4 = F (.95; ∞, 8) = 2.93, F5 = F (.95; 3, 8) = 4.07, F6 =
F (.95; 8, 3) = 8.85, G1 = .6154, G2 = .4845, G3 = −.1409, G4 = −1.5134,
HL = 9.042, HU = 95.444, 10.1068−9.042, 10.1068+95.444, 1.065 ≤ στ2 ≤ 105.551
26.21. a.
eijk :
k
1
2
3
j=1
.1667
−.0333
−.1333
i=1
j=2
j=3
.0667
.0333
−.1333 −.1667
.0667
.1333
j=4
−.0333
.1667
−.1333
k
1
2
3
j=1
.0333
.1333
−.1667
i=2
j=2
j=3
j=4
.0333 −.0667 −.2000
−.1667 −.0667
.2000
.1333
.1333
.0000
k
1
2
3
j=1
.0000
.1000
−.1000
i=3
j=2
j=3
.1667 −.1333
.0667 −.0333
−.2333
.1667
j=4
.0667
−.2333
.1667
k
1
2
3
j=1
−.1333
.1667
−.0333
i=4
j=2
j=3
j=4
−.0333
.1667 −.0333
.1667 −.1333
.1667
−.1333 −.0333 −.1333
j=1
.0333
.1333
−.1667
i=5
j=2
j=3
.1000
.1333
.1000 −.0667
−.2000 −.0667
j=4
.2000
−.2000
.0000
k
1
2
3
26-8
r = .981
b.
H0 : all σ 2 {²j(i) } are equal (i = 1, ..., 5), Ha : not all σ 2 {²j(i) } are equal.
H ∗ = .100833/.014167 = 7.117, H(.99; 5, 3) = 151.
If H ∗ ≤ 151 conclude H0 , otherwise Ha . Conclude H0 .
26.22. a.
Source
SS
Batches
10.6843
Barrels, within batches
.6508
Determinations, within barrels 1.0067
Total
12.3418
b.
df
MS
4 2.67108
15 .04339
40 .02517
59
H0 : στ2 = 0, Ha : στ2 > 0. F ∗ = 2.67108/.04339 = 61.56, F (.99; 4, 15) = 4.89.
If F ∗ ≤ 4.89 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
H0 : σ 2 = 0, Ha : σ 2 > 0. F ∗ = .04339/.02517 = 1.724, F (.99; 15, 40) = 2.52.
If F ∗ ≤ 2.52 conclude H0 , otherwise Ha . Conclude H0 . P -value = .085
d.
Ȳ... = 2.9117, s{Ȳ... } = .21099, t(.995; 4) = 4.604,
2.9117 ± 4.604(.21099), 1.94 ≤ µ.. ≤ 3.88
e.
σ̂ 2τ = .2190, σ̂ 2 = .0061, σ̂ 2η = .0252
f.
c1 = .08333, c2 = −.08333, M S1 = 2.67108, M S2 = .04339, df1 = 4, df2 = 15,
F1 = F (.975; 4, ∞) = 2.79, F2 = F (.975; 15, ∞) = 1.83, F3 = F (.975; ∞, 4) =
8.26, F4 = F (.975; ∞, 15) = 2.40, F5 = F (.975; 4, 15) = 3.80, F6 = F (.975; 15, 4) =
8.66, G1 = .6416, G2 = .4536, G3 = .1082, G4 = −1.0925, HL = .1432,
HU = 1.6157, .2190 − .1432, .2190 + 1.6157, .076 ≤ στ2 ≤ 1.835
PPP
26.23.
(Yijk − Ȳ... )2 =
=
PPP
PPP
[(Ȳi.. − Ȳ... ) + (Ȳij. − Ȳi.. ) + (Yijk − Ȳij. )]2
[(Ȳi.. − Ȳ... )2 + (Ȳij. − Ȳi.. )2 + (Yijk − Ȳij. )2 + 2(Ȳi.. − Ȳ... )(Ȳij. − Ȳi.. )
+2(Ȳi.. − Ȳ... )(Yijk − Ȳij. ) + 2(Ȳij. − Ȳi.. )(Yijk − Ȳij. )]
= bn
P
(Ȳi.. − Ȳ... )2 + n
PP
(Ȳij. − Ȳi.. )2 +
PPP
(Yijk − Ȳij. )2
All cross products equal zero by arguments similar to that given in Section 16.8.
P
26.24.
26.25. a.
b.
26.26.
PP
P
P
2
Y.j.2
Yij.
Y.j.2
Y2
Yi..2
Y2
SSB + SSAB =
− ... +
−
−
+ ...
na
nab
n
nb
na
nab
PP 2
P 2
Yij.
Yi..
−
= SSB(A)
=
n
nb
σβ2 σ 2
+
b
bn
2
σ
σ2
σ 2 {Ȳ... } = σ 2 {µ.. + β̄ .(.) + ²̄.(..) } = β +
ab abn
[M SB(A) − M SE]/n
σ 2 {Ȳi.. } = σ 2 {µ.. + αi + β̄ .(i) + ²̄.(i.) } =
σ 2 {Ȳi.. } = σ 2 {µ.. + τi + ²̄.(i) + η̄ .(i.) }
26-9
ση2
mσ 2 + ση2
σ2
=
+
=
n
mn
mn
σ 2 {Ȳ... } = σ 2 {µ.. + τ̄ . + ²̄.(.) + η̄ .(..) }
26.27.
ση2
στ2 σ 2
=
+
+
r
rn rnm
½
¾
n
o
ση2 + mσ 2 + nmστ2
M ST R
2
=
= σ 2 {Ȳ... }
E s {Ȳ... } = E
rnm
rnm
Ã
2
σ2
2
σ 2 {Ȳ1j.. − Ȳ2j.. } =
σβγ
+
+ σγ2
c
n
26.28.
df =
26.29. a.
b.
!
[bM SBC(A) + M SC(A) − M SE]2
[bM SBC(A)]2
[M SC(A)]2
(M SE)2
+
+
a(b − 1)(c − 1)
a(c − 1)
abc(n − 1)
Yijk = µ.. + αi + βj(i) + ²k(ij) , βj(i) and ²k(ij) random
eijk :
k
1
2
j=1
−.040
.040
k j=1
1
.025
2 −.025
r = .938
i=1
j=2 j=3 j=4
.045
.020 −.035
−.045 −.020
.035
k
1
2
j=1
.035
−.035
i=2
j=2 j=3 j=4
−.045 −.025
.040
.045
.025 −.040
i=3
j=2 j=3 j=4
.040 −.035 −.060
−.040
.035
.060
26.30. a.
b.
Source
SS
df
A (lever press rate)
.89306 2
D(A) (rats within A) .12019 9
Error
.03555 12
Total
1.04880 23
H0 : α1 = α2 = α3 = 0, Ha : not all αi
MS
.44653
.01335
.00296
equal zero.
∗
F = .44653/.01335 = 33.448, F (.95; 2, 9) = 4.26.
If F ∗ ≤ 4.26 conclude H0 , otherwise Ha . Conclude Ha . P -value = .0001
c.
H0 : σβ2 = 0, Ha : σβ2 > 0. F ∗ = .01335/.00296 = 4.510, F (.95; 9, 12) = 2.80.
If F ∗ ≤ 2.80 conclude H0 , otherwise Ha . Conclude Ha . P -value = .009
d.
Ȳ1.. = .53500, Ȳ2.. = .77375, Ȳ3.. = 1.00750, L̂1 = Ȳ1.. − Ȳ2.. = −.23875, L̂2 =
Ȳ1.. − Ȳ3.. = −.47250, L̂3 = Ȳ2.. − Ȳ3.. = −.23375, s{L̂i } = .0578 (i = 1, 2, 3),
q(.90; 3, 9) = 3.32, T = 2.3476
−.23875 ± 2.3476(.0578)
−.47250 ± 2.3476(.0578)
−.23375 ± 2.3476(.0578)
−.374 ≤ L1 ≤ −.103
−.608 ≤ L2 ≤ −.337
−.369 ≤ L3 ≤ −.098
26-10
e.
26.31. a.
b.
σ̂ 2β = .005195, c1 = .5, c2 = −.5, M S1 = .013354, M S2 = .002963, df1 = 9, df2 =
12, F1 = F (.95; 9, ∞) = 1.88, F2 = F (.95; 12, ∞) = 1.75, F3 = F (.95; ∞, 9) =
2.71, F4 = F (.95; ∞, 12) = 2.30, F5 = F (.95; 9, 12) = 2.80, F6 = F (.95; 12, 9) =
3.07, G1 = .4681, G2 = .4286, G3 = −.05996, G4 = −.1210, HL = .003589,
HU = .01138, .005195 − .003589, .005195 + .01138, .00161 ≤ σβ2 ≤ .0166
Yijk = µ.. + τi + ²j(i) + ηk(ij) , ²j(i) and ηk(ij) random
eijk :
k
1
2
k
1
2
j=1
−.035
.035
i=1
j=2 j=3 j=4
−.030 −.030 −.025
.030
.030
.025
j=1
−.050
.050
i=3
j=2 j=3 j=4
−.025 −.035
.045
.025
.035 −.045
k
1
2
j=1
.020
−.020
i=2
j=2 j=3 j=4
.030 −.035 −.020
−.030
.035
.020
r = .940
c.
H0 : all σ 2 {²j(i) } are equal (i = 1, 2, 3), Ha : not all σ 2 {²j(i) } are equal.
Ye1 = 1.9075, Ye2 = 2.2200, Ye3 = 2.4075, M ST R = .001431, M SE = .004204,
∗
FBF
= .001431/.004204 = .34, F (.99; 2, 9) = 8.02.
∗
If FBF
≤ 8.02 conclude H0 , otherwise Ha . Conclude H0 .
26.32. a.
Source
SS
df
Treatments (lever press rates) 1.013125 2
Experimental error
.182025 9
Observational error
.025900 12
Total
1.221050 23
b.
MS
.50656
.02023
.00216
H0 : all τ1 = τ2 = τ3 = 0, Ha : not all τi equal zero.
F ∗ = .50656/.02023 = 25.040, F (.99; 2, 9) = 8.02.
If F ∗ ≤ 8.02 conclude H0 , otherwise Ha . Conclude Ha . P -value = .0002
c.
H0 : σ 2 = 0, Ha : σ 2 > 0. F ∗ = .02023/.00216 = 9.366, F (.99; 9, 12) = 4.39.
If F ∗ ≤ 4.39 conclude H0 , otherwise Ha . Conclude Ha . P -value = .0003
d.
Ȳ1.. = 1.88750, Ȳ2.. = 2.21875, Ȳ3.. = 2.38125, L̂1 = Ȳ1.. − Ȳ2.. = −.33125, L̂2 =
Ȳ1.. − Ȳ3.. = −.49375, L̂3 = Ȳ2.. − Ȳ3.. = −.16250, s{L̂i } = .071116 (i = 1, 2, 3),
q(.95; 3, 9) = 3.95, T = 2.793
−.33125 ± 2.793(.071116)
−.49375 ± 2.793(.071116)
−.16250 ± 2.793(.071116)
f.
−.530 ≤ L1 ≤ −.133
−.692 ≤ L2 ≤ −.295
−.361 ≤ L3 ≤ .036
For σ 2 : σ̂ 2 = .00904, c1 = .5, c2 = −.5, M S1 = .020225, M S2 = .0021583,
df1 = 9, df2 = 12, F1 = F (.95; 9, ∞) = 1.88, F2 = F (.95; 12, ∞) = 1.75, F3 =
26-11
F (.95; ∞, 9) = 2.71, F4 = F (.95; ∞, 12) = 2.30, F5 = F (.95; 9, 12) = 2.80, F6 =
F (.95; 12, 9) = 3.07, G1 = .4681, G2 = .4286, G3 = −.05996, G4 = −.1210,
HL = .00487, HU = .0173, .00904 − .00487, .00904 + .0173, .0042 ≤ σ 2 ≤ .0263
For ση2 : σ̂ 2η = .00216, df = 12, χ2 (.05; 12) = 5.23, χ2 (.95; 12) = 21.03,
.0012 =
12(.00216)
12(.00216)
≤ ση2 ≤
= .0050
21.03
5.23
26-12
Chapter 27
REPEATED MEASURES AND
RELATED DESIGNS
27.3. a.
eij :
i
j=1
j=2
1
2.5556 −1.9444
2 −.1111
.3889
3 −2.9444
5.5556
4 −2.7778
.7222
5 −2.4444 −3.9444
6 −.1111
3.3889
7 −.9444 −2.4444
8
2.3889 −2.1111
9 −.6111 −5.1111
10
1.5556
1.0556
11
1.3889
.8889
12
2.0556
3.5556
j=3
−3.7778
−.4444
3.7222
−2.1111
1.2222
−.4444
−1.2778
4.0556
.0556
−3.7778
.0556
2.7222
j=4
2.4722
2.8056
−3.0278
1.1389
1.4722
−3.9944
−2.0278
−1.6944
−1.6944
3.4722
1.3056
−1.0278
j=5
−2.7778
−3.4444
−2.2778
1.8889
1.2222
.5556
2.7222
.0556
3.0556
−.7778
1.0556
−1.2778
j=6
3.4722
.8056
−1.0278
1.1389
2.4722
−.1944
3.9722
−2.6944
4.3056
−1.5278
−4.6944
−6.0278
r = .995
d.
H0 : D = 0, Ha : D 6= 0. SST R.S = 467.3889, SST R.S ∗ = 8.7643, SSRem∗ =
458.6246, F ∗ = (8.7643/1) ÷ (458.6246/54) = 1.032, F (.995; 1, 54) = 8.567. If
F ∗ ≤ 8.567 conclude H0 , otherwise Ha . Conclude H0 . P -value = .31
27.4. a.
Source
Subjects
Doses
Error
Total
SS
df
MS
1, 197.4444 11
108.8586
5, 826.2778 5 1, 165.2556
467.3889 55
8.4980
7, 491.1111 71
b.
H0 : all τj equal zero (j = 1, ..., 6), Ha : not all τj equal zero. F ∗ = 1, 165.2556/8.4980 =
137.12, F (.99; 5, 55) = 3.37. If F ∗ ≤ 3.37 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
c.
Ȳ.1 = 14.6667, Ȳ.2 = 19.1667, Ȳ.3 = 23.0000, Ȳ.4 = 28.7500, Ȳ.5 = 35.0000, Ȳ.6 =
40.7500, L̂1 = Ȳ.1 − Ȳ.2 = −4.5000, L̂2 = Ȳ.2 − Ȳ.3 = −3.8333, L̂3 = Ȳ.3 − Ȳ.4 =
27-1
−5.7500, L̂4 = Ȳ.4 − Ȳ.5 = −6.2500, L̂5 = Ȳ.5 − Ȳ.6 = −5.7500, s{L̂i } = 1.1901
(i = 1, ..., 5), B = t(.995; 55) = 2.668
−4.5000 ± 2.668(1.1901)
−3.8333 ± 2.668(1.1901)
−5.7500 ± 2.668(1.1901)
−6.2500 ± 2.668(1.1901)
−5.7500 ± 2.668(1.1901)
d.
27.5. a.
−7.6752 ≤ L1
−7.0085 ≤ L2
−8.9252 ≤ L3
−9.4252 ≤ L4
−8.9252 ≤ L5
≤ −1.3248
≤ −.6581
≤ −2.5748
≤ −3.0748
≤ −2.5748
Ê = 2.83
Yij = µ.. + ρ1 Xij1 + ρ2 Xij2 + ρ3 Xij3 + ρ4 Xij4 + ρ5 Xij5 + ρ6 Xij6 + ρ7 Xij7
+ρ8 Xij8 + ρ9 Xij9 + ρ10 Xij10 + ρ11 Xij11 + γ1 xij + γ2 x2ij + γ3 x3ij + ²ij
Xij1
1 if experimental unit from block 1
−1 if experimental unit from block 12
=
0 otherwise
Xij2 , ..., Xij11 are defined similarly
xij =
−.97
−.77
−.57
−.07
.43
1.93
b.
if
if
if
if
if
if
experimental
experimental
experimental
experimental
experimental
experimental
unit
unit
unit
unit
unit
unit
received
received
received
received
received
received
treatment
treatment
treatment
treatment
treatment
treatment
1
2
3
4
5
6
Ŷ = 30.3903 + 3.7778X1 + 4.4444X2 + .2778X3 − 2.8889X4 − 5.2222X5
+3.4444X6 − 4.7222X7 + 2.9444X8 + 3.9444X9 − 8.2222X10
+1.9444X11 + 11.5329x − 4.0297x2 + .4353x3
c.
eij :
i
1
2
3
4
5
6
7
8
9
10
11
12
d.
j=1
2.2076
−.4591
−3.2924
−3.1257
−2.7924
−.4591
−1.2934
2.0409
−.9591
1.2076
1.0409
1.7076
j=2
−1.6998
.6335
5.8002
.9668
−3.6998
3.6335
−2.1998
−1.8665
−4.8665
1.3002
1.1335
3.8002
j=3
−3.2045
.1288
4.2955
−1.5378
1.7955
.1288
−.7045
4.6288
.6288
−3.2045
.6288
3.2955
j=4
1.6591
1.9925
−3.8409
.3258
.6591
−4.0075
−2.8409
−2.5075
−2.5075
2.6591
.4925
−1.8409
j=5
−2.4167
−3.0834
−1.9167
2.2499
1.5833
.9166
3.0833
.4166
3.4166
−.4167
1.4166
−.9167
j=6
3.4543
.7877
−1.0457
1.1210
2.4543
−.2123
3.9543
−2.7123
4.2877
−1.5457
−4.7123
−6.0457
H0 : γ3 = 0, Ha : γ3 6= 0. SSE(F ) = 483.0053, SSE(R) = 484.8980,
F ∗ = (1.8927/1) ÷ (483.0053/57) = .223, F (.95; 1, 57) = 4.01.
If F ∗ ≤ 4.01 conclude H0 , otherwise Ha . Conclude H0 . P -value = .64
27-2
27.6. a.
eij :
i
1
2
3
4
5
6
7
8
j=1
j=2
−1.2792 −.2417
−.8458
.6917
.6208
.0583
.5542
.1917
.5208 −.3417
−.1458
.3917
.9875 −.7750
−.4125
.0250
j=3
1.5208
.1542
−.6792
−.7458
−.1792
−.2458
−.2125
.3875
r = .992
d.
H0 : D = 0, Ha : D 6= 0. SST R.S = 9.5725, SST R.S ∗ = 2.9410, SSRem∗ =
6.6315, F ∗ = (2.9410/1) ÷ (6.6315/13) = 5.765, F (.99; 1, 13) = 9.07. If F ∗ ≤ 9.07
conclude H0 , otherwise Ha . Conclude H0 . P -value = .032
27.7. a.
Source
SS
df
MS
Stores 745.1850 7 106.4550
Prices
67.4808 2 33.7404
Error
9.5725 14
.68375
Total
822.2383 23
b.
H0 : all τj equal zero (j = 1, 2, 3), Ha : not all τj equal zero. F ∗ = 33.7404/.68375 =
49.346, F (.95; 2, 14) = 3.739. If F ∗ ≤ 3.739 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
c.
Ȳ.1 = 55.4375, Ȳ.2 = 53.6000, Ȳ.3 = 51.3375, L̂1 = Ȳ.1 − Ȳ.2 = 1.8375, L̂2 =
Ȳ.1 − Ȳ.3 = 4.1000, L̂3 = Ȳ.2 − Ȳ.3 = 2.2625, s{L̂i } = .413446 (i = 1, 2, 3),
q(.95; 3, 14) = 3.70, T = 2.616
1.8375 ± 2.616(.413446)
4.1000 ± 2.616(.413446)
2.2625 ± 2.616(.413446)
d.
27.8.
.756 ≤ L1 ≤ 2.919
3.018 ≤ L2 ≤ 5.182
1.181 ≤ L3 ≤ 3.344
Ê = 48.08
H0 : all τj equal zero (j = 1, ..., 6), Ha : not all τj equal zero.
M ST R = 39.8583, M ST R.S = .2883, FR∗ = 39.8583/.2883 = 138.24,
F (.99; 5, 25) = 3.855. If FR∗ ≤ 3.855 conclude H0 , otherwise Ha . Conclude Ha .
27.9.
H0 : all τj equal zero (j = 1, 2, 3), Ha : not all τj equal zero. M ST R = 8,
M ST R.S = 0, FR∗ = 8/0. Note: Nonparametric F test results in SST R.S = 0
and therefore should not be used.
27.10. a.
H0 : all τj equal zero (j = 1, ..., 5), Ha : not all τj equal zero.
M ST R = 15.8500, M ST R.S = 1.0167, FR∗ = 15.8500/1.0167 = 15.59, F (.95; 4, 36) =
2.63. If FR∗ ≤ 2.63 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
27-3
b.
R̄.1 = 4.0, R̄.2 = 1.4, R̄.3 = 2.1, R̄.4 = 3.1, R̄.5 = 4.4, B = z(.995) = 2.576,
B[r(r + 1)/6n]1/2 = 1.82
Group 1: B, C, D
Group 2: A, D, E
c.
27.11. a.
W = .634
eijk :
i
1
2
3
4
5
6
j=1
k=1
k=2
−1.9167
1.9167
−.4167
.4167
1.5833 −1.5833
.0833 −.0833
1.0833 −1.0833
−.4167
.4167
j=2
k=1
k=2
−2.3333
2.3333
.6667 −.6667
−.3333
.3333
.6667 −.6667
−.3333
.3333
1.6667 −1.6667
r = .994
27.12. a.
Source
SS
df
MS
A (incentive stimulus) 975.38 1 975.38
S(A)
148.75 10 14.875
B (problem type)
513.37 1 513.37
AB interactions
155.04 1 155.04
B.S(A)(Error)
34.08 10 3.408
Total
1826.63 23
b.
Ȳ.11 = 12.667, Ȳ.12 = 16.833, Ȳ.21 = 20.333, Ȳ.22 = 34.667
c.
H0 : all (αβ)jk equal zero, Ha : not all (αβ)jk equal zero.
F ∗ = 155.04/3.408 = 45.49, F (.95; 1, 10) = 4.96.
d.
If F ∗ ≤ 4.96 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
√
T = q(.95; 2, 10)/ 2 = 2.227, s2 {D̂} = 2(3.408)/6 = 1.136, s{D̂} = 1.0658
−4.17 ± 2.227(1.0658)
−14.33 ± 2.227(1.0658)
e.
dfadj =
−6.54 ≤ L1 ≤ −1.80
−16.70 ≤ L2 ≤ −11.96
√
[34.08 + 148.75]2
= 14.35, T = q(.95; 2, 14)/ 2 = 2.143
2
2
34.08 /10 + 148.75 /10
M S(Within Treatments)=(34.08+148.75)/20=9.1415
s2 {D̂} = 2(9.1415)/6 = 3.0472, s{D̂} = 1.7456
−7.67 ± 2.143(1.7456)
−17.83 ± 2.143(1.7456)
27.13. a.
−11.41 ≤ L1 ≤ −3.93
−21.57 ≤ L2 ≤ −14.09
eijk :
27-4
j=1
j=2
i=1
i=2
i=3
i=4
i=1
i=2
i=3
i=4
k=1
9.250
−11.750
7.750
−5.250
3.625
15.375
−8.375
−10.625
k=2
−8.750
−2.750
−5.250
16.750
−3.125
6.625
−3.125
−.375
k=3
1.250
15.250
5.750
−22.250
−13.875
7.875
−3.875
9.875
k=4
−1.750
−.750
−8.250
10.750
13.375
−29.875
15.375
1.125
r = .981
27.14. a.
H0 : σ 2 {ρi(1) } = σ 2 {ρi(2) }, Ha : σ 2 {ρi(1) } 6= σ 2 {ρi(2) }.
SSS(A1 ) = 1, 478, 757.00, SSS(A2 ) = 1, 525, 262.25,
H ∗ = (1, 525, 262.25/3) ÷ (1, 478, 757.00/3) = 1.03, H(.99; 2, 3) = 47.5.
If H ∗ ≤ 47.5 conclude H0 , otherwise Ha . Conclude H0 .
b.
H0 : σ 2 {²1jk } = σ 2 {²2jk }, Ha : σ 2 {²1jk } 6= σ 2 {²2jk }.
SSB.S(A1 ) = 1, 653.00, SSB.S(A2 ) = 2, 172.25,
H ∗ = (2, 172.25/9) ÷ (1, 653.00/9) = 1.31, H(.99; 2, 9) = 6.54.
If H ∗ ≤ 6.54 conclude H0 , otherwise Ha . Conclude H0 .
27.15. a.
Source
SS
df
MS
A (type display)
266, 085.1250 1 266, 085.1250
S(A)
3, 004, 019.2500 6 500, 669.8750
B (time)
53, 321.6250 3 17, 773.8750
AB interactions
690.6250 3
230.2083
Error
3, 825.2500 18
212.5139
Total
3, 327, 941.8750 31
b.
Ȳ.11 = 681.500, Ȳ.12 = 696.500, Ȳ.13 = 671.500, Ȳ.14 = 785.500,
Ȳ.21 = 508.500, Ȳ.22 = 512.250, Ȳ.23 = 496.000, Ȳ.24 = 588.750
c.
H0 : all (αβ)jk equal zero, Ha : not all (αβ)jk equal zero.
F ∗ = 230.2083/212.5139 = 1.08, F (.975; 3, 18) = 3.95.
If F ∗ ≤ 3.95 conclude H0 , otherwise Ha . Conclude H0 . P -value = .38
d.
H0 : α1 = α2 = 0, Ha : not both αj equal zero.
F ∗ = 266, 085.1250/500, 669.8750 = .53, F (.975; 1, 6) = 8.81.
If F ∗ ≤ 8.81 conclude H0 , otherwise Ha . Conclude H0 . P -value = .49
H0 : all βk equal zero (k = 1, ..., 4), Ha : not all βk equal zero.
F ∗ = 17, 773.8750/212.5139 = 83.636, F (.975; 3, 18) = 3.95.
If F ∗ ≤ 3.95 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
27-5
e.
Ȳ.1. = 708.750, Ȳ.2. = 526.375, Ȳ..1 = 595.000, Ȳ..2 = 604.375, Ȳ..3 = 583.750,
Ȳ..4 = 687.125, L̂1 = 182.375, L̂2 = −9.375, L̂3 = 20.625, L̂4 = −103.375,
s{L̂1 } = 250.1674, s{L̂i } = 7.2889 (i = 2, 3, 4), B1 = t(.9875; 6) = 2.969, Bi =
t(.9875; 18) = 2.445 (i = 2, 3, 4)
182.375 ± 2.969(250.1674)
−9.375 ± 2.445(7.2889)
20.625 ± 2.445(7.2889)
−103.375 ± 2.445(7.2889)
27.16. a.
−560.372 ≤ L1 ≤ 925.122
−27.196 ≤ L2 ≤ 8.446
2.804 ≤ L3 ≤ 38.446
−121.196 ≤ L4 ≤ −85.554
eijk :
i
1
2
3
4
5
6
j=1
k=1
k=2
−.05833
.05833
−.05833
.05833
−.03333
.03333
−.00833
.00833
.21667 −.21667
−.05833
.05833
j=2
k=1
k=2
.05833 −.05833
.05833 −.05833
.03333 −.03333
.00833 −.00833
−.21667
.21667
.05833 −.05833
r = .9685
27.17. a.
Source
SS
df
MS
Subjects
1.0533 5
.2107
A (problem) 16.6667 1 16.6667
B (model)
72.1067 1 72.1067
AB
3.6817 1 3.6817
AS
.5983 5
.1197
BS
.1783 5
.0357
ABS
.2333 5
.0467
Total
94.5183 23
b.
Ȳ.11 = 3.367, Ȳ.12 = 7.617, Ȳ.21 = 2.483, Ȳ.22 = 5.167
c.
H0 : all (αβ)jk equal zero, Ha : not all (αβ)jk equal zero.
F ∗ = 3.6817/.0467 = 78.84, F (.99; 1, 5) = 16.3.
If F ∗ ≤ 16.3 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
d.
L̂1 = 4.250, L̂2 = 2.684, L̂3 = −1.566,
s{L̂i } = .1248 (i = 1, 2), s{L̂3 } = .1765, B = t(.9917; 5) = 3.538
4.250 ± 3.538(.1248)
2.684 ± 3.538(.1248)
−1.566 ± 3.538(.1765)
27.18. a.
3.808 ≤ L1 ≤ 4.692
2.242 ≤ L2 ≤ 3.126
−2.190 ≤ L3 ≤ −.942
eijk :
27-6
i
1
2
3
4
5
6
7
8
9
10
j=1
k=1 k=2
−.045
.045
−.120
.120
.080 −.080
−.045
.045
.080 −.080
.055 −.055
.030 −.030
−.045
.045
.055 −.055
−.045
.045
j=2
k=1 k=2
.045 −.045
.120 −.120
−.080
.080
.045 −.045
−.080
.080
−.055
.055
−.030
.030
.045 −.045
−.055
.055
.045 −.045
r = .973
27.19. a.
Source
Subjects
A
B
AB
AS
BS
ABS
Total
SS
df
MS
154.579 9 17.175
3.025 1 3.025
14.449 1 11.449
.001 1
.001
2.035 9
.226
5.061 9
.562
.169 9
.019
176.319 39
b.
Ȳ.11 = 3.93, Ȳ.12 = 5.01, Ȳ.21 = 4.49, Ȳ.22 = 5.55
c.
H0 : all (αβ)jk equal zero, Ha : not all (αβ)jk equal zero.
F ∗ = .001/.019 = .05, F (.995; 1, 9) = 13.6.
If F ∗ ≤ 13.6 conclude H0 , otherwise Ha . Conclude H0 . P -value = .82
d.
H0 : α1 = α2 = 0, Ha : not both αj equal zero.
F ∗ = 3.025/.226 = 13.38, F (.95; 1, 9) = 5.12.
If F ∗ ≤ 13.6 conclude H0 , otherwise Ha . Conclude Ha . P -value = .005
H0 : β1 = β2 = 0, Ha : not both βk equal zero.
F ∗ = 11.449/.562 = 20.36, F (.95; 1, 9) = 5.12.
If F ∗ ≤ 13.6 conclude H0 , otherwise Ha . Conclude Ha . P -value = .001
e.
L̂1 = .56, L̂2 = 1.08, L̂3 = −.52, L̂4 = 1.62,
s{L̂i } = .0613 (i = 1, ..., 4), B = t(.99375; 9) = 3.11
.56 ± 3.11(.0613)
1.08 ± 3.11(.0613)
−.52 ± 3.11(.0613)
1.62 ± 3.11(.0613)
27.20. a.
.37 ≤ L1 ≤ .75
.89 ≤ L2 ≤ 1.27
−.71 ≤ L3 ≤ −.33
1.43 ≤ L4 ≤ 1.81
eijk :
27-7
i
1
j
1
2
2
1
2
3
1
2
4
1
2
5
1
2
r = .981
k=1
−.6
−1.7
.4
1.3
−.6
.3
.4
−.2
.4
.3
k=2
.6
1.7
−.4
−1.3
.6
−.3
−.4
.2
−.4
−.3
27.21. a.
Source
SS
Whole plots
Irrigation method (A) 1, 394.45
Whole-plot error
837.60
Split plots
Fertilizer (B)
AB Interactions
Split-plot error
Total
df
MS
1 1, 394.45
8
104.70
68.45 1
.05 1
12.00 8
2, 312.55 19
68.45
.05
1.50
b.
Ȳ.11 = 35.4, Ȳ.21 = 52.2, Ȳ.12 = 39.2, Ȳ.22 = 55.8
c.
H0 : all (αβ)jk equal zero, Ha : not all (αβ)jk equal zero. F ∗ = .05/1.50 = .033,
F (.95; 1, 8) = 5.32. If F ∗ ≤ 5.32 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .86
d.
H0 : α1 = α2 = 0, Ha : not both αj equal zero. F ∗ = 1, 394.45/104.70 = 13.32,
F (.95; 1, 8) = 5.32. If F ∗ ≤ 5.32 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .006
H0 : β1 = β2 = 0, Ha : not both βk equal zero. F ∗ = 68.45/1.50 = 45.63,
F (.95; 1, 8) = 5.32. If F ∗ ≤ 5.32 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .0001
e.
Ȳ.1. = 37.3, Ȳ.2. = 54.0, Ȳ..1 = 43.8, Ȳ..2 = 47.5, L̂1 = −16.7, L̂2 = −3.7, s{L̂1 } =
4.5760, s{L̂2 } = .5477, B1 = t(.975; 8) = 2.306, B2 = t(.975; 8) = 2.306
−16.7 ± 2.306(4.5760)
−3.7 ± 2.306(.5477)
−27.252 ≤ L1 ≤ −6.148
−4.963 ≤ L2 ≤ −2.437
27.22.
XX
(Yij − Ȳ.. )2 =
=
=
XX
XX
XX
[(Yij − Ȳi. ) + (Ȳi. − Ȳ.. )]2
(Yij − Ȳi. )2 +
XX
(Yij − Ȳi. )2 + r
27-8
X
(Ȳi. − Ȳ.. )2 + 2
(Ȳi. − Ȳ.. )2
XX
(Yij − Ȳi. )(Ȳi. − Ȳ.. )
Cross-product term equals zero by argument similar to that given in Section 16.5.
27.23.
j
1
2
3
4
5
6
j0
27.3333 20.8788 23.0909 19.1818
29.4242 23.4545 14.3182
30.9091 14.7273
18.3864
16.7273
12.6364
18.2727
11.1818
17.4545
17.0909
11.0455
16.4545
15.0227
16.8182
27.8409
j0
27.24.
1
29.6084 33.0114 34.0598
j 2
37.5886 38.7000
3
40.6255
27.25. a.
b.
Yij = µ.. + ρi + τj + ²(ij)
eij :
i
1
2
3
4
5
6
7
8
9
10
11
12
j=1
.02083
.00083
.00083
.04083
−.03167
.01833
−.00167
−.00167
−.02417
−.00417
.03333
−.05167
j=2
−.00917
−.00917
.00083
.02083
.04833
−.03167
−.00167
.00833
−.00417
−.04417
−.01667
.03833
j=3
.00833
−.00167
.00833
−.02167
−.04417
.05583
−.02417
−.01417
.02333
.02333
−.03917
.02583
j=4
−.02000
.01000
−.01000
−.04000
.02750
−.04250
.02750
.00750
.00500
.02500
.02250
−.01250
r = .994
27.26. a.
Source
Subjects
Dosage
Error
Total
b.
SS
df
1.80012 11
.72615 3
.03220 33
2.55847 47
MS
.163647
.242050
.000976
H0 : all τj equal zero (j = 1, ..., 4), Ha : not all τj equal zero.
F ∗ = .242050/.000976 = 248.0, F (.95; 3, 33) = 2.89.
If F ∗ ≤ 2.89 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
Ȳ.1 = 1.03833, Ȳ.2 = 1.05833, Ȳ.3 = 1.06083, Ȳ.4 = .76917, L̂1 = Ȳ.1 − Ȳ.2 =
−.02000, L̂2 = Ȳ.2 − Ȳ.3 = −.00250, L̂3 = Ȳ.3 − Ȳ.4 = .29166, s{L̂i } = .01275
(i = 1, 2, 3), B = t(.983; 33) = 2.22
27-9
−.02000 ± 2.22(.01275)
−.00250 ± 2.22(.01275)
.29166 ± 2.22(.01275)
d.
−.048 ≤ L1 ≤ .008
−.031 ≤ L2 ≤ .026
.263 ≤ L3 ≤ .320
Yij = µ.. + ρ1 Xij1 + ρ2 Xij2 + ρ3 Xij3 + ρ4 Xij4 + ρ5 Xij5 + ρ6 Xij6
+ρ7 Xij7 + ρ8 Xij8 + ρ9 Xij9 + ρ10 Xij10 + ρ11 Xij11 + γ1 xij + γ2 x2ij + ²ij
Xij1
1 if experimental unit from subject 1
= −1 if experimental unit from subject 12
0 otherwise
Xij2 , ..., Xij11 are defined similarly
−.825
if
−.325 if
xij =
.175 if
.975 if
experimental
experimental
experimental
experimental
unit
unit
unit
unit
received
received
received
received
treatment
treatment
treatment
treatment
1
2
3
4
Ŷ = 1.06647 − .24917X1 − .26917X2 − .23917X3 − .12917X4 + .02333X5 − .09667X6
−.05667X7 +.13333X8 +.18583X9 +.21583X10 +.15833X11 −.11341x−.19192x2
e.
eij :
i
j=1
j=2
1
.02976 −.03389
2
.00976 −.03389
3
.00976 −.02389
4
.04976 −.00389
5 −.02274
.02361
6
.02726 −.05639
7
.00726 −.02639
8
.00726 −.01639
9 −.01524 −.02889
10
.00476 −.06889
11
.04226 −.04139
12 −.04274
.01361
f.
j=3
.02842
.01842
.02842
−.00158
−.02408
.07592
−.00408
.00592
.04342
.04342
−.01908
.04592
j=4
−.02429
.00571
−.01429
−.04429
.02321
−.04679
.02321
.00321
.00071
.02071
.01821
−.01679
H0 : γ2 = 0, Ha : γ2 6= 0. SSE(F ) = .0456, SSE(R) = .2816,
F ∗ = (.2360/1) ÷ (.0456/34) = 175.96, F (.99; 1, 34) = 7.44.
If F ∗ ≤ 7.44 conclude H0 , otherwise Ha . Conclude Ha .
Note: The subscript for subjects here is l instead of the usual i and the subscripts
for factors A, B, and C are i, j, and k, respectively.
27.27.
a.
Yijklm = µ.... + αi + βj + γk + ρl(ik) + (αβ)ij + (αγ)ik + (βγ)jk + (αβγ)ijk + ²m(ijkl)
b.
r = .990
27.28. a.
27-10
Source
SS
df
MS
A (initial lever press rate)
7.99586
2 3.99793
B (dosage level)
25.90210
3 8.63403
C (reinforcement schedule) 59.74172
1 59.74172
AB interactions
.35167
6
.05861
AC interactions
.09465
2
.04733
BC interactions
12.36104
3 4.12035
ABC interactions
.37040
6
.06173
S(AC) (rats, within AC)
1.64179 18
.09121
Error
.36711 150
.00245
Total
108.82634 191
E{M SA}
E{M SB}
E{M SC}
E{M SS(AC)}
E{M SAB}
E{M SAC}
E{M SBC}
E{M SABC}
E{M SE}
b.
=
=
=
=
=
=
=
=
=
P
64 αi2 /2 + 8σρ2 + σ 2
P
48 βj2 /3 + σ 2
P
96 γk2 /1 + 8σρ2 + σ 2
8σρ2 + σ 2
PP
16
(αβ)2ij /6 + σ 2
PP
32
(αγ)2ik /2 + 8σρ2 + σ 2
PP
24
(βγ)2jk /3 + σ 2
PPP
8
(αβγ)2ijk /6 + σ 2
σ2
H0 : all (αβγ)ijk equal zero, Ha : not all (αβγ)ijk equal zero.
F ∗ = .06173/.00245 = 25.196, F (.99; 6, 150) = 2.92.
If F ∗ ≤ 2.92 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
Ȳijk.. :
k
1
i
1
2
3
j=1
j=2
j=3
.81375 .82375 .83625
1.05375 1.06625 1.05625
1.25500 1.25625 1.27125
j=4
.53500
.77375
1.00750
2
1 2.15125 2.33625 1.88750
2 2.59250 2.58375 2.21875
3 3.04750 2.75125 2.38125
.88125
1.01250
1.29250
27.29. a.
FR∗
0
.25
1.00
1.60
7.00
Undefined
b.
P (FR∗ )
12/216
90/216
36/216
36/216
36/216
6/216
F (.90; 2, 4) = 4.32, P (FR∗ ≤ 7.00) = .972, P (FR∗ ≤ 1.60) = .806
27-11
27-12
Chapter 28
BALANCED INCOMPLETE
BLOCK, LATIN SQUARE, AND
RELATED DESIGNS
28.3.
One such design, for which nb = 3, n = 2, and np = 1:
1
1
2
28.4.
2
3
3
For r = 7, rb = 5, a BIBD exists for nb =
7!
= 21.
5!(7 − 5)!
Since nb rb = nr, n = 21(5)/7 = 15.
Since np (r − 1) = n(rb − 1), np = 15(5 − 1)/(7 − 1) = 10.
28.5.
For r = 8, rb = 3, a BIBD exists for nb =
8!
= 56.
3!(8 − 3)!
Since nb rb = nr, n = 56(3)/8 = 21.
Since np (r − 1) = n(rb − 1), np = 21(3 − 1)/(8 − 1) = 6.
28.6.
eij :
i j=1 j=2 j=3
1
−.704
2
.222
3
.556
4 −.481
5
−.926
6
7
8
.222
9 −.111
10
−.222
11
1.444
12
.370
.926 −1.296
j=4
.185
−.111
j=5
j=6
j=7
j=8
.519
−.111
.444
.741
−1.000
−.259
−.519
.481
.296
.037
.630
.815
−.667
.444
.667
.444
−.667
28-1
j=9
.259
−.556
−.222
−.778
−1.074
r = .990
28.7. a.
µ̂.. = 19.36, τ̂ 1 = .33, τ̂ 2 = −2.22, τ̂ 3 = −6.00, τ̂ 4 = −12.89, τ̂ 5 = 6.11, τ̂ 6 = 3.56,
τ̂ 7 = 1.22, τ̂ 8 = −.22, τ̂ 9 = 10.11.
µ̂.1 = 19.69, µ̂.2 = 17.14, µ̂.3 = 13.36, µ̂.4 = 6.47, µ̂.5 = 25.47, µ̂.6 = 22.92,
µ̂.7 = 20.58, µ̂.8 = 19.14, µ̂.9 = 29.47.
b.
H0 : all τj equal zero (j = 1, 2, . . . , 8), Ha : not all τj equal zero. SSE(F ) = 14.519,
SSE(R) = 1097.33, F ∗ = (1082.811/8) ÷ (14.519/16) = 149.2, F (.95; 8, 16) =
2.59. If F ∗ ≤ 2.59 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+
c.
H0 : all ρi equal zero (j = 1, 2, . . . , 11), Ha : not all ρi equal zero. SSE(F ) = 14.519,
SSE(R) = 25.25, F ∗ = (10.731/11) ÷ (14.519/16) = 1.08, F (.95; 11, 16) = 2.46.
If F ∗ ≤ 2.46 conclude H0 , otherwise Ha . Conclude H0 . P -value = .43
d.
µ̂.5 = 25.47, s2 (µ̂.5 ) = s2 (µ̂.. ) + s2 (τˆ5 ) = (.02778 + .29630).907 = .2939, B =
t(.975; 16) = 2.120, 25.47 ± 2.120(.542), 24.32 ≤ µ.5 ≤ 26.62
e.
28-2
95% C.I.
µ.1 − µ.2
µ.1 − µ.3
µ.1 − µ.4
µ.1 − µ.5
µ.1 − µ.6
µ.1 − µ.7
µ.1 − µ.8
µ.1 − µ.9
µ.2 − µ.3
µ.2 − µ.4
µ.2 − µ.5
µ.2 − µ.6
µ.2 − µ.7
µ.2 − µ.8
µ.2 − µ.9
µ.3 − µ.4
µ.3 − µ.5
µ.3 − µ.6
µ.3 − µ.7
µ.3 − µ.8
µ.3 − µ.9
µ.4 − µ.5
µ.4 − µ.6
µ.4 − µ.7
µ.4 − µ.8
µ.4 − µ.9
µ.5 − µ.6
µ.5 − µ.7
µ.5 − µ.8
µ.5 − µ.9
µ.6 − µ.7
µ.6 − µ.8
µ.6 − µ.9
µ.7 − µ.8
µ.7 − µ.9
µ.8 − µ.9
28.8. a.
lower
−.21
3.57
10.46
−8.54
−5.99
−3.66
−2.21
−12.54
1.01
7.90
−11.10
−8.54
−6.21
−4.77
−15.10
4.12
−14.88
−12.32
−9.99
−8.54
−18.88
−21.77
−19.21
−16.88
−15.43
−25.77
−.21
2.12
3.57
−6.77
−.43
1.01
−9.32
−1.32
−11.66
−13.10
center
2.56
6.33
13.22
−5.78
−3.22
−.89
.56
−9.78
3.78
10.67
−8.33
−5.78
−3.44
−2.00
−12.33
6.89
−12.11
−9.56
−7.22
−5.78
−16.11
−19.00
−16.44
−14.11
−12.67
−23.00
2.56
4.89
6.33
−4.00
2.33
3.78
−6.56
1.44
−8.89
−10.33
upper
5.32
9.10
15.99
−3.01
−.46
1.88
3.32
−7.01
6.54
13.43
−5.57
−3.01
−.68
.77
−9.57
9.66
−9.35
−6.79
−4.46
−3.01
−13.35
−16.23
−13.68
−11.35
−9.90
−20.23
5.32
7.66
9.10
−1.23
5.10
6.54
−3.79
4.21
−6.12
−7.57
eij :
i
1
2
3
4
j=1
13.2083
−7.9167
−5.2917
j=2
8.8333
4.7083
−13.5417
j=3
−22.0417
−1.5417
23.5833
r = .995
28-3
j=4
3.2083
6.8333
−10.0417
28.9. a.
µ̂.. = 297.667, τ̂ 1 = −45.375, τ̂ 2 = −41.000, τ̂ 3 = 30.875, τ̂ 4 = 55.500
µ̂.1 = 252.292, µ̂.2 = 256.667, µ̂.3 = 328.542, µ̂.4 = 353.167
b.
H0 : τ1 = τ2 = τ3 = 0, Ha : not all τj equal zero. SSE(F ) = 1750.9, SSE(R) =
22480, F ∗ = (20729.1/3) ÷ (1750.9/5) = 19.73, F (.95; 3, 5) = 5.41. If F ∗ ≤ 5.41
conclude H0 , otherwise Ha . Conclude Ha . P -value = .003
c.
H0 : ρ1 = ρ2 = ρ3 = 0, Ha : not all ρi equal zero. SSE(F ) = 14.519, SSE(R) =
22789, F ∗ = (21038.1/3) ÷ (1750.9/5) = 20.03, F (.95; 3, 5) = 5.41. If F ∗ ≤ 5.41
conclude H0 , otherwise Ha . Conclude Ha . P -value = .003
d.
µ̂.1 = 252.292, s2 (µ̂.1 ) = s2 (µ̂.. ) + s2 (τˆ1 ) = (.08333 + .28125)350.2 = 127.68,
B = t(.975; 5) = 2.571, 252.292 ± 2.571(11.30), 223.240 ≤ µ.1 ≤ 281.344
e.
95% C.I.
µ.1 − µ.2
µ.1 − µ.3
µ.1 − µ.4
µ.2 − µ.3
µ.2 − µ.4
µ.3 − µ.4
28.10.
lower
center
−64.19
−4.375
−136.07 −76.250
−160.69 −100.875
−131.70
−71.87
−156.30
−96.50
−84.44
−24.63
upper
55.44
−16.43
−41.06
−12.06
−36.68
35.19
r = 4, and rb = 3, dfe = 4n − 4 − 4n/3 + 1 = 8n/3 − 3.
Since np = n(3 − 1)/(4 − 1) = 2n/3, σ 2 {D̂j } = 2σ 2 (3)/(4np ) = 9σ 2 /(4n)
s
9σ 2
1
T σ{D̂j } = √ q[.95; 4, 8n/3 − 3]
4n
2
For σ 2 = 2.0 and T σ{D̂j } ≤ 1.5, so we need to iterate to find n so that
n ≥ q 2 [.95; 4, 8n/3 − 3]
We iteratively find n ≥ 15. Since design 2 in Table 28.1 has n = 3, we require
that design 2 be repeated 5 times. Thus, n = 15, and nb = 20.
28.11.
r = 5, and rb = 4, dfe = 5n − 5 − 5n/4 + 1 = 15n/4 − 4.
Since np = n(4 − 1)/(5 − 1) = 3n/4, σ 2 {D̂j } = 2σ 2 (4)/(5np ) = 32σ 2 /(15n)
s
32σ 2
1
T σ{D̂j } = √ q[.90; 5, 15n/4 − 4]
15n
2
2
For σ = 1.5 and T σ{D̂j } ≤ 1.25, so we need to iterate to find n so that
n ≥ 1.024q 2 [.90; 5, 15n/4 − 4]
We iteratively find n ≥ 14. Since design 5 in Table 28.1 has n = 4, we require
that design 2 be repeated 4 times. Thus, n = 16, and nb = 20.
28.14.
eijk :
i
1
2
3
4
j=1
−.1375
−.0125
.1375
.0125
j=2
j=3
.0875 −.0125
−.0125
.1625
−.0875 −.0625
.0125 −.0875
j=4
.0625
−.1375
.0125
.0625
28-4
r = .986
28.15. a.
Ȳ..1 = 1.725, Ȳ..2 = 1.900, Ȳ..3 = 2.175, Ȳ..4 = 2.425
b.
Source
SS
df
MS
Rows (sales volumes) 5.98187 3 1.99396
Columns (locations)
.12188 3 .04062
Treatments (prices)
1.13688 3 .37896
Error
.11875 6 .01979
Total
7.35938 15
H0 : all τk equal zero (k = 1, ..., 4), Ha : not all τk equal zero. F ∗ = .37896/.01979 =
19.149, F (.95; 3, 6) = 4.76. If F ∗ ≤ 4.76 conclude H0 , otherwise Ha . Conclude
Ha . P -value = .002
c.
L̂1 = Ȳ..1 − Ȳ..2 = −.175, L̂2 = Ȳ..1 − Ȳ..3 = −.450, L̂3 = Ȳ..1 − Ȳ..4 = −.700,
L̂4 = Ȳ..2 − Ȳ..3 = −.275, L̂5 = Ȳ..2 − Ȳ..4 = −.525, L̂6 = Ȳ..3 − Ȳ..4 = −.250,
s{L̂i } = .09947 (i = 1, ..., 6), q(.90; 4, 6) = 4.07, T = 2.8779
−.175 ± 2.8779(.09947)
−.450 ± 2.8779(.09947)
−.700 ± 2.8779(.09947)
−.275 ± 2.8779(.09947)
−.525 ± 2.8779(.09947)
−.250 ± 2.8779(.09947)
−.461 ≤ L1
−.736 ≤ L2
−.986 ≤ L3
−.561 ≤ L4
−.811 ≤ L5
−.536 ≤ L6
28.16. a.
Ê1 = 21.1617, Ê2 = 1.2631, Ê3 = 25.9390
28.17.
eijk :
i j=1
1 −.88
2
.32
3
.52
4 −.68
5
.72
r = .993
28.18. a.
j=2
−.68
.12
−.68
1.92
−.68
j=3
.92
−.28
−1.08
.52
−.08
≤ .111
≤ −.164
≤ −.414
≤ .011
≤ −.239
≤ .036
j=4 j=5
.32
.32
.92 −1.08
.12
1.12
−.08 −1.68
−1.28
1.32
Ȳ..1 = 7.0, Ȳ..2 = 7.4, Ȳ..3 = 15.0, Ȳ..4 = 19.0, Ȳ..5 = 13.4
b.
Source
SS
df
MS
Rows (executives)
220.16 4 55.040
Columns (months)
10.96 4
2.740
Treatments (reports) 527.36 4 131.840
Error
19.28 12
1.607
Total
777.76 24
H0 : all τk equal zero (k = 1, ..., 5), Ha : not all τk equal zero. F ∗ = 131.840/1.607 =
82.04, F (.99; 4, 12) = 5.41. If F ∗ ≤ 5.41 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
28-5
c.
L̂1 = Ȳ..1 − Ȳ..2 = −.4, L̂2 = Ȳ..1 − Ȳ..3 = −8.0, L̂3 = Ȳ..1 − Ȳ..4 = −12.0,
L̂4 = Ȳ..1 − Ȳ..5 = −6.4, L̂5 = Ȳ..2 − Ȳ..3 = −7.6, L̂6 = Ȳ..2 − Ȳ..4 = −11.6,
L̂7 = Ȳ..2 − Ȳ..5 = −6.0, L̂8 = Ȳ..3 − Ȳ..4 = −4.0, L̂9 = Ȳ..3 − Ȳ..5 = 1.6, L̂10 =
Ȳ..4 − Ȳ..5 = 5.6, s{L̂i } = .8017 (i = 1, ..., 10), q(.95; 5, 12) = 4.51, T = 3.189
−.4 ± 3.189(.8017)
−8.0 ± 3.189(.8017)
−12.0 ± 3.189(.8017)
−6.4 ± 3.189(.8017)
−7.6 ± 3.189(.8017)
−11.6 ± 3.189(.8017)
−6.0 ± 3.189(.8017)
−4.0 ± 3.189(.8017)
1.6 ± 3.189(.8017)
5.6 ± 3.189(.8017)
−2.96 ≤ L1 ≤ 2.16
−10.56 ≤ L2 ≤ −5.44
−14.56 ≤ L3 ≤ −9.44
−8.96 ≤ L4 ≤ −3.84
−10.16 ≤ L5 ≤ −5.04
−14.16 ≤ L6 ≤ −9.04
−8.56 ≤ L7 ≤ −3.44
−6.56 ≤ L8 ≤ −1.44
−.96 ≤ L9 ≤ 4.16
3.04 ≤ L10 ≤ 8.16
28.19. a.
Ê1 = 6.66, Ê2 = 1.14, Ê3 = 7.65
28.20.
φ = 3.399, 1 − β ∼
= .99
28.21.
φ = 2.202, 1 − β ∼
= .69
28.22.
eijkl :
i
1
2
3
4
j=1
.01625
.00625
−.00875
−.01375
j=2
−.01875
.01875
−.03125
.03125
j=3
.01625
−.05375
.03375
.00375
j=4
−.01375
.02875
.00625
−.02125
r = .980
28.23. a.
Yijkl = µ... + ρi + κj + αk + βl + (αβ)kl + ²(ijkl)
b.
Source
SS
df
Rows (subjects)
.03462 3
Columns (periods) .00592 3
Treatments
.43333 3
X
.22801 1
Y
.19581 1
XY interactions
.00951 1
Error
.00904 6
Total
.48291 15
MS
.01154
.00197
.14444
.22801
.19581
.00951
.00151
H0 : all (αβ)kl equal zero, Ha : not all (αβ)kl equal zero. F ∗ = .00951/.00151 =
6.298, F (.90; 1, 6) = 3.78. If F ∗ ≤ 3.78 conclude H0 , otherwise Ha . Conclude Ha .
P -value = .046
c.
Ȳ..1 = .0050, Ȳ..2 = .1950, Ȳ..3 = .1775, Ȳ..4 = .4650, L̂ = −.0975, s{L̂} = .03886,
t(.95; 6) = 1.943, −.0975 ± 1.943(.03886), −.1730 ≤ L ≤ −.0220
28-6
28.24. a.
Yijk = µ... + ρ1 Xijk1 + ρ2 Xijk2 + ρ3 Xijk3 + κ1 Xijk4 + κ2 Xijk5
+κ3 Xijk6 + τ1 Xijk7 + τ2 Xijk8 + τ3 Xijk9 + ²(ijk)
Xijk1
1 if experimental unit from row blocking class 1
= −1 if experimental unit from row blocking class 4
0 otherwise
Xijk2 and Xijk3 are defined similarly
Xijk4
1 if experimental unit from column blocking class 1
= −1 if experimental unit from column blocking class 4
0 otherwise
Xijk5 and Xijk6 are defined similarly
Xijk7
1 if experimental unit received treatment 1
−1 if experimental unit received treatment 4
=
0 otherwise
Xijk8 and Xijk9 are defined similarly
b.
Full model:
Ŷ = 2.05625 − .70625X1 − .45625X2 + .34375X3 + .14375X4
−.05625X5 − .00625X6 − .33125X7 − .15625X8 + .11875X9
SSE(F ) = .1188
Reduced model:
Ŷ = 2.05625 − .70625X1 − .45625X2 + .34375X3 + .14375X4 − .05625X5 − .00625X6
SSE(R) = 1.2556
H0 : all τk equal zero (k = 1, 2, 3), Ha : not all τk equal zero. F ∗ = (1.1368/3) ÷
(.1188/6) = 19.138, F (.95; 3, 6) = 4.76. If F ∗ ≤ 4.76 conclude H0 , otherwise Ha .
Conclude Ha .
c.
L̂ = τ̂ 3 − (−τ̂ 1 − τ̂ 2 − τ̂ 3 ) = 2τ̂ 3 + τ̂ 1 + τ̂ 2 = −.250, s2 {τ̂ i } = .00371 (i = 1, 2, 3),
s{τ̂ 1 , τ̂ 2 } = s{τ̂ 1 , τ̂ 3 } = s{τ̂ 2 , τ̂ 3 } = −.00124, s{L̂} = .09930, t(.975; 6) = 2.447,
−.250 ± 2.447(.09930), −.493 ≤ L ≤ −.007
d.
(i) Full model:
Ŷ = 2.02917 − .67917X1 − .53750X2 + .37083X3 + .17083X4 − .02917X5
−.08750X6 − .30417X7 − .23750X8 + .14583X9
SSE(F ) = .0483
Reduced model:
Ŷ = 2.05556 − .70556X1 − .45833X2 + .34444X3 + .14444X4 − .05556X5 − .00833X6
SSE(R) = 1.2556
H0 : all τk equal zero (k = 1, 2, 3), Ha : not all τk equal zero. F ∗ = (1.2073/3) ÷
(.0483/5) = 41.66, F (.95; 3, 5) = 5.41. If F ∗ ≤ 5.41 conclude H0 , otherwise Ha .
Conclude Ha .
28-7
(ii) L̂ = τ̂ 1 − τ̂ 2 = −.06667, s2 {τ̂ 1 } = .00191, s2 {τ̂ 2 } = .00272, s{τ̂ 1 , τ̂ 2 } =
−.00091, s{L̂} = .0803, t(.975; 5) = 2.571, −.06667 ± 2.571(.0803), −.273 ≤ L ≤
.140
28.25. a.
Full model:
Ŷ = 12.54286 + 1.91429X1 − 3.54286X2 + 3.25714X3 − 3.28571X4 + 1.11429X5
−.34286X6 − .94286X7 − .74286X8 − 5.54286X9
−5.14286X10 + 3.11329X11 + 6.71429X12
SSE(F ) = 12.6286
Reduced model:
Ŷ = 11.96471 + .44706X1 − 2.96471X2 + 3.83529X3 − 3.55294X4 − .35294X5
+.23529X6 − .36471X7 − .16471X8
SSE(R) = 494.2353
H0 : all τk equal zero (k = 1, ..., 4), Ha : not all τk equal zero. F ∗ = (481.6067/4) ÷
(12.6286/10) = 95.340, F (.99; 4, 10) = 5.99. If F ∗ ≤ 5.99 conclude H0 , otherwise
Ha . Conclude Ha .
b.
28.26.
L̂ = τ̂ 4 − τ̂ 1 = 12.25715, s2 {τ̂ 1 } = .20927, s2 {τ̂ 4 } = .28144, s{τ̂ 1 , τ̂ 4 } = −.06134,
s{L̂} = .7832, t(.995; 10) = 3.169, 12.25715 ± 3.169(.7832), 9.775 ≤ L ≤ 14.739
eijkm :
i
1
2
3
4
m
1
2
1
2
1
2
1
2
j=1
−1.9375
1.0625
−3.6875
2.3125
−4.0625
2.9375
−.3125
3.6875
j=2
−1.5625
3.4375
6.0625
.0625
1.1875
−4.8125
.3125
−4.6875
j=3
.6875
−2.3125
−1.1875
−5.1875
−.6875
4.3125
.1875
4.1875
j=4
1.3125
−.6875
3.8125
−2.1875
4.0625
−2.9375
−2.6875
−.6875
r = .990
28.27. a.
Yijklm = µ... + ρi + κj + αk + βl + (αβ)kl + ²m(ijkl)
b.
Source
SS
df
Rows (ages)
658.09375 3
Columns (education levels)
18.34375 3
Treatments
1, 251.34375 3
Volumes
399.03125 1
Products
850.78125 1
Volume-product interactions
1.53125 1
Error
285.43750 22
Total
2, 213.21875 31
28-8
MS
219.36458
6.11458
417.11458
399.03125
850.78125
1.53125
12.97443
H0 : all (αβ)kl equal zero, Ha : not all (αβ)kl equal zero. F ∗ = 1.53125/12.97443 =
.118, F (.99; 1, 22) = 7.95. If F ∗ ≤ 7.95 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .73
c.
H0 : α1 = α2 = 0, Ha : not both α1 and α2 equal zero. F ∗ = 399.03125/12.97443 =
30.755, F (.99; 1, 22) = 7.95. If F ∗ ≤ 7.95 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : β1 = β2 = 0, Ha : not both β1 and β2 equal zero. F ∗ = 850.78125/12.97443 =
65.574, F (.99; 1, 22) = 7.95. If F ∗ ≤ 7.95 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
d.
Ȳ..1. = 61.750, Ȳ..2. = 69.250, Ȳ..3. = 72.500, Ȳ..4. = 79.125,
µ..1 + µ..3 µ..2 + µ..4
L1 =
−
2
2
µ..1 + µ..2 µ..3 + µ..4
L2 =
−
2
2
L̂1 = −7.0625, L̂2 = −10.3125, s{L̂1 } = s{L̂2 } = 1.2735,
B = t(.9875; 22) = 2.4055
−7.0625 ± 2.4055(1.2735)
−10.3125 ± 2.4055(1.2735)
28.28.
−10.126 ≤ L1 ≤ −3.999
−13.376 ≤ L2 ≤ −7.249
eijkm :
i=1
i=2
i=3
m=1
m=2
m=3
m=1
m=2
m=3
m=1
m=2
m=3
j=1
4.3704
−3.6296
−2.2963
−.9630
1.0370
2.0370
−3.5185
.1481
2.8148
j=2
−2.7407
1.2593
3.5926
−1.1852
−1.1852
1.8148
−.8519
3.8148
−4.5185
j=3
−1.6296
2.3704
−1.2963
2.1481
.1481
−3.8519
4.3704
−3.9630
1.7037
r = .986
28.29. a.
Source
SS
df
MS
Patterns
14.2963 2
7.1481
Order positions
1, 803.6296 2
901.8148
Questionnaires
3, 472.0741 2 1, 736.0370
Subjects (within patterns)
159.5556 6
26.5926
Error
194.9630 14
13.9259
Total
5, 644.5185 26
H0 : all ρi equal zero (i = 1, 2, 3), Ha : not all ρi equal zero. F ∗ = 7.1481/26.5926 =
.269, F (.95; 2, 6) = 5.14. If F ∗ ≤ 5.14 conclude H0 , otherwise Ha . Conclude H0 .
P -value = .77
28-9
H0 : all κj equal zero (j = 1, 2, 3), Ha : not all κj equal zero. F ∗ = 901.8148/13.9259 =
64.758, F (.95; 2, 14) = 3.74. If F ∗ ≤ 3.74 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
H0 : all τk equal zero (k = 1, 2, 3), Ha : not all τk equal zero. F ∗ = 1, 736.0370/13.9259 =
126.66, F (.95; 2, 14) = 3.74. If F ∗ ≤ 3.74 conclude H0 , otherwise Ha . Conclude
Ha . P -value = 0+
b.
Ȳ..1. = 22.3333, Ȳ..2. = 22.4444, Ȳ..3. = 46.4444, L̂1 = Ȳ..1. − Ȳ..2. = −.1111,
L̂2 = Ȳ..1. − Ȳ..3. = −24.1111, L̂3 = Ȳ..2. − Ȳ..3. = −24.0000, s{L̂i } = 1.75916
(i = 1, 2, 3), q(.90; 3, 14) = 3.16, T = 2.234
−.1111 ± 2.234(1.75916)
−24.1111 ± 2.234(1.75916)
−24.0000 ± 2.234(1.75916)
−4.0411 ≤ µ..1 − µ..2 ≤ 3.8189
−28.0411 ≤ µ..1 − µ..3 ≤ −20.1811
−27.9300 ≤ µ..2 − µ..3 ≤ −20.0700
28-10
Chapter 29
EXPLORATORY EXPERIMENTS –
TWO-LEVEL FACTORIAL AND
FRACTIONAL FACTORIAL
DESIGNS
29.1.
Yi = β0 Xi0 + β1 Xil + β2 Xi2 + β3 Xi3 + β4 Xi4 + β12 Xi12 + β13 Xi13
+β14 Xi14 + β23 Xi23 + β24 Xi24 + β34 Xi34 + β123 Xi123
+β124 Xi124 + β134 Xi134 + β234 Xi234 + β1234 Xi1234 + ²i
6, 4, 1
29.2.
Fractional factorial designs can be used.
29.3. a.
Six factors, two levels, 64 trials
b.
29.4. a.
b.
No
Seven factors, two levels, 8 trials; no
Yes, no
29.5.
X0
1
1
1
1
1
1
1
1
X1
−1
1
−1
1
−1
1
−1
1
X2
−1
−1
1
1
−1
−1
1
1
X3
−1
−1
−1
−1
1
1
1
1
X12 X13 X23 X123
1
1
1
−1
−1 −1
1
1
−1
1 −1
1
1 −1 −1
−1
1 −1 −1
1
−1
1 −1
−1
−1 −1
1
−1
1
1
1
1
29-1
0
X X =
29.6. a.
8
0
0
0
0
0
0
0
0
8
0
0
0
0
0
0
0
0
8
0
0
0
0
0
0
0
0
8
0
0
0
0
0
0
0
0
8
0
0
0
0
0
0
0
0
8
0
0
0
0
0
0
0
0
8
0
0
0
0
0
0
0
0
8
= 8I = nT I
σ 2 {b1 } = σ 2 /nT = 52 /64 = .391. Yes, yes
b.
z(.975) = 1.96, nT = [1.96(5)/(.5)]2 = 384.16, 384.16/64 = 6 replicates
29.7. a.
Yi = β0 Xi0 + β1 Xi1 + · · · + β5 Xi5 + β12 Xi12 + · · · + β45 Xi45 + β123 Xi123
+ · · · +β345 Xi345 + β1234 Xi1234 + · · · + β2345 Xi2345 + β12345 Xi12345 + ²i
Coef.
b0
b1
b2
b3
b4
b5
b12
b13
29.8. a.
bq
6.853
1.606
−.099
1.258
−1.151
−1.338
−.033
.455
Coef.
b14
b15
b23
b24
b25
b34
b35
b45
bq
−.239
.611
−.134
−.127
−.045
−.311
.912
−.198
Coef.
b123
b124
b125
b134
b135
b145
b234
b235
bq
.070
.020
−.118
−.378
−.138
−.183
.233
.055
Coef.
b245
b345
b1234
b1235
b1245
b1345
b2345
b12345
bq
.076
−.576
.062
.323
.357
−.122
−.292
.043
Yi = β0 Xi0 + β1 Xi1 + · · · + β5 Xi5 + β12 Xi12 + · · · + β45 Xi45 + ²i
Coef.
b0
b1
b2
b3
b4
b5
b12
b13
bq
6.853
1.606
−.099
1.258
−1.151
−1.338
−.033
.455
P -value
.000
.689
.000
.000
.000
.892
.080
Coef.
b14
b15
b23
b24
b25
b34
b35
b45
bq
P -value
−.239
.340
.611
.023
−.134
.589
−.127
.610
−.045
.855
−.311
.219
.912
.002
−.198
.426
b.
H0 : Normal, Ha : not normal. r = .983. If r ≥ .9656 conclude H0 , otherwise Ha .
Conclude H0 .
c.
H0 : βq = 0, Ha : βq 6= 0. s{bq } = .2432. If P -value ≥ .0034 conclude H0 , otherwise
Ha . Active effects (see part a): β1 , β3 , β4 , β5 , β35
29.9. a.
Yi = β0 + β1 Xi1 + · · · + β4 Xi4 + β12 Xi12 + · · · + β34 Xi34 + β123 Xi123
+ · · · +β234 Xi234 + β1234 Xi1234 + ²i
29-2
Coef.
b0
b1
b2
b3
b4
b12
b13
b14
bq
3.7784
−.3113
−.0062
−.1463
.0837
.0050
.0400
.0025
P -value
.020
.903
.083
.204
.922
.468
.961
Coef.
b23
b24
b34
b123
b124
b134
b234
b1234
bq
P -value
−.0925
.176
.0125
.807
−.2175
.040
−.0087
.865
.0538
.354
−.0363
.505
−.0138
.788
.0050
.922
(Note: P -values based on M SP E; see part d.)
d.
q
H0 : βq = 0, Ha : βq 6= 0. M SP E = .0324, s{bq } = .0324/16 = .0450. If P -value
≥ .05 conclude H0 , otherwise Ha . Active effects (see part a): β1 , β34
29.10. a.
Coef.
b0
b1
b3
b4
b34
bq
3.778
−.3112
−.1462
.0838
−.2175
P -value
.000
.006
.084
.000
b.
H0 : Normal, Ha : not normal. r = .970. If r ≥ .9485 conclude H0 , otherwise Ha .
Conclude H0 .
c.
H0 : βq = 0, Ha : βq 6= 0. s{bq } = .0449. If P -value ≥ .01 conclude H0 , otherwise
Ha . Active effects (see part a): β1 , β3 , β34 .
d.
H0 : No lack of fit, Ha : lack of fit. SSLF = SSE − SSP E = .45248 − .25617 =
.19631. F ∗ = [.19631/4] ÷ (.25617/10) = 1.92, F (.95; 4, 10) = 3.48. If F ∗ ≤ 3.48
conclude H0 , otherwise Ha . Conclude H0 .
e.
Set X1 , X3 , X4 at high levels to minimize failure rate.
29.11. a.
b.
0 = −234, resolution = III
0 = −234, 1 = −1234, 2 = −34, 3 = −24, 4 = −23, 12 = −134,
13 = −124, 14 = −123
29.12. a.
X1
−1
1
−1
1
−1
1
−1
1
X2
−1
−1
1
1
−1
−1
1
1
X3
−1
−1
−1
−1
1
1
1
1
X4
−1
1
1
−1
1
−1
−1
1
Resolution = IV
29-3
b.
29.13.
For example, dropping X1 and arranging in standard order:
X2 X3 X4
−1 −1 −1
1 −1 −1
−1
1 −1
1
1 −1
−1 −1
1
1 −1
1
−1
1
1
1
1
1
No
29.14.
X1
1
−1
−1
1
1
−1
−1
1
Yes;
X2 X3 X4
−1 −1 −1
1 −1 −1
−1
1 −1
1
1 −1
−1 −1
1
1 −1
1
−1
1
1
1
1
1
use 0 = 1234 for resolution IV.
29.15.
Defining relation: 0 = 123 = 245 = 1345
Confounding scheme:
0 = 123 = 245 = 1345
1 = 23
= 1245 = 345
2 = 13
= 45
= 12345
3 = 12
= 2345 = 145
4 = 1234 = 25
= 135
5 = 1235 = 24
= 134
14 = 234 = 125 = 35
15 = 235 = 124 = 34
Resolution = III, no
29.16.
Defining relation: 0 = −145 = −234 = 1235
Confounding scheme:
0 = −145 = −234 = 1235
1 = −45
= −1234 = 235
2 = −1245 = −34
= 135
3 = −1345 = −24
= 125
4 = −15
= −23
= 12345
5 = −14
= −2345 = 123
12 = −245 = −134 = 35
13 = −345 = −124 = 25
29-4
No
29.17.
Defining relation: 0 = 124 = 135 = 2345 = 236 = 1346 = 1256 = 456
X1
−1
1
−1
1
−1
1
−1
1
X2
−1
−1
1
1
−1
−1
1
1
X3
−1
−1
−1
−1
1
1
1
1
X4
1
−1
−1
1
1
−1
−1
1
X5
1
−1
1
−1
−1
1
−1
1
X6
1
1
−1
−1
−1
−1
1
1
Resolution = III
29.18. a.
Defining relation: 0 = 1235 = 2346 = 1247 = 1456 = 3457 = 1367 = 2567,
resolution = IV, no
b.
Omitting four-factor and higher-order interactions:
1
2
3
4
5
6
7
12
13
14
15
16
17
26
c.
=
=
=
=
=
=
=
=
=
=
=
=
=
=
235
135
125
127
123
137
124
35
25
27
23
37
24
34
=
=
=
=
=
=
=
=
=
=
=
=
=
=
247
147
167
156
146
145
136
47
67
56
46
45
36
57
=
=
=
=
=
=
=
367
346
246
236
267
234
256
=
=
=
=
=
=
=
456
567
457
357
347
257
345
Yi = β0 Xi0 + β1 Xi1 + · · · + β7 Xi7 + β12 Xi12 + β13 Xi13 + β14 Xi14
+β15 Xi15 + β16 Xi16 + β17 Xi17 + β26 Xi26 + ²i
Coef.
b0
b1
b2
b3
b4
e.
bq
8.028
.127
.003
.021
−2.077
Coef.
b5
b6
b7
b12
b13
bq
.724
−.467
−.766
.354
−.066
Coef.
b14
b15
b16
b17
b26
bq
−.316
.318
.117
.021
−.182
H0 : β12 = · · · = β17 = β26 = 0, Ha : not all βq = 0. F ∗ = (6.046/7) ÷ (.1958/1) =
4.41, F (.99; 7, 1) = 5, 928. If F ∗ ≤ 5, 928 conclude H0 , otherwise Ha . Conclude
H0 .
29.19. a.
29-5
Coef.
b0
b1
b2
b3
bq
8.028
.127
.003
.021
P -value
.581
.989
.928
Coef.
b4
b5
b6
b7
bq
−2.077
.724
−.467
−.766
P -value
.000
.011
.067
.008
b.
H0 : Case i not an outlier, Ha : case i an outlier (i = 3, 14). t3 = 2.70, t14 = −4.09,
t(.99844; 7) = 4.41. If | ti |≤ 4.41 conclude H0 , otherwise Ha . Conclude H0 for
both cases.
c.
H0 : Normal, Ha : not normal. r = .938. If r ≥ .929 conclude H0 , otherwise Ha .
Conclude H0 .
d.
H0 : βq = 0, Ha : βq 6= 0. s{bq } = .2208. If P -value ≥ .02 conclude H0 , otherwise
Ha . Active effects (see part a): β4 , β5 , β7
e.
Set X4 = −1, X5 = 1, X7 = −1 to maximize extraction.
29.20. a.
Yi = β0 Xi0 + β1 Xi1 + · · · + β9 Xi9 + ²i
Coef.
b0
b1
b2
b3
b4
d.
29.21. a.
bq
P -value
70.11
13.52
.060
−.99
.870
1.32
.829
2.36
.701
Coef.
b5
b6
b7
b8
b9
bq
13.49
.12
−21.58
−4.07
3.07
P -value
.060
.984
.010
.512
.618
H0 : βq = 0, Ha : βq 6= 0. s{bq } = 5.841. If P -value ≥ .10 conclude H0 , otherwise
Ha . Active effects (see part a): β1 , β5 , β7 .
b0 = 70.11, b1 = 13.52, b5 = 13.49, b7 = −21.58
b.
H0 : Normal, Ha : not normal. r = .951. If r ≥ .941 conclude H0 , otherwise Ha .
Conclude H0 .
c.
H0 : No lack of fit, Ha : lack of fit. SSLF = SSE−SSP E = 3, 824−1, 068 = 2, 756,
F ∗ = (2, 756/4) ÷ (1, 068/8) = 5.16, F (.95; 4, 8) = 3.84. If F ∗ ≤ 3.84 conclude
H0 , otherwise Ha . Conclude Ha .
29.22. a.
Yi = β0 Xi0 + β1 Xi1 + β5 Xi5 + β7 Xi7 + β15 Xi15 + β17 Xi17
+β57 Xi57 + β157 Xi157 + ²i
Coef.
b0
b1
b5
b7
bq
70.11
13.52
13.49
−21.58
P -value
.000
.000
.000
Coef.
b15
b17
b57
b157
bq
P -value
11.68
.004
−1.32
.660
5.83
.078
.12
.968
H0 : βq = 0, Ha : βq =
6 0. s{bq } = 2.889. If P -value ≥ .01 conclude H0 , otherwise
Ha . Active effects: β1 , β5 , β7 , β15
29.23. a.
Defining relation: 0 = 134
Confounding scheme:
29-6
0
1
2
3
=
=
=
=
134
34
1234
14
4
12
23
24
= 13
= 234
= 124
= 123
Yes. Defining relation 0 = 1234 would yield a resolution IV design.
b.
Yi = β0 Xi0 + β1 Xi1 + β2 Xi2 + β3 Xi3 + β4 Xi4 + β12 Xi12
+β23 Xi23 + β24 Xi24 + ²i
Coef.
b0
b1
b2
b3
bq
747.50
−207.25
−17.00
108.00
Coef.
b0
b1
b2
b3
b4
bq
747.50
−207.25
−17.00
108.00
88.25
Coef.
b4
b12
b23
b24
bq
88.25
−24.75
−29.00
−18.75
29.24. a.
P -value
.003
.538
.022
.037
H0 : βq = 0, Ha : βq =
6 0. s{bq } = 24.53. If P -value ≥ .05 conclude H0 , otherwise
Ha . Active effects: β1 , β3 , β4
b.
29.25.
Set X1 = −1, X3 = 1, X4 = 1 to maximize defect-free moldings.
Confounding scheme for design:
0
1
2
3
4
5
23
25
=
=
=
=
=
=
=
=
124 = 135 = 2345
24
= 35
= 12345
14
= 1235 = 345
1234 = 15
= 245
12
= 1345 = 235
1245 = 13
= 234
134 = 125 = 45
145 = 123 = 34
Design:
Block
1
1
1
1
2
2
2
2
X1
−1
1
−1
1
−1
1
−1
1
X2
1
1
−1
−1
−1
−1
1
1
X3
−1
−1
1
1
−1
−1
1
1
X4
−1
1
1
−1
1
−1
−1
1
X5
1
−1
−1
1
1
−1
−1
1
29-7
= Block effect
29.26. b.
The seven block effects are confounded with the following interaction terms: β135 ,
β146 , β236 , β245 , β1234 , β1256 , β3456
No, no
c.
Yi = β0 Xi0 + β1 Xi1 + · · · + β6 Xi6 + β12 Xi12 + · · · + β56 Xi56 + β123 Xi123
+ · · · +β456 Xi456 + β1235 Xi1235 + · · · + β2456 Xi2456 + β12345 Xi12345
+ · · · +β23456 Xi23456 + β123456 Xi123456 + α1 Zi1 + · · · + α7 Zi7 + ²i
where α1 , ..., α7 are the block effects
Coef.
b0
b1
b2
b3
b4
b5
b6
b12
b13
b14
b15
b16
b23
b24
b25
b26
bq
63.922
2.297
5.797
2.172
2.359
2.828
2.922
.547
−.266
−.203
−.797
−.141
−.641
−1.141
.891
.047
Coef.
b0
b1
b2
b3
b4
b5
b6
b12
b13
b14
b15
b16
b23
b24
b25
bq
63.922
2.297
5.797
2.172
2.359
2.828
2.922
.547
−.266
−.203
−.797
−.141
−.641
−1.141
.891
Coef.
b34
b35
b36
b45
b46
b56
b123
b124
b125
b126
b134
b136
b145
b156
b234
b235
bq
.297
.266
.984
−.422
−.141
.516
.422
.172
1.391
.984
.297
−.641
−.109
−.547
.234
.266
Coef.
b246
b256
b345
b346
b356
b456
b1235
b1236
b1245
b1246
b1345
b1346
b1356
b1456
b2345
b2346
bq
−.391
.078
−.672
.734
−.734
−.234
.578
.922
.453
.109
−.797
.547
−1.109
−.109
.328
−.578
Coef.
b2356
b2456
b12345
b12346
b12356
b12456
b13456
b23456
b123456
Block 1
Block 2
Block 3
Block 4
Block 5
Block 6
Block 7
bq
.766
.203
−.297
−.391
−.734
−.422
−.109
.203
.016
−4.172
−.422
1.203
6.703
−.797
−1.047
−9.547
29.27. a.
b.
P -value
.000
.000
.001
.000
.000
.000
.346
.645
.725
.172
.807
.270
.054
.128
Coef.
b26
b34
b35
b36
b45
b46
b56
Block
Block
Block
Block
Block
Block
Block
1
2
3
4
5
6
7
bq
.047
.297
.266
.984
−.422
−.141
.516
−4.172
−.422
1.203
6.703
−.797
−1.047
−9.547
P -value
.935
.607
.645
.094
.466
.807
.373
.009
.782
.432
.000
.602
.494
.000
H0 : Normal, Ha : not normal. r = .989. If r ≥ .9812 conclude H0 , otherwise Ha .
Conclude H0 .
29-8
c.
H0 : βq = 0, Ha : βq 6= 0. s{α̂i } = 1.513 for block effects, s{bq } = .5719 for factor
effects. If P -value ≥ .01 conclude H0 , otherwise Ha . Active effects (see part a):
Block effects 1, 4, 7, all main effects
29.28. a.
See Problem 29.27a for estimated factor and block effects. (These do not change
with subset model.)
b.
Maximum team effectiveness is accomplished by setting each factor at its high
level.
c.
Ŷh = 82.297, s{pred} = 4.857, t(.975; 50) = 2.009, 82.297 ± 2.009(4.857), 72.54 ≤
Yh(new) ≤ 92.05
29.29. a.
b.
Defining relation: 0 = 12345, resolution = V
Yi = β0 Xi0 + β1 Xi1 + · · · + β5 Xi5 + β12 Xi12 + · · · + β35 Xi35 + α1 Zi1 + ²i
where α1 is the block effect
Coef.
Coef.
bq
b0
113.18
b14
b1
26.69
b15
b2
−10.94
b23
b3
5.69
b24
b4
4.44
b25
b5
14.69
b34
b12
21.94
b35
b13
.56
Block effect
d.
bq
−1.44
−2.94
1.44
5.19
2.94
−3.44
−.94
2.27
H0 : α1 = 0, Ha : α1 6= 0. s{α̂1 } = 3.673, t∗ = 2.27/3.673 = .62, (.975; 6) = 2.447.
If | t∗ |≤ 2.447 conclude H0 , otherwise Ha . Conclude H0 .
e.
Coef.
b0
b1
b2
b3
b4
b5
b12
b13
bq
113.18
26.69
−10.94
5.69
4.44
14.69
21.94
.56
P -value
.000
.011
.094
.169
.003
.001
.847
Coef.
b14
b15
b23
b24
b25
b34
b35
bq
P -value
−1.44
.625
−2.94
.336
1.44
.625
5.19
.119
2.94
.336
−3.44
.268
−.94
.748
(Note: P -values based on M SP E; see part f.)
H0 : No lack of fit, Ha : lack of fit. SSLF = SSE − SSP E = 1, 894.4 − 609.5 =
1, 284.9, F ∗ = (1, 284.9/2)÷(609.5/5) = 5.270, F (.95; 2, 5) = 5.786. If F ∗ ≤ 5.786
conclude H0 , otherwise Ha . Conclude H0 .
f.
q
H0 : βq = 0, Ha : βq 6= 0. M SP E = 121.90, s{bq } = 121.90/16 = 2.760. If
P -value ≥ .025 conclude H0 , otherwise Ha . Active effects (see part a): β1 , β2 , β5 ,
β12
29.30. a.
29-9
Coef.
b0
b1
b2
bq
113.18
26.69
−10.94
Coef.
b5
b12
bq
14.69
21.94
b.
H0 : Normal, Ha : not normal. r = .961. If r ≥ .954 conclude H0 , otherwise Ha .
Conclude H0
c.
Set factors 1, 2, 5 at their high levels to maximize whippability.
d.
Ŷh = 165.56, s{Ŷh } = 8.03, t(.975; 17) = 2.110, 165.56 ± 2.110(8.03), 148.62 ≤
E{Yh } ≤ 182.50
29.31. a.
i
1
2
3
s2i 1.244 1.299 1.103
loge s2i
.218 .261 .098
i
s2i
loge s2i
4
5
6
7
8
.992 1.966 1.916 1.589 1.576
−.008 .676 .650 .463 .455
9
10
11
12
13
14
2.201 1.818 1.901 1.547 1.681 1.020
.789 .598 .642 .437 .520 .020
15
16
1.033 1.151
.033 .141
b.
ds2 = .3746−.0553X −.0919X −.0048X +.0229X +.0289X +.0021X −
log
i1
i2
i3
i4
i12
i13
e i
.0432Xi14 −.0048Xi23 +.0077Xi24 −.2142Xi34 +.0493Xi123 +.0453Xi124 −.0016Xi134 −
.0024Xi234 + .0284Xi1234
X34 appears to be active.
c.
v̂i = 1.17395 (for i = 1, 2, 3, 4, 13, 14, 15, 16)
v̂i = 1.80173 (for i = 5, . . . , 12)
d.
Ŷi = 3.7082 − .3754Xi1
e.
From location model: X1 = +1; and from location model: (X3 , X4 ) = (−1, −1) or
(+1, +1)
From dispersion model: sˆ2 = exp(.3746 − .2142) = 1.17395,
f.
and a 95% P.I. is (exp(.0453), exp(.2755)), or (1.0463, 1.3172).
g.
Md
SE = 1.17395 + 3.3332 = 12.284
29.32. a.
i
1
s2i
.0164
loge s2i −4.109
2
.0173
−4.058
3
4
.0804
.1100
−2.521 −2.207
5
6
.0010
.0079
−6.949 −4.838
7
.0953
−2.351
8
.1134
−2.176
b.
ds2 = −3.651 + .331X + 1.337X − .427X − .275X − .209X + .240X +
log
i1
i2
i3
i4
i12
i13
e i
.477Xi14 .
X2 appears to be active.
c.
v̂i = .00682 (for i = 1, 2, 5, 6)
v̂i = .0989 (for i = 3, 4, 7, 8)
d.
Ŷi = 7.5800 + .0772Xi1
29-10
e.
f.
From the location model: X1 = +1; from the dispersion model: X2 = −1
From dispersion model: sˆ2 = exp(−3.651 + 1.337(−1)) = .006819,
and a 95% P.I. is (exp(−6.169), exp(−3.808)), or (.0021, .0222).
g.
Md
SE = .00682 + (8 − 7.657)2 = .124
From (2.51), SSR(Xq ) = b2q
29.33.
SSR(Xq ) = b2q
P
P
Xiq2 = b2q
(Xiq − X̄q )2 . For coding in (29.2a), X̄q = 0. Then:
n
T
P
(±1)2 = nT b2q
i=1
29.34. a.
E{β̂ 1 }
b.
Let:
= E{(X01 X1 )−1 X01 Y}
= (X01 X1 )−1 X01 E{Y}
= (X01 X1 )−1 X01 (X1 β 1 + X2 β 2 )
= β 1 + (X01 X1 )−1 X01 X2 β 2 = β 1 + Aβ 2
β1 =
Then:
X1 =
and:
A=
β0
β1
β2
β3
β12
β 2 = β13
β23
1
1 −1 −1
1 −1
1 −1
1 −1 −1
1
1
1
1
1
0
0
0
1
0
0
1
0
0
1
0
0
X2 =
The results follow from E{b1 } = β 1 + Aβ 2 .
29-11
−1 −1
1
−1
1 −1
1 −1 −1
1
1
1
29-12
Chapter 30
RESPONSE SURFACE
METHODOLOGY
Second block:
30.2.
X1 X2 X3 X4
2
0
0
0
−2
0
0
0
0
2
0
0
0 −2
0
0
0
0
2
0
0
0 −2
0
0
0
0
2
0
0
0 −2
Any number of center points may be added to the second block.
30.7. a.
21
b.
5, 5, 10
c.
21, 27
30.8.
30-1
X1
−1
1
1
−1
1
−1
−1
1
1
−1
−1
1
−1
1
1
−1
X2
−1
1
−1
1
−1
1
−1
1
−1
1
−1
1
−1
1
−1
1
X3
−1
−1
1
1
−1
−1
1
1
−1
−1
1
1
−1
−1
1
1
X4
−1
−1
−1
−1
1
1
1
1
−1
−1
−1
−1
1
1
1
1
X1 X2 X3
2
0
0
−2
0
0
0
2
0
0 −2
0
0
0
2
0
0 −2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
X5
−1
−1
−1
−1
−1
−1
−1
−1
1
1
1
1
1
1
1
1
30.9.
No, base design is resolution III.
30.10.
α = [29−3 (1)/(1)]1/4 = 2.828
X4
0
0
0
0
0
0
2
−2
0
0
0
0
0
X5
0
0
0
0
0
0
0
0
2
−2
0
0
0
30.11. b.
Coef.
b0
b1
b2
b3
b12
d.
Coef.
b13
b23
b11
b22
b33
bq
P -value
1.868
.190
.007
.195
.006
−.120
.039
.162
.020
bq
−.038
−.062
.228
−.047
.028
P -value
.471
.251
.044
.602
.757
H0 : βq = 0, Ha : βq 6= 0. s{bq } = .0431 (for linear effects), s{bq } = .0481
(for interaction effects), s{bq } = .0849 (for quadratic effects). If P -value ≥ .05
conclude H0 , otherwise Ha . Active effects (see part b): β1 , β2 , β3 , β12 , β11
30.12. a.
Coef.
b0
b1
b2
b.
bq
1.860
.190
.195
Coef.
b3
b12
b11
bq
−.120
.162
.220
H0 : Normal, Ha : not normal. r = .947. If r ≥ .938 conclude H0 , otherwise Ha .
Conclude H0 .
30.13. a.
Coef.
b0
b1
b2
bq
189.750
28.247
−.772
Coef.
b12
b11
b22
bq
13.750
−18.128
−6.875
30-2
c.
H0 : No lack of fit, Ha : lack of fit. SSLF = SSE − SSP E = 978.9 − 230.75 =
748.15, F ∗ = (748.15/3) ÷ (230.75/3) = 3.24, F (.99; 3, 3) = 29.5. If F ∗ ≤ 29.5
conclude H0 , otherwise Ha . Conclude H0 .
e. (1.22, 1.16)
f. Ŷh = 206.54, s{Ŷh } = 13.70, t(.975; 6) = 2.447, 206.54 ± 2.447(13.70), 173.0 ≤
E{Yh } ≤ 240.1.
30.14. a.
Corner Points:
X1
X2
−.707 −.707
.707 −.707
−.707
.707
.707
.707
Design Matrix:
X1
X2
−.707
−.707
.707
−.707
−.707
.707
.707
.707
−1
0
1
0
0
−1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
b.
0
−1
(X X) =
.125
0
0 −.125 −.125 0
0
.250 0
0
0
0
0
0 .250
0
0
0
−.125 0
0
.5
0
0
−.125 0
0
0
.5
0
0
0
0
0
0
1
30.15. a.
X1
−1
1
−1
1
−1.414
1.414
0
0
X2
−1
−1
1
1
0
0
−1.414
1.414
X1
0
0
0
0
0
n0 = 5
30-3
X2
0
0
0
0
0
b.
Variance function:
.20 − .075X12 − .075X22 + .14375X14 + .14375X24 + .2875X12 X22
30.16. a.
"
b∗ =
−2.077
.724
#
s = 2.200
b.
t
X1
X2
1.5 −1.416 .494
2.5 −2.361 .823
3.5 −3.304 1.152
30.17. a.
13.519
∗
b = 13.494
−21.581
s = 28.820
b.
t
X1
−1 −.469
−2 −.938
−3 −1.407
X2
X3
−.468 .749
−.936 1.498
−1.404 2.246
30.18. a.
Design
1
2
3
Variance Function
.6788 − .5116X + .1710X 2 − .02264X 3 + .001029X 4
.5266 − .4048X + .1475X 2 − .02012X 3 + .000914X 4
.6615 − .4504X + .1393X 2 − .01788X 3 + .0008129X 4
Design
1
2
3
V̄
.1993
.2037
.1869
b.
Design 3 preferred
c.
Comparison
EV
Design 1 relative to design 2 1.02
Design 1 relative to design 3 .94
Design 2 relative to design 3 .92
1/.94 = 1.06 times
d.
Design
1
2
3
| (X0 X)−1 |
6.35057 x 10−7
4.70419 x 10−7
5.57533 x 10−7
30-4
Design 2 preferred
e.
Comparison
ED
Design 1 relative to design 2 .90
Design 1 relative to design 3 .96
Design 2 relative to design 3 1.06
1/.90 = 1.11 times
30.19. a. Design 2 is D-optimal
b. Design 3 is V -optimal
30.20. a.
c.
d.
e.
f.
Irregular
Design
V̄
1
.5235
2
.8962
Design 1 preferred
EV = 1.712, 1/1.712 = .584 times
Design | (X0 X)−1 |
1
.393
2
1.567
Design 1 preferred
ED = .794, 1/.794 = 1.26 times
D-optimal design: | (X0 X)−1 |= .2998
X1
X2 Number of Replicates
−1
−1
1
−.5 −1
1
−.5
0
1
.25
0
1
0
1
2
1
1
1
.5
1
2
No
b. V -optimal design: V̄ = .4765
X1
X2 Number of Replicates
−1
−1
1
−.5
−1
1
−.5
0
1
.25
0
1
0
1
1
.5
1
1
1
1
1
0
.25
2
30.21. a.
30-5
No
30-6
Appendix D: RULES FOR
DEVELOPING ANOVA MODELS
AND TABLES FOR BALANCED
DESIGNS
D.1.
αi
βj
(αβ)ij
²k(ij)
D.2.
i
R
a
1
a
1
1
j
R
b
b
1
1
1
k
R
n
n
n
n
1
Variance
σα2
σβ2
2
σαβ
σ2
E{MSA} E{MSB}
i
j
bn
0
0
an
n
n
1
1
E{MSAB} E{MSE}
ij
(ij)k
0
0
0
0
n
0
1
1
a.
Model
Term
Coefficient
Symbolic
Product
Term to
be Squared
Degrees of
Freedom
αi
βj
γk
(αβ)ij
(αγ)ik
(βγ)jk
(αβγ)ijk
bcn
acn
abn
cn
bn
an
n
i−1
j−1
k−1
ij − i − j + 1
ik − i − k + 1
jk − j − k + 1
ijk − ij − ik
−jk + i + j + k − 1
ijkm − ijk
Ȳi··· − Ȳ····
Ȳ·j·· − Ȳ····
Ȳ··k· − Ȳ····
Ȳij·· − Ȳi··· − Ȳ·j·· + Ȳ····
Ȳi·k· − Ȳi··· − Ȳ··k· + Ȳ····
Ȳ·jk· − Ȳ·j·· − Ȳ··k· + Ȳ····
Ȳijk· − Ȳij·· − Ȳi·k· − Ȳ·jk·
+Ȳi··· + Ȳ·j·· + Ȳ··k· − Ȳ····
Yijkm − Ȳijk
a−1
b−1
c−1
(a − 1)(b − 1)
(a − 1)(c − 1)
(b − 1)(c − 1)
(a − 1)(b − 1)(c − 1)
Yijkm − Ȳ····
abcn − 1
²m(ijk)
1
Total
D.1
abc(n − 1)
b.
αi
βj
γk
(αβ)ij
(αγ)ik
(βγ)jk
(αβγ)ijk
²m(ijk)
D.3.
i
j
k
m
R
a
1
a
a
1
1
a
1
1
R
b
b
1
b
1
b
1
1
1
R
c
c
c
1
c
1
1
1
1
R
n
n
n
n
n
n
n
n
1
Variance
σα2
σβ2
σγ2
2
σαβ
2
σαγ
2
σβγ
2
σαβγ
σ2
A
i
bcn
0
0
nc
nb
0
n
1
B
j
0
nac
0
nc
0
na
n
1
Expected Mean Square of −−
C
AB AC BC ABC
k
ij
ik
jk
ijk
0
0
0
0
0
0
0
0
0
0
abn
0
0
0
0
0
nc
0
0
0
nb
0
nb
0
0
na
0
0
na
0
n
n
n
n
n
1
1
1
1
1
E
(ijk)m
0
0
0
0
0
0
0
1
a. See Problem D.2a.
b.
αi
βj
γk
(αβ)ij
(αγ)ik
(βγ)jk
(αβγ)ijk
²m(ijk)
D.4.
i
j
k
m
F
a
0
a
a
1
1
a
1
1
R
b
b
1
b
1
b
1
1
1
R
c
c
c
1
c
1
1
1
1
R
n
n
n
n
n
n
n
n
1
Vari- A
ance
i
2
σα
bcn
σβ2
0
2
σγ
0
2
σαβ
nc
2
σαγ
nb
2
σβγ
0
2
σαβγ
n
σ2
1
B
j
0
nac
0
0
0
na
0
1
Expected Mean Square of −−
C AB AC BC ABC
k
ij
ik
jk
ijk
0
0
0
0
0
0
0
0
0
0
abn
0
0
0
0
0
nc
0
0
0
0
0
nb
0
0
na
0
0
na
0
0
n
n
0
n
1
1
1
1
1
a.
Model Symbolic
Term Product
βj
αi(j)
²k(ij)
j−1
ij − j
ijk − ij
Sum of Squares
P
an (Ȳ·j· − Ȳ··· )2
PP
n
(Ȳ − Ȳ·j· )2
P P P ij·
(Yijk − Ȳij· )2
PPP
Total
(Yijk − Ȳ··· )2
Degrees of
Freedom
b−1
b(a − 1)
ab(n − 1)
abn − 1
b.
βj
αi(j)
²k(ij)
j
F
b
0
1
1
i
R
a
a
1
1
k
R
n
n
n
1
Vari- E{MSB} E{MSA(B)}
ance
j
i(j)
2
an
0
σβ
n
n
σα2
σ2
1
1
D.2
E{MSE}
(ij)k
0
0
1
E
(ijk)m
0
0
0
0
0
0
0
1
P
E{MSB} =
an
βj2
b−1
2
+ nσα2 + σ 2
E{MSE} = σ
E{MSA(B)} = nσα2 + σ 2
c. MSA(B)
D.5.
a.
Model Coef- Symbolic
Term ficient Product
ρi
τj
Error
r
nb
i−1
j−1
Term to be
Squared
Degrees of
Freedom
Ȳi· − Ȳ··
Ȳ·j − Ȳ··
Remainder =
Yij − Ȳi· − Ȳ·j + Ȳ··
n−1
r−1
Remainder =
(r − 1)(nb − 1)
Yij − Ȳ··
rnb − 1
Total
b.
ρi
τj
²(ij)
D.6.
i
F
nb
0
nb
1
j
F
r
r
0
1
Vari- E{MSBL}
ance
i
2
σρ
r
στ2
0
σ2
1
E{MSTR} E{MSE}
j
(ij)
0
0
nb
0
1
1
a. See Problem D.5a.
b.
ρi
τj
²(ij)
D.7.
i
F
nb
0
nb
1
j
R
r
r
1
1
Vari- E{MSBL} E{MSTR}
ance
i
j
2
σρ
r
0
στ2
0
nb
σ2
1
1
E{MSE}
(ij)
0
0
1
a. See Problem D.5a.
b.
ρi
τj
²(ij)
i
R
nb
1
nb
1
j
F
r
r
0
1
Vari- E{MSBL}
ance
i
2
r
σρ
0
στ2
σ2
1
E{MSTR} E{MSE}
j
(ij)
0
0
nb
0
1
1
D.3
D.8.
a.
Model Coef- Symbolic
Term ficient Product
ρi
τj
²(ij)
rm
nb m
m
i−1
j−1
ηk(ij)
1
ijk − ij
Term to be Squared
Degrees of
Freedom
Ȳi·· − Ȳ···
Ȳ·j· − Ȳ···
Remainder =
Ȳij· − Ȳi·· − Ȳ·j· + Ȳ···
Yijk − Ȳij·
nb − 1
r−1
Remainder =
(nb − 1)(r − 1)
nb r(m − 1)
Yijk − Ȳ···
nb rm − 1
Total
P
SSBL = rm (Ȳi·· − Ȳ··· )2
PP
SSEE = m
(Ȳ − Ȳi·· − Ȳ·j· + Ȳ··· )2
P ij·
SSTR = nb m (Ȳ·j· − Ȳ··· )2
PPP
SSOE =
(Yijk − Ȳij· )2
b.
ρi
τj
²(ij)
ηk(ij)
i
R
nb
1
nb
1
1
j
F
r
r
0
1
1
k
R
m
m
m
m
1
Variance
σρ2
στ2
σ2
ση2
E{MSBL}
i
rm
0
m
1
E{M ST R}
j
0
nb m
m
1
2
2
2
E{MSBL} = rmσP
ρ + mσ + ση
E{MSTR} =
D.9.
nb m
r−1
τj2
E{MSEE}
(ij)
0
0
m
1
E{MSOE}
(ij)k
0
0
0
1
E{MSEE} = mσ 2 + ση2
+ mσ 2 + ση2
E{MSOE} = ση2
a.
Model
Term
Symbolic
Product
Sum of Squares
Degrees of
Freedom
αi
βj(k)
γk
(αγ)ik
(αβ)ij(k)
²m(ijk)
i−1
jk − k
k−1
ik − i − k + 1
ijk − ik − jk + k
ijkm − ijk
bcn (Ȳi··· − Ȳ···· )2
PP
an
(Ȳ·jk· − Ȳ··k· )2
P
abn (Ȳ··k· − Ȳ···· )2
PP
bn
(Ȳ − Ȳi··· − Ȳ··k· + Ȳ···· )2
P P P i·k·
n
(Ȳ − Ȳi·k· − Ȳ·jk· + Ȳ··k· )2
P P P P ijk·
(Yijkm − Ȳijk· )2
a−1
c(b − 1)
c−1
(a − 1)(c − 1)
(a − 1)(b − 1)c
abc(n − 1)
Total
P
PPPP
(Yijkm − Ȳ···· )2
D.4
abcn − 1
b.
i
αi
βj(k)
γk
(αγ)ik
(αβ)ij(k)
²m(ijk)
j
F
a
0
a
a
0
0
1
k
R
b
b
1
b
b
1
1
bcn
P
R
c
c
1
1
1
1
1
m
R
n
n
n
n
n
n
1
Expected Mean Square of −−
Vari- A B(C) C AC AB(C)
E
ance
i
j(k)
k
ik
ij(k) m(ijk)
σα2
bcn
0
0
0
0
0
σβ2
0
an
an
0
0
0
2
σγ
0
0
abn
0
0
0
2
σαγ
bn
0
0
bn
0
0
2
σαβ
n
0
0
n
n
0
2
σ
1
1
1
1
1
1
α2
2
2
E{MSA} = a−1 i + bnσαγ
+ nσαβ
+ σ2
E{MSB(C)} = anσβ2 + σ 2
2
E{MSC} = anσβ2 + abnσγ2 + σ
E{MSAB(C)} = nσαβ
+ σ2
2
2
E{MSAC} = bnσαγ
+ nσαβ
+ σ2
E{MSE} = σ 2
D.10. eijk :
k
k
k
k
k
k
k
k
k
i=1
i=2
i=3
=1
=2
=3
=1
=2
=3
=1
=2
=3
j=1
−2.3333
1.6667
.6667
−.3333
1.6667
−1.3333
−1.6667
1.3333
.3333
j=2
j=3
.3333 −1.6667
−1.6667
.3333
1.3333
1.3333
2.3333 −1.0000
−.6667
.0000
−1.6667
1.0000
−1.6667 −1.3333
1.3333
.6667
.3333
.6667
r = .981
D.11.
a.
Source
SS
df
MS
Blocks
520.963 2 260.4815
Treatments
103.185 2 51.5925
Experimental error
5.259 4
1.3148
Observation error
45.333 18
2.5185
Total
674.741 26
b. H0 : all τj equal zero (j = 1, 2, 3), Ha : not all τj equal zero. F ∗ = 51.5925/1.3148 =
39.24, F (.95; 2, 4) = 6.94. If F ∗ ≤ 6.94 conclude H0 , otherwise Ha . Conclude Ha .
P −value = .002.
c. Ȳ·1· = 24.77778, Ȳ·2· = 20.00000, Ȳ·3· = 22.66667, L̂1 = Ȳ·1· − Ȳ·2· = 4.77778,
L̂2 = Ȳ·1· − Ȳ·3· = 2.11111, L̂3 = Ȳ·2· − Ȳ·3· = −2.66667, s{L̂i } = .5405 (i = 1, 2, 3),
q(.90; 3, 4) = 3.98, T = 2.8143
4.77778 ± 2.8143(.5405)
D.5
3.257 ≤ L1 ≤ 6.299
2.11111 ± 2.8143(.5405)
−2.66667 ± 2.8143(.5405)
.590 ≤ L2 ≤ 3.632
− 4.188 ≤ L3 ≤ −1.146
d. σ̂ 2 = 0, σ̂ 2η = 2.5185
D.12.
a.
Model
Term
ρi
αj
βk
(αβ)jk
Error
Coefficient
ab
sb
sa
s
Symbolic
Product
i−1
j−1
k−1
jk − j − k + 1
Term to
Be Squared
Ȳi·· − Ȳ···
Ȳ·j· − Ȳ···
Ȳ··k − Ȳ···
Ȳ·jk − Ȳ·j· − Ȳ··k + Ȳ···
Remainder =
Yijk − Ȳi·· − Ȳ·jk + Ȳ···
Degrees of
Freedom
s−1
a−1
b−1
(a − 1)(b − 1)
Remainder =
(s − 1)(ab − 1)
Yijk − Ȳ···
abs − 1
Total
b.
i
D.13.
j
F
a
a
0
a
0
1
k
F
b
b
b
0
0
1
Expected Mean Square of −−
Vari- S A B AB
Rem
ance i
j k
jk
(ijk)
σρ2
ab 0 0
0
0
2
σα
0 sb 0
0
0
σβ2
0 0 sa
0
0
2
σαβ
0 0 0
s
0
2
σ
1 1 1
1
1
ρi
αj
βk
(αβ)jk
²(ijk)
R
s
1
s
s
s
1
Model
Term
Coefficient
Symbolic
Product
Term to
Be Squared
Degrees of
Freedom
αj
βk
(αβ)jk
ρi(j)
Error
bs
as
s
b
j−1
k−1
jk − j − k + 1
ij − j
Ȳ·j· − Ȳ···
Ȳ··k − Ȳ···
Ȳ·jk − Ȳ·j· − Ȳ··k + Ȳ···
Ȳij· − Ȳ·j·
Remainder =
Yijk − Ȳ·jk − Ȳij· + Ȳ·j·
a−1
b−1
(a − 1)(b − 1)
a(s − 1)
Remainder =
a(s − 1)(b − 1)
Yijk − Ȳ···
abs − 1
a.
Total
D.6
b.
αj
βk
(αβ)jk
ρi(j)
²(ijk)
j
k
i
F
a
0
a
0
1
1
F
b
b
0
0
b
1
R
s
s
s
s
1
1
Expected Mean Square of −−
Vari- A B AB S(A)
Rem
ance j k
jk
i(j)
(ijk)
2
σα
bs 0
0
0
0
σβ2
0 as
0
0
0
2
σαβ
0 0
s
0
0
2
σρ
b 0
0
b
0
σ2
1 1
1
1
1
D.14. Note: The subscript for subjects here is l instead of the usual i and the subscripts for
factor A, B, and C are i, j, and k, respectively.
a.
Yijklm = µ···· + αi + βj + γk + ρl(ik) + (αβ)ij + (αγ)ik + (βγ)jk
+(αβγ)ijk + ²m(ijkl)
b.
Model
Term
Coefficient
Symbolic
Product
αi
βj
γk
ρl(ik)
(αβ)ij
(αγ)ik
(βγ)jk
(αβγ)ijk
bcrn
acrn
abrn
abcn
crn
brn
arn
rn
i−1
j−1
k−1
ikl − ik
ij − i − j + 1
ik − i − k + 1
jk − j − k + 1
ijk − ik − jk
−ij + i + j + k − 1
Error
Term to
Be Squared
Ȳi···· − Ȳ·····
Ȳ·j··· − Ȳ·····
Ȳ··k·· − Ȳ·····
Ȳi·kl· − Ȳi·k··
Ȳij··· − Ȳi···· − Ȳ·j··· + Ȳ·····
Ȳi·k·· − Ȳi···· − Ȳ··k·· + Ȳ·····
Ȳ·jk·· − Ȳ·j··· − Ȳ··k·· + Ȳ·····
Ȳijk·· − Ȳi·k·· − Ȳ·jk·· − Ȳij···
+Ȳi···· + Ȳ·j··· + Ȳ··k·· − Ȳ·····
Remainder =
Yijklm − Ȳi·kl· − Ȳijk·· + Ȳi·k··
Total
Yijklm − Ȳ·····
SSA = bcrn
SSB = acrn
SSC = abrn
etc.
D.7
X
(Ȳi···· − Ȳ····· )2
X
X
(Ȳ·j··· − Ȳ····· )2
(Ȳ··k·· − Ȳ····· )2
Degrees of
Freedom
a−1
b−1
c−1
ac(r − 1)
(a − 1)(b − 1)
(a − 1)(c − 1)
(b − 1)(c − 1)
(a − 1)(b − 1)(c − 1)
Remainder =
abcrn − acr − abc + ac
abcrn − 1
c.
αi
βj
γk
ρl(ik)
(αβ)ij
(αγ)ik
(βγ)jk
(αβγ)ijk
²m(ijkl)
αi
βj
γk
ρl(ik)
(αβ)ij
(αγ)ik
(βγ)jk
(αβγ)ijk
²m(ijkl)
i
F
a=3
0
a
a
1
0
0
a
0
1
A
i
bcrn
0
0
bn
0
0
0
0
1
j
F
b=4
b
0
b
b
0
b
0
0
1
B
j
0
acrn
0
0
0
0
0
0
1
k
F
c=2
c
c
0
1
c
0
0
0
1
l
R
r=4
r
r
r
1
r
r
r
r
1
m
R
n=2
n
n
n
n
n
n
n
n
1
Variance
σα2
σβ2
σγ2
σρ2
2
σαβ
2
σαγ
2
σβγ
2
σαβγ
σ2
Expected Mean Square of −−
C
S(AC) AB AC BC ABC
k
l(ik)
ij
ik
jk
ijk
0
0
0
0
0
0
0
0
0
0
0
0
abrn
0
0
0
0
0
bn
bn
0
bn
0
0
0
0
crn
0
0
0
0
0
0
brn
0
0
0
0
0
0
arn
0
0
0
0
0
0
rn
1
1
1
1
1
1
E{MSA} = 64
X αi2
2
+ 8σρ2 + σ 2
X βj2
+ σ2
3
X γk2
E{MSC} = 96
+ 8σρ2 + σ 2
1
E{MSS(AC)} = 8σρ2 + σ 2
E{MSB} = 48
E{MSAC} =
E{MSBC} =
E{MSABC} =
E{MSE} =
X X (αβ)2ij
+ σ2
6
X X (αγ)2ik
32
+ 8σρ2 + σ 2
2
X X (βγ)2jk
24
+ σ2
3
X X X (αβγ)2ijk
8
+ σ2
6
σ2
E{MSAB} = 16
D.8
E
m(ijkl)
0
0
0
0
0
0
0
0
1
D.15.
ρi
κj
τk
²(ijk)
i
F
r
0
r
r
1
j
F
r
r
0
r
1
k
F
r
r
r
0
1
Vari- E{MSROW}
ance
i
2
σρ
r
σκ2
0
στ2
0
2
σ
1
E{MSCOL}
j
0
r
0
1
E{MSTR}
k
0
0
r
1
E{MSE}
(ijk)
0
0
0
1
D.16.
ρi
κj
τk
²m(ijk)
i
F
r
0
r
r
1
j
F
r
r
0
r
1
k
F
r
r
r
0
1
m
R
n
n
n
n
1
Vari- E{MSROW}
ance
i
2
σρ
rn
0
σκ2
2
στ
0
2
σ
1
P
rn ρ2i
E{MSROW} = σ +
r−1
P
rn κ2j
2
E{MSCOL} = σ +
r−1
2
E{MSCOL} E{MSTR}
j
k
0
0
rn
0
0
rn
1
1
P
rn τk2
E{MSTR} = σ +
r−1
2
E{MSRem} = σ 2
D.17.
ρi
κj
τk
ηm(i)
²(ijkm)
i
j
k
m
F
r
0
r
r
1
1
F
r
r
0
r
r
1
F
r
r
r
0
r
1
R
n
n
n
n
1
1
Variance
σρ2
σκ2
στ2
ση2
σ2
Expected Mean Square of −−
P O T R S(P )
Rem
i
j
k
m(i) (ijkm)
rn 0
0
0
0
0 rn 0
0
0
0
0 rn
0
0
r
0
0
r
0
1
1
1
1
1
D.9
E{MSRem}
m(ijk)
0
0
0
1
Source Exif Data:
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