CHAPTER 3: Solutions to Selected Exercises
3.1
a.
Scatterplot of Salary vs GPA
b. The straight line appearance of the plot in Part a indicates that a simple linear regression
model should be appropriate.
3.2
a.
 y| x 4.00   0  1 (4.00) is the mean of the population of all potential starting salaries for
marketing graduates having a 4.00 GPA.
b.
 y| x 2.50   0  1 (2.50) is the mean of the population of all potential starting salaries
for marketing graduates having a 2.50 GPA.
c.
1 = the change in mean starting salary associated with a one point increase in the grade
point average.
d.
 0 = the mean starting salary for marketing graduates with a grade point average of 0.00.
The interpretation of  0 fails to make practical sense because it requires that a marketing
graduate have a grade point average of 0.00.
3.3
e.
All factors other than the grade point average. For example, extra-curricular activities and
overall GPA.
a.
b1 =5.70657 may be interpreted as follows:
An increase of one point in the GPA corresponds to an increase of $5,706.57 in the
average starting salary.
13
b0  14.8156 clearly has no practical interpretation because it indicates a mean starting
salary of $14,815.60 when the GPA=0.000.
b.
yˆ  14.8156  5.70657(3.25)  33.362
($33,362)
c.
SS xy  xi y i 
xi yi 
7
21.57 226.8  10.5040
 709.372 
7
2
xi 
SS xx  xi2 
7
21.57 2  1.8407
 68.3071 7
SS xy 10.5040
b1 

 5.7065
SS xx
1.8407
x
xi 21.57

 3.0814
7
7
y
y i 226.8

 32.4
7
7
b0  y  b1 x  32.4  (5.7065)(3.0814)  14.816
Note: For more accurate estimate carry more decimal places.
3.4
a.
s2 
SSE 1.438

 .2876
n2 72
s  s 2  .2876  .5363
14
b.
SSE  y i2  [b0 y i  b1xi y i ]
 7409.7  [(14.8156)( 226.8)  (5.70657)(709.372)]
 1.44
Note: For more accurate estimate carrying more decimal places, starting in Exercise
3.3. Round-off-error can be surprisingly large.
3.5
The straight line appearance of the data plot indicates that a simple linear regression model
should be appropriate.
3.6
a.
It is the mean of the service times required when the number of copiers is 4.
b.
It is the mean of the service times required when the number of copiers is 6.
c.
The slope parameter equals the change in the mean service time that is associated with each
additional copier serviced.
d.
The intercept is the mean service time when there are no copiers. It could make practical
sense if it requires service time to be 0 when no copies are serviced.
e.
All factors other than the number of copiers serviced.
a.
b1  24.6022 may be interpreted as follows: An increase of one copier serviced
3.7
corresponds to an increase of 24.6 minutes in the mean time required.
b0  11.464 clearly has no practical meaning because it indicates that the mean service
time is 11.464 minutes for 0 copiers.
b.
yˆ  11.4641  24.6022(4)  109.873
c.
.
15
(109.9 minutes)
SS xy  xi y i 
xi yi 
11
431184  809.6364
 5438 
11
SS xx  xi2 
xi 2
11
(43) 2
 201 
 32.9091
11
SS xy 809.6364
b1 

 24.6022
SS xx
32.9091
xi 43
y 1184

 3.9091
y i 
 107.6364
11 11
11
11
b0  107.6364  (24.6022)(3.9091)  11.464
x
3.8
a.
s2 
SSE 191.70166

 21.3002
n2
11  2
s  s 2  21.3002  4.6152
b.
SSE  y i2  [b0 y i  b1xi y i ]
 147552  [(11.4641)(1184)  (24.6022)(5438)]
 191.7
(Carry more decimal places in 3.7c and here for more accuracy)
3.9
A straight line appears to be a reasonable approximation for relating the average demand to price
difference.
3.10
a.
Mean demand when price difference is .10.
b.
Mean demand when price difference is -.05.
c.
Change in mean demand per dollar increase in price difference.
d.
Mean demand when price difference = 0; yes
e.
Factors other than price difference, such as amount and type of advertising.
a.
b1  2.66522 has the following interpretation:
3.11
For each increase in the price difference (fresh price - ind. price) of one dollar, the mean
demand for Fresh increases by 266, 522 bottles.
b0  7.81409 has the following interpretation:
16
When there is no price difference between fresh price and average industry price, mean
demand for the large bottle of Fresh is 781,409 bottles.
b.
yˆ  7.814088  2.665214(.10)  8.081
c.
yˆ  b0  b1 x
8.5 = 7.81409 + 2.6652x
x
d.
.68591
 .257, or about 26 cents
2.6652
s2 
SSE 2.8059

 .10021
n  2 30  2
s  s 2  .10021  .3166
3.12
3.13
a.
b.
Since the relationship between direct labor cost (y) and batch size (x) has a straight
appearance, a simple linear regression model is appropriate.
a.
Mean labor cost when batch size = 60.
b.
Mean labor cost when batch size = 30.
c.
Change in mean labor cost per unit increase in batch size.
d.
Mean labor cost when batch size = 0. This could make sense if mean labor cost is 0 when
batch size is 0.
e.
Factors other than batch size; answers will vary.
17
3.14
a.
SS xy  xi y i 
xi yi 
n
(548)(5,782)
 365,027 
 100,982.33
12
xi 2
SS xx  xi2 
n
5482  9,952.667
 34,978 
12
SS xy 100,982.32
b1 

 10.14626
SS xx
9,952.667
 5782 
 548 
b0  y  b1 x  
  10.14626
  18.4875
 12 
 12 
b.
b1 is the estimated increase in mean labor cost ($1014.63) for every 1 unit increase in the
batch size.
b0 is the estimated mean labor cost (18.4880) when batch size = 0; no.
3.15
c.
yˆ  18.4880  10.1463x
d.
yˆ  18.4880  10.1463(60)  627.266
a.
s2 
b.
SSE  y i2  [b0 y i  b1xi y i ]
= 3,811,300  [(18.48751)(5,782)+(10.14626)(365,027)]
SSE
747

 74.7
n  2 12  2
s  s 2  74.7  8.643
= 746 (Carry more decimal places in 3.14 Part a for more accuracy)
18
3.16
a.
b0 =14.816
b1 =5.7066
b.
SSE = 1.438
s 2  .288
c.
sb1  .3953
t = 14.44
s=.536321
t  b1 / sb1  5.7066 / .3953  14.44
d.

t[.5025
]  2.571
Reject H 0 , strong evidence of a significant linear relationship between x
and y.
e.

t[.5005
]  4.032
Reject H 0 , very strong evidence of a significant linear relationship
between x and y.
f.
p-value = .000
Reject at all  , extremely strong evidence of a significant linear
relationship between x and y.
g.

95% CI : [b1  t.5025
 sbi ]  5.7066  (2.571)(.3953)  [4.690, 6.723]
We are 95% confident that the mean starting salary increases by between $4690 and $6723
for each 1.0 increase in GPA.
h.

99% CI : [b1  t.5005
 sbi ]  5.7066  (4.032)(.3953)  [4.113, 7.300]
We are 99% confident that the mean starting salary increases by between $4113 and $7300
for each 1.0 increase in GPA.
i.
sbb  1.235
t  12.00
t  b0 / sb0  14.816 / 1.235  12.00
j.
k.
p-value = .000 Reject at all  , Extremely strong evidence that the y-intercept is
significant.
sbi 
s
SS xx

.5363
1.8407
 .3953
1
x2
1 3.0814

 .5363

 1.235
n SS xx
7
1.8407
2
sb0  s
3.17
a.
bo  11.46403
b1  24.60221
19
b.
SSE  191.70166
c.
sb1  .80451
s 2  21.30019
s  4.61521
t  30.58
t  b1 / sb1  24.60221 / .80451  30.58
d.
e.

t .9025
  2.262

t.9005
  3.250
x and y .
Reject H 0 , strong evidence of a linear relationship between x and y .
Reject H 0 , very strong evidence of a linear relationship between
f.
p-value = .000
Reject at all  , extremely strong evidence of a linear relationship
between x and y .
g.
[24.60221  2.262.80451]  [22.782, 26.422]
h.
[24.60221  3.250.80451]  [21.988, 27.217]
i.
sb0  3.43903
t  3.33
t  b0 / sb0  11.46403 / 3.43903  3.33
j.
p-value = .0087 Reject at all  except .001, very strong evidence of a linear relationship
between x and y .
k.
sb1 
s
SS xx

4.61521
32.9091
 .80451
1
x2
1 3.9091

 4.61521

 3.439
n SS xx
11 32.9091
2
sb0  s
3.18
a.
b0  7.81409, b1  2.6652
b.
SSE  2.806, s 2  .100, s  .316561
c.
sb1  .2585, t  10.31
d.

t .28
025  2.048
Reject H 0 , strong evidence of a linear relationship
e.

t.28
005  2.763
Reject H 0 , very strong evidence of a linear relationship
t  b1
sb 1
 2.6652 / .2585  10.31
20
f.
p-value = 0.000 < .001. Reject H 0 , extremely strong evidence of linear relationship.
g.
[2.6652  2.048(.2585)]  [2.136,3.194]
h.
[2.6652  2.763(.2585)]  [1.951,3.379]
i.
sb0  .07988, t  97.82
j.
p-value = 0.000 < .001; reject H 0 .
k.
s
sb1 
SS xx

.316561
1.49967
 2585
1
x2
1 .2133

 .316561

 .07988
n SS xx
30 1.49967
2
sb0  s
3.19
a.
b0  18.48751, b1  10.14626
b.
SSE  746.76238, s 2  74.67624, s  8.64154
c.
sb1  .08662, t  117.13
d.

t .10
025  2.228
Reject H 0 , strong evidence of a linear relationship
e.

t.10
005  3.168
Reject H 0 , very strong evidence of a linear relationship
t  b1 / sb1  10.14626 / .08662  117.13
f.
p-value = 0.000; Reject H 0 at each value of  , extremely strong evidence of linear
relationship.
g.
[10.1463  2.228(.08662)]  [9.953,10.339]
h.
[10.1463  3.169(.08662)]  [9.872,10.421]
i.
sb0  4.67658, t  3.95
j.
p-value = 0.003; fail to reject H 0 at  = .001. Reject H 0 at all other values of  .
k.
sb1 
s
SS xx

8.64154
9952.667
 .086621
21
1
x2
1 45.667 

 8.64154

 4.6766
n SS xx
12 9952.667
2
sb0  s
3.20
a.
33.362, [32.813, 33.910]
b.
33.362, [31.878, 34.846]
c.
1 3.25  3.0814 
 .1583
Distance Value = 
7
1.8407
[33.362  2.571.5363 .1583 ]  [32.813, 33.911]
2
[33.362  2.571.5363 1  .1583 ]  [31.878, 34.846]
3.21
3.22
a.
109.873, [106.7207, 113.0252]
b.
109.873, [98.9671, 120.7788]
c.
113 minutes
a.
8.0806, [7.9479, 8.2133]
b.
8.0806, [7.4187, 8.7425]
c.
See graph with Exercise 3.22. A vertical line at Pricedif = 0.1 will cross the curves at the
points that correspond to the values for 95% CI (Part a) and 95% PI (Part b).
d.
s D.V .  .0648
s  .316561
2
 .0648 
D.V .  
  .0419
 .316561 
99% C.I . is [8.0806  2.763(.316561) .0419 ]
 [8.0806  .1790]  [7.9016, 8.2596]
99% P.I . is [8.0806  2.763.316561 1.0419 ]
 [8.0806  .8928]  [7.1878, 8.9734]
e.
a) 8.4804, [8.3604, 8.6004]
b) 8.4804, [7.8209, 9.1398]
c) Use vertical line at Pricedif = .25
d) s D.V .  .0586
s  .316561
22
2
 .0586 
D.V .  
  .0343
 .316561 
99% C.I . is [8.4804  2.763(.316561) .0343 ]
 [8.4804  .1620]
 [8.3184, 8.6424]
99% P.I . is [8.4804  2.763(.316561) 1  .0343 ]
 [8.4804  .8895]
 [7.5909, 9.3699]
3.23
a.
627.2630, [621.0544, 633.4717]
b.
627.2630, [607.0322, 647.4939]
c.
s D.V .  2.7865
s  8.64154
2
 2.7865 
D.V .  
  .1040
 8.64154 
99% C.I . is [627.2630  3.169(8.64154) .1040 ]
 [627.2630  8.8314]
 [618.4316, 636.0944]
99% P.I . is [627.2630  3.169(8.64154) 1  .1040
 [627.2630  28.7738]
 [598.4892, 656.0368]
3.24
a.
61.380; 1.438; 59.942; r 2 =.977, r =.988
97% of the total variation in the starting salaries can be explained by the linear
relationship between the starting salaries and GPA.
b.
t
r n2
1 r
2

.988 7  2
1  .977
 14.58

5 
Since t.5025
  2.571, and t .005  4.032,
(Difference from 14.44 is round-off error. Need
r 2 with more decimal places.)
we can reject H 0 at  =.05 and  =.01
3.25
a.
20,111; 191.70166; 19,919; r 2 =.9905, r = .9952
99.05% the total variation in service time can be explained by the linear relationship
between service time and the number of copiers serviced.
23
3.26
3.27
3.28
t
r n2

.9952 11  2
b.
 30.63 (Difference from 30.58 is round-off error)
1  .9905
1 r2
9)
9 
t(.025
  2.262, and t .005  3.250
Reject H 0 :  =0 at  =.05 and  =.01
a.
13.459; 2.806; 10.653; .792; .890
79.2% of the total variation in demand can be explained by the linear relationship
between demand and price difference.
b.
t
a.
1,025,340; 746.76238; 1,024,593; .9993; .9996
99.96% of the total variation in direct labor costs are explained by the linear relationship
between direct labor costs and batch size.
b.
t
(Difference from 117.13 is round-off error)
 119.48
1  .9993


t.10
t.10
025  2.228
005  3.169
Reject H 0 at  =.05 and  =.01.
a.
F = 59.942 /(1.438 / 5) = 208.42 (approximately 208.39, round-off error)
b.
F.05  6.61
.890 30  2
(Difference from 10.31 is round-off error)
 10.33
1  .792


t.28
t.28
025  2.048
005  2.763
.9996 12  2
numerator df  1, denominato r df  5
Since 208.39 > 6.61, reject H 0 with strong evidence of a linear relationship between x
and y.
c.
F.01  16.26
numerator df  1, denominato r df  5
Since 208.39 > 16.26, reject H 0 with very strong evidence of a significant relationship
between x and y.
d.
p-value = .000; Reject H 0 at all levels of  , extremely strong evidence of a significant
relationship between x and y.
e.
t 2  (14.44) 2  208.51 (approximately equals F = 208.39, round-off error)
f.
t     2.571
2
5
.025
2
1,5 
 6.61  F.05

24
3.29
a.
F =19919 / (191.70166/9) = 935.16 (approximately 935.15, round-off error)
b.
F.05  5.12
numerator df  1, denominato r df  9
Since 935.15 > 5.12, reject H 0 with strong evidence of a linear relationship between x
and y.
c.
F.01  10.56
numerator df  1, denominato r df  9
Since 935.149 > 10.56, reject H 0 with very strong evidence of a linear relationship
between x and y.
d.
p-value = less than .001; Reject H 0 at all levels of  , extremely strong evidence of a
linear relationship between x and y.
e.
t 2  (30.58) 2  935.14 (approximately equals F = 935.15)
t   
2
11
.025
3.30
1, 11
 (2.262) 2  5.12  F.05

a.
F = 10.653 / (2.806 / 28) = 106.30
b.
F.05  4.20 , reject H 0 (df numerator  1,
df denominato r  28) . Strong
evidence of a linear relationship between x and y.
c.
F.01  7.64 , reject H 0 (df numerator  1,
df denominato r  28) . Very strong
evidence of a linear relationship between x and y.
d.
e.
p-value = .000 and is less than .001, reject H 0 . Extremely strong evidence of a linear
relationship between x and y.
10.312
 106.30
t     2.048
2
28
.025
3.31
a.
F = 13,720.5
b.
F.05  4.95
2
1, 28)
 4.19  4.20  F[.(05
]
df numerator  1, df denominato r  10
Since 13,720.5 > 4.95, reject H 0 at   .05
c.
F.01  10.04
df numerator  1, df denominato r  10
Since 13,720.5 > 10.04, reject H 0 at
 = .01
25
d.
p-value = .000; reject H 0 because .000 < .001, extremely strong evidence of a linear
relationship.
e.
(117.13) 2 =13,719.4 (approximately equals 13,720.5, round-off error)
t     2.228
2
10
.025
3.32
a.
b.
3.33
3.34
2
1,10
 4.96  4.95  F.05

Using Figure 3.21, there does seem to be a negative relationship between temperature and
o-ring failure.
The temperature of 31 o was outside the experimental region.
a.
Yes, diet and type of exercise. While this evidence definitely indicates a potential link
between smoking and lung cancer, a well-designed study will take into account other
factors.
b.
The two slopes appear to differ. A statistical test is needed to show if there is in fact a
statistical difference. It appears that the incidence of lung cancer increases more rapidly
when one increases amount of smoking at low levels of smoking than when a heavy
smoker does a similar increase.
a.
Yes, there is a linear relationship at   .0002 because the p-value for H 0 : 1  0 vs.
H a : 1  0 is .000196916. There is extremely strong statistical evidence.
b.
b1  35.2877
95% C.I. for  1 is [19.2, 51.3]
Thus, we are 95% confident that a 1% increase in the percentage of minority population
corresponds to an increase of between 19 and 51 in the mean number of residents per
branch bank.
3.35
yˆ  2.0572  6.3545(1/ 5)  3.3281
3.36
Minitab Output
26
a.
When x = 15, yˆ  10.004
And 95% C. I. for mean market return rate is [8.494, 11.514]
b.
When x = 15, yˆ  10.004 and 95% P. I. for market return rate of this individual stock is
[-0.310, 20.318]
27