Chap 17
17.6 a
Scatter Diagram
30
Test scores
25
20
15
10
5
0
0
20
40
60
80
Lengths
b b1 
s xy
s 2x
=
51 .86
= .2675, b 0  y  b1x = 13.80 – .2675(38.00) = 3.635
193 .9
Regression line: ŷ = 3.635 + .2675x (Excel: ŷ = 3.636 + .2675x)
c b1 = .2675; for each additional second of commercial, the memory test score increases on average by
.2675. b0 = 3.64 is the y-intercept.
17.8 a
Scatter Diagram
200
Income
150
100
50
0
0
5
10
15
20
25
Education
b b1 
s xy
s 2x
=
46 .02
 4.138 , b 0  y  b1x = 78.13 – 4.138(13.17) =23.63.
11 .12
Regression line: ŷ = 23.63 + 4.138x (Excel: ŷ = 23.63 + 4.137x)
c The slope coefficient tells us that for each additional year of education income increases on average by
$4.138 thousand ($4,138). The y-intercept has no meaning.
17.16 a b1 
s xy
s 2x
=
10 .78
 .3039 , b 0  y  b1x = 17.20 – (–.3039)(11.33) = 20.64.
35 .47
Regression line: ŷ = 20.64 – .3039x (Excel: ŷ = 20.64 – .3038x)
b The slope indicates that for each additional one percentage point increase in the vacancy rate rents on
average decrease by $.3039. The y-intercept is 20.64.
17.18 b1 
s xy
s 2x
=
.8258
 .0514 , b 0  y  b1x = 93.89 –.0514(79.47) = 89.81.
16 .07
Regression line: ŷ = 89.81 + .0514x (Excel: ŷ = 89.81 + .0514x)
17.98 a b1 
s xy
s 2x
=
936 .82
 2.47 b 0  y  b1x = 395.21 – 2.47(113.35) = 115.24.
378 .77
Regression line: ŷ = 115.24 + 2.47x (Excel: ŷ = 114.85 + 2.47x)
b b1 = 2.47; for each additional month of age, repair costs increase on average by $2.47.
b0 = 114.85 is the y-intercept.
c R2 
s 2xy
s 2x s 2y
=
(936 .82)2
 .5659 (Excel: R 2 = .5659) 56.59% of the variation in repair costs s
(378 .77 )( 4,094 .79)
explained by the variation in ages.
H0 :   0
17.104
H1 :   0
Rejection region: t  t , n  2  t .05, 428  1.645 or
r
s xy
s xs y
tr

n2
1  r2
255 ,877
(99 .11)( 2,152 ,602 ,614 )
 (.5540 )
430  2
1  (.5540 ) 2
 .5540 (Excel: .5540)
 13 .77 (Excel: t = 13.77, p–value = 0). There is enough evidence of a
positive linear relationship. The theory appears to be valid.
18.8
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
A
SUMMARY OUTPUT
B
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
C
D
E
F
0.8415
0.7081
0.7021
213.7
100
ANOVA
df
Regression
Residual
Total
Intercept
Space
Water
SS
10,744,454
4,429,664
15,174,118
2
97
99
Coefficients Standard Error
576.8
514.0
90.61
6.48
9.66
2.41
MS
5,372,227
45,667
F
Significance F
117.6
0.0000
t Stat
P-value
1.12
0.2646
13.99
0.0000
4.00
0.0001
a The regression equation is ŷ = 576.8 + 90.61x 1 + 9.66x 2
b The coefficient of determination is R 2 = .7081; 70.81% of the variation in electricity
consumption is explained by the model. The model fits reasonably well.
H 0 : 1   2  0
c
H1 : At least one  i is not equal to zero
F = 117.6, p-value = 0. There is enough evidence to conclude that the model is valid.
d&e
A
B
C
1 Prediction Interval
2
3
Consumption
4
5 Predicted value
8175
6
7 Prediction Interval
8 Lower limit
7748
9 Upper limit
8601
10
11 Interval Estimate of Expected Value
12 Lower limit
8127
13 Upper limit
8222
D
e We predict that the house will consume between 7748 and 8601 units of electricity.
f We estimate that the average house will consume between 8127 and 8222 units of
electricity.
18.10a
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
b
A
SUMMARY OUTPUT
B
C
D
E
F
Regression Statistics
Multiple R
0.8608
R Square
0.7411
Adjusted R Square
0.7301
Standard Error
2.66
Observations
100
ANOVA
df
Regression
Residual
Total
Intercept
Mother
Father
Gmothers
Gfathers
SS
4
95
99
1930
674
2604
Coefficients Standard Error
3.24
5.42
0.451
0.0545
0.411
0.0498
0.0166
0.0661
0.0869
0.0657
MS
482.38
7.10
F
Significance F
67.97
0.0000
t Stat
P-value
0.60
0.5512
8.27
0.0000
8.26
0.0000
0.25
0.8028
1.32
0.1890
H 0 : 1   2  3  0
H1 : At least one  i is not equal to zero
F = 67.97, p-value = 0. There is enough evidence to conclude that the model is valid.
c b1 = .451; for each one year increase in the mother's age the customer's age increases on
average by .451 provided the other variables are constant (which may not be possible
because of the multicollinearity).
b2 = .411; for each one year increase in the father's age the customer's age increases on
average by .411 provided the other variables are constant.
b3 = .0166; for each one year increase in the grandmothers' mean age the customer's age
increases on average by .0166 provided the other variables are constant.
b4 = .0869; for each one year increase in the grandfathers' mean age the customer's age
increases on average by .0869 provided the other variables are constant.
H 0 : i  0
H 1 : i  0
Mothers: t = 8.27, p-value = 0
Fathers: t = 8.26, p-value = 0
Grandmothers: t = .25, p-value .8028
Grandfathers: t = 1.32, p-value = .1890
The ages of mothers and fathers are linearly related to the ages of their children. The
other two variables are not.
d
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
Prediction Interval
C
D
Longvity
Predicted value
71.43
Prediction Interval
Lower limit
Upper limit
65.54
77.31
Interval Estimate of Expected Value
Lower limit
68.85
Upper limit
74.00
The man is predicted to live to an age between 65.54 and 77.31
g
A
B
C
1 Prediction Interval
2
3
Longvity
4
5 Predicted value
71.71
6
7 Prediction Interval
8 Lower limit
65.65
9 Upper limit
77.77
10
11 Interval Estimate of Expected Value
12 Lower limit
68.75
13 Upper limit
74.66
D
The mean longevity is estimated to fall between 68.75 and 74.66.
18.12
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
A
SUMMARY OUTPUT
B
C
D
E
F
Regression Statistics
Multiple R
0.8984
R Square
0.8072
Adjusted R Square
0.7990
Standard Error
7.07
Observations
50
ANOVA
df
Regression
Residual
Total
SS
2
47
49
Coefficients Standard Error
-28.43
6.89
0.604
0.0557
0.374
0.0847
Intercept
Boxes
Weight
MS
4,916
49.97
9,832
2,349
12,181
F
Significance F
98.37
0.0000
t Stat
P-value
-4.13
0.0001
10.85
0.0000
4.42
0.0001
a ŷ = –28.43 + .604x 1 + .374x 2
b s  = 7.07 and R 2 = .8072; the model fits well.
c b1 = .604; for each one additional box, the amount of time to unload increases on
average by .604 minutes provided the weight is constant.
b2 = .374; for each additional hundred pounds the amount of time to unload increases on
average by .374 minutes provided the number of boxes is constant.
H 0 : i  0
H1 :  i
0
Boxes: t = 10.85, p-value = 0
Weight: t = 4.42, p-value = .0001
Both variables are linearly related to time to unload.
d&e
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
Prediction Interval
C
D
Time
Predicted value
50.70
Prediction Interval
Lower limit
Upper limit
35.16
66.24
Interval Estimate of Expected Value
Lower limit
44.43
Upper limit
56.96
d It is predicted that the truck will be unloaded in a time between 35.16 and 66.24
minutes.
e The mean time to unload the trucks is estimated to lie between 44.43 and 56.96 minutes
18.40
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
b
A
SUMMARY OUTPUT
B
C
D
E
F
Regression Statistics
Multiple R
0.6882
R Square
0.4736
Adjusted R Square
0.4134
Standard Error
2,644
Observations
40
ANOVA
df
Regression
Residual
Total
Intercept
Size
Apartments
Age
Floors
4
35
39
SS
MS
220,130,124 55,032,531
244,690,939 6,991,170
464,821,063
Coefficients Standard Error
1,433
2,093
-14.55
20.70
113.0
24.01
-50.10
98.81
-223.8
171.1
F
7.87
Significance F
0.0001
t Stat
P-value
0.68
0.4980
-0.70
0.4866
4.70
0.0000
-0.51
0.6153
-1.31
0.1994
H 0 : 1  2  3  4  0
H1 : At least one  i is not equal to zero
F = 7.87, p-value = .0001. There is enough evidence to conclude that the model is valid.
The regression equation for Exercise 17.12 is ŷ = 4040 + 44.97x. The addition of the new
variables changes the coefficients of the regression line in Exercise 17.12.
19.4a First–order model: a Demand =  0 + 1 Price+ 
Second–order model: a Demand =  0 + 1 Price +  2 Price 2 + 
First–order model:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
A
SUMMARY OUTPUT
B
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
C
D
E
F
0.9249
0.8553
0.8473
13.29
20
ANOVA
df
Regression
Residual
Total
Intercept
Price
Second–order model:
1
18
19
SS
18,798
3,179
21,977
Coefficients Standard Error
453.6
15.18
-68.91
6.68
MS
18,798
176.6
F
Significance F
106.44
0.0000
t Stat
P-value
29.87
0.0000
-10.32
0.0000
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
A
SUMMARY OUTPUT
B
C
D
E
F
Regression Statistics
Multiple R
0.9862
R Square
0.9726
Adjusted R Square
0.9693
Standard Error
5.96
Observations
20
ANOVA
df
Regression
Residual
Total
Intercept
Price
Price-sq
2
17
19
SS
21,374
603
21,977
Coefficients Standard Error
766.9
37.40
-359.1
34.19
64.55
7.58
MS
10,687
35.49
F
Significance F
301.15
0.0000
t Stat
P-value
20.50
0.0000
-10.50
0.0000
8.52
0.0000
c The second order model fits better because its standard error of estimate is 5.96,
whereas that of the first–order models is 13.29
d ŷ .= 766.9 –359.1(2.95) + 64.55(2.95) 2 = 269.3
19.8a
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
A
SUMMARY OUTPUT
B
C
D
E
F
Regression Statistics
Multiple R
0.9255
R Square
0.8566
Adjusted R Square
0.8362
Standard Error
5.20
Observations
25
ANOVA
df
Regression
Residual
Total
Intercept
Temperature
Currency
Temp-Curr
3
21
24
SS
3398.7
568.8
3967.4
Coefficients Standard Error
260.7
162.3
-3.32
2.09
-164.3
667.1
3.64
8.54
MS
1132.9
27.08
F
Significance F
41.83
0.0000
t Stat
P-value
1.61
0.1230
-1.59
0.1270
-0.25
0.8078
0.43
0.6741
b
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
A
SUMMARY OUTPUT
C
B
D
E
F
Regression Statistics
0.9312
Multiple R
0.8671
R Square
0.8322
Adjusted R Square
5.27
Standard Error
25
Observations
ANOVA
df
5
19
24
Regression
Residual
Total
Intercept
Temperature
Currency
Temp-sq
Curr-sq
Temp-Curr
SS
3440.3
527.1
3967.4
Coefficients Standard Error
283.8
274.8
6.88
-1.72
888.5
-828.6
0.0475
-0.0024
1718.5
2054.0
10.57
-0.870
MS
688.1
27.74
Significance F
F
0.0000
24.80
P-value
t Stat
0.3449
0.97
0.8053
-0.25
0.3627
-0.93
0.9608
-0.05
0.2467
1.20
0.9353
-0.08
c Both models fit equally well. The standard errors of estimate and coefficients of
determination are quite similar.
19.16a
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
b
A
SUMMARY OUTPUT
B
Regression Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
Observations
C
D
E
F
0.8368
0.7002
0.6659
810.8
40
ANOVA
df
Regression
Residual
Total
4
35
39
SS
MS
53,729,535 13,432,384
23,007,438
657,355
76,736,973
Coefficients Standard Error
3490
469.2
0.369
0.078
1623
492.5
733.5
394.4
-765.5
484.7
Intercept
Yest Att
I1
I2
I3
t Stat
7.44
4.73
3.30
1.86
-1.58
F
Significance F
20.43
0.0000
P-value
0.0000
0.0000
0.0023
0.0713
0.1232
H 0 : 1   2   3   4  0
H1 : At least on  i is not equal to 0
F = 20.43, p-value = 0. There is enough evidence to infer that the model is valid.
c
H 0 : i  0
H1 :  i
0
I 2 : t = 1.86, p-value = .0713
I 3 : t = –1.58, p-value = .1232
Weather is not a factor in attendance.
d
H0 : 2  0
H1 :  2 > 0
t = 3.30, p-value = .0023/2 = .0012. There is sufficient evidence to infer that weekend
attendance is larger than weekday attendance.
19.22a
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
A
SUMMARY OUTPUT
B
C
D
E
F
F
Significance F
0.0000
Regression Statistics
Multiple R
0.5125
R Square
0.2626
Adjusted R Square
0.2234
Standard Error
5866
Observations
100
ANOVA
df
Regression
Residual
Total
5
94
99
Coefficients Standard Error
30,523
2,358
-108.9
77.58
63.95
33.86
2591
1,287
-3714
1,347
-1260
221.5
Intercept
Pct PT
Pct U
Av Shift
UM Rel
Absent
b
SS
MS
1,151,889,624 230,377,925
3,234,297,164 34,407,417
4,386,186,788
6.70
t Stat
P-value
12.95
0.0000
-1.40
0.1635
1.89
0.0620
2.01
0.0470
-2.76
0.0070
-5.69
0.0000
H0 : 4  0
H1 :  4
0
t = 2.01, p-value = .0470. There is enough evidence to infer that the availability of
shiftwork affects absenteeism.
c
H 0 : 5  0
H1 : 5
 0
t = –2.76, p-value =.0070. There is enough evidence to infer that in organizations where
the union–management relationship is good absenteeism is lower.
19.40a
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
H
I
J
Results of stepwise regression
K
L
M
N
Step 1 - Entering variable: Absent
Summary measures
Multiple R
R-Square
Adj R-Square
StErr of Est
0.3989
0.1591
0.1505
6134.7729
ANOVA Table
Source
Explained
Unexplained
df
SS
MS
1 697913636.0400 697913636.0400
98 3688273152.0000 37635440.3265
F
18.5441
p-value
0.0000
Regression coefficients
Constant
Absent
Coefficient
28516.9941
-790.9393
Std Err
1298.6729
183.6711
t-value
21.9586
-4.3063
Change
0.0520
0.0442
0.0363
-132.6689
% Change
%13.0
%27.8
%24.1
-%2.2
p-value
0.0000
0.0000
Step 2 - Entering variable: UM_Rel
Summary measures
Multiple R
R-Square
Adj R-Square
StErr of Est
0.4509
0.2033
0.1869
6002.1040
ANOVA Table
Source
Explained
Unexplained
df
SS
MS
2 891737380.0400 445868690.0200
97 3494449408.0000 36025251.6289
F
12.3766
p-value
0.0000
Regression coefficients
Constant
Absent
UM_Rel
Coefficient
31636.3125
-967.8824
-3150.9519
Std Err
1850.1073
195.2204
1358.4437
t-value
17.0997
-4.9579
-2.3195
p-value
0.0000
0.0000
0.0225
b In the stepwise regression equation only the number of days absent and union–
management relations were statistically significant.
c The three variables that were not statistically significant and one that was borderline
were excluded by the stepwise regression process.
19.48a Depletion = 0 + 1 Temperature +  2 PH–level + 3 PH–level 2 +  4 I 4 +  5 I 5 + 
where
I1 = 1 if mainly cloudy
I1 = 0 otherwise
I 2 = 1 if sunny
I 2 = 0 otherwise
b
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
c
A
SUMMARY OUTPUT
B
C
D
E
F
Regression Statistics
Multiple R
0.8085
R Square
0.6537
Adjusted R Square
0.6452
Standard Error
4.14
Observations
210
ANOVA
df
Regression
Residual
Total
Intercept
Temperature
PH Level
PH-sq
I1
I2
SS
5
204
209
6596
3495
10091
Coefficients Standard Error
1003
55.12
0.194
0.029
-265.6
14.75
17.76
0.983
-1.07
0.700
1.16
0.700
MS
1319
17.13
F
Significance F
77.00
0.0000
t Stat
P-value
18.19
0.0000
6.78
0.0000
-18.01
0.0000
18.07
0.0000
-1.53
0.1282
1.65
0.0997
H 0 : 1   2   3   4   5  0
H1 : At least on  i is not equal to 0
F = 77.00, p-value = 0. There is enough evidence to infer that the model is valid.
d
H 0 : 1  0
H1 : 1 > 0
t = 6.78, p-value = 0. There is enough evidence to infer that higher temperatures deplete
chlorine more quickly.
e
H 0 : 3  0
H1 : 3 > 0
t = 18.07, p-value = 0. There is enough evidence to infer that there is a quadratic
relationship between chlorine depletion and PH level.
f
H 0 : i  0
H1 :  i  0
I1 : t = –1.53, p-value = .1282. There is not enough evidence to infer that chlorine
depletion differs between mainly cloudy days and partly sunny days.
I 2 : t = 1.65, p-value = .0997. There is not enough evidence to infer that chlorine
depletion differs between sunny days and partly sunny days.
Weather is not a factor in chlorine depletion.