CHAPTER 4: Solutions to Selected Exercises
4.1
a.
The plot of price versus size has a straight-line appearance and thus the model
y   0  1 x1   is appropriate. The plot of price versus rating has a straight-line
appearance. The model y   0  1 x2   is appropriate. Combining these two models,
we obtain the model y   0  1 x1   2 x2   .
b.
 y| x
 20, x 2  9
is the mean (or average) of the sales prices of all houses having 20 hundred
(that is, 2000) square feet and a niceness rating of 9.
c.
 0 = mean sales price of all houses having 0 square feet and 0 niceness rating –
meaningless.
1 = change in mean sales price associated with each increase in house size of 100 square
feet, when niceness rating stays constant.
 2 = change in mean sales price associated with each increase in niceness rating of 1,
when house size remains constant.
d.
4.2
The error term represents all factors other than the square footage and the niceness rating.
One such factor is the ability and effort of the real estate agent listing the house.
The plots suggest a linear relationship between each of the independent variables and hours.
 0 = meaningless
1 = The change in mean monthly labor hours associated with an increase of one x-ray exposure,
when the other independent variables are held constant.
 2 - The change in mean monthly labor hours associated with an increase of one in the number
of monthly bed days, when the other independent variables are held constant.
 3 = The change in mean monthly labor hours associated with a one day increase in the average
length of stay, when the other independent variables are held constant.
The error term represents all factors other than x1 , x2 , and x3 . Such factors could involve patient
load, size of hospital, etc.
4.3
a.
b0  29.347, b1  5.6128, b2  3.8344
b0 = meaningless
b1 = 5.6128 implies that we estimate that mean sales price increases by $5,612.80 for
each increase of 100 square feet in house size, when the niceness rating stays constant.
b2 = 3.8344 implies that we estimate that mean sales price increases by $3,834.40 for
each increase in niceness rating of 1, when the square footage remains constant.
4.4
b.
172.28. From yˆ  29.347  5.6128(20)  3.8344(8)
a.
b0 =1523.38924
b1 =.05299
b2 =.97848
28
b3 = -320.95083
b.
16,065 is computed by
yˆ  1523.38924  .05299(56194)  .97848(14077.88)  320.95083(6.89)
c.
y  yˆ  17,207.31  16,065  1,142.31 hours
Actual hours exceeds predicted hours by 1,142.31
4.5
SSE
73.6
73.6


 10.5; s  10.5  3.242
n  (k  1) 10  (2  1)
7
a.
SSE  73.6; s 2 
b.
Total variation = 7447.5
Unexplained variation = 73.6
Explained variation = 7374
c.
R2 
7374
 .99
7447.5
k  n  1 


R 2   R2 

n  1  n  (k  1) 



=  .99 
2  10  1 


10  1  10  (2  1) 
= .987
d.
F(model) =
=
e.
(Explained variation ) / k
( Unexplaine d variation ))/(n - (k  1))
7374 / 2
7374 / 2

 350.87
73.6 /(10  (2  1)) 73.6 / 7
Based on 2 and 7 degrees of freedom, F.05 = 4.74. Since F(model) = 350.87 > 4.74, we
reject H 0 : 1   2  0 by setting  =.05.
f.
Based on 2 and 7 degrees of freedom, F.01 = 9.55. Since F (model) = 350.87 > 9.55, we
reject H 0 : 1   2  0 by setting  = .01.
g.
p-value = 0.000 (which means less than.001). Since this p-value is less than  = .10,
.05, .01, and .001, we have extremely strong evidence that H 0 : 1   2  0 is false.
That is, we have extremely strong evidence that at least one of x1 and x2 is significantly
related to y.
4.6
a.
SSE = 4,913,399
29
s2 
SSE
4,913,399

 377,954
n  (k  1) 17  (3  1)
s  377,954  614.77942
b.
Total Variation = 494,712,540
Unexplained Variation = 4,913,399
Explained Variation = 489,799,142
c.
R2 
489,799,142
 .9901
494,712,540
k  n  1  
3  17  1 

   .9901 
  .9878
R 2   R2 


n  1  n  (k  1)  
17  1  17  (3  1) 

d.
F (model) =
(Explained Variation) / k
489,799,142 / 3

 431.97
(Unexplain ed Variation) /(n-(k  1 )) 4,913,399 /(17  (3  1))
e.
Based on 3 and 13 degrees of freedom, F.05 = 3.41
F (model) = 431.97 > F.05 =3.41. Reject H 0 : 1   2  0 at  =.05
f.
Based on 3 and 13 degrees of freedom, F.01 =4.35.
F (model) = 431.97 > F.01 =4.35. Reject H 0 : 1   2  0 at  =.01
g.
p-value < .0001. Reject H 0 at   .001 . We have extremely strong evidence that at
least one of x1 , x2 , and x3 is significantly related to y.
4.7
We first consider the intercept  0
a.
b0 = 29.347, sb0 =4.891, t = 6.00
where t = b0 / sb0  29.347 / 4.891  6.00
b.
We reject H 0 :  0  0 (and conclude that the intercept is significant) with   .05 if

| t |  t.705 / 2  t.7025


Since t .7025
 = 2.365 (with n – (k + 1) = 10 – (2 + 1) = 7 degrees of freedom), we have

t = 6.00 > t .7025
 = 2.365.
We reject H 0 :  0  0 with   .05 and conclude that the intercept is significant at the
.05 level.
30

c. We reject H 0 :  0  0 with   .01 if |t| > t .701 / 2  t .7005


Since t .7005
 = 3.499, we have t = 6.00 > t .005 = 3.499.
We reject H 0 :  0  0 with   .01 and conclude that the intercept is significant at the
.01 level.
d. The Minitab output tells us that the p-value for testing H 0 :  0  0 is 0.000. Since this pvalue is less than each given value of  , we reject H 0 :  0  0 at each of these values of  .
We can conclude that the intercept  0 is significant at the .10, .05, .01, and the .001 levels of
significance.
e. A 95% confidence interval for  0 is

[b0  t n / 2k1 sb0 ]  [b0  t .7025
 s b0 ]
= [29.347  2.365 (4.891)]
= [17.780, 40.914]
This interval has no practical interpretation since  0 is meaningless.
f.
A 99% confidence interval for  0 is

[b0  t.7005
 sb0 ]  [29.347  3.499(4.891)]
= [12.233, 46.461]
We next consider 1 .
a.
b1  5.6128, sb1  .2285, t  24.56
where t  b1 / sb1 = 5.6128 / .2285 = 24.56
b., c., and d.:
We reject H 0 : 1  0 (and conclude that the independent variable x1 is significant) at
level of significance  if | t | t7 / 2 (based on n – (k + 1) = 10 – 3 = 7 d.f.)

7 
7 
For  = .05, t7 / 2  t.7025
  2.365 , and for  = .01, t / 2  t.005  3.499.

Since t = 24.56 > t.7025
  2.365 , we reject H 0 : 1  0 with  = .05.

Since t = 24.56 > t.7005
  3.499 , we reject H 0 : 1  0 with  = .01.
Further, the Minitab output tells us that the p-value related to testing H 0 : 1  0 is 0.000.
Since this p-value is less than each given value of  , we reject H 0 at each of these
values of  (.10, .05, .01, and .001).
31
The rejection points and p-value tell us to reject H 0 : 1  0 with  = .10,  = .05,  =
.01, and  = .001. We conclude that the independent variable x1 (home size) is
significant at the .10, .05, .01, and .001 levels of significance.
e. and f.:
95% interval for 1 :

[b1  t.7025
sbi ]  [5.6128  2.365(.2285)]
 [5.072,6.153]
99% interval for 1 :

[b1  t.7005
sbi ]  [5.6128  3.499(.2285)]
 [4.813,6.412]
For instance, we are 95% confident that the mean sales price increases by between $5072
and $6153 for each increase of 100 square feet in home size, when the rating stays
constant.
Last, we consider  2 .
a.
b2  3.8344, sb2  .4332, t  8.85
where t  b2 / sb2 = 3.8344 / .4332 = 8.85
b., c. and d.:
We reject H 0 :  2  0 (and conclude that the independent variable x2 is significant) at
level of significance  if | t | t7 / 2 .
7 
7 

For  = .05, t7 / 2  t.7025
  2.365 , and for  = .01, t / 2  t.005  3.499.

Since t = 8.85 > t.7025
  2.365 , we reject H 0 :  2  0 with  = .05.

Since t = 8.85 > t.7005
  3.499 , we reject H 0 :  2  0 with  = .01.
Further, the Minitab output tells us that the p-value related to testing H 0 :  2  0 is
0.000. Since this p-value is less than each given value of  , we reject H 0 at each of
these values of  (.10, .05, .01, and .001).
The rejection points and p-value tell us to reject H 0 :  2  0 with  = .10,  = .05,  =
.01, and  = .001.
We conclude that the independent variable x2 (niceness rating) is significant at the .10,
.05, .01, and .001 levels of significance.
e. and f.:
95% interval for  2 :

[b2  t.7025
sb2 ]  [3.8344  2.365(.4332)]
 [2.810,4.860]
32
99% interval for  2 :

[b2  t.7005
sb2 ]  [3.8344  3.499(.4332)]
 [2.319,5.350]
For instance, we are 95% confident that the mean sales price increases by between $2810
and $4860 for each increase of one rating point, when the home size stays constant.
4.8
Same process as Exercise 4.7
y   0  1 x1   2 x2   3 x3  
n  (k  1)  17  (3  1)  13

t.13
005  3.012

t.13
025  2.160
H 0 : 0  0
t
1523.38924
 1.94 . Do not reject H 0 at   .05,  .01
786.89772
.05299
 2.64 . Reject at   .05 , not .01
.02009
.97848
t
 9.31 . Reject at   .05,  .01
H 0 : 2  0
.10515
 320.95083
t
 2.10 . Do not reject at   .05 , .01
H 0 : 3  0
153.1922
p-value for testing H 0 : 1  0 is .0205. Reject at   .05 .
H 0 : 1  0
t
H 0 :  2  0 is <.0001. Reject at  =.001.
H 0 : 3  0 is .0583. Do not reject at   .05
but can reject a   .10 .
95% C.I. [b j  2.160sb j ]
99% C.I. [b j  3.012sb j ]
4.9
a.
Point estimate is yˆ  172.28 ($172,280)
95% confidence interval is [168.56, 175.99]
b.
Point prediction is yˆ  172.28
95% prediction interval is [163.76, 180.80]
g. Stdev Fit = s Distance value  1.57
This implies that Distance value = (1.57 /s) 2
= (1.57 / 3.242) 2
= 0.2345
The 99% confidence interval for mean sales price is

[ yˆ  t[.7005
] s Distance value ]
= [172.28  3.499 (1.57)]
= [172.28  5.49]
= [166.79, 177.77]
33
The 99% prediction interval for an individual sales price is

[ yˆ  t[.7005
] s 1  Distance value ]
= [172.28  3.499 (3.242) 1 0.2345 ]
= [172.28  12.60]
= [159.68, 184.88]
4.10
y = 17,207.31 is within the P.I. interval [14,511, 17,618]. There is no statistical evidence to say
the labor hours are unusually high or low for this hospital.
4.11
ŷ =30,626 + 3.893 (28,000) – 29,607 (1.56) + 86.52 (1821.7) = 251,056.564
4.12
For a house of a given square footage and age, each additional room adds $6321.78 to the price of
the home when the number of bedrooms stays constant, while adding a bedroom deducts
$11,103.16 from the price when the number of rooms stays constant.
4.13
a.
The straight line appearance of the plot of y versus x1 suggests that the model
y   0  1 x1  
might appropriately relate y to x1 . The possibly quadratic appearance of the plot of y
versus x2 suggests that the model
y   0  1 x2   2 x22  
might appropriately relate y to x2 . Combining these models, we obtain the model
y   0  1 x1   2 x2  3 x22  
which might appropriately relate y to x1 and x2 .
b.
The p-value related to F(model) is 0.000. Since this p-value is less than .001, we have
extremely strong evidence that the model is significant.
The p-values related to x1, x2 , and x22 are, respectively, 0.000, 0.000, and 0.006. Since
each of these p-values is less than .01, we have very strong evidence that each of x1, x2 ,
and x22 is significant (important).
c.
The point prediction is ŷ = 171.222 ($171,222)
The 95% prediction interval is [166.365, 176.079]
We are 95% confident that the sales price for an individual house with 2,000 square feet
and a “niceness rating” of 8 will be between $166,365 and $176,079.
4.14
a.
b.
The plots have a quadratic appearance.
ŷ = 35.0261
(1)
95% C.I. = [34.4997, 35.5525]
ŷ = 35.0261
(2)
95% P.I. = [35.5954, 36.4568]
4.15
a.
The p-value for x1x2 is .014. Since this p-value is less than .05, we have strong evidence
that x1x2 is important.
34
b.
171.751, [168.835, 174.666]
The length of this interval is 174.666 – 168.835 = 5.83
The length of the interval for the model in Figure 4.27 is 176.079 – 166.36 = 9.72
Hence, the interaction model is giving us a more accurate estimate for the sales price of a
house with 2,000 square feet and a rating of 8.
a,
yˆ  27.438  5.0813x1  7.2899(2)  .5311(2)2  .11473x1 (2)
 27.438  5.0813x1  14.5798  2.1244  .22946 x1
4.16
 39.8934  5.31076 x1
when x1 = 13
ŷ = 39.8934 + 5.31076 (13)
when x1 = 22
ŷ = 39.8934 + 5.31076 (22)
= 39.8934 + 116.83672
= 156.73012
= 39.8934 + 69.03988
= 108.93328
b.
yˆ  27.438  5.0813x1  7.2899(8)  .5311(8)2  .11473x1 (8)
 27.438  5.0813x1  58.3192  33.9904  .91784 x1
 51.7668  5.99914 x1
when x1 = 13
ŷ = 51.7668 + 5.99914 (13)
when x1 = 22
ŷ = 51.7668 + 5.99914 (22)
= 51.7668 + 131.98108
= 183.74788
= 51.7668 + 77.98882
= 129.75562
c.
One can see from the slopes in the two equations that the slope of 5.99914 when x2  8
is somewhat larger than the slope of 5.31076 when x2  2 . The graph also shows that
the line for x2  8 rises a little faster. Thus we estimate that as square feet (home size)
increases, the mean sales price increases faster when the rating is 8 than when the rating
is 2.
4.17
ŷ = 6.0599; 95% P.I. = [3.7578, 8.3620]; 95% confident the actual profit for a future
construction project with a contract size of $480,000 and a supervisor with 6 years experience
will be between $375,780 and $836,200.
4.18
a.
yˆ  19.3050  1.4866 x1  6.375152  .7522 x12  1.7171x1 2
 6.562  1.9476 x1  .7522 x12
35
When x1  3 , yˆ  6.562  1.94763  .75223  5.635.
2
When x1  4, yˆ  6.562  1.94764  .75224   2.3172 .
2
When x1  5, yˆ  6.562  1.94765  .75225  2.505.
2
Plot of ŷ against x1 (for x1 = 3, 4, and 5) when x2  2
b.
yˆ  19.3050  1.4866 x1  6.3715(4)  .7522 x12  1.7171x1 (4)
=  6.181  5.3818 x1  .7522 x12
When x1 =3, yˆ  6.181  5.3818(3)  .7522(3) 2  3.1946.
When x1 =4, yˆ  6.181  5.3818(4)  .7522(4) 2  3.311.
When x1 =5, yˆ  6.181  5.3818(5)  .7522(5) 2  1.923.
Plot of ŷ against x1 (for x1 = 3, 4, and 5) when x2  4
c.
yˆ  19.3050  1.4866 x1  6.37156  .7522 x12  1.7171x1 6
 18.924  8.816 x1  .7522 x12
When x1 =3, yˆ  18.924  8.816(3)  .7522(3) 2  .7542.
When x1 =4, yˆ  18.924  8.816(4)  .7522(4) 2  4.3048.
When x1 =5, yˆ  18.924  8.816(5)  .7522(5) 2  6.351.
36
Plot of ŷ against x1 (for x1 = 3, 4, and 5) when x2  6
Plot of ŷ against x1 (for x1 = 3, 4, and 5)
These plots suggest that:
For a low level of supervisor experience ( x2  2 ), profit decreases substantially as the
contract size ( x1 ) increases.
For a moderate level of supervisor experience ( x2  4 ), profit decreases when the
contract size ( x1 ) becomes large.
For a high level of supervisor experience ( x2  6 ), profit increases as the contract size
( x1 ) increases.
These results suggest that inexperienced supervisors should be assigned to smaller
contracts, while the most experienced supervisors should be assigned to the largest
37
contracts. Medium sized contracts can be assigned to supervisors with medium
experience levels with slightly better results than would be obtained when an
inexperienced supervisor is assigned, and the slightly worse results than would be
obtained when a more experienced supervisor is assigned.
 
If we set 6.562 + 1.9476 x1*  .7522 x1*
2
 
 6.181  5.3818 x1*  .7522 x1*
2
we find
12.743 = 3.4342 x or x  3.71 . (See part a when x2  2 . See part b when x2  4 .)
*
1
*
1
If we set 6.562 + 1.9476 x1* - .7522( x1* ) = -18.924 + 8.816 x1* - .7522 ( x1* ) 2 we find
25.486 = 6.8684 x1* or x1* = 3.71. (See part a when x2  2 . See part c when x2  6 .)
Therefore, all three curves meet at a point corresponding to a contract size of x1* = 3.71.
(i)
The curve when x2  2 . Supervisors with 2 years of experience should be
assigned to contracts of size less than x1* = 3.71.
(ii)
The curve when x2  6 . Supervisors with 6 years of experience should be
assigned to contracts of size greater than x1* = 3.71.
(iii)
Contract sizes “near” x1* = 3.71 should be assigned to supervisors with 4 years of
experience.
4.19
The effect on attendance from a promotion for a day game if all other variables remain the same
is (Promotion = 1, Daygame = 1)
4,745 (1) + 5,059 (1) (1) – 4,690 (1) (Weekend) + 696.5 (1) (Rival)
The effect on attendance from a promotion for a night game if all other variables remain the same
is (Promotion = 1, Daygame = 0)
4,745 (1) + 5,059 (1) (0) – 4,690 (1) (Weekend) + 696.5 (1) (Rival)
Thus, the impact of promotion on attendance is greater by 5,059 for a day game. For example, a
promotion for a day game on a weekday against a rival is expected to increase average attendance
by 4,745 + 5,059 – 0 + 696.5 = 10,500.5. For a night game on a weekday against a rival the
increase would be 4745 + 0 – 0 + 696.5 = 5,441.5 (i.e. 5,059 less). By a similar argument,
promotion on a weekend would decrease the increase in attendance by 4,690 compared to
promotion on a weekday.
4.20
a.
The lines relating the size of the firm and average months to adoptions are parallel,
indicating no interaction of size and type of firm. Also, the two lines are different.
b.
 2 equals the difference between the mean innovation adoption times of stock companies
and mutual companies.
c.
p-value is less than .001; Reject H 0 at both levels of  .
 2 is significant at both levels of  .
95% C.I. for  2 : [4.9770, 11.1339]; 95% confident that for any given size of insurance
firm, the mean adoption time for an insurance innovation is between 4.9770 and 11.1339
months longer if the firm is a stock company rather than a mutual company.
h. No interaction.
4.21
a.
B  B  M 0  T 0  B
38
M  B  M 1  T 0  B  M
T   B  M 0  T 1   B  T
b.
F = 184.57, p-value < .001:
Reject H 0 ; conclude the means are not equal, that is, at least one is different.
c.
M  B  B  M  (B )  M
T   B   B   T  (  B )   T
M  T  B  M  (B  T )  M  T
bM  21.4
bT  - 4.3
bM  bT  21.4  (4.3)  25.7
95% C.I. for M : [21.4  (2.131) 1.433]  [18.346, 24.454]
95% C.I. for T : [-4.300  (2.131) 1.433]  [-7.354, - 1.246]
d.
77.20, [75.040, 79.360], [71.486, 82.914]
e.
[25.700  2.131 (1.433)] = [22.646, 28.754]
For  M , p - value  .001. Hence,  M  T   M  0. Or since 0 is not in 95% C.I.
for  M ,  M  T   M  0.
4.22
a.
b5  .21369
b6  .38178
b6  b5  .38178  .21369  .16809
[.21369  2.069.06215]  [.0851, .3423]
95% C.I. for  5
95% C.I. for  6
[.38178  2.069.06125]  [.2551, .5085]
Both p-values < .01, 5 and  6 are significant.
b.
2
yˆ  25.61270  9.0568.20  6.57676.50  .584446.50
 1.15648.206.50  .381781
 8.5005
95% C.I. for mean demand: [8.4037, 8.5977]
We are 95% confident that the mean demand for all sales periods when the price
difference is .20, the advertising expenditure is 6.50, and campaign C is used will be
between 840,370 and 859,770.
95% P.I. for individual demand: [8.2132, 8.7881]
We are 95% confident that the actual demand in a particular sales period when the price
difference is .20, the advertising expenditure is 6.50, and campaign C is used will be
between 821,322 bottles and 878,813 bottles.
c.
[.0363, .2999]; p-value = .0147  6 significant at  = .05.
39
95% C.I. for  6 : [.16809  2.069 (.06371)] = [.0363, .2999]
4.23
a.
d ,a,C   d ,a, A  [ 0  1d   2 a   3 a 2   4 da   5 (0)   6 (1)   7 a(0)   8 a1]
 [  0   1 d   2 a   3 a 2   4 da   5 (0)   6 (0)   7 a(0)   8 a0]
 6  8a
  .9351  .2035 (6.2)  .3266
  .9351  .2035 (6.6)  .408
d ,a,C   d ,a,B  [0  1d  2a  3a 2  4da  5 (0)  6 (1)  7 a(0)  8a1]
 [  0   1 d   2 a   3 a 2   4 da   5 (1)   6 (0)   7 a(1)   8 a0 
 6  8a  5  7a
 6 - 5  8a  7a
  .9351  (.4807)  .2035 (6.2)  .1072 (6.2)  .14266
  .9351  (.4807)  .2035 (6.6)  .1072 (6.6)  .18118
Both differences increased with the larger value of a.
b.
yˆ  8.5118 (851,180 bottles)
95% P. I.
[8.2249, 8.7988] length = 8.7988 – 8.2249 = .5739
For P.I. in Exercise 4.22 (Figure 4.36),
Length = 8.7881 - 8.2132 = .5739
The intervals are essentially the same length. This is not surprising since the p-values for
the two additional interaction terms both exceed  = .10.
4.24
To test H 0 : 5   6  0 in Model 2, we note that Model 2 is the complete model, and therefore k
= 6. If H 0 : 5   6  0 is true, Model 2 reduces to Model 1, which is the reduced model, and
therefore k – g = 2. It follows that
SSE R  SSEC 1.0644  .3936 6708
.3354
kg
2
F

 2 
 19.599.
SSEC
.3936
.3936 .017113
23
23
n  (k  1)
Based on 2 and 23 degrees of freedom F.05 = 3.42 and F.01 = 5.66. Since F = 19.599 is greater
than F.05 and F.01 , we reject H 0 : 5   6  0 at  = .05 and  = .01. We have seen in
Exercise 4.22 that 5  d , a , B   d , a , A and  6  d , a ,C   d , a , A. Therefore, if 5  6  0 ,
it follows that d , a , B   d , a , A  0 and d , a ,C   d , a , A  0. That is, H 0 : 5   6  0 is
equivalent to H 0 : d , a , A  d , a , B   d , a ,C . If H 0 is true, the mean demand does not differ
with type of advertising. We have shown that at  = .01, the mean demand for at least one type
of advertising is different from the other two means when price difference and advertising
expenditures remain the same.
4.25
Model 3 – complete
Model 1 – reduced
40
H 0 : 5   6   7  8  0
1.0644  .3518
4
F
 10.634
.3518
21
F.05 = 2.84 based on 4 and 21 degrees of freedom.
F.01 = 4.37 based on 4 and 21 degrees of freedom.
Since 10.634 > 4.37, reject H 0 at  = .05 and .01.
The type of advertising has an impact on mean demand.
4.26
Model 3 – complete
Model 2 – reduced
H 0 :  7  8  0
.3936  .3518
2
F
 1.248
.3518
21
F.05 = 3.47 based on 2 and 21 degrees of freedom.
Since 1.248 < 3.47, do not reject H 0 at  = .05 and .01.
The effect of the type of advertising on demand is not altered by amount spent on advertising.
4.27
4.28
pˆ (4.5) 
a.
e 3.74561.1109( 4.5) 
3.5024
 .7779
3.7561.1109( 4.5)  
4.5024
1 e
group 0:
pˆ (85,82) 
e 56.17.4833(85).1652(82) 
.2317
 .1761
56.17.4833(85) .1652(82)  
1.2137
1 e
41