Chapter 13
For all exercises in Chapters 13, 14, and 24 we employed the F-test of two variances at the 5%
significance level to decide which one of the equal-variances or unequal-variances t-test and
estimator of the difference between two means to use to solve the problem. Additionally, for exercises
that compare two populations and are accompanied by data files, our answers were derived by
defining the sample from population 1 as the data stored in the first column (often column A in Excel
and column 1 in Minitab). The data stored in the second column represent the sample from population
2. Paired differences were defined as the difference between the variable in the first column minus the
variable in the second column.
1 1
1 
 1
  = 55  76.79
13.1 ( x1  x2 )  t / 2 s 2p    = (524 – 469)  2.009 18,261
 25 25 
 n1 n2 
1 1
1 
 1
  = 55  160.35
13.2 ( x1  x2 )  t / 2 s 2p    = (524 – 469)  2.009 79,637 
25
25
n
n


2
 1
13.3 The interval widens.
1 1
1 
 1

13.4 ( x1  x2 )  t / 2 s 2p    = (524 – 469)  1.972 18,261
 = 55  37.69
 100 100 
 n1 n2 
13.5 The interval narrows.
13.6 to 13.12
H 0 : (1   2 ) = 0
H 1 : (1   2 )
0
13.6 Rejection region: t  t  / 2,n 1  t .025, 22  -2.074 or t  t  / 2,n 1  t .025, 22  2.074
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
=
(74  71)  0
1
1
275   
 12 12 
= .44, p-value = .6617 (Excel). There is not enough
evidence to infer that the population means differ.
209
13.7 Rejection region: t  t  / 2,n 1  t .025, 22  -2.074 or t  t  / 2,n 1  t .025, 22  2.074
t
( x1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



=
(74  71)  0
1 1
41,652   
 12 12 
= .036, p-value = .9716 (Excel). There is not enough
evidence to infer that the population means differ.
13.8 The value of the test statistic decreases and the p-value increases.
13.9 Rejection region: t  t  / 2,  t .025, 289  -1.960 or t  t  / 2,   t .025, 289  1.960
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(74  71)  0
 324 225 



 150 150 
= 1.57, p-value = .1179 (Excel). There is not enough
evidence to infer that the population means differ.
13.10 The value of the test statistic increases and the p-value decreases.
13.11 Rejection region: t  t  / 2,n 1  t .025, 22  -2.074 or t  t  / 2,n 1  t .025, 22  2.074
t
( x1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



=
(76  71)  0
1 1
275   
 12 12 
= .74, p-value = .4676 (Excel). There is not enough
evidence to infer that the population means differ.
13.12 The value of the test statistic increases and the p-value decreases.
 s2 s2 
 324 49 
  = 3  4.60
13.13 ( x1  x2 )  t / 2  1  2  = (63 –60)  1.671 
n n 
 50 45 
2
 1
 s2 s2 
 1681 225 

13.14 ( x1  x2 )  t / 2  1  2  = (63 –60)  1.671 
 = 3  10.38
n n 
45 
 50
2
 1
13.15 The interval widens.
 s2 s2 
 324 49 
  = 3  3.22
13.16 ( x1  x2 )  t / 2  1  2  = (63 –60)  1.671 
n n 
 100 90 
2
 1
210
13.17 The interval narrows.
13.18 to 13.24
H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
13.18 Rejection region: t  t ,  t .05, 200  1.653
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(412  405 )  0
 16,384 2,916 



150 
 150
= .62, p-value = .2689 (Excel). There is not enough
evidence to infer that 1 is greater than  2 .
13.19 Rejection region: t  t ,  t .05, 289  1.645
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(412  405 )  0
 961 676 



 150 150 
= 2.12, p-value = .0175 (Excel). There is enough evidence
to infer that 1 is greater than  2 .
13.20 The value of the test statistic increases and the p-value decreases.
13.21 Rejection region: t  t ,  t .05, 26  1.706
t
( x1  x 2 )  (1   2 )





 n1 n 2 


s12
=
s 22
(412  405 )  0
 16,384 2,916 



20 
 20
= .23, p-value = .4118 (Excel). There is not enough
evidence to infer that 1 is greater than  2 .
13.22 The value of the test statistic decreases and the p-value increases.
13.23 Rejection region: t  t ,  t .05, 200  1.653
t
( x1  x 2 )  (1   2 )
 s12 s 22 



n

 1 n2 
=
(409  405 )  0
 16,384 2,916 



150 
 150
= .35, p-value = .3624 (Excel). There is not enough
evidence to infer that 1 is greater than  2 .
13.24 The value of the test statistic decreases and the p-value increases.
211
13.25 a Equal-variances t-test degrees of freedom = 28, Unequal-variances t-test degrees of freedom
=26.4
b Equal-variances t-test degrees of freedom = 24, Unequal-variances t-test degrees of freedom = 10.7
c Equal-variances t-test degrees of freedom = 98 , Unequal-variances t-test degrees of freedom =91.2
d Equal-variances t-test degrees of freedom = 103, Unequal-variances t-test degrees of freedom = 78.5
13.26 a In all cases the equal-variances t-test degrees of freedom is greater than the unequal-variances
t-test degrees of freedom.
13.27
H 0 : (1   2 ) = 0
H 1 : (1   2 )
0
Rejection region: t  t  / 2,  t .025,14  2.145 or t  t  / 2,  t .025,14  2.145
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 


 n1 n 2 
=
(5.88  5.13)  0
1 1
2.27   
8 8
= 1.00, p-value = .3361 (Excel). There is not enough
evidence to conclude that the population means differ.
13.28
H 0 : (1   2 ) = 0
H 1 : (1   2 )
0
Rejection region: t  t  / 2,  t .025,10  2.228 or t  t  / 2,  t .025,10  2.228
t
( x1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



=
(7.83  8.50 )  0
1 1
9.03  
6 6
= -.38, p- value = .7089 (Excel). There is not enough
evidence to conclude that the population means differ.
13.29a
H 0 : (1   2 ) = 0
H 1 : (1   2 )
0
Rejection region: t  t  / 2,  t .025,380  1.960 or t  t  / 2,  t .025,380  1.960
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
=
(99 .30  95 .77 )  0
1 
 1
565 


 165 217 
= 1.43, p-value = .1522 (Excel). There is not enough
evidence to infer that the population means differ.
212
1 1
1 
 1

b ( x1  x2 )  t / 2 s 2p    = (99.30 –95.77)  1.960 565 
 = 3.53  4.81;
 165 217 
 n1 n2 
LCL = -1.28, UCL = 8.34
c The two populations must be normally distributed.
d The histograms (not shown here) are bell shaped.
13.30a
H 0 : (1   2 ) = 0
H 1 : (1   2 ) > 0
Rejection region: t  t ,  t .10, 203  1.286
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
=
(21 .51  19 .76 )  0
1 
 1
20 .33
 
121
84


= 2.72, p-value = .0036 (Excel). There is enough
evidence to infer that the mean of population 1 is greater than the mean of population 2.
1 1
1 
 1
  = 1.75  1.06;
b ( x1  x2 )  t / 2 s 2p    = (21.51 –19.76)  1.653 20 .33
121
84
n
n


2
 1
LCL = .69, UCL = 2.81
c The two populations must be normally distributed.
d The histograms are bell shaped.
13.31
H 0 : (1   2 ) = 0
H 1 : (1   2 ) < 0
Rejection region: t  t ,  t .01, 46  2.412
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(250 .40  259 .80 )  0
 175 .12 1893 .75 



40 
 40
= -1.31, p-value = .0988 (Excel). There is not
enough evidence to infer that the mean of population 2 is less than the mean of population 1.
13.32
H 0 : (1   2 ) = 0
H 1 : (1   2 ) < 0
Rejection region: t  t ,  t .10, 485  1.28
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(72 .93  73 .99 )  0
 26 .32 241 .66 



400 
 400
= -1.30, p-value = .0974 (Excel). There is enough
evidence to infer that the mean of population 1 is less than the mean of population 2.
213
13.33a
H 0 : (1   2 ) = 0
H 1 : (1   2 ) > 0
Rejection region: t  t ,  t .05,38  1.684
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 


 n1 n 2 
=
(36 .93  31 .36 )  0
1 
1
13 .70  
 15 25 
= 4.61, p-value = 0. There is enough evidence to infer
that Tastee is superior.
1 1
1 
1
b ( x1  x2 )  t / 2 s 2p    = (36.93 – 31.36)  2.021 13 .70    = 5.57  2.44;
 15 25 
 n1 n2 
LCL = 3.13, UCL = 8.01
c The histograms are somewhat bell shaped. The weight gains may be normally distributed.
13.34
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Rejection region: t  t ,  t .05,94  1.662
t
( x1  x 2 )  (1   2 )

s 2p 
1
1

 n1 n 2



=
(19 .02  21 .85)  0
1 
 1
33 .41  
48
48


= -2.40, p-value = .0092 (Excel). There is enough
evidence to infer that taking vitamin and mineral supplements daily increases the body's immune
system?
13.35 a H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Rejection region: t  t  / 2,  t .025, 449  1.960 or t  t  / 2,  t .025, 449  1.960
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(58 .99  52 .96 )  0
 946 .97 1876 .44 



250 
 250
= 1.79, p-value = .0737 (Excel). There is not
enough evidence to conclude that a difference in mean listening times exist between the two
populations.
 s2 s2 
 946 .97 1876 .44 

b ( x1  x2 )  t / 2  1  2  = (58.99 –52.96)  1.960 
 = 6.03  6.59;
n n 
250 
 250
2
 1
LCL = -.56, UCL = 12.62
c The histograms are bell shaped.
214
13.36
H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
Rejection region: t  t ,  t .05, 220  1.645
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 


 n1 n 2 
(3.19  4.35  0
=
1 
 1
14 .18
 
 124 98 
= -2.27, p-value = .0122 (Excel). There is enough
evidence to infer that the number of visits to health professionals grew between 1997 and 1998.
13.37a
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Rejection region: t  t  / 2,  t .025,198  1.972 or t  t  / 2,  t .025,198  1.972
t
( x1  x 2 )  (1   2 )

s 2p 
1
1

 n1 n 2



(10 .23  9.66 )  0
=
1 
 1
8.30


100
100


= 1.41, p-value = .1606 (Excel). There is not enough
evidence to infer that male and female drivers differ.
1 1
1 
 1

b ( x1  x2 )  t / 2 s 2p    = (10.23 – 9.66)  1.972 8.30 
 = .57  .80; LCL = -.23,
100
100
n
n


2
 1
UCL = 1.38
c The histograms are bell shaped.
13.38
H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
Rejection region: t  t ,  t .05,56  1.671
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 


 n1 n 2 
=
(115 .50  110 .32 )  0
1 
 1
492 .28  
 30 28 
= .89, p-value = .1891 (Excel). There is not enough
evidence to retain supplier A—switch to supplier B.
13.39a H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
Rejection region: t  t  ,  t .01, 42  2.423
215
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(13 .52  9.92 )  0
 5.76 13 .16 



25 
 25
= 4.14, p-value = .0001 (Excel). There is enough
evidence to conclude that exercise is more effective.
 s2 s2 
 5.76 13 .16 

b ( x1  x2 )  t / 2  1  2  = (13.52 – 9.92)  2.021 
 = 3.60  1.76; LCL = 1.84,
n n 
25 
 25
2
 1
UCL = 5.36
c The histograms are bell shaped.
13.40a
H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
Rejection region: t  t  / 2,  t .005, 29  2.756 or t  t  / 2,  t .005, 29  2.756
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(74 .91  53 .40 )  0
 601 .90 73 .69 



15 
 23
= 3.86, p-value = .0006 (Excel). There is enough
evidence to conclude that the two packages differ in the amount of time needed to learn how to use
them.
 s2 s2 
 601 .90 73 .69 

b ( x1  x2 )  t / 2  1  2  = (74.91 –53.40)  2.045 
 = 21.51  11.40;
n n 
15 
 23
2
 1
LCL = 10.11, UCL = 32.92
c The amount of time is required to be normally distributed.
d The histograms are somewhat bell shaped.
13.41a
H 0 : (1   2 ) = 0
H 1 : (1   2 ) < 0
Rejection region: t  t ,  t .01, 277  2.326
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(5.02  7.80 )  0
 1.94 9.57 



 200 200 
= -11.61, p-value = 0. There is enough evidence to infer
that the amount of time wasted in unsuccessful firms exceeds that of successful firms.
 s2 s2 
 1.94 9.57 

b ( x1  x2 )  t / 2  1  2  = (5.02 – 7.80)  1.960 
 = -2.78  .47;
n n 
 200 200 
2
 1
LCL = -3.25, UCL = -2.31. Workers in unsuccessful companies waste on average between 2.31 and
3.25 hours per week more than workers in successful companies.
216
13.42
H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
Rejection region: t  t ,  t .05, 268  1.645
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 


 n1 n 2 
(.646  .601)  0
=
1 
 1
.00243 


 125 145 
= 7.44, p-value = 0. There is enough evidence to
conclude that the reaction time of drivers using cell phones is slower that for non-cell phone users.
13.43
H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
Rejection region: t  t  / 2,  t .025,183  1.973 or t  t / 2,  t .025,183  1.973
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
(.654  .662 )  0
=
1 
 1
.00217   
95
90


= -1.21, p-value = .2268 (Excel). There is not enough
evidence to infer that the type of discussion affects reaction times.
13.44
H 0 : (1   2 ) = 0
H 1 : (1   2 ) > 0
Rejection region: t  t ,  t .05,143  1.656
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 


n
n
2 
 1
=
(6.18  5.94 )  0
1
 1
2.57   
 64 81 
= .87, p-value = .1917 (Excel). There is not enough
evidence to infer that people spend more time researching for a financial planner than they do for a
stock broker.
13.45a
H 0 : (1   2 ) = 0
H 1 : (1   2 ) < 0
Rejection region: t  t ,  t .05,373  1.645
t
( x1  x 2 )  (1   2 )
 s12 s 22 



 n1 n 2 


=
(63 .71  66 .80 )  0
 34 .82 46 .90 



202 
 173
= -4.69, p-value = 0. There is enough evidence to infer
that students without textbooks outperform those with textbooks.
217
13.46
H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
Rejection region: t  t  / 2,  t .025, 413  1.960 or t  t  / 2,  t .025, 413  1.960
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 


 n1 n 2 
(149 .85  154 .43)  0
=
= -2.05, p-value = .0407 (Excel). There is enough
1 
 1
516 .40


 213 202 
evidence to conclude that there are differences in service times between the two chains.
13.47
H 0 : (1   2 ) = 0
H 1 : (1   2 )  0
Rejection region: t  t  / 2,  t .025,168  1.975 or t  t  / 2,  t .025,168  1.975
t
( x1  x 2 )  (1   2 )

s 2p 
1
1

 n1 n 2



=
(53 .05  51 .67 )  0
1
 1
11 .44  
79
91


= 2.65, p-value = .0087 (Excel). There is enough
evidence to conclude that the two types of specialties differ in the time devoted to patients.
13.48
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Rejection region: t  t / 2,   t.025,190  1.973 or t  t / 2,   t.025,190  1.973
t
( x1  x2 )  (1   2 )


  
n

 1 n2 
s12
=
s22
(130 .93  126 .14 )  0
 1023 .36 675 .85 



100 
 100
= 1.16, p-value = .2467 (Excel). There is not
enough evidence to infer that differences exist between the two types of customers.
13.49
H 0 : (1   2 ) = 0
H1 : (1   2 ) > 0
Rejection region: t  t,   t.05,38  1.684
t
( x1  x2 )  (1   2 )

s 2p 
1
1 
 
n
n
2 
 1
=
(73 .60  69 .20 )  0
1 
 1
235 .11  
20
20


= .91, p-value = .1849 (Excel). There is not enough
evidence to infer that the new design tire lasts longer than the existing design tire.
218
13.50
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
Rejection region: t  t,  t.05,531  1.645
t
( x1  x2 )  (1   2 )
 s12 s22 
  
n

 1 n2 
(237 .99  251 .99 )  0
=
 149 .91 220 .25 



279 
 263
= -12.00, p-value = 0. There is enough evidence to
infer that British golfers play golf in less time than do American golfers.
13.51
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
Rejection region: t  t,   t.05,53  1.676
t
( x1  x2 )  (1   2 )
1
1 
s 2p   
 n1 n2 
=
(6,345  6,358 )  0
1
 1
4,010   
 28 33 
= -.84, p-value = .2010 (Excel). There is not enough
evidence to conclude that British courses are shorter than American courses.
13.52
H 0 : (1   2 ) = 0
H 1 : (1   2 ) < 0
Rejection region: t  t,   t.05,53  1.676
t
( x1  x2 )  (1   2 )


  
n

 1 n2 
s12
s22
=
(7,137  9,304 )  0
 38,051 110 ,151 



33 
 28
= -31.61, p-value = 0. There is enough evidence to
conclude that the total distance of American golf courses is greater than that of British courses.
13.53 The data are observational. Experimental data could be produced by randomly assigning babies
to either Tastee or the competitor’s product.
13.54 More affluent mothers use Tastee and babies with more affluent mothers gain weight faster.
13.55 The data are experimental.
13.56a The data are observational.
b Randomly assign students to use either of the software packages.
c Better students tend to choose Program B and better students learn how to use computer software
more quickly.
219
13.57a Let students select the section they wish to attend and compare test results.
b Randomly assign students to either section and compare test results.
13.58a Randomly select finance and marketing MBA graduates and determine their starting salaries.
b Randomly assign some MBA students to major in finance and others to major in marketing.
Compare starting salaries after they graduate.
c Better students may be attracted to finance and better students draw higher starting salaries.
13.59a The data are observational because to obtain experimental data would entail randomly
assigning some people to smoke and others not to smoke.
b It is possible that some people smoke because of a genetic defect (Genetics have been associated
with alcoholism.), which may also be linked to lung cancer.
c In our society the experiment described in part a is impossible.
13.60
H0 : D = 0
H1 :  D > 0
Rejection region: t  t,   t.01, 4  3.747
t
xD   D
s D / nD

3.8  0
3.11 / 5
 2.73, p-value = .0263 (Excel). There is not enough evidence to infer that
the mean of population 1 exceeds the mean of population 2.
13.61
H0 : D = 0
H1 :  D  0
Rejection region: t  t / 2,   t.05,9  1.833 or t  t / 2,   t.05,9  1.833
t
xD   D
s D / nD

 1.30  0
2.16 / 10
 -1.90, p-value = .0898 (Excel). There is enough evidence to conclude
that the two population means differ
13.62 a xD  t / 2
sD
nD
=  2.10  1.833
2.23
 2.10  1.29 ; LCL = -3.39, UCL = -.81
10
b The mean difference is estimated to lie between -3.39 and -.81. This type of estimate is correct 90%
of the time.
220
13.63a
H0 : D = 0
H1 :  D  0
Rejection region: t  t / 2,  t.025,7  2.365 or t  t / 2,  t.025,7  2.365
t
xD   D

s D / nD
.75  0
.71 / 8
 2.99, p-value = .0199 (Excel). There is enough evidence to infer that the
population means differ.
b The matched pairs experiment reduced the amount of variation and as a result the p-value decreased.
13.64a
H0 : D = 0
H1 :  D  0
Rejection region: t  t / 2,  t.025,5  2.571 or t  t / 2,  t.025,5  2.571
t
xD   D

s D / nD
 .67  0
5.16 / 6
 -.32, p-value = .7646 (Excel). There is not enough evidence at the 5%
significance level to infer that the population means differ.
b The variable used to match the pairs was not strongly related to the variable being tested. As a
consequence the matched pairs experiment did not reduce the variation. The smaller number of
degrees of freedom produced a larger p-value.
13.65a
H0 : D = 0
H1 :  D  0
Rejection region: t  t ,  t.05,11  1.796
t
xD   D

s D / nD
 2.08  0
3.58 / 12
 -2.01, p-value = .0344 (Excel). There is enough evidence to infer that the
new fertilizer is better.
b x D  t / 2
sD
nD
=  2.08  2.201
3.58
 2.08  2.27 ; LCL = -4.35, UCL = .19
12
c The differences are required to be normally distributed
d The data are experimental.
e The data are experimental.
f The experimental design should be independent samples.
13.66 a H 0 :  D = 0
H1 :  D  0
Rejection region: t  t ,  t.05,11  1.796
221
t
xD   D
3.08  0

s D / nD
5.88 / 12
 1.81, p-value = .0484 (Excel). There is enough evidence to infer that
companies with exercise programs have lower medical expenses.
b x D  t / 2
sD
= 3.08  2.201
nD
5.88
 3.08  3.74 ; LCL = -.66, UCL = 6.82
12
c Yes because medical expenses will vary by the month of the year.
13.67a
H0 : D = 0
H1 :  D  0
Rejection region: t  t,  t.05,149  1.656
t
xD   D
12 .4  0

s D / nD
99 .1 / 150
 1.53, p-value = .0638 (Excel). There is not enough evidence to infer that
mortgage payments have increases in the past 5 years.
13.68
H0 : D = 0
H1 :  D  0
Rejection region: t  t / 2,  t.025, 49  2.009 or t  t / 2,  t.025, 49  2.009
t
xD   D

s D / nD
 1.16  0
2.22 / 50
 -3.70, p-value = .0006 (Excel). There is enough evidence to infer that
waiters and waitresses earn different amounts in tips.
13.69a xD  t / 2
b
sD
nD
= 19.75  1.684
30 .63
 19 .75  8.16 ; LCL = 11.59, UCL = 27.91
40
H0 : D = 0
H1 :  D  0
Rejection region: t  t,   t.05,39  1.684
t
xD   D
s D / nD

19 .75  0
30 .63 / 40
 4.08, p-value = .0001 (Excel). There is enough evidence to conclude
that companies that advertise in the Yellow Pages have higher sales than companies that do not.
c The histogram of the differences is bell shaped.
d No, because we expect a great deal of variation between stores.
222
13.70
H0 : D = 0
H1 :  D  0
Rejection region: t  t,  t.05,99  1.660
t
xD   D

s D / nD
 3.6  0
5.91 / 100
 -6.09, p-value = 0. There is enough evidence to infer that the new drug
is effective.
13.71
H0 : D = 0
H1 :  D  0
Rejection region: t  t / 2,  t.025, 44  2.014 or t  t / 2,  t.025, 44  2.014
t
xD   D

s D / nD
 42 .94  0
317 .16 / 45
 -.91, p-value = .3686 (Excel). There is not enough evidence to infer
men and women spend different amounts on health care.
13.72
H0 : D = 0
H1 :  D  0
Rejection region: t  t,  t.05,169  1.654
t
xD   D
s D / nD

 183 .35  0
1568 .94 / 170
 -1.52, p-value = .0647 (Excel). There is not enough to infer stock
holdings have decreased.
13.73
H0 : D = 0
H1 :  D > 0
Rejection region: t  t,  t.05,37  1.690
t
xD   D
s D / nD

.0422  0
.1634 / 38
 t = 1.59, p-value = .0597 (Excel). There is not enough evidence to
conclude that ratios are higher this year.
13.74
H0 : D = 0
H1 :  D > 0
Rejection region: t  t,  t.05,54  1.676
223
t
xD   D

s D / nD
520 .85  0
1854 .92 / 55
 2.08, p-value = .0210 (Excel). There is enough evidence to infer that
company 1’s calculated tax payable is higher than company 2’s.
13.75
H0 : D = 0
H1 :  D > 0
Rejection region: t  t,  t.05,19  1.729
t
xD   D

s D / nD
4.55  0
7.22 / 20
 2.82, p-value = .0055 (Excel). There is enough evidence to that the new
design tire lasts longer than the existing design.
13.76 The matched pairs experiment reduced the variation caused by different drivers.
13.77
H0 : D = 0
H1 :  D > 0
Rejection region: t  t,  t.05, 24  1.711
t
xD   D
s D / nD

4587  0
22 ,851 / 25
 1.00, p-value = .1628 (Excel). There is not enough evidence to infer
that finance majors attract higher salary offers than do marketing majors.
13.78
H 0 : 12 / 22  1
H1 : 12 / 22  1
Rejection region: F  F / 2,1,2  F.025,29,29  2.09 or
F  F1 / 2,1 ,2  1 / F / 2,2 ,1  1 / F.025,29,29  1 / 2.09  .48
F = s12 / s 22 = 350/700 =.50, p-value = .0669 (Excel). There is not enough evidence to conclude that
the population variances differ.
13.79 Rejection region: F  F / 2,1,2  F.025,14,14  2.95 or
F  F1 / 2,1 ,  2  1 / F / 2,  2 , 1  1 / F.025,14,14  1 / 2.95  .34
F = s12 / s 22 = 350/700 =.50, p-value = .3357 (Excel). There is not enough evidence to conclude that
the population variances differ.
224
13.80 The value of the test statistic is unchanged. The p-value increases.
 s2 
1
 28  1
13.81 LCL =  12 
=  
= .366, UCL =
s F
 2   / 2,1,2  19  4.03
 s12 
  F / 2,  ,  =  28 4.03 = 5.94
2 1
 s2 
 19 
 2
 s2 
1
 28  1
13.82 LCL =  12 
=  
= .649, UCL =
s F
 2   / 2,1,2  19  2.27
 s12 
  F / 2,  ,  =  28 2.27 = 3.35
2 1
 s2 
 19 
 2
13.83 The interval widens.
13.84
H 0 : 12 / 22  1
H1 : 12 / 22  1
Rejection region: F  F,1,2  F.05,19,19  2.16
F = s12 / s 22 = 60/25 = 2.40, p-value = .0318 (Excel). There is enough evidence to infer that variance of
population 1 is greater than the variance of population 2.
13.85
H 0 : 12 / 22 = 1
H1 : 12 / 22  1
Rejection region: F  F / 2,1,2  F.05,7,9  3.29 or
F  F1 / 2,1 , 2  1 / F / 2,  2 , 1  1 / F.05,9,7  1 / 3.68  .27
F = s12 / s 22 = 10.27/39.16 = .26, p- value = .0914 (Excel). There is enough evidence to infer that the
population variances differ.
13.86
H 0 : 12 / 22 = 1
H1 : 12 / 22  1
Rejection region: F  F / 2,1,2  F.025,9,9  4.03 or
F  F1 / 2,1 , 2  1 / F / 2,  2 , 1  1 / F.025,9,9  1 / 4.03  .25
F = s12 / s 22 = 163.88/233.38 = .70, p-value = .6068 (Excel). There is no evidence to conclude that the
population variances differ.
225
13.87
H 0 : 12 / 22 = 1
H1 : 12 / 22 < 1
Rejection region: F  F1, 1 ,  2  1 / F,  2 ,1  1 / F.01,125,129  1 / 1.53  .65
F = s12 / s 22 = 15,800/18,734 =.84, p-value = .1690 (Excel). There is not enough evidence to infer that
the variance of population 1 is less that the variance of population 2.
 s2 
 15,800  1
1

13.88 LCL =  12 
= 
= .590, UCL =
s F
 2   / 2,1,2  18,734  1.43
13.89
 s12 
  F / 2,  ,  =
2 1
 s2 
 2
 15,800 

1.43 = 1.21
 18,734 
H 0 : 12 / 22 = 1
H1 : 12 / 22  1
Rejection region: F  F / 2,1,2  F.025,24,24  2.27 or
F  F1 / 2,1 , 2  1 / F / 2,  2 , 1  1 / F.025,24,24  1 / 2.27  .44
F = s12 / s 22 =5.76/13.16 = .438, p-value = .0482 (Excel). There is enough evidence to conclude that the
population variances are not equal.
13.90
H 0 : 12 / 22 = 1
H1 : 12 / 22  1
Rejection region: F  F / 2,1,2  F.05,78,90  1.43 or
F  F1 / 2,1 , 2  1 / F / 2,  2 ,1  1 / F.05,90,78  1 / 1.53  .65
F = s12 / s 22 = 9.38/13.22 = .71, p-value = .1214 (Excel). There is not enough evidence to conclude that
the population variances are not equal.
13.91
H 0 : 12 / 22 = 1
H1 : 12 / 22  1
Rejection region: F  F / 2,1,2  F.025,99,99  1.43 or
F  F1 / 2,1 ,  2  1 / F / 2,  2 , 1  1 / F.025,99,99  1 / 1.43  .70
F = s12 / s 22 = 1023.36/675.85= 1.51, p-value = .0402 (Excel). There is enough evidence to conclude
that the population variances are not equal.
226
13.92
H 0 : 12 / 22 = 1
H1 : 12 / 22 < 1
Rejection region: F  F1, 1 ,  2  1 / F,  2 , 1  1 / F.05,51,51  1 / 1.53  .65
F = s12 / s 22 = .026/.087 = .30, p-value = 0 (Excel). There is enough evidence to infer that portfolio 2 is
riskier than portfolio 1.
13.93
H 0 : 12 / 22 = 1
H1 : 12 / 22  1
Rejection region: F  F / 2,1,2  F.05,99,99  1.35 or
F  F1, 1 ,  2  1 / F,  2 , 1  1 / F.05,99,99  1 / 1.35  .74
F = s12 / s 22 = 3.35/10.95 = .31, p-value = 0 (Excel). There is enough evidence to conclude that the
population variances differ.
13.94 to 13.100
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z   z / 2   z.025 = -1.96 or z  z / 2  z.025 = 1.96
13.94 z 
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.45  .40 )
1 
 1
.425 (1  .425 )


100
100


= .72, p-value = 2P(Z > .72) = 2(.5 -
.2642) = .4716.
13.95 z 
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
n
n
2 
 1
=
(.45  .40 )
1 
 1
.425 (1  .425 )


 400 400 
= 1.43, p-value = 2P(Z > 1.43)
= 2(.5 - .4236) = .1528.
13.96 The p-value decreases.
13.97 z 
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.95  .90 )
1 
 1
.925 (1  .925 )


 100 100 
= 2(.5 - .4099) = .1802.
227
= 1.34, p-value = 2P(Z > 1.34)
13.98 The p-value decreases.
13.99 z 
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.05  .10 )
1 
 1
.075 (1  .075 )


 100 100 
= -1.34, p-value = 2P(Z < -1.34)
= 2(.5 - .4099) = .1802.
13.100 The p-value deceases.
13.101 ( pˆ1  pˆ 2 )  z / 2
.18(1  .18) .22 (1  .22 )
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )
= (.18-.22)  1.645


n1
n2
100
100
= -.040  .093
13.102 ( pˆ1  pˆ 2 )  z / 2
.48(1  .48) .52 (1  .52 )
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )
= (.48-.52)  1.645


n1
n2
100
100
= -.040  .116
13.103 The interval widens.
13.104
H 0 : ( p1  p2 ) = 0
H1 : ( p1  p2 )  0
Rejection region: z  z  z.05 = 1.645
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.071  .064 )
1 
 1
.068 (1  .068 )


 1604 1109 
= .71, p-value = P(Z > .71) = .5 - .2611
= .2389. There is not enough evidence to infer that the claim is false.
13.105a H 0 : ( p1  p 2 ) = 0
H1 : ( p1  p2 )  0
Rejection region: z  z  z.05 = 1.645
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
n
n
2 
 1
=
(.56  .46 )
1 
 1
.518 (1  .518 )


 1100 800 
= 4.31, p-value = 0. There is enough
evidence to infer that the leader’s popularity has decreased.
228
H 0 : ( p1  p2 ) = .05
b
H1 : ( p1  p2 )  .05
Rejection region: z  z  z.05 = 1.645
z
( pˆ 1  pˆ 2 )  ( p1  p2 )
=
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )

n1
n2
(.56  .46 )  .05
.56 (1  .56 ) .46 (1  .46 )

1100
800
= 2.16, p-value = P(Z > 2.16) = .5 -
.4846 = .0154. There is enough evidence to infer that the leader’s popularity has decreased by more
than 5%.
c ( pˆ1  pˆ 2 )  z / 2
.56 (1  .56 ) .46 (1  .46 )
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )
= (.56  .46 )  1.96
= .10


n1
n2
1100
800
 .045; LCL = .055, UCL = .145; we estimate that the leader’s popularity has fallen by between 5.5%
and 14.5%.
13.106
H 0 : ( p1  p2 ) = -.08
H 1 : ( p1  p 2 )  -.08
Rejection region: z   z   z.01 = -2.33
z
( pˆ 1  pˆ 2 )  ( p1  p2 )
=
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )

n1
n2
(.11  .28)  (.08)
.11(1  .11) .28(1  .28)

300
300
= -2.85, p-value =P(Z < -2.85) = .5 -
.4978 = .0022. There is enough evidence to conclude that management should adopt process 1.
13.107
H 0 : ( p1  p2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z   z   z.05 = -1.645
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.093  .115 )
1 
 1
.104 (1  .104 )


 6281 6281 
= -4.04, p-value = 0. There is enough
evidence to infer that Plavix is effective.
13.108
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z   z / 2   z.05 = -1.96 or z  z / 2  z.05 = 1.96
229
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.0995  .1297 )
1 
 1
.1132 (1  .1132 )


 382 316 
= -1.26, p-value = 2P(Z < -1.26)
= 2(.5 - .3962) = .2076. There is not enough evidence to infer differences between the two sources.
13.109
H 0 : ( p1  p 2 ) = 0
H1 : ( p1  p 2 ) < 0
Rejection region: z   z   z.05 = -1.645
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.5702  .6365 )
1 
 1
.6042 (1  .109 )


577
608


= -2.33, p-value = P(Z < -2.33) = .5 - .4901
= .0099. There is enough evidence to infer that New Yorkers are more likely to respond no.
13.110
H 0 : ( p1  p2 ) = 0
H 1 : ( p1  p 2 ) < 0
Rejection region: z   z   z.05 = -1.645
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.2462  .2691 )
1 
 1
.2579 (1  .2579 )


 662 695 
= -.96, p-value = P(Z < -.96) = .5 - .3315
= .1685. There is not enough evidence to infer that the proportion of smokers has decreased in the last
10 years.
13.111
H 0 : ( p1  p2 ) = 0
H 1 : ( p1  p 2 ) > 0
Rejection region: z  z  z.10 = 1.28
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.2632  .0741 )
1 
 1
.11(1  .11) 

 38 162 
= 3.35, p-value = 0. There is enough evidence to
conclude that smokers have a higher incidence of heart diseases than nonsmokers.
b ( pˆ1  pˆ 2 )  z / 2
 1.645
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )

= (.2632-0741)
n1
n2
.2632 (1  .2632 ) .0741 (1  .0741 )
=.1891  .1222; LCL = .0669, UCL = .3113

38
162
230
13.112a H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z  z.10 = 1.28
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.62  .52 )
1 
 1
.56 (1  .56 )


 400 500 
= 3.01, p-value = P(Z > 3.01) = .5 - .4987 =
.0013. There is enough evidence to infer that there has been a decrease in belief in the greenhouse
effect.
b ( pˆ1  pˆ 2 )  z / 2
.62 (1  .62 ) .52 (1  .52 )
pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )
= (.62-.52)  1.645
=.10


n1
n2
400
500
 .0543; LCL = .0457 and UCL = .1543; the change in the public’s opinion is estimated to lie
between 4.57 and 15.43%
13.113
H 0 : ( p1  p2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z  z.05 = 1.645
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.3585  .3420 )
1 
 1
.3504 (1  .3504 )


 477 462 
= .53, p-value = P(Z > .53) = .5 - .2019
= .2981.There is not enough evidence to infer that the use of illicit drugs in the United States has
increased in the past decade.
13.114
H 0 : ( p1  p2 ) = -.02
H 1 : ( p1  p 2 )  -.02
Rejection region: z   z   z.05 = -1.645
z
( pˆ 1  pˆ 2 )  ( p1  p2 )
=
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )

n1
n2
(.055  .11)  (.02 )
.055 (1  .055 ) .11(1  .11)

200
200
= -1.28, p-value = P(Z < -1.28)
= .5 - .3997= .1003. There is not enough evidence to choose machine A.
13.115
H 0 : ( p1  p2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z  z.05 = 1.645
231
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.1385  .0905 )
1 
 1
.1035 (1  .1035 )


 231 619 
z = 2.05, p-value = P(Z > 2.05) = .5 -
.4798 = .0202. There is enough evidence to conclude that health conscious adults are more likely to
buy Special X.
13.116
H 0 : ( p1  p2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z  z.05 = 1.645
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.1561  .0921 )
1 
 1
.1377 (1  .1377 )


 378 152 
= 1.93, p-value = P(Z > 1.93) = .5 -
.4732 = .0268. There is enough evidence to infer that low-income individuals are more likely to use
the company’s services.
13.117
H 0 : ( p1  p2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z   z   z.05 = -1.645
z
( pˆ 1  pˆ 2 )
1
1 
pˆ (1  pˆ )  
 n1 n2 
=
(.0095  .0172 )
1 
 1
.01335 (1  .01335 )


11000
11000


= -5.00, p-value = 0. There is
enough evidence to infer that aspirin is effective in reducing the incidence of heart attacks.
The answers to the chapter review exercises were produced by Excel only.
13.118 H 0 : (1   2 ) = 0
H1 : (1   2 )  0
232
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Oat bran
10.01
19.64
120
19.74
0
238
1.56
0.0602
1.6513
0.1204
1.9700
Other
9.12
19.84
120
t = 1.56, p-value = .1204. There is not enough evidence to infer that oat bran is different from other
cereals in terms of cholesterol reduction?
13.119
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
$100-Limit $3000-Limit
0.5234
0.5551
491
490
0
-1.00
0.1598
1.2816
0.3196
1.6449
z = -1.00, p-value = .1598. There is not enough evidence to infer that the dealer at the more expensive
table is cheating.
13.120
H 0 : (1   2 ) = 0
H1 : (1   2 ) > 0
233
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
During
5746.07
167289
15
184373
0
37
2.65
0.0059
1.6871
0.0119
2.0262
Before
5372.13
194772
24
t = 2.65, p-value = .0059. There is enough evidence to conclude that the campaign is successful.
13.121 Gross sales must increase by 50/.20 = $250 to pay for ads.
H 0 : (1   2 ) = 250
H1 : (1   2 ) > 250
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
During
5746.07
167289
15
184373
250
37
0.88
0.1931
1.6871
0.3862
2.0262
Before
5372.13
194772
24
t = .88, p-value = .1931. There is not enough evidence to conclude that the ads are profitable.
13.122
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
234
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Group 1 Group 2
4.94
9.48
11.19
20.29
68
193
0
158
-8.73
0.0000
1.6546
0.0000
1.9751
t = -8.73, p-value = 0. There is enough evidence to conclude that men and women who suffer heart
attacks vacation less than those who do not suffer heart attacks.
13.123a H 0 :  D = 0
H1 :  D  0
t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Uninsulated
775.53
48106.3
15
0.9988
0
14
16.92
0.0000
1.3450
0.0000
1.7613
Insulated
718.13
51464.7
15
t = 16.92, p-value = 0. There is enough evidence to conclude that heating costs for insulated homes is
less than that for uninsulated homes.
b
t-Estimate: Mean
Mean
Standard Deviation
LCL
UCL
Difference
57.40
13.14
50.12
64.68
235
LCL = 50.12, UCL = 64.68
c Differences are required to be normally distributed.
13.124
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Vendor Delivered
19.50
20.03
14.35
14.97
205
155
14.62
0
358
-1.29
0.0996
1.6491
0.1993
1.9666
t = -1.29, p-value = .1993. There is no evidence of a difference in reading time between the two
groups.
13.125
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
Last Year This Year
0.6758
0.7539
327
382
0
-2.30
0.0106
1.6449
0.0212
1.9600
z = -2.30, p-value = .0106. There is enough evidence to infer an increase in seatbelt use.
13.126a H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
236
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
5 Years This Year
32.42
33.72
36.92
45.52
200
200
41.22
0
398
-2.02
0.0218
1.2837
0.0436
1.6487
t = -2.02, p-value = .0218. There is enough evidence to infer that housing cost a percentage of total
income has increased.
b The histograms are be bell shaped.
13.127a H 0 : (1   2 ) = 0
H1 : (1   2 )  0
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Male
Female
39.75
49.00
803.88
733.16
20
20
768.52
0
38
-1.06
0.1490
1.3042
0.2980
1.6860
t = -1.06, p-value = .2980. There is not enough evidence to conclude that men and women differ in the
amount of time spent reading magazines.
b
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
237
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Variable 1 Variable 2
33.10
56.84
278.69
1047.81
21
19
0
26
-2.87
0.0040
1.3150
0.0080
1.7056
t = -2.87, p-value = .0040. There is enough evidence to conclude that high-income individuals devote
more time to reading magazines than do low-income individuals.
13.128a H 0 :  D = 0
H1 :  D  0
t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Female
55.68
105.64
25
0.9553
0
24
-1.13
0.1355
1.3178
0.2710
1.7109
Male
56.40
116.75
25
t = -1.13, p-value = .2710. There is no evidence to infer that gender is a factor.
b A large variation within each gender group was expected.
c The histogram of the differences is somewhat bell shaped.
3.129
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
238
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
This Year 3 Years Ago
0.4351
0.3558
393
385
0
2.2605
0.0119
1.2816
0.0238
1.6449
z = 2.26, p-value = .0119. There is enough evidence to infer that Americans have become more
distrustful of television and newspaper reporting this year than they were three years ago.
13.130
H 0 : (1   2 ) = 25
H1 : (1   2 ) > 25
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
A Nondefectives B Nondefectives
230.13
200.92
79.51
59.04
24
24
69.27
25
46
1.75
0.0433
1.6787
0.0865
2.0129
t = 1.75, p-value = .0433. There is enough evidence to conclude that machine A should be purchased.
13.131
H 0 : ( p1  p2 ) = 0
H1 : ( p1  p2 )  0
The totals in columns A through D are 5788, 265, 5154, and 332, respectively.
z-Test of the Difference Between Two Proportions (Case 1)
Sample proportion
Sample size
Alpha
Sample 1 Sample 2
0.0458
0.0644
5788
5154
0.05
z Stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
239
-4.28
0.0000
1.6449
0.0000
1.9600
z = -4.28, p-value = 0. There is enough evidence to infer that the defective rate differs between the
two machines.
13.132
H0 : D = 0
H1 :  D  0
Dry Cleaner
t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Dry C Before Dry C After
168.00
165.50
351.38
321.96
14
14
0.8590
0
13
0.96
0.1780
1.7709
0.3559
2.1604
t = .96, p-value = .1780. There is not enough evidence to conclude that the dry cleaner sales have
decreased.
Doughnut shop
t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Donut Before
308.14
809.67
14
0.8640
0
13
3.24
0.0032
1.7709
0.0065
2.1604
Donut After
295.29
812.07
14
t = 3.24, p-value = .0032. There is enough evidence to conclude that the doughnut shop sales have
decreased.
240
Convenience store
t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Convenience Before Convenience After
374.64
348.14
2270.40
2941.82
14
14
0.9731
0
13
7.34
0.0000
1.7709
0.0000
2.1604
t = 7.34, p-value = 0. There is enough evidence to conclude that the convenience store sales have
decreased.
13.133a H 0 : (1   2 ) = 0
H1 : (1   2 )  0
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
20-Year-old 40-year old
125.74
129.93
31.90
31.95
26
24
31.92
0
48
-2.62
0.0059
1.2994
0.0119
1.6772
t = -2.62, p-value = .0059. There is enough evidence to infer that 40-year-old men have more iron in
their bodies than do 20-year-old men.
b
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
241
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
20-year old 40-year old
134.02
141.11
36.15
39.47
26
24
37.74
0
48
-4.08
0.0001
1.2994
0.0002
1.6772
t = -4.08, p-value = .0001. There is enough evidence to infer that 40-year-old women have more iron
in their bodies than do 20-year-old women.
13.134a H 0 : ( p1  p 2 ) = 0
H1 : ( p1  p2 )  0
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
Depressed Not Depressed
0.2879
0.2004
132
1058
0
2.33
0.0100
2.3263
0.0200
2.5758
z = 2.33, p-value = .0100. There is enough evidence to infer that men who are clinically depressed are
more likely to die from heart diseases.
b No, we cannot establish a causal relationship.
13.135
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
242
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Men
Women
5.56
5.49
28.76
31.10
306
290
29.90
0
594
0.15
0.4422
1.6474
0.8844
1.9640
t = .15, p-value = .8844. There is no evidence of a difference in job tenures between men and women.
13.136
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Group 1 Group 2
7.46
8.46
25.06
12.98
50
50
0
89
-1.14
0.1288
1.6622
0.2575
1.9870
t = -1.14, p-value = .1288. There is not enough evidence to conclude that people who exercise
moderately more frequently lose weight faster
13.137
H0 : D = 0
H1 :  D  0
243
t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Group 1 Group 2
7.53
8.57
29.77
43.37
50
50
0.8885
0
49
-2.40
0.0100
1.6766
0.0201
2.0096
t = -2.40, p-value = .0100. There is enough evidence to conclude that people who exercise moderately
more frequently lose weight faster
13.138
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
Special K Other
0.635
0.530
200
200
0
2.13
0.0166
1.6449
0.0332
1.96
z = 2.13, p-value = .0166. There is enough evidence to conclude that Special K buyers like the ad
more than non-buyers.
13.139
H 0 : ( p1  p 2 ) = 0
H1 : ( p1  p2 )  0
244
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
Special K Other
0.575
0.515
200
200
0
1.20
0.1141
1.6449
0.2282
1.96
z = 1.20, p-value = .1141. There is not enough evidence to conclude that Special K buyers are more
likely to think the ad is effective.
13.140
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Small space Large space
1245.68
1915.84
23811.89
65566.31
25
25
0
39
-11.21
0.0000
1.3036
0.0000
1.6849
t = -11.21, p-value = 0. There is enough evidence to infer that students write in such a way as to fill
the allotted space.
13.141
H 0 : (1  2 ) = 0
H1 : (1   2 )  0
245
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Computer No Computer
69933
48246
63359040
101588525
89
61
0
109
14.07
0.0000
1.2894
0.0000
1.6590
t = 14.07, p-value = 0. There is enough evidence to conclude that single-person businesses that use a
PC earn more.
13.142
H 0 : ( p1  p 2 ) = 0
H1 : ( p1  p2 )  0
z-Test: Two Proportions
New
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
Older
0.948
250
0
1.26
0.1037
1.2816
0.2074
1.6449
0.92
250
z = 1.26, p-value = .1037. There is not enough evidence to conclude that the new company is better.
13.143
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
246
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Happy
Unhappy
4.92
2.33
10.24
6.67
61
42
8.79
0
101
4.35
0.0000
1.6601
0.0000
1.9837
t = 4.35, p-value = 0. There is enough evidence to infer that people in happy marriages have a greater
reduction in blood pressure.
13.144
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Teenagers 20-to-30
18.18
14.30
357.32
130.79
176
154
0
293
2.28
0.0115
1.6501
0.0230
1.9681
t = 2.28, p-value = .0115.There is enough evidence to infer that teenagers see more movies than do
twenty to thirty year olds.
Case 13.1 For ACT 241 and ACT 242 we test 0 high school courses versus 1 course and then 0 versus
2 high school courses. In each instance we test the following.
H 0 : (1   2 )  0
H1 : (1   2 )  0
247
ACT 241 0 versus 1
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
ACT 241-0 ACT 241-1
73.49
74.54
45.62
67.91
296
24
47.23
0
318
-0.72
0.2364
1.6497
0.4728
1.9675
t = -.72, p-value .2364; there is not enough evidence to infer that students with 1 high school
accounting course outperform students with no high school accounting in ACT 241.
ACT 241 0 versus 2
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
ACT 241-0 ACT 241-2
73.49
80.26
45.62
26.72
296
54
0
90
-8.40
0.0000
1.6620
0.0000
1.9867
t = -8.40, p-value = 0; there is enough evidence to infer that students with 2 high school accounting
courses outperform students with no high school accounting in ACT 241.
248
ACT 242 0 versus 1
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
ACT242-0 ACT 242-1
76.89
77.60
15.09
9.97
210
15
14.77
0
223
-0.70
0.2437
1.6517
0.4875
1.9707
t = -.70, p-value = .2437; there is not enough evidence to infer that students with 1 high school
accounting course outperform students with no high school accounting in ACT 242.
ACT 242 0 versus 2
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
ACT242-0 ACT 242-2
76.89
78.92
15.09
14.12
210
50
14.90
0
258
-3.35
0.0005
1.6508
0.0009
1.9692
t = -3.35, p-value = .0005; there is enough evidence to infer that students with 2 high school
accounting course outperform students with no high school accounting in ACT 242.
Overall Conclusion
The statistical evidence suggests that students with 2 high school accounting courses should be
exempted from both ACT 241 and ACT 242. No other exemptions should be allowed.
249