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Hints for Selected Exercises
Hints for Selected Exercises
CHAPTER 1 Practice Exercises 1. Enrollment by College 4000 3500
Students
3000 2500 2000 1500 1000 500 0 Ag
Bus
E.S.
L.A.
Sc
S.Sc.
2. Mean = 10.9, median = 12.5, std dev = 5.3427, variance = 28.54444. 3. Mean = 2, std dev = 2.94. 4. (a) Ages of Students 30 25
Students
20 15 10 5 0 20224 25229 30234 35239 40244 45249 50254 Age
Statistical Methods, DOI: 10.1016/B978-0-12-374970-3.00022-6 © 2010, Elsevier Inc. All rights reserved.
769
Hints for Selected Exercises
(b) Mean = 32.65, stddev = 6.91. Used 100 − 1 in calculation of std dev. If using midpoints 22.5, 27.5, etc., mean = 33.15. 5. (a) There is no consistent trend in CPI. (b) Mean = 2.6467, std dev = 0.5604, median = 2.8. (c) l. fence = 1.25, u. fence = 4.05, no outliers. 3.5 3.0 CPI
770
2.5 2.0 1.5 1992
1996
2000 Year
2004
2008
Exercises 1. (a) Mean = 17, median = 17, variance = 30.57, range = 22, interquartilerange = 7. (c) The stem and leaf reveals a heavier concentration of data in the range 10–19, piling up around the mean and median, indicative of a symmetrical distribution. 3. (a) FUTURE: Mean = − 0.20848, median = − 0.3, variance = 0.601018. INDEX: Mean = − 0.14935, median = − 0.155, variance = 1.770753. (b) Yes. The plot shows that as the futures contract increases, the NYSE Composite index also tends to increase. 5. (a) DAYS: Mean = 15.85366, median = 17, variance = 24.324. TEMP: Mean = 39.34756, median = 40, variance = 11.15191. (b) From the scatterplot, there appears to be no definitive relationship between the average temperature and the number of rainy January days. 7. The strongest relationship exists between DFOOT, the diameter of the tree at one foot above ground level and HT, the total height of the tree. One would expect that as the base of a tree increases in diameter the tree would increase in height as well.
Hints for Selected Exercises
9. (a) The mean is larger than (to the right of) the median, indicating a distribution skewed to the right. Yes, both the stem and leaf plot and the box plot reveal the skewness of the distribution. (b) The outliers 955 and 1160 may have resulted from younger patients. (c) Approximately 75% or 38 of the 51 patients were in remission for less than one year. 11. 12
Cost
11
10
9
8 1994 1996 1998 2000 2002 2004 2006 2008 Year
13. (a) Plot shows initial doses for drug G are much lower than for drug A. (b) For a given drug, there is not much relation between half life and initial dose. (c) Drug A: mean = 9.209, std dev = 1.142; Drug G: mean = 2.668, std dev = 0.440. This supports the conclusion in part (a).
CHAPTER 2 Practice Exercises 1. (a) P(Both) = (0.4)(0.3) = 0.12 (b) P(Neither) = (0.6)(0.7) = 0.42 (c) P(At Least One) = (0.4) + (0.3) − (0.12) = 0.58 2. (a) P(A) = 0.2 P(B) = 0.3 (b) P(A and B) = 0 (c) P(A or B) = (0.2) + (0.3) = 0.50 3. (a) μ = 1.0 σ 2 = 1.5 (b) 0.03125
771
772
Hints for Selected Exercises
4. $1,450 5. (a) (0.1587)(0.5793) = 0.0919 (b) 0.6461 (c) (0.5)(0.5) = 0.25
Exercises Y 1. (a)
| $0
$10
$1,000
$10,000
$50,000
148,969 1000 25 5 1 150,000 150,000 150,000 150,000 150,000 (b) μ = $0.90 (c) Expected net winnings are −$0.10; therefore a purchase is not worthwhile from a strictly monetary point of view. (d) σ = $142.00 p(Y ) |
3. (a) μ = 1 σ 2 = 0.8 (b) Yes 5. Arrangement I: P(system fail) = 0.0001999 Arrangement II: P(system fail) = 0.000396 7. (a) 0.1586 (b) 0.3108 (c) 16.4 9. 0.0571 11. 39 13. 0.5762 15. μ = 75 σ 2 = (11.72)2 = 137.36 17. (a) 4.35 × 10−8 (b) 1.233 × 10−5 (c) 0.01484 19. (a) A: 0.939, B: 0.921 (b) A: 0.322, B: 0.097 (c) Plan B is more expensive but has a smaller probability of missing an increase in the true error rate. 21. (a) 119.2 (b) 0.4972 (TI84: 0.4950) (c) 0.3734 (TI84: 0.3711) 23. About 5%, using F distribution with 5 and 5 df.
Hints for Selected Exercises
CHAPTER 3 Practice Exercises 1. (158.4, 175.6) 2. Z =−0.91 3. n ≈ 62, n ≈ 246 4. z = 3.40
Exercises 3. (a) (b) (c) (d)
β = 0.8300 β = 0.9428 When α = 0.05, β = 0.1492; when α = 0.01, β = 0.3372 When α = 0.05, β = 0.0877; when α = 0.01, β = 0.2514
5. (a) z =−4.38 (b) p value 0 7. (a) α = 0.0256 (b) β = 0.8704 9. (a) E = 1.31 (b) (78.3, 80.9) (c) n = 246 11. z = −22 p value 0 13. z = −4.0 15. n = 9604 17. (a) 0.0314 (b) 0.4231 19. (a) (2473, 2767) (b) About 138 or 139 21. (a) 0.1087 (b) Binomial. 0.7486
CHAPTER 4 Practice Exercises 1. (a) 1.7709 (b) 2.4786
773
774
Hints for Selected Exercises
(c) (d) (e) (f) (g)
1.3968 37.566 13.362 24.433 2.088
2. (a) (7.2, 10.8) (b) (3.0, 5.3) 3. (a) t = −4.59 (b) X 2 = 26 4. z = 0.0 5. n = 9220
Exercises 1. t = 3.99 3. χ 2 = 92.17 5. (a) z = −0.67 7. z = 0.65 9. t = −1.08 11. χ 2 = 23.96 15. Normality assumption violated 17. (1.596, 2.155) 19. z = 2.67 21. Type 1: χ 2 = 50.32, Type 2: χ 2 = 102.13 Type 3: χ 2 = 58.88, Type 4: χ 2 = 24.84 23. (0.376, 0.624) 25. z = 1.195, p value = 0.116. 27. (a) X 2 = 2.18, 11 df. 29. (a) Duval z = 2.49 (b) Putnam p value = 2 × (0.2539)
CHAPTER 5 Practice Exercises 1. z = 1.75 2. t = −2.44 3. (−9.92, −1.68)
Hints for Selected Exercises
4. t = 1.94 5. z = 1.84
Exercises 1. t = 1.223 3. t = 1.479 5. t = 3.136 7. z = 0 9. F = 1.502 11. t = −1.1798 13. t = −0.8862 15. (a) z = −0.56 17. (a) z = −4.766 (b) t = −4.04 19. z = 2.556
CHAPTER 6 Exercises 1. (a) F = 15.32 3. (a) F = 19.04 (b) Est(σs2 ) = 629.94 Est(σ 2 ) = 139.65 5. (a) F = 53.55 (b) Control: F = 143.19 MFG: F = 5.09 ADD: F = 63.59 (c) LIN: F = 195.61 QUAD: F = 13.60 LOF: F = 2.489 7. F = 15.04 9. F = 17.99 11. CD: F = 58.0 PB: F = 13.82 13. F = 30.40 15. F = 6.21
775
776
Hints for Selected Exercises
17. ANOM limits 45.96, 61.59 19. (a) 28 × 0.05 = 1.4 (b) Bonferroni’s method, each test using α = 0.05/28 = 0.0018 (c) No, each p value > 0.0018
CHAPTER 7 Exercises 1. (a) (b) (c) (d)
μˆ y|x = 2.8 − 0.5X 3.8, 3.3, 2.8, 2.3, 1.8 0.2,−0.3, 0.2, −0.3, 0.2 t = −5.0
3. (a) μˆ y|x = 14.396 + 0.765X (d) 77.1328 5. (a) μˆ y|x = −71.451 + 1.209X (b) 1.117 7. (b) μˆ y|x = −97.924 + 2.001X 9. (a) μˆ y|x = 7.526 + 0.225X 11. Relationship is not linear. 15. (b) 20 × (−0.03 ± 2.1009 × 0.0142) = −1.196 to −0.034 (c) r 2 = 0.198
CHAPTER 8 Exercises 3. μˆ y1|x = 700.62 − 1.526X 1 + 175.984X 2 − 6.697X 3. μˆ y 2|x = −5.611 + 0.668X 1 − 1.235X 2 + 0.073X 3, but residual plot suggests multiplicative model. 5. (a) μˆ y|x = −379.248 + 170.220DBH + 1.900HEIGHT + 8.146AGE −1192.868 GRAV, but residual plot suggests multiplicative model. 7. (a) μˆ y|x = 219.275 + 77.725X . (b) μˆ y|x = 178.078 + 93.106X − 0.729X 2 . 9. (a) COOL: μˆ y|x = −2.638 + 0.439WIDTH + 0.110HEIGHT. WARM: μˆ y|x − 2.117 + 0.207WIDTH + 0.118HEIGHT.
Hints for Selected Exercises
(b) COOL: μˆ y|x = −4.597 + 1.571LWIDTH + 0.747LHEIGHT. WARM: μˆ y|x = −4.421 + 1.669LWIDTH + 0.209LHEIGHT. 11. μˆ y|x = −10.305 + 0.378AGE + 2.294SEX + 0.179COLLEGE + 0.293INCOME. 13. μˆ y|x = 104.906 − 6.682AGE + 0.636SQFT − 0.403SD + 0.098UNTS + 65.000GAR + 33.051CP − 9.977SS + 14.327FIT. 15. F (2, 46) = 0.79 17. (a) 16
B
14 G
Fit
12 10
G
8 6 4
B 0
1
2
3
4
5
X2
(b) For low values of X2, girls tend to score higher than boys. But for high values of X2, boys tend to score higher than girls. (c) Girls–boys: β1 + 3β3 (d) 5 − 1.5 × 3 = 0.5 19. (a) Model 1: SSR = 3.2, SSE = 42.578, F (2, 97) = 3.65 Model 2: SSR = 8.70, SSE = 37.08, F (4, 95) = 5.57 (b) t(95) = −3.29, significant (c) (−0.897, −0.222) (d) F (2, 95) = 7.05
Chapter 9 Exercises 1.
Source
df
F Value
A T A∗T
1 3 3
85.88 4.41 9.48
To compare control, Ho : 3μ.c − μ.m − μ.n − μ.q = 0
777
778
Hints for Selected Exercises
3.
Source
df
F Value
A C A*C
2 2 4
14.23 6.89 3.18
μˆ y|x = −11.13 + 6.508A − 0.633A2 + 6.303C − 0.732C 2 − 0.597AC Lack of Fit F = 0.485 5.
7.
Source
df
F Value
FUNGICID CONCENTR FUNGICID∗CONCENTR
2 1 2
6.57 21.50 4.91
Source
df
F Value
TEMPR CLEAN TEMPR∗ CLEAN
4 4 16
323.05 1233.17 86.09
μˆ y|x = −8.035 + 36.275T − 12.460T 2 −30.952C + 23.787C2 + 17.654 TC 9.
11.
Source
df
F Value
TEMPL TEMPSQ DAY TEMPL∗ DAY TEMPSQ∗ DAY
1 1 1 1 1
4.28 2.78 3.95 1.46 23.58
Source
df
F Value
GRAIN PREP GRAIN∗PREP
2 2 4
2.93 7.33 1.85
13. (a)
Source MODEL TEST DISTRACTION TEST*DIST. ERROR CORRECTED
df
SS
MS
F
5 2 1 2 22 27
92 15 69 17 82 174
18.4 7.5 69 8.5 3.727
4.94 2.01 18.5 2.28
Hints for Selected Exercises
(b) At α = 0.05, the only significant effect is the main effect for distraction. Scores in one Distraction condition were consistently higher than in the other condition, averaged over all versions of the test. But without the cell means, we don’t know which distraction condition was higher. (c) The SS for the main and interaction effects don’t sum to the Model SS. 15. (a) 7
L H
Mean out. satis.
6 5 4
H 3
L H
2
L 1 Eq
Btr Outcome
Wrs
(b) The plot and test statistics are consistent in showing a very strong main effect of Outcome, with participants showing the highest satisfaction when Outcome is perceived as Equal. There is a weak main effect for Cog. Busyness, with a tendency for those with the Low Busyness to have less satisfaction. However, this tendency seems strongest in the Better Outcome category, leading to a significant interaction. (c) 6 × 5/2 = 15 (d) Yes, as noted in (b), the difference between the High and Low Busyness categories is most pronounced in the Better outcome category. There is less of a difference in the Equal and Worse categories. This is consistent with the significant interaction. 17. Reduced model SSE = 542.822, df = 88, F (16, 72) = 0.96
Chapter 10 Exercises 1.
Source
df
TRT 2 EXP 1 TRT∗EXP 2 TOTAL 29
SS
F Value
2.766 1.580 2.130
1.30
779
780
Hints for Selected Exercises
3. (b) Variety: F (2, 6) = [6.231/2] ÷ [1.011/6] = 18.49, where error uses SS of REP*VAR. NIT: F (1, 3) = 0.00 Variety*NIT: F (2, 6) = 14.81 5. (b) Salt: F (3, 8) = 6.95; Day: F (3, 24) = 347.16; Salt*Day: F (9, 24) = 49.13 7. (a)
Source
df
SS
REP 4 LIGHT 2 REP∗ LIGHT (A) 8 LEAF 4 LIGHT∗LEAF 8 ERROR (B) 48
F Value
1.033 1.947 1.264 0.885 0.302 1.073
6.16 9.90 1.69
(b) μˆ y|x = 1.781 + 0.013LIGHT − 0.0001LIGHT 2 + 0.267LEAF − 0.047LEAF 2 + 0.0005LIGHT∗ LEAF (c) F = 3.33 9. SURFACE: Source
df
SS
F Value
TRT(Shade) COLOR TRT∗COLOR
3 1 3
73.541 0.003 3.176
16.61 0.00 0.72
RECTAL: Source
df
SS
F Value
TRT(Shade) COLOR TRT∗COLOR
3 1 3
1.253 0.062 0.521
1.79 0.26 0.75
13. WEIGHT: F = 16.54 LENGTH: F = 8.31 RELWT: F = 27.91
CHAPTER 11 Exercises 1.
Source
df
SS
F Value
STAGE WWT
2 1
289.82 394.08
12.59 34.24
Hints for Selected Exercises
3. For 52 df, t-table value √ is 2.0066 from Excel (a) 4.058 ± 2.0066 2.664 = 4.058 √ ± 3.275 (b) (4.058 − (−0.957)) ± 2.0066 2.664 − 2(1.095) + 2.504 = 5.014 ± 3.463 √ (c) (4.058 − 4(−0.773)) ± 2.0066 2.92 = 0.966 ± 3.429 5. (a) Mean = 12 − 12 = 0 (b) Variance = 0.0037, SD = 0.0608 7. (a) ANOVA: Source
df
SS
F Value
PAVE TREAD PAVE∗ TREAD
2 2 4
216.774 203.676 22.154
43.67 41.03 2.23
DUMMY VARIABLE:
(b)
Source
df
SS
F Value
PAVE TREAD PAVE∗ TREAD
2 2 4
233.584 212.463 6.699
47.06 42.80 0.67
Source
df
SS
F Value
PAVE TREAD
2 1
232.818 219.062
52.31 98.44
(c) μˆ y|x = 26.194 + 28.660FRICT + 1.374TREAD 9. (a) Source
df
SS
F Value
MEDIUM TIME MEDIUM∗ TIME
2 3 6
3137.392 1514.468 514.574
50.97 16.40 2.79
11. (a) gas = 112.727 + 2.258oil + error (c) DW = 0.744, p value for positive correlation < .0001 13. (a) Source
df
SS
F Value
SIZE TYPE SIZE∗ TYPE
1 1 1
913381.32 85.55 461.12
3881.24 0.36 1.96
781
782
Hints for Selected Exercises
17. (a) pooled t(22) = 0.74 (b) Sex: F (1, 21) = 7.22 19. (a) SOUTH has positive coefficient with small p value.
CHAPTER 12 Exercises 1.
(i) (0.67, 0.75) (ii) X 2 = 3.23
3. X 2 = 0.275 5. X 2 = 4.21 7. X 2 = 3.306 9. X 22 = 7.66 11. (a) X 2 = 26.25 13. Net_type: X 2 = 8.65, Size: X 2 = 29.99 15. Fisher’s exact test two-tailed p = 0.6834
CHAPTER 13 Exercises 1. (b) Race dummy variable has X 2 = 29.99 (c) Likelihood ratio test for interaction parameters has X 2 = 4.685 with 2df 3. Fitted ln(ODDS) = −4.0478 + 0.0569V, likelihood ratio X 2 = 6.836 5. (c) Fitted ln(ODDS) = −3.207 − 1.214Sex + 1.051(A2) + 1.268(A3) + 1.1183(A4), where Sex = 0 for males, 1 for females, and A2, A3, A4 are dummy variables for Age group that use reference cell coding with < 18 as baseline. 7. Poisson regression with log(AWTL) as offset variable, design coded as 0 for design A, 1 for design B. β = 0.7784, X 2 = 22.96
CHAPTER 14 Exercises 1. (a) T (+) = 9 (b) t = −2.178
Hints for Selected Exercises
3. T = 81 5. (a) T (+) = 2.0 (b) t = −2.281 7. T ∗ = 0.486 9. H = 22.68 11. T ∗ = 1.448 13. (a) Cons. Staples: T (−) = 0 Financial: T (−) = 0 (b) Mann-Whitney: T = 89
783
CHAPTER 1 Practice Exercises 1. Enrollment by College 4000 3500
Students
3000 2500 2000 1500 1000 500 0 Ag
Bus
E.S.
L.A.
Sc
S.Sc.
2. Mean = 10.9, median = 12.5, std dev = 5.3427, variance = 28.54444. 3. Mean = 2, std dev = 2.94. 4. (a) Ages of Students 30 25
Students
20 15 10 5 0 20224 25229 30234 35239 40244 45249 50254 Age
Statistical Methods, DOI: 10.1016/B978-0-12-374970-3.00022-6 © 2010, Elsevier Inc. All rights reserved.
769
Hints for Selected Exercises
(b) Mean = 32.65, stddev = 6.91. Used 100 − 1 in calculation of std dev. If using midpoints 22.5, 27.5, etc., mean = 33.15. 5. (a) There is no consistent trend in CPI. (b) Mean = 2.6467, std dev = 0.5604, median = 2.8. (c) l. fence = 1.25, u. fence = 4.05, no outliers. 3.5 3.0 CPI
770
2.5 2.0 1.5 1992
1996
2000 Year
2004
2008
Exercises 1. (a) Mean = 17, median = 17, variance = 30.57, range = 22, interquartilerange = 7. (c) The stem and leaf reveals a heavier concentration of data in the range 10–19, piling up around the mean and median, indicative of a symmetrical distribution. 3. (a) FUTURE: Mean = − 0.20848, median = − 0.3, variance = 0.601018. INDEX: Mean = − 0.14935, median = − 0.155, variance = 1.770753. (b) Yes. The plot shows that as the futures contract increases, the NYSE Composite index also tends to increase. 5. (a) DAYS: Mean = 15.85366, median = 17, variance = 24.324. TEMP: Mean = 39.34756, median = 40, variance = 11.15191. (b) From the scatterplot, there appears to be no definitive relationship between the average temperature and the number of rainy January days. 7. The strongest relationship exists between DFOOT, the diameter of the tree at one foot above ground level and HT, the total height of the tree. One would expect that as the base of a tree increases in diameter the tree would increase in height as well.
Hints for Selected Exercises
9. (a) The mean is larger than (to the right of) the median, indicating a distribution skewed to the right. Yes, both the stem and leaf plot and the box plot reveal the skewness of the distribution. (b) The outliers 955 and 1160 may have resulted from younger patients. (c) Approximately 75% or 38 of the 51 patients were in remission for less than one year. 11. 12
Cost
11
10
9
8 1994 1996 1998 2000 2002 2004 2006 2008 Year
13. (a) Plot shows initial doses for drug G are much lower than for drug A. (b) For a given drug, there is not much relation between half life and initial dose. (c) Drug A: mean = 9.209, std dev = 1.142; Drug G: mean = 2.668, std dev = 0.440. This supports the conclusion in part (a).
CHAPTER 2 Practice Exercises 1. (a) P(Both) = (0.4)(0.3) = 0.12 (b) P(Neither) = (0.6)(0.7) = 0.42 (c) P(At Least One) = (0.4) + (0.3) − (0.12) = 0.58 2. (a) P(A) = 0.2 P(B) = 0.3 (b) P(A and B) = 0 (c) P(A or B) = (0.2) + (0.3) = 0.50 3. (a) μ = 1.0 σ 2 = 1.5 (b) 0.03125
771
772
Hints for Selected Exercises
4. $1,450 5. (a) (0.1587)(0.5793) = 0.0919 (b) 0.6461 (c) (0.5)(0.5) = 0.25
Exercises Y 1. (a)
| $0
$10
$1,000
$10,000
$50,000
148,969 1000 25 5 1 150,000 150,000 150,000 150,000 150,000 (b) μ = $0.90 (c) Expected net winnings are −$0.10; therefore a purchase is not worthwhile from a strictly monetary point of view. (d) σ = $142.00 p(Y ) |
3. (a) μ = 1 σ 2 = 0.8 (b) Yes 5. Arrangement I: P(system fail) = 0.0001999 Arrangement II: P(system fail) = 0.000396 7. (a) 0.1586 (b) 0.3108 (c) 16.4 9. 0.0571 11. 39 13. 0.5762 15. μ = 75 σ 2 = (11.72)2 = 137.36 17. (a) 4.35 × 10−8 (b) 1.233 × 10−5 (c) 0.01484 19. (a) A: 0.939, B: 0.921 (b) A: 0.322, B: 0.097 (c) Plan B is more expensive but has a smaller probability of missing an increase in the true error rate. 21. (a) 119.2 (b) 0.4972 (TI84: 0.4950) (c) 0.3734 (TI84: 0.3711) 23. About 5%, using F distribution with 5 and 5 df.
Hints for Selected Exercises
CHAPTER 3 Practice Exercises 1. (158.4, 175.6) 2. Z =−0.91 3. n ≈ 62, n ≈ 246 4. z = 3.40
Exercises 3. (a) (b) (c) (d)
β = 0.8300 β = 0.9428 When α = 0.05, β = 0.1492; when α = 0.01, β = 0.3372 When α = 0.05, β = 0.0877; when α = 0.01, β = 0.2514
5. (a) z =−4.38 (b) p value 0 7. (a) α = 0.0256 (b) β = 0.8704 9. (a) E = 1.31 (b) (78.3, 80.9) (c) n = 246 11. z = −22 p value 0 13. z = −4.0 15. n = 9604 17. (a) 0.0314 (b) 0.4231 19. (a) (2473, 2767) (b) About 138 or 139 21. (a) 0.1087 (b) Binomial. 0.7486
CHAPTER 4 Practice Exercises 1. (a) 1.7709 (b) 2.4786
773
774
Hints for Selected Exercises
(c) (d) (e) (f) (g)
1.3968 37.566 13.362 24.433 2.088
2. (a) (7.2, 10.8) (b) (3.0, 5.3) 3. (a) t = −4.59 (b) X 2 = 26 4. z = 0.0 5. n = 9220
Exercises 1. t = 3.99 3. χ 2 = 92.17 5. (a) z = −0.67 7. z = 0.65 9. t = −1.08 11. χ 2 = 23.96 15. Normality assumption violated 17. (1.596, 2.155) 19. z = 2.67 21. Type 1: χ 2 = 50.32, Type 2: χ 2 = 102.13 Type 3: χ 2 = 58.88, Type 4: χ 2 = 24.84 23. (0.376, 0.624) 25. z = 1.195, p value = 0.116. 27. (a) X 2 = 2.18, 11 df. 29. (a) Duval z = 2.49 (b) Putnam p value = 2 × (0.2539)
CHAPTER 5 Practice Exercises 1. z = 1.75 2. t = −2.44 3. (−9.92, −1.68)
Hints for Selected Exercises
4. t = 1.94 5. z = 1.84
Exercises 1. t = 1.223 3. t = 1.479 5. t = 3.136 7. z = 0 9. F = 1.502 11. t = −1.1798 13. t = −0.8862 15. (a) z = −0.56 17. (a) z = −4.766 (b) t = −4.04 19. z = 2.556
CHAPTER 6 Exercises 1. (a) F = 15.32 3. (a) F = 19.04 (b) Est(σs2 ) = 629.94 Est(σ 2 ) = 139.65 5. (a) F = 53.55 (b) Control: F = 143.19 MFG: F = 5.09 ADD: F = 63.59 (c) LIN: F = 195.61 QUAD: F = 13.60 LOF: F = 2.489 7. F = 15.04 9. F = 17.99 11. CD: F = 58.0 PB: F = 13.82 13. F = 30.40 15. F = 6.21
775
776
Hints for Selected Exercises
17. ANOM limits 45.96, 61.59 19. (a) 28 × 0.05 = 1.4 (b) Bonferroni’s method, each test using α = 0.05/28 = 0.0018 (c) No, each p value > 0.0018
CHAPTER 7 Exercises 1. (a) (b) (c) (d)
μˆ y|x = 2.8 − 0.5X 3.8, 3.3, 2.8, 2.3, 1.8 0.2,−0.3, 0.2, −0.3, 0.2 t = −5.0
3. (a) μˆ y|x = 14.396 + 0.765X (d) 77.1328 5. (a) μˆ y|x = −71.451 + 1.209X (b) 1.117 7. (b) μˆ y|x = −97.924 + 2.001X 9. (a) μˆ y|x = 7.526 + 0.225X 11. Relationship is not linear. 15. (b) 20 × (−0.03 ± 2.1009 × 0.0142) = −1.196 to −0.034 (c) r 2 = 0.198
CHAPTER 8 Exercises 3. μˆ y1|x = 700.62 − 1.526X 1 + 175.984X 2 − 6.697X 3. μˆ y 2|x = −5.611 + 0.668X 1 − 1.235X 2 + 0.073X 3, but residual plot suggests multiplicative model. 5. (a) μˆ y|x = −379.248 + 170.220DBH + 1.900HEIGHT + 8.146AGE −1192.868 GRAV, but residual plot suggests multiplicative model. 7. (a) μˆ y|x = 219.275 + 77.725X . (b) μˆ y|x = 178.078 + 93.106X − 0.729X 2 . 9. (a) COOL: μˆ y|x = −2.638 + 0.439WIDTH + 0.110HEIGHT. WARM: μˆ y|x − 2.117 + 0.207WIDTH + 0.118HEIGHT.
Hints for Selected Exercises
(b) COOL: μˆ y|x = −4.597 + 1.571LWIDTH + 0.747LHEIGHT. WARM: μˆ y|x = −4.421 + 1.669LWIDTH + 0.209LHEIGHT. 11. μˆ y|x = −10.305 + 0.378AGE + 2.294SEX + 0.179COLLEGE + 0.293INCOME. 13. μˆ y|x = 104.906 − 6.682AGE + 0.636SQFT − 0.403SD + 0.098UNTS + 65.000GAR + 33.051CP − 9.977SS + 14.327FIT. 15. F (2, 46) = 0.79 17. (a) 16
B
14 G
Fit
12 10
G
8 6 4
B 0
1
2
3
4
5
X2
(b) For low values of X2, girls tend to score higher than boys. But for high values of X2, boys tend to score higher than girls. (c) Girls–boys: β1 + 3β3 (d) 5 − 1.5 × 3 = 0.5 19. (a) Model 1: SSR = 3.2, SSE = 42.578, F (2, 97) = 3.65 Model 2: SSR = 8.70, SSE = 37.08, F (4, 95) = 5.57 (b) t(95) = −3.29, significant (c) (−0.897, −0.222) (d) F (2, 95) = 7.05
Chapter 9 Exercises 1.
Source
df
F Value
A T A∗T
1 3 3
85.88 4.41 9.48
To compare control, Ho : 3μ.c − μ.m − μ.n − μ.q = 0
777
778
Hints for Selected Exercises
3.
Source
df
F Value
A C A*C
2 2 4
14.23 6.89 3.18
μˆ y|x = −11.13 + 6.508A − 0.633A2 + 6.303C − 0.732C 2 − 0.597AC Lack of Fit F = 0.485 5.
7.
Source
df
F Value
FUNGICID CONCENTR FUNGICID∗CONCENTR
2 1 2
6.57 21.50 4.91
Source
df
F Value
TEMPR CLEAN TEMPR∗ CLEAN
4 4 16
323.05 1233.17 86.09
μˆ y|x = −8.035 + 36.275T − 12.460T 2 −30.952C + 23.787C2 + 17.654 TC 9.
11.
Source
df
F Value
TEMPL TEMPSQ DAY TEMPL∗ DAY TEMPSQ∗ DAY
1 1 1 1 1
4.28 2.78 3.95 1.46 23.58
Source
df
F Value
GRAIN PREP GRAIN∗PREP
2 2 4
2.93 7.33 1.85
13. (a)
Source MODEL TEST DISTRACTION TEST*DIST. ERROR CORRECTED
df
SS
MS
F
5 2 1 2 22 27
92 15 69 17 82 174
18.4 7.5 69 8.5 3.727
4.94 2.01 18.5 2.28
Hints for Selected Exercises
(b) At α = 0.05, the only significant effect is the main effect for distraction. Scores in one Distraction condition were consistently higher than in the other condition, averaged over all versions of the test. But without the cell means, we don’t know which distraction condition was higher. (c) The SS for the main and interaction effects don’t sum to the Model SS. 15. (a) 7
L H
Mean out. satis.
6 5 4
H 3
L H
2
L 1 Eq
Btr Outcome
Wrs
(b) The plot and test statistics are consistent in showing a very strong main effect of Outcome, with participants showing the highest satisfaction when Outcome is perceived as Equal. There is a weak main effect for Cog. Busyness, with a tendency for those with the Low Busyness to have less satisfaction. However, this tendency seems strongest in the Better Outcome category, leading to a significant interaction. (c) 6 × 5/2 = 15 (d) Yes, as noted in (b), the difference between the High and Low Busyness categories is most pronounced in the Better outcome category. There is less of a difference in the Equal and Worse categories. This is consistent with the significant interaction. 17. Reduced model SSE = 542.822, df = 88, F (16, 72) = 0.96
Chapter 10 Exercises 1.
Source
df
TRT 2 EXP 1 TRT∗EXP 2 TOTAL 29
SS
F Value
2.766 1.580 2.130
1.30
779
780
Hints for Selected Exercises
3. (b) Variety: F (2, 6) = [6.231/2] ÷ [1.011/6] = 18.49, where error uses SS of REP*VAR. NIT: F (1, 3) = 0.00 Variety*NIT: F (2, 6) = 14.81 5. (b) Salt: F (3, 8) = 6.95; Day: F (3, 24) = 347.16; Salt*Day: F (9, 24) = 49.13 7. (a)
Source
df
SS
REP 4 LIGHT 2 REP∗ LIGHT (A) 8 LEAF 4 LIGHT∗LEAF 8 ERROR (B) 48
F Value
1.033 1.947 1.264 0.885 0.302 1.073
6.16 9.90 1.69
(b) μˆ y|x = 1.781 + 0.013LIGHT − 0.0001LIGHT 2 + 0.267LEAF − 0.047LEAF 2 + 0.0005LIGHT∗ LEAF (c) F = 3.33 9. SURFACE: Source
df
SS
F Value
TRT(Shade) COLOR TRT∗COLOR
3 1 3
73.541 0.003 3.176
16.61 0.00 0.72
RECTAL: Source
df
SS
F Value
TRT(Shade) COLOR TRT∗COLOR
3 1 3
1.253 0.062 0.521
1.79 0.26 0.75
13. WEIGHT: F = 16.54 LENGTH: F = 8.31 RELWT: F = 27.91
CHAPTER 11 Exercises 1.
Source
df
SS
F Value
STAGE WWT
2 1
289.82 394.08
12.59 34.24
Hints for Selected Exercises
3. For 52 df, t-table value √ is 2.0066 from Excel (a) 4.058 ± 2.0066 2.664 = 4.058 √ ± 3.275 (b) (4.058 − (−0.957)) ± 2.0066 2.664 − 2(1.095) + 2.504 = 5.014 ± 3.463 √ (c) (4.058 − 4(−0.773)) ± 2.0066 2.92 = 0.966 ± 3.429 5. (a) Mean = 12 − 12 = 0 (b) Variance = 0.0037, SD = 0.0608 7. (a) ANOVA: Source
df
SS
F Value
PAVE TREAD PAVE∗ TREAD
2 2 4
216.774 203.676 22.154
43.67 41.03 2.23
DUMMY VARIABLE:
(b)
Source
df
SS
F Value
PAVE TREAD PAVE∗ TREAD
2 2 4
233.584 212.463 6.699
47.06 42.80 0.67
Source
df
SS
F Value
PAVE TREAD
2 1
232.818 219.062
52.31 98.44
(c) μˆ y|x = 26.194 + 28.660FRICT + 1.374TREAD 9. (a) Source
df
SS
F Value
MEDIUM TIME MEDIUM∗ TIME
2 3 6
3137.392 1514.468 514.574
50.97 16.40 2.79
11. (a) gas = 112.727 + 2.258oil + error (c) DW = 0.744, p value for positive correlation < .0001 13. (a) Source
df
SS
F Value
SIZE TYPE SIZE∗ TYPE
1 1 1
913381.32 85.55 461.12
3881.24 0.36 1.96
781
782
Hints for Selected Exercises
17. (a) pooled t(22) = 0.74 (b) Sex: F (1, 21) = 7.22 19. (a) SOUTH has positive coefficient with small p value.
CHAPTER 12 Exercises 1.
(i) (0.67, 0.75) (ii) X 2 = 3.23
3. X 2 = 0.275 5. X 2 = 4.21 7. X 2 = 3.306 9. X 22 = 7.66 11. (a) X 2 = 26.25 13. Net_type: X 2 = 8.65, Size: X 2 = 29.99 15. Fisher’s exact test two-tailed p = 0.6834
CHAPTER 13 Exercises 1. (b) Race dummy variable has X 2 = 29.99 (c) Likelihood ratio test for interaction parameters has X 2 = 4.685 with 2df 3. Fitted ln(ODDS) = −4.0478 + 0.0569V, likelihood ratio X 2 = 6.836 5. (c) Fitted ln(ODDS) = −3.207 − 1.214Sex + 1.051(A2) + 1.268(A3) + 1.1183(A4), where Sex = 0 for males, 1 for females, and A2, A3, A4 are dummy variables for Age group that use reference cell coding with < 18 as baseline. 7. Poisson regression with log(AWTL) as offset variable, design coded as 0 for design A, 1 for design B. β = 0.7784, X 2 = 22.96
CHAPTER 14 Exercises 1. (a) T (+) = 9 (b) t = −2.178
Hints for Selected Exercises
3. T = 81 5. (a) T (+) = 2.0 (b) t = −2.281 7. T ∗ = 0.486 9. H = 22.68 11. T ∗ = 1.448 13. (a) Cons. Staples: T (−) = 0 Financial: T (−) = 0 (b) Mann-Whitney: T = 89
783