Table D.12 Percentage Points of the Studentized Range, q(p, v), Upper 1%
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Appendix D Useful Statistical Tables
777
Table D.12 (Continued)
p
ν
1
12
13
14
15
260.0
266.2
271.8
277.0
16
281.8
17
18
19
20
286.3
290.0
294.3
298.0
2
33.40
34.13
34.81
35.43
36.00
36.53
37.03
37.50
37.95
3
17.53
17.89
18.22
18.52
18.81
19.07
19.32
19.55
19.77
4
12.84
13.09
13.32
13.53
13.73
13.91
14.08
14.24
14.40
5
10.70
10.89
11.08
11.24
11.40
11.55
11.68
11.81
11.93
6
9.48
9.65
9.81
9.95
10.08
10.21
10.32
10.43
10.54
7
8.71
8.86
9.00
9.12
9.24
9.35
9.46
9.55
9.65
8
8.18
8.31
8.44
8.55
8.66
8.76
8.85
8.94
9.03
9
7.78
7.91
8.03
8.13
8.23
8.33
8.41
8.49
8.57
10
7.49
7.60
7.71
7.81
7.91
7.99
8.08
8.15
8.23
11
7.25
7.36
7.46
7.56
7.65
7.73
7.81
7.88
7.95
12
7.06
7.17
7.26
7.36
7.44
7.52
7.59
7.66
7.73
13
6.90
7.01
7.10
7.19
7.27
7.35
7.42
7.48
7.55
14
6.77
6.87
6.96
7.05
7.13
7.20
7.27
7.33
7.39
15
6.66
6.76
6.84
6.93
7.00
7.07
7.14
7.20
7.26
16
6.56
6.66
6.74
6.82
6.90
6.97
7.03
7.09
7.15
17
6.48
6.57
6.66
6.73
6.81
6.87
6.94
7.00
7.05
18
6.41
6.50
6.58
6.65
6.72
6.79
6.85
6.91
6.97
19
6.34
6.43
6.51
6.58
6.65
6.72
6.78
6.84
6.89
20
6.28
6.37
6.45
6.52
6.59
6.65
6.71
6.77
6.82
24
6.11
6.19
6.26
6.33
6.39
6.45
6.51
6.56
6.61
30
5.93
6.01
6.08
6.14
6.20
6.26
6.31
6.36
6.41
40
5.76
5.83
5.90
5.96
6.02
6.07
6.12
6.16
6.21
60
5.60
5.67
5.73
5.78
5.84
5.89
5.93
5.97
6.01
120
5.44
5.50
5.56
5.61
5.66
5.71
5.75
5.79
5.83
∞
5.29
5.35
5.40
5.45
5.49
5.54
5.57
5.61
5.65
Source: Biometrika Tables for Statisticians, Vol. 1, 3rd ed., edited by E. S. Pearson and H. O. Hartley (Cambridge
University Press, 1966). Reproduced by permission of Professor E. S. Pearson and the Biometrika Trustees.
Appendix
File Layouts for Case
Study Data Sets
E
Case Study 1: Legal Advertising—Does It Pay?
LEGALADV (n = 48 observations)
Variable
Type
Description
MONTH
Numeric
Month
TOTADVEXP
Numeric
Advertising expenditure (thousands of dollars)
NEWPI
Numeric
Number of new personal injury cases
NEWWC
Numeric
Number of new worker’s compensation cases
ADVEXP6
Numeric
Cumulative advertising expenditure over previous 6 months (thousands of dollars)
Case Study 2: Modeling the Sales Prices of Properties in Four Neighborhoods
TAMSALES4 (n = 176 observations)
Variable
Type
Description
SALES
Numeric
Sales price (thousands of dollars)
LAND
Numeric
Land value (thousands of dollars)
IMP
Numeric
Value of improvements (thousands of dollars)
NBHD
Character
Neighborhood (HYDEPARK, DAVISISLES, CHEVAL, HUNTERSGREEN)
TAMSALES8 (n = 350 observations)
Variable
Type
Description
SALES
Numeric
Sales price (thousands of dollars)
LAND
Numeric
Land value (thousands of dollars)
IMP
Numeric
Value of improvements (thousands of dollars)
NBHD
Character
Neighborhood (HYDEPARK, DAVISISLES, CHEVAL, HUNTERSGREEN,
AVILA, CRLLWOODVILL, TAMPAPALMS, TOWN&CNTRY)
778
Appendix E File Layouts for Case Study Sets
Case Study 3: Deregulation of the Intrastate Trucking Industry
TRUCKING (n = 134 observations)
Variable
Type
Description
PRICPTM
Numeric
Price charged per ton-mile (dollars)
DISTANCE
Numeric
Distance traveled (hundreds of miles)
WEIGHT
Numeric
Weight of product shipped (thousands of pounds)
PCTLOAD
Numeric
Weight as a percentage of truck load capacity
ORIGIN
Character
City of origin (JAX or MIA)
MARKET
Character
Size of market destination (LARGE or SMALL)
DEREG
Character
Deregulation in effect (YES or NO)
PRODUCT
Numeric
Product classiﬁcation (100, 150, or 200)
LNPRICE
Numeric
Natural logarithm of price charged
TRUCKING4 (n = 448 observations)
Variable
Type
Description
PRICPTM
Numeric
Price charged per ton-mile (dollars)
DISTANCE
Numeric
Distance traveled (hundreds of miles)
WEIGHT
Numeric
Weight of product shipped (thousands of pounds)
PCTLOAD
Numeric
Weight as a percentage of truck load capacity
ORIGIN
Character
City of origin (JAX or MIA)
MARKET
Character
Size of market destination (LARGE or SMALL)
DEREG
Character
Deregulation in effect (YES or NO)
PRODUCT
Numeric
Product classiﬁcation (100, 150, or 200)
CARRIER
Character
Florida trucking carrier (A, B, C, or D)
LNPRICE
Numeric
Natural logarithm of price charged
Case Study 4: An Analysis of Rain Levels in California
CALIRAIN (n = 30 observations)
Variable
Type
Description
STATION
Numeric
Station number
NAME
Character
Name of meteorological station
PRECIP
Numeric
Average annual precipitation (inches)
ALTITUDE
Numeric
Altitude (feet)
LATITUDE
Numeric
Latitude (degrees)
DISTANCE
Numeric
Distance to Paciﬁc Ocean (miles)
SHADOW
Character
Slope face (W = westward or L = leeward)
779
780 Appendix E File Layouts for Case Study Sets
Case Study 5: An Investigation of Factors Affecting the Sales Price of
Condominium Units Sold at Public Auction
CONDO (n = 209 observations)
Variable
Type
Description
PRICE
Numeric
Price paid (hundreds of dollars)
FLOOR
Numeric
Floor height (1, 2, 3, . . . , 8)
DISTELEV
Numeric
Distance from elevator (1, 2, 3, . . . , 15)
VIEW
Numeric
View (1 = Ocean, 0 = Bay)
ENDUNIT
Numeric
End unit (1 = end, 0 = not)
FURNISH
Numeric
Furnished (1 = furnish, 0 = not)
METHOD
Character
Sales method (A = auction, F = ﬁxed price)
Case Study 7: Reluctance to Transmit Bad News: The MUM Effect
MUM (n = 40 observations)
Variable
Type
Description
FEEDBACK
Numeric
Latency to feedback (seconds)
SUBJECT
Character
Subject visibility (NOTVIS or VISIBLE)
CONFED
Character
Confederate success (FAILURE or SUCCESS)
ANSWERS TO SELECTED EXERCISES
Chapter 1
1.1 (a) Quantitative (b) Qualitative (c) Quantitative
(d) Qualitative (e) Quantitative (f) Quantitative
1.3 (a) earthquakes (b) type of ground motion–qualitative;
magnitude and acceleration–quantitative
1.5 (a) Qualitative (b) Qualitative (c) Quantitative
(d) Quantitative (e) Quantitative (f) Quantitative
(g) Quantitative (h) Qualitative
1.7 (a) Population–all decision makers; sample–155 volunteer students; variables–emotional state and whether or not
to repair a very old car (b) subjects in the guilty-state group
are less likely to repair an old car
1.9 (a) Amateur boxers (b) massage or rest group–
qualitative; heart rate and blood lactate level–both quantitative (c) no difference in mean heart rates of two groups
of boxers (d) no
1.11 (a) Population–all adults in Tennessee; sample–575
study participants (b) years of education–quantitative;
insomnia–qualitative (c) less educated adults are more
likely to have chronic insomnia
1.13 (a) Black–.203; White–.637; Sumatran–.017; Javan–
.003; Indian–.140 (c) .84; .16
1.15 (a) pie chart (b) type of ﬁrearms owned (c) riﬂe
(33 %), shotgun (21%), and revolver (20 %)
1.17 (d) public wells (40 %); private wells (21 %)
1.19 (b) yes
1.21 yes
1.23 (b) .98
1.25 (a) 2.12; average magnitude for the aftershocks is 2.12
(b) 6.7; difference between the largest and smallest magnitude is 6.7 (c) .66; about 95 % of the magnitudes fall in
the interval mean ± 2(std. dev.) = (.8, 3.44) (d) μ = mean;
σ = standard deviation
y = 94.91,
1.27 (a)
(c) .976; yes
s = 4.83 (b)
(85.25,
104.57)
1.29 (a) 1.47; average daily ammonia concentration is
1.47 ppm (b) .0640; about 95 % of the daily ammonia
concentration levels fall within y ± 2s = (1.34, 1.60) ppm
(c) morning
1.31 (a) (−111, 149) (b) (−91, 105) (c) SAT-Math
1.33 (a) .9544 (b) .1587 (c) .1587 (d) .8185 (e) .1498
(f) .9974
1.35 (a) .9406 (b) .9406 (c) .1140
1.37 (a) .2690 (b) .2611 (c) (13.6, 62.2)
1.39 .8664
1.41 standard deviation for means of n = 12 measurements
is smaller
1.43 (a) .10; .0141 (b) Central Limit Theorem (c) .017
1.45 (a) 1.13 ± .67 (b) yes
1.47 (a) 99.6 (b) (97.4, 101.8) (c) 95 % conﬁdent that the
true mean Mach rating score is between 97.4 and 101.8
(d) yes
1.49 (17.1, 20.9); 99 % conﬁdent that the true mean quality
of the methodology is between 17.1 and 20.9 Wong Scale
points
1.51 (a) null hypothesis (b) alternative hypothesis
(c) rejecting H0 when H0 is true (d) accepting H0 when
H0 is false (e) probability of Type I error (f) probability
of Type II error (g) observed signiﬁcance level
1.53 (a) .025 (b) .05 (c) .005 (d) .0985 (e) .10 (f) .01
1.55 (a) H0 : μ = 15, Ha : μ < 15 (b) concluding that the
average level of mercury uptake is less than 15 ppm, when
the average level is equal to 15 ppm (c) concluding that the
average level of mercury uptake is equal to 15 ppm, when
the average level is less than 15 ppm
1.57 (a) z = −3.72, reject H0
1.59 (b) p = .0214, reject H0 at α = .05 (c) p ≈ 0, reject
H0 at α = .05
1.61 t = 7.83, reject H0
1.63 Approx. normal (for large n) with mean μy 1 −y 2 =
(μ1 − μ2 ) and standard deviation
σy 1 −y 2 =
σ12
σ2
+ 2
n1
n2
1.65 t = 1.08, fail to reject H0 ; insufﬁcient evidence to
conclude that the mean ratings for the two groups differ
1.67 t = 2.68, reject H0 ; sufﬁcient evidence to conclude that
the mean performance level in the rudeness condition is less
than the mean in the control group
1.69 (a) t = −7.24, reject H0 (b) t = −.50, fail to reject H0
(c) t = −1.25, fail to reject H0
1.71 (a) each participant acted as a speaker and an audience member (b) μd = μspeaker − μaudience (c) No; need
sample statistics for differences (d) reject H0 : μd = 0
1.73 (a) t = −2.97, p-value = .016, fail to reject H0 (b) no
(c) t = .57, p-value = .58, fail to reject H0 ; no (d) t = 3.23,
p-value = .01, fail to reject H0 ; no
2
/σH2 = 1
1.75 F = 1.30, fail to reject H0 : σDM
2
1.77 F = 1.16, fail to reject H0 : σN /σB2 = 1
1.79 F = 5.87, reject H0 : σ12 /σ32 = 1
1.81 (a) 5, 21.5, 4.637 (b) 16.75, 36.25, 6.021 (c) 4.857,
29.81, 5.460 (d) 4, 0, 0
1.83 (a) z = −4 (b) z = .5 (c) z = 0 (d) z = 6
1.85 (a) all men and women (b) 300 people from
Gainesville, FL (c) inferential (d) qualitative
1.87 Relative frequencies: Burnished (.159), Monochrome
(.550), Slipped (.066), Paint-curvilinear (.017), Paintgeometric (.197), Paint-natural (.005), Cycladic (.005), Conical (.002)
781
782
Answers to Selected Exercises
1.89 (c) Eclipse
1.91 (a) .3085 (b) .1587 (c) .1359 (d) .6915 (e) 0
(f) .9938
1.93 (a) .26 (b) .086 (c) y = 7.43, s = .82; (5.79, 9.06); 95 %
(Empirical Rule) (d) y = 1.22, s = 5.11; (−9.00, 11.44); 95 %
(Empirical Rule)
1.95 (a) .3156 (b) .1894
1.97 (a) no, p ≈ 0 (b) μ and/or σ differ from stated values
1.99 (.61, 2.13); 99 % conﬁdent that mean weight of dry
seeds in the crops of all spinifex pigeons is between 0.61 and
2.13 grams.
1.101 (a) 293; 119.78 (c) .0158
1.103 t = 1.21, p-value = .257, fail to reject H0 : μ = 2,550
1.105 (a) z = −2.64, reject H0 : μNever − μRepeat = 0
1.107 (a) no; z = 1.19, fail to reject H0 : μMales − μFemales = 0
(b) .5 ± .690 (d) p = .2340 (d) F = 1.196, reject H0 :
2
2
/σFemales
=1
σMales
1.109 (a) yes; z = −6.02, p-value = 0, reject H0 : μS −
2
/σS2 = 1
μTRI = 0 (b) no; F = 1.41, fail to reject H0 : σTRI
Chapter 3
3.3 (a) β0 = 2; β1 = 2 (b) β0 = 4; β1 = 1 (c) β0 = −2;
β1 = 4 (d) β0 = −4; β1 = −1
3.5 (a) β1 = 2; β0 = 3 (b) β1 = 1; β0 = 1 (c) β1 = 3;
β0 = −2 (d) β1 = 5; β0 = 0 (e) β1 = −2; β0 = 4
3.7 (a) βˆ0 = 2; βˆ1 = −1.2
3.9 (a) yes (b) positive (c) line is based on sample data
3.11 (a) no (b) no (c) yes (d) negative (e) yes; winners
tend to punish less than non-winners
3.13 (a) yˆ = 6.25 − .0023x (c) 5.56
3.15 yˆ = .5704 + .0264x; since x = 0 is nonsensical, no
practical interpretation of βˆ0 = .5704; for each one-position
increase in order, estimated recall proportion increases by
βˆ1 = .0264
3.17 (a) .0313 (b) .1769
3.19 .2
(b) yˆ = 119.9 + .3456x
3.21 (a) y = β0 + β1 x + ε
(d) 635.2 (e) yˆ ± 1270.4
3.23 (a) yes, t = 6.71 (b) yes, t = −5.20
3.25 (−.0039, −.0008)
3.27 (a) t = 6.29, reject H0 (b) .88 ± .24 (c) no evidence
to say slope differs from 1
3.29 yes, t = 2.86
3.31 yes; t = 3.02, p-value = .009/2 = .0045, reject H0 :
β1 = 0 in favor of Ha : β1 > 0
3.33 (a) βˆ0 = .5151, βˆ1 = .000021 (b) yes, p-value =
.008/2 = .004 (c) Denver’s elevation is very high (d) βˆ0 =
.5154, βˆ1 = .000020 (b) no, p-value = .332/2 = .166
3.35 (a) .9583; .9183 (b) −.9487; .90
3.37 positive
3.39 (a) y = β0 + β1 x + ε (b) moderate positive linear
relationship between RMP and SET ratings (c) positive
(d) reject H0 at α = .05 (e) .4624
3.41 (b) .1998; .0032; .3832; .0864; .9006 (c) reject H0 ; fail
to reject H0 ; reject H0 ; fail to reject H0 ; reject H0
3.43 (a) fail to reject H0 : ρ = 0 at α = .05 (c) .25 (d) fail
to reject H0 : ρ = 0 at α = .05 (f) .0144
3.45 r = .5702; r 2 = .3251
3.47 (a) Negative (b) r = −.0895
3.49 (a) 4.06; .23 (b) 10.6 ± .233 (c) 8.9 ± .319
(d) 12.3 ± .319 (e) wider (f) 12.3 ± 1.046
3.51 (a) prediction interval for y with x = 10 (b) conﬁdence
interval for E(y) with x = 10
3.53 for x = 220, 95% conﬁdence interval for E(y) is
(5.65, 5.84)
3.55 (a) (3.17, 3.85) (b) (2.00, 5.02)
3.57 (0, 1535.3)
3.59 yˆ = 9470.5 + .192x; t = 12.69, p-value = 0, reject
H0 : β1 = 0; r 2 = .712; 2s = 1724; 95% PI for y when
x = 12, 000: (10,033, 13,508)
3.61 yˆ = 62.5 − .35x; t = −12.56, p-value = 0, reject
H0 : β1 = 0; r 2 = .07; 2s = 24.6; 95% PI for y when x = 50:
(21.1, 69.3)
3.63 (a) yˆ = −9.2667x (b) 12.8667; 3.2167; 1.7935
(c) yes; t = −28.30 (d) −9.267 ± .909 (e) −9.267 ± .909
(f) −9.267 ± 5.061
3.65 (a) yˆ = 5.364x (b) yes; t = 25.28 (c) 18.77 ± 6.30
3.67 (a) yˆ = 51.18x (b) yes; t = 154.56 (c) yˆ =
1, 855.45 + 47.07x; yes, t = 93.36 (d) y = β0 + β1 x + ε
3.69 (a) y = β0 + β1 x + ε; negative (b) no
3.71 yˆ = −.00105 + .00321x; t = 7.43, reject H0 : β1 = 0;
r 2 = .81; 2s = .014; 95% PI for y when x = 8: (.0085,. 0408)
3.73 100(r 2 )% of sample variation in ESLR score can be
explained by x (SG, SR, or ER score) in linear model
3.75 yˆ = −3.05 + .108x; t = 4.00, reject H0 : β1 = 0; r 2 = .57;
2s = 2.18
3.77 (b) yˆ = 78.52 − .24x (d) t = −2.31, fail to reject H0
(e) observation #5 is an outlier (f) yˆ = 139.76 − .45x; t =
−15.35, reject H0
3.79 (a) y = β0 + β1 x + ε (b) β1 > 0 (d) no; t = −3.12,
fail to reject H0
3.81 (a) yes; t = 3.05, reject H0 : ρ = 0 (b) mother,
number of friends; t = −2.43 (d) only testing for linear
relationships
3.83 (a) yˆ = 14, 012 − 1, 783x; no (at α = .05), t = −.98
(b) yˆ = 19, 680 − 3, 887x; yes (at α = .05), t = −2.29
(c) Predicting outside range of x (1.52–4.11)
Chapter 4
4.1 df = n − (number of independent variables + 1)
4.3 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 (b) 8% of the sample variation in frequency of marijuana is explained by
Answers to Selected Exercises
the model (c) reject H0 : β1 = β2 = β3 = 0 (d) reject
H0 : β1 = 0 (e) fail to reject H0 : β2 = 0 (f) fail to reject
H0 : β3 = 0
4.5 (a) yˆ = 3.70 + .34x1 + .49x2 + .72x3 + 1.14x4 + 1.51x5 +
.26x6 − .14x7 − .10x8 − .10x9 (b) t = −1.00, fail to reject
H0 (c) 1.51 ± .098
4.7 (a) for every 1-unit increase in proportion of block with
low density (x1 ), population density will increase by 2; for
every 1-unit increase in proportion of block with high density
(x2 ), population density will increase by 5 (b) 68.6% of the
sample variation in population density is explained by the
model (c) H0 : β1 = β2 = 0 (d) F = 133.27 (e) reject H0
4.9 (a) yˆ = 1.81231 + .10875x1 + .00017x2 (b) for every
1-mile increase in road length (x1 ), number of crashes
increase by .109; for every one-vehicle increase in AADT
(x2 ), number of crashes increase by .00017 (c) .109 ± .082
(d) .00017 ± .00008 (e) yˆ = 1.20785 + .06343x1 + .00056x2 ;
for every 1-mile increase in road length (x1 ), number of
crashes increase by .063; for every one-vehicle increase
in AADT (x2 ), number of crashes increase by .00056;
.063 ± .046; .00056 ± .00031
4.11 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 (b) yˆ =
21,087.95 + 108.45x1 + 557.91x2 − 340.17x3 + 85.68x4
(d) t = 1.22, p-value = .238, fail to reject H0 (e) R 2 = .912,
Ra2 = .894; Ra2 (f) F = 51.72, reject H0
4.13 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 + β5 x5
(b) yˆ = 13,614 + .09x1 − 9.20x2 + 14.40x3 + .35x4 − .85x5
(d) 458.8 ; ≈ 95% of sample heat rates fall within 917.6 kJ/kwhr of model predicted values (e) .917; 91.7% of the sample
variation in heat rate is explained by the model (f) yes,
F = 147.3, p-value ≈ 0
4.15 (a) F = 1.056; do not reject H0 (b) .05; 5% of the
sample variation in IQ is explained by the model
4.17 (a) 36.2% of the sample variation in active caring score
is explained by the model (b) F = 5.11, reject H0
4.21 (a) (2.68, 5.82) (b) (−3.04, 11.54)
4.23 95% PI for y when x1 = 23.755, x2 = 90.662, x3 = 25.0:
(24.03, 440.64)
4.25 (−1.233, 1.038)
4.27 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x1 x2 (b) linear relationship between number of defects and turntable speed
depends on blade position (c) β3 < 0
4.29 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x1 x2 (b) linear relationship between negative feelings score and number ahead
in line depends on number behind in line (c) fail to reject
H0 : β3 = 0 (d) β1 > 0; β2 < 0
4.31 (a) E(y) = β0 +β1 x1 + β2 x2 + β3 x3 + β4 x1 x3 + β5 x2 x3
(b) yˆ = 10,845 − 1280.0x1 + 217.4x2 − 1549.2x3 − 11.0x1 x3
+ 19.98x2 x3 (c) t = −.93, p-value = .355, fail to reject H0
(d) t = 1.78, p-value = .076, fail to reject H0 (e) no interaction
4.33 (a) slope depends on x2 and x5 (b) F = 5.505, R 2 =
.6792, s = .505; yes
4.35 (a) E(y) = β0 + β1 x1 + β2 x12 (b) β2 (c) negative
783
4.37 (a) yes (b) t = 2.69, p-value = .031, reject H0
4.39 (a) E(y) = β0 + β1 x + β2 x 2 (b) positive (c) no;
E(y) = β0 + β1 x1
4.41 (a) curvilinear trend (b) yˆ = 1.01 − 1.17x + .29x 2
(c) yes; t = 7.36, p-value ≈ 0, reject H0
4.43 (a) E(y) = β0 + β1 x1 + β2 x2 (b) E(y) = β0 + β1 x1 +
β2 x2 + β3 x3 + β4 x4
4.45 (a) E(y) = β0 + β1 x1 , where x = {1 if A, 0 if B}
(b) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 , where x1 = {1 if A, 0
if not}, x2 = {1 if B, 0 if not}, x3 = {1 if C, 0 if not};
β0 = μD , β1 = μA − μD , β2 = μB − μD , β3 = μC − μD
4.47 (c) Parallel lines
4.49 (b) second-order (c) different shapes (d) yes
(e) shift curves along the x1 -axis
4.51 (a) method: x1 = {1 if manual, 0 if automated}; soil:
x2 ={1 if clay, 0 if not}, x3 = {1 if gravel, 0 if not};
slope: x4 ={1 if east, 0 if not}, x5 = {1 if south, 0 if not},
x6 = {1 if west, 0 if not}, x7 = {1 if southeast, 0 if not}
(b) E(y) = β0 + β1 x1 ; β0 = μAuto , β1 = μManual − μAuto
(c) E(y) = β0 +β1 x2 + β2 x3 ; β0 = μSand , β1 = μClay − μSand ,
β2 = μGravel − μSand (d) E(y) = β0 + β1 x4 + β2 x5 + β3 x6 +
β4 x7 ; β0 = μSW , β1 = μE − μSW , β2 = μS − μSW , β3 = μW −
μSW , β4 = μSE − μSW
4.53 (a) E(y) = β0 + β1 x1 + β2 x2 , where x1 = {1 if Full
solution, 0 if not}, x2 = {1 if Check ﬁgures, 0 if
not} (b) β1 (c) yˆ = 2.433 − .483x1 + .287x2 (c) F = .45,
p-value = .637, fail to reject H0
4.55 (a) reject H0 : β1 = β2 = . . . β12 = 0; sufﬁcient evidence to indicate the model is adequate for predicting
log of card price (b) fail to reject H0 : β1 = 0 (c) reject
H0 : β3 = 0 (d) E[ln(y)] = β0 + β1 x4 + β2 x5 + β3 x6 + . . . +
β9 x12 + β10 x4 x5 + β11 x4 x6 + . . . + β17 x4 x12
4.57 (a) Model 1: t = 2.58, reject H0 ; Model 2: reject
H0 : β1 = 0 (t = 3.32), reject H0 : β2 = 0 (t = 6.47), reject
H0 : β3 = 0 (t = −4.77), do not reject H0 : β4 = 0 (t = .24);
Model 3: reject H0 : β1 = 0 (t = 3.21), reject H0 : β2 = 0
(t = 5.24), reject H0 : β3 = 0 (t = −4.00), do not reject
H0 : β4 = 0 (t = 2.28), do not reject H0 : β5 = 0 (t = .014)
(c) Model 2
4.59 (a) yˆ = 80.22 + 156.5x1 − 42.3x12 + 272.84x2 +
760.1x1 x2 + 47.0x12 x2 (b) yes, F = 417.05, p-value ≈ 0
(c) no; fail to reject H0 : β2 = 0 and H0 : β5 = 0
4.61 Nested models: a and b, a and d, a and e, b and c, b and
d, b and e, c and e, d and e
4.63 (a) 10.1% (55.5%) of the sample variation in aggression
score is explained by Model 1 (Model 2) (b) H0 : β5 = β6 =
β7 = β8 = 0 (c) yes (d) reject H0 (e) E(y) = β0 + β1 x1
+ β2 x2 + β3 x3 + β4 x4 + β5 x5 + β6 x6 + β7 x7 + β8 x8 + β9 x5 x6
+ β10 x5 x7 + β11 x5 x8 + β12 x6 x7 + β13 x6 x8 + β14 x7 x8 (f) fail
to reject H0 : β9 = β10 = . . . = β14 = 0
4.65 (a) E(y) = β0 + β1 x1 + β2 x2 + . . . + β11 x11 (b) E(y) =
β0 + β1 x1 + β2 x2 + . . . + β11 x11 + β12 x1 x9 + β13 x1 x10 +
β14 x1 x11 + β15 x2 x9 + β16 x2 x10 + β17 x2 x11 + . . . + β35 x8 x11
(c) Test H0 : β12 = β13 = β14 = . . . = β35 = 0
784
Answers to Selected Exercises
4.67 (a) E(y) = β0 + β1 x1 + β2 x2 + . . . + β10 x10 (b) H0 : β3 =
β4 = . . . = β10 = 0 (c) reject H0 (e) 14 ± 5.88 (f) yes
(g) E(y) = β0 + β1 x1 + β2 x2 + . . . + β10 x10 + β11 x2 x1 + β12
x2 x3 + . . . + β19 x2 x10 (h) H0 : β11 = β12 = . . . = β19 = 0;
partial (nested model) F -test
4.69 (a) multiple t-tests result in an increased Type I error
rate (b) H0 : β2 = β5 = 0 (c) fail to reject H0
4.71 (a) yes; t = 5.96, reject H0 (b) t = .01, do not reject
H0 (c) t = 1.91, reject H0
4.73 (a) estimate of β1 (b) prediction equation less reliable
for values of x’s outside the range of the sample data
4.75 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 + β5 x5
(b) F = 34.47, reject H0 (c) E(y) = β0 + β1 x1 + β2 x2 +
(d) 60.3% of the sample variation in GSI
. . . + β7 x7
is explained by the model (e) reject H0 for both
4.77 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x1 x2 (b) p-value =
.02, reject H0 (c) impact of intensity on test score depends
on treatment and is measured by (β2 + β3 x1 ), not by β2 alone
4.79 (a) Negative (b) F = 1.60, fail to reject H0 (c) F =
1.61, fail to reject H0
4.81 yes; t = 4.20, p-value = .004
4.83 (a) E(y) = β0 + β1 x1 + β2 x2 where x1 = {1 if Communist, 0 if not}, x2 = {1 if Democratic, 0 if not} (b) β0 =
μDictator , β1 = μCommunist − μDictator , β2 = μDemocratic
− μDictator
4.85 no; income (x1 ) not signiﬁcant, air carrier dummies
(x3 –x30 ) are signiﬁcant
4.87 (a) quantitative (b) quantitative (c) qualitative
(d) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 , where x1 =
{1 if Benzene, 0 if not}, x2 ={1 if Toluene, 0 if not},
x3 = {1 if Chloroform, 0 if not}, x4 ={1 if Methanol,
0 if not} (e) β0 = μA , β1 = μB − μA , β2 = μT − μA , β3 =
μC − μA , β4 = μM − μA (f) F -test with H0 : β1 = β2 =
β3 = β4 = 0
4.89 (a) H0 : β5 = β6 = β7 = β8 = 0; males: F > 2.37;
females: F > 2.45 (c) reject H0 for both
Chapter 5
5.1 (a) Qualitative (b) Quantitative (c) Quantitative
5.3 Gender: qualitative; Testimony: qualitative
5.5 (a) Quantitative (b) Quantitative (c) Qualitative
(d) Qualitative
(e) Qualitative
(f) Quantitative
(g) Qualitative (h) Qualitative (i) Qualitative (j) Quantitative (k) Qualitative
5.7 (a) (i) 1; (ii) 3; (iii) 1; (iv) 2 (b) (i) E(y) = β0 + β1 x;
(ii) E(y) = β0 + β1 x + β2 x 2 + β3 x 3 ; (iii) E(y) = β0 + β1 x;
(iv) E(y) = β0 + β1 x + β2 x 2 (c) (i) β1 > 0; (ii) β3 > 0;
(iii) β1 < 0; (iv) β2 < 0
5.9 E(y) = β0 + β1 x + β2 x 2 + β3 x 3
5.11 E(y) = β0 + β1 x + β2 x 2
5.15 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x1 x2 + β4 x12 + β5 x22
(b) E(y) = β0 + β1 x1 + β2 x2 (c) E(y) = β0 + β1 x1 + β2 x2
+ β3 x1 x2 (d) β1 + β3 x2 (e) β2 + β3 x1
5.17 (a) E(y) = β0 + β1 x1 + β2 x12 + β3 x2 + β4 x22 + β5 x4 +
β6 x42 + β7 x1 x2 + β8 x1 x4 + β9 x2 x4
(b) yˆ = 10, 283.2 +
276.8x1 + .990x12 + 3325.2x2 + 266.6x22 + 1301.3x4 + 40.22x42
+ 41.98x1 x2 + 15.98x1 x4 + 207.4x2 x4 ; yes, F = 613.278,
reject H0 (c) F = 108.43, reject H0
5.19 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x1 x2 + β4 x12 + β5 x22
(b)
yˆ = 15,583 + .078x1 − 523x2 + .0044x1 x2 − .0000002x12
+ 8.84x22 (c) F = 93.55, reject H0 (f) graphs have similar
shape
5.21 (a) u = (x − 85.1)/14.81 (b) −.668, .446, 1.026, −1.411,
−.223, 1.695, −.527, −.338 (c) .9967 (d) .376 (e) yˆ =
110.953 + 14.377u + 7.425u2
5.23 (a) .975, .928, .987 (b) u = (x − 5.5)/3.03 (c) 0, .923,
0; yes
5.25 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 , where x1 = {1 if
low, 0 if not}, x2 = {1 if moderate, 0 if not}, x3 = {1 if high,
0 if not} (b) β0 = μNone , β1 = μLow − μNone , β2 = μMod −
μNone , β3 = μHigh − μNone (c) F -test of H0 : β1 = β2 =
β3 = 0
5.27 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x1 x2 + β5 x1 x3 ,
where x1 = {1 if manual, 0 if automated}, x2 = {1 if clay, 0
if not}, x3 = {1 if gravel, 0 if not} (b) μAutomated/Sand (c) β0 +
β1 + β2 + β4 (d) β1
5.29 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x1 x2 , where x1 =
{1 if common, 0 if ambiguous}, x2 = {1 if low, 0 if high}
(b) βˆ0 = 6.1, βˆ1 = .2, βˆ2 = 11.9, βˆ3 = −10.4 (c) t-test for
H0 : β3 = 0
5.31 (a) E(y) = β0 + β1 x1 + β2 x12 + β3 x2 + β4 x3 + β5 x1 x2
+ β6 x1 x3 + β7 x12 x2 + β8 x12 x3 , where x1 = level of bullying,
x2 = {1 if low, 0 if not}, x3 = {1 if neutral, 0 if not} (b) β0 +
25β1 + 625β2 + β3 + 25β5 + 625β7 (c) nested F -test of
H0 : β2 = β7 = β8 = 0 (d) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 +
β4 x1 x2 + β5 x1 x3 (e) low: β1 + β4 ; neutral: β1 + β5 ; high: β1
5.33 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 + β5 x5 +
β6 x6 + β7 x7 + β8 x1 x2 + β9 x1 x3 + β10 x1 x4 + β11 x1 x5 +
β12 x1 x6 + β13 x1 x7 + β14 x2 x4 + β15 x2 x5 + β16 x2 x6 + β17 x2 x7 +
β18 x3 x4 + β19 x3 x5 + β20 x3 x6 + β21 x3 x7 + β22 x1 x2 x4 +
β23 x1 x2 x5 + β24 x1 x2 x6 + β25 x1 x2 x7 + β26 x1 x3 x4 + β27 x1 x3 x5
+ β28 x1 x3 x6 + β29 x1 x3 x7 , where x1 = {1 if manual, 0 if
automated}, x2 = {1 if clay, 0 if not}, x3 = {1 if gravel, 0 if not},
x4 = {1 if East, 0 if not}, x5 = {1 if South, 0 if not}, x6 = {1 if
West, 0 if not}, x7 = {1 if Southeast, 0 if not} (b) μAutomated
/Sand/SW (c)β0 + β1 + β2 + β4 + β8 + β10 + β14 + β22 (d) β1
(e) β8 = β9 = β22 = β23 = β24 = β25 = β26 = β27 = β28 = β29 = 0
5.35 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x1 x2 (c) β1 + β3
(d) no; F = .26, p-value = .857, do not reject H0 (e) E(y) =
β0 + β1 x1 + β2 x12 + β3 x2 + β4 x1 x2 + β5 x12 x2
5.37 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 (b) E(y) = β0 +
β1 x1 + β2 x2 + β3 x3 + β4 x1 x2 + β5 x1 x3 (c) AL: β1 ; TDS:
β1 + β4 ; FE: β1 + β5 (d) nested F -test of H0 : β4 = β5 = 0
5.39 (a) Qualitative (b) Quantitative (c) Quantitative
5.41 (a) Quantitative (b) Quantitative (c) Qualitative (d) Qualitative (e) Qualitative (f) Qualitative
(g) Quantitative (h) Qualitative
Answers to Selected Exercises
5.43 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 (b) μBA − μN
(c) μE − μN (d) μLAS − μN (e) μJ − μN (f) E(y) = (β0 +
β5 ) + β1 x1 + β2 x2 + β3 x3 + β4 x4 (g) μBA − μN (h) μE − μN
(i) μLAS − μN (j) μJ − μN (k) μF − μM l. Reject
H0 : β5 = 0; gender has an effect
5.45 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 , where x1 = {1 if
boy, 0 if girl}, x2 = {1 if youngest third, 0 if not}, x3 = {1 if
middle third, 0 if not} (b) β0 = μGirls/Oldest , β1 = μBoys −
μGirls , β2 = μYoungest − μOldest , β3 = μMiddle − μOldest
(c) E(y) = β0 +β1 x1 +β2 x2 +β3 x3 + β4 x1 x2 + β5 x1 x3 (d) .21,
−.05, .06, −.03, .11, .20 (e) nested F -test of H0 : β4 = β5 = 0
5.47 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 , where x1 =
{1 if P1 , 0 if P2 }, x2 = {1 if L1 , 0 if not}, x3 = {1 if L2 , 0 if not},
x4 = {1 if L3 , 0 if not} (b) 8; E(y) = β0 + β1 x1 + β2 x2 +
β3 x3 + β4 x4 + β5 x1 x2 + β6 x1 x3 + β7 x1 x4 (c) F = 2.33, do
not reject H0
5.49 (a) E(y) = β0 + β1 x1 + β2 x2 , where x1 = years of
education, x2 = {1 if Certiﬁcate, 0 if not} (b) E(y) = β0 +
β1 x1 + β2 x2 + β3 x1 x2 (c) E(y) = β0 + β1 x1 + β2 x12 + β3 x2 +
β4 x1 x2 + β5 x12 x2
5.51 (a) F = 8.79, p-value = .0096; reject H0 (b) DF-2:
2.14; blended: 4.865; adv. timing: 7.815
5.53 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 , where x1 = {1 if
Low, 0 if not}, x2 = {1 if Moderate, 0 if not}, x3 = {1 if producer, 0 if consumer} (b)β0 = μHigh/Cons , β1 = μLow − μHigh ,
β2 = μMod − μHigh , β3 = μProd − μCons (c) E(y) = β0 +
β1 x1 + β2 x2 + β3 x3 + β4 x1 x3 + β5 x2 x3 (d) β0 = μHigh/Cons ,
β1 = (μLow − μHigh ) for consumers, β2 = (μMod − μHigh )
for consumers, β3 = (μProd − μCons ) for high, β4 = (μLow −
μHigh )Prod − (μLow − μHigh )Cons , β5 = (μMod − μHigh )Prod −
(μMod − μHigh )Cons , (e) nested F -test of H0 : β4 = β5 = 0
5.55 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 (b) t-test of
H0 : β3 = 0 vs. Ha : β3 < 0 (c) E(y) = β0 + β1 x1 +
β2 x2 + β3 x1 x2 + β4 x12 + β5 x22 + β6 x3 + β7 x1 x3 +
β8 x2 x3 + β9 x1 x2 x3 + β10 x12 x3 + β11 x22 x3 (d) nested F -test
of H0 : β1 = β3 = β4 = β7 = β9 = β10 = 0
Chapter 6
6.1 (a) x2 (b) yes (c) ﬁt all possible 2-variable models,
E(y) = β0 + β1 x2 + β2 xj
6.3 (a) 11 (b) 10 (c) 1 (d) E(y) = β0 + β1 x11 + β2 x4 +
β3 x2 + β4 x7 + β5 x10 + β6 x1 + β7 x9 + β8 x3 (e) 67.7% of
sample variation in satisfaction is explained by the model
(f) no interactions or higher-order terms tested
6.5 (a) x5 , x6 , and x4 (b) no (c) E(y) = β0 + β1 x4 +
β2 x5 + β3 x6 + β4 x4 x5 + β5 x4 x6 + β6 x5 x6 (d) nested F -test
of Ho : β4 = β5 = β6 = 0 (e) consider interaction and
higher-order terms
6.7 (a) (i) 4; (ii) 6; (iii) 4; (iv) 1 (b) (i) .213, 193.8, 2.5,
10,507; (ii) .247, 189.1, 2.3, 10,494; (iii) .266, 188.2, 3.1,
10,489; (iv) .268, 191.7, 5.0, 10,710 (d) x2 , x3 , x4
6.9 yes; DOT estimate and low-bid-estimate ratio
6.11 Stepwise: well depth and percent of land allocated to
industry
785
Chapter 7
7.1 model less reliable
7.3 (a) x = ln(p) (b) yes, t = −15.89 (c) (924.5, 975.5)
7.5 (a) no (b) no
7.7 (a) no (b) yes
7.9 Unable to test model adequacy since df(Error) =
n−3=0
7.11 (a) yˆ = 2.74 + .80x1 ; yes, t = 15.92 (b) yˆ = 1.66 +
12.40x2 ; yes, t = 11.76 (c) yˆ = −11.80 + 25.07x3 ; yes,
t = 2.51 (d) yes
7.13 (a) multicollinearity (b) no, β3 not estimable
7.15 no multicollinearity, use DOTEST and LBERATIO
7.17 Two levels each; n > 4
7.19 yes, high VIFs for Inlet-temp, Airﬂow and Power;
include only one of these 3 variables
7.21 (a) .0025; no (b) .434; no (c) no (d) yˆ = −45.154 +
3.097x1 + 1.032x2 , F = 39,222.34, reject H0 : β1 = β2 = 0;
R 2 = .9998 (e) –.8998; high correlation (f) no
7.23 df(Error) = 0, s 2 undeﬁned, no test of model
adequacy
Chapter 8
8.1 (a) yˆ = 2.588 + .541x (b) −.406, −.206, −.047, .053, .112,
.212, .271, .471, .330, .230, −.411, −.611 (c) Yes; needs
curvature
8.3 (a) yˆ = 40.35 − .207x (b) −4.64, −3.94, −1.83, .57, 2.58,
1.28, 4.69, 4.09, 4.39, 2.79, .50, 1.10, −6.09, −5.49 (c) Yes;
needs curvature (d) yˆ = −1051 + 66.19x − 1.006x 2 ; yes,
t = −11.80
8.5 yˆ = 30, 856 − 191.57x; yes, quadratic trend; yes
8.7 (a) −389, −178, 496, . . . , 651 (b) No trends (c) No
trends (d) No trends (e) No trends
8.9 Yes; assumption of constant variance appears satisﬁed;
add curvature
8.11 Yes
8.13 (a) Yes; assumption of equal variances violated
√
(b) Use transformation y* = y
8.15 (a) yˆ = .94 − .214x (b) 0, .02, −.026, .034, .088, −.112,
−.058, .002, .036, .016 (c) Unequal variances (d) Use
√
ˆ = 1.307 − .2496x; yes
transformation y* = sin−1 y (e) y*
8.17 (a) Lagos: −.223 (b) yes
8.19 No; remove outliers or normalizing transformation
8.21 Residuals are approximately normal
8.23 No outliers
8.25 Observations #8 and #3 are inﬂuential
8.27 No outliers
8.29 Several, including wells 4567 and 7893; both inﬂuential;
possibly delete
8.31 Inﬂated t-statistics for testing model parameters
8.33 (b) Model adequate at α = .05 for all banks except bank
5 (c) Reject H0 (two-tailed at α = .05) for banks 2, 5; fail
786
Answers to Selected Exercises
to reject H0 for banks 4, 6, 8; test inconclusive for banks 1, 3,
7, 9
8.35 (a) Yes; residual correlation (b) d = .058, reject H0
(c) Normal errors
8.37 (a) yˆ = 1668.44 + 105.83t; yes, t = 2.11, reject H0
(b) Yes (c) d = .845, reject H0
8.39 (a) Misspeciﬁed model; quadratic term missing
(b) Unequal variances (c) Outlier (d) Unequal variances
(e) Nonnormal errors
8.41 Assumptions reasonably satisﬁed
8.43 Assumptions reasonably satisﬁed
8.45 (a) yˆ = −3.94 + .082x (b) R 2 = .372; F = 2.96, p-value
= .146, model not useful (d) Obs. for horse #3 is outside 2s
interval (e) Yes; R 2 = .970; F = 130.71, p-value = 0
8.47 Possible violation of constant variance assumption
8.49 Violation of constant variance assumption; variancestabilizing transformation ln(y)
Chapter 9
9.1 (a) E(y) = β0 + β1 x1 + β2 (x1 − 15)x2 , where x1 = x
and x2 = {1 if x1 > 15, 0 if not} (b) x ≤ 15 : y-intercept
= β0 , slope = β1 ; x > 15: y-intercept = β0 − 15β2 , slope =
β1 + β2 (c) t-test for H0 : β2 = 0
9.3 (a) E(y) = β0 + β1 x1 + β2 (x1 − 320)x2 + β3 x2 , where
x1 = x and x2 = {1 if x1 > 320, 0 if not} (b) x ≤ 320:
y-intercept = β0 , slope = β1 ; x > 320: y-intercept = β0 −
320β2 + β3 , slope = β1 + β2 (c) nested F -test for H0 : β2 =
β3 = 0
9.5 (a) 4 and 7 (b) E(y) = β0 + β1 x1 + β2 (x1 − 4)x2 +
β3 (x1 − 7)x3 , where x1 = x, x2 = {1 if x1 > 4, 0 if not}, x3 =
{1 if x1 > 7, 0 if not} (c) x ≤ 4 : β1 ; 4 < x ≤ 7 : (β1 + β2 );
x > 7: (β1 + β2 + β3 ) (d) for every 1-point increase
in performance over range x ≤ 4, satisfaction increases
5.05 units
9.7 (a) yes; 3.55 (b) E(y) = β0 + β1 x1 + β2 (x1 − 3.55)x2 ,
where x1 = load and x2 = {1 if load > 3.55, 0 if not}
(c) y=
ˆ 2.22 + .529x1 + 2.63(x1 − 3.55)x2 ; R 2 = .994, F = 1371,
p-value = 0, reject H0
9.9 135.1 ± 74.1
9.11 1.93 ± 16.45
9.13 (a) yˆ = −2.03 + 6.06x (b) t = 10.35, reject H0
(c) 1.985 ± .687
9.15 (a) yˆ = − 3.367 + .194x; yes, t = 4.52 (p-value = .0006)
(b) Residuals: −1.03, 6.97, −5.03, −2.03, 1.97, −6.73, 3.27,
−5.73, 9.27, −1.73, −9.43, 12.57, −8.43, 2.57, 3.57; as speed
increases, variation tends to increase (c) wi = 1/xi2 ; x =
100: 20.7; x = 150: 44.3; x = 200: 84.3 (d) y=
ˆ − 3.057 +
.192x; s = .044
9.17 (a) yˆ = 140.6 − .67x; assumption violated (b) w =
1/(x)
¯ 2
9.19 violation of constant error variance assumption; violation of assumption of normal errors; predicted y is not
bounded between 0 and 1
9.21 (a) β1 = change in P(hired) for every 1-year increase in
higher education; β2 = change in P(hired) for every 1-year
increase in experience; β3 = P(hired)Males - P(hired)Females
(b) yˆ = −.5279 + .0750x1 + .0747x2 + .3912x3 (c) F =
21.79, reject H0 (d) yes; t = 4.01 (e) (−.097, .089)
9.23 (a) P(Maryland nappe) (b) π ∗ = β0 + β1 x, where
π ∗ = ln{π/(1 − π )} (c) change in log-odds of a Maryland nappe for every 1-degree increase in FIA (d) exp
(β0 + 80β1 )/{1 − exp(β0 + 80β1 )}
9.25 (a) χ 2 = 20.43, reject H0 (b) Yes; χ 2 = 4.63
(c) (.00048, .40027)
9.27 (a) P(landfast ice) (b) π ∗ = β0 + β1 x1 + β2 x2
+ β3 x3 , where π ∗ = ln{π/(1 − π )} (c) πˆ ∗ = .30 + 4.13x1 +
47.12x2 − 31.14x3 (d) χ 2 = 70.45, reject H0 (e) π ∗ = β0
+ β1 x1 + β2 x2 + β3 x3 + β4 x1 x2 + β5 x1 x3 + β6 x2 x3
(f) πˆ ∗ = 6.10−3.00x1 + 10.56x2 − 39.69x3 + 50.49x1 x2 − 6.14
x1 x3 + 56.24x2 x3 (g) χ 2 = 32.19, reject H0 ; interaction
model is better
Chapter 10
10.1 (a) yes; yes (b) 366.3, 335.8, 310.8, 284.0, 261.5, 237.0,
202.3, 176.8, 155.5 (c) yes (d) 81.9 (e) 113.5 (f) Quarter 1: 94.2; Quarter 2: 107.8
10.3 (a) Moving average: 16.2; Exponential smoothing: 107;
Holt-Winters: 35.9 (b) Moving average: 16.2; Exponential
smoothing: 109.4; Holt–Winters: 43.9 (c) Moving average
10.5 (a) Yes (b) forecast = 231.7 (c) forecast = 205.4
(d) forecast = 232.8
10.7 (a) yes (b) 2006: 495; 2007: 540 ; 2008: 585 (d) 2006:
435.6 ; 2007: 435.6; 2008: 435.6 (e) 2006: 480.85; 2007: 518.1;
2008: 555.35 (f) moving average most accurate
10.9 (a) βˆ0 = 4.36: estimated price is $4.36 in 1990; βˆ1 = .455:
for each additional year, price increases by $.455 (b) t =
8.43, p-value = 0, reject H0 (c) 2008: (9.742, 15.357); 2009:
(10.151, 15.858) (d) extrapolation; no account for cyclical
trends
10.11 (a) E(yt ) = β0 + β1 t + β2 Q1 + β3 Q2 + β4 Q3 (b) yˆ t =
119.85 + 16.51t + 262.34Q1 + 222.83Q2 + 105.51Q3 ; F =
117.82, p-value = 0, reject H0 (c) independent error
(d) Q1 : 728.95, (662.8, 795.1); Q2 : 705.95, (639.8 772.1); Q3 :
605.15, (539.0, 671.3); Q4 : 516.115, (450.0, 582.3)
10.13 (a)yes (b) yˆ t = 39.49 + 19.13t − 1.315t 2 (d) (−31.25,
48.97)
10.15 (a) no, t = −1.39 (b) 2003.48 (c) no, t = −1.61
(d) 1901.81
10.17 (a) 0, 0, 0, .5, 0, 0, 0, .25, 0, 0, 0, .125, 0, 0, 0, .0625, 0, 0,
0, .03125 (b) .5, .25, .125, .0625, .03125, .0156, . . .
10.19 Rt = φ1 Rt−1 + φ2 Rt−2 + φ3 Rt−3 + φ4 Rt−4 + εt
10.21 (a) E(yt ) = β0 +β1 x1t + β2 x2t + β3 x3t + β4 t (b)E(yt ) =
β0 + β1 x1t + β2 x2t + β3 x3t + β4 t + β5 x1t t + β6 x2t t + β7 x3t t
(c) Rt = φRt−1 + εt
10.23 (a) E(yt ) = β0 + β1 [cos(2π /365)t] +β2 [sin(2π /365)t]
(c) E(yt ) = β0 +β1 [cos(2π /365)t] + β2 [sin(2π /365)t] +β3 t +
β4 t[cos(2π /365)t] +β5 t[sin(2π /365)t] (d) no; Rt = φRt−1 +εt
10.25 (a) yt = β0 + β1 t + φRt−1 + εt (b) yˆ t = 11, 374 +
160.23t + .3743Rˆ t−1 (d) R 2 = .9874, s = 115.13
Answers to Selected Exercises
10.27 (a) yes, upward trend (b) yt = β0 + β1 t + β2 t 2 +
φRt−1 + εt (c) yˆ t = 263.14 + 1.145t + .056t 2 + .792Rˆ t−1 ;
R 2 = .747; t = 3.66, p-value = .0004, reject H0
10.29 (a) F49 = 336.91; F50 = 323.41; F51 = 309.46 (b) t =
49: 336.91 ± 6.48; t =50: 323.41 ± 8.34; t = 51: 309.46 ± 9.36
10.35 (a) 2136.2 (b) (1404.3, 3249.7) (c) 1944; (1301.8,
2902.9)
10.37 (a) yes; possibly (b) 64 (d) 86.6 (e) 73.5
(f) yt = β0 + β1 t + β2 S1 + β3 S2 + β4 S3 + φRt−1 + εt , where
S1 = {1 if Jan/Feb/Mar, 0 if not}, S2 = {1 if Apr/May/Jun,
0 if not}, S3 = {1 if Jul/Aug/Sep, 0 if not} (g) yˆ t = 101.69 −
1.47t + 13.32S1 + 29.94S2 + 32.45S3 − .543Rˆ t−1 (h)96.1 ±11
10.39 (a) yes, curvilinear trend (b) yt = β0 + β1 t + β2 t 2 +
εt (c) yˆ t = 189.03 + 1.38t − .041t 2 (e) yes (f) Durbin–
Watson test; d =.96, reject H0 (g) yt = β0 + β1 t + β2 t 2 +
φRt−1 + εt ; (h) yˆ t = 188.7 + 1.39t − .04t 2 + .456Rˆ t−1
10.41 (a) E(yt ) = β0 + β1 t + β2 t 2 + β3 x,where x = {1 if Jan–
Apr, 0 if not} (b) add interaction terms
10.43 (a) μPost − μPre (b) μPre (c) −.55 (d) 2.53
Chapter 11
11.1 (a) Noise (variability) and volume (n) (b) remove
noise from an extraneous source of variation
11.3 (a) cockatiel (b) yes; completely randomized design
(c) experimental group (d) 1,2,3 (e) 3 (f) total consumption (g) E(y) = β0 + β1 x1 + β2 x2 , where x1 = {1 if group 1,
0 if not}, x2 = {1 if group 2, 0 if not}
11.5 (a) yB1 = β0 + β2 + β4 + εB1 ; yB2 = β0 + β2 + β5 + εB2 ;
. . . ; yB,10 = β0 + β2 + εB,10 ; y¯B = β0 + β2 + (β4 + β5 + · · · +
yD1 = β0 + β4 + εD1 ; yD2 = β0 + β5 +
β12 )/10 + ε¯ B (b)
y¯D = β0 + (β4 + β5 + · · · +
εD2 ; . . . ; yD,10 = β0 + εD,10 ;
β12 )/10 + ε¯ D
11.7 ability to investigate factor interaction
11.9 (a) students (b) yes; factorial design (c) class standing and type of preparation (d) class standing: low, medium,
high; type of preparation: review session and practice exam
(e) (low, review), (medium, review), (high, review), (low,
practice),(medium, practice), (high, practice) (f) ﬁnal exam
score
11.11 (a) training method, practice session, task consistency (b) task consistency is QL, others are QN
(c) 48; (CC/1/100), (AC/1/100), (CC/2/100), (AC/2/100),
. . . ,(CC/6/33), (AC/6/33)
11.13 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x1 x2 + β5 x1 x3 ,
where x1 is dummy variable for QL factor A; x2, x3
are dummy variables for QL factor B (b) E(y) = β0
+ β1 x1 + β2 x2 + β3 x3 + β4 x4 + β5 x5 + β6 x1 x2 + β7 x1 x3
+ β8 x1 x4 + β9 x1 x5 + β10 x2 x4 + β11 x2 x5 + β12 x3 x4 + β13 x3 x5 +
β14 x1 x2 x4 + β15 x1 x2 x5 + β16 x1 x3 x4 + β17 x1 x3 x5 , where x1 =
QN factor A; x2 , x3 are dummy variables for QL factor B;
x4 , x5 are dummy variables for QL factor C
11.15 Cannot investigate factor interaction
11.17 11
11.19 sample size (n) and standard deviation of estimator
787
11.21 Step 4
11.23 8 treatments: A1 B1 , A1 B2 , A1 B3 , A1 B4 , A2 B1 , A2 B2 ,
A2 B3 , A2 B4
11.25 E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 + β5 x5 ; 10
11.27 (a) Sex and weight (b) Both qualitative (c) 4 ;
(ML), (MH), (FL), and (FH)
Chapter 12
12.3 (a) E(y) = β0 + β1 x, where x = {1 if treatment 1, 0 if
treatment 2} (b) yˆ = 10.667 − 1.524x; t = −1.775, fail to
reject H0
12.5 (a) t = −1.78; do not reject H0 (c) Two-tailed
12.7 (a) completely randomized design (b) colonies 3, 6,
9 and 12; energy expended (c) H0 : μ1 = μ2 = μ3 = μ4
(d) Reject H0
12.9 (a) H0 : μ1 = μ2 = μ3 (b) E(y) = β0 + β1 x1 + β2 x2
(c) Reject H0 (d) Fail to reject H0
12.11 (a)
Source
df
SS
MS
F
Solvent
Error
2
29
3.3054
1.9553
1.6527
0.0674
24.51
Total
31
5.2607
(b) F = 24.51, reject H0
12.13 F = 7.69, reject H0
12.15 F = 9.97, reject H0
12.17 (a) H0 : μ1 = μ2 = μ3
(b)
Source
df
SS
MS
F
Level
Error
2
72
6.643
527.357
3.322
7.324
0.45
Total
74
534.000
(c) fail to reject H0
12.19 (a) same subjects used in parts A and B (b) response
= WTP; treatments = A and B; blocks: subjects (c) treatments: H0 : μA = μB
12.21 No evidence of a difference among the three plant
session means; F = .019
12.23 (a) same genes examined across all three conditions (b) H0 : μFull-Dark = μTR-Light = μTR-Dark (c) F =
5.33, reject H0
12.25 (a) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 + β4 x4 + β5 x5 +
· · · + β10 x10 , where x1 , x2 and x3 are dummy variables for
interventions (treatments), x4 , x5, . . . , x10 are dummy variables for boxers (blocks) (b) E(y) = β0 + β4 x4 + β5 x5 +
· · · + β10 x10 , where x4 , x5, . . . , x10 are dummy variables for
boxers (blocks) (c) E(y) = β0 + β1 x1 + β2 x2 + β3 x3 , where
x1 , x2 and x3 are dummy variables for interventions (treatments)
12.27 Yes, F = 34.12