Chapter 13
13.1a Equal-variances estimator
1
1 
 = (524 – 469)  2.009
( x1  x 2 )  t  / 2 s 2p  
n
n
2
 1
 (25  1)129 2  (25  1)131 2


25  25  2

 1
  1 
 25 25 

 (25  1)255 2  (25  1)260 2


25  25  2

 1
  1 
 25 25 

= 55  73.87
b Equal-variances estimator
1
1 
 = (524 – 469)  2.009
( x1  x 2 )  t  / 2 s 2p  
n
n
2
 1
= 55  146.33
c The interval widens.
d Equal-variances estimator
1
1 
 = (524 – 469)  1.972
( x1  x 2 )  t  / 2 s 2p  
 n1 n 2 
 (100  1)129 2  (100  1)131 2


100  100  2

 1
1 


 100 100 

= 55  36.26
e The interval narrows.
13.2
H 0 : (1   2 ) = 0
H1 : (1   2 )
0
a Equal-variances test statistic
Rejection region: t   t  / 2,   t .025, 22  –2.074 or t  t  / 2,  t .025, 22  2.074
t
( x 1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



=
(74  71)  0
 (12  1)18 2  (12  1)16 2


12  12  2

 1 1 
  
 12 12 

= .43, p-value = .6703. There is not
enough evidence to infer that the population means differ.
b Equal-variances test statistic
Rejection region: t   t  / 2,   t .025, 22  –2.074 or t  t  / 2,  t .025, 22  2.074
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1 


n
n
2 
 1
=
(74  71)  0
 (12  1)210 2  (12  1)198 2


12  12  2

 1 1 
  
 12 12 

enough evidence to infer that the population means differ.
c The value of the test statistic decreases and the p-value increases.
269
= .04, p-value = .9716. There is not
d Equal-variances test statistic
Rejection region: t   t  / 2,   t .025, 298  –1.960 or t  t  / 2,  t .025, 298  1.960
t
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
=
(74  71)  0
 (150  1)18 2  (150  1)16 2


150  150  2

 1
1 


 150 150 

= 1.53, p-value = .1282. There is
not enough evidence to infer that the population means differ.
e The value of the test statistic increases and the p-value decreases.
f Rejection region: t   t  / 2,   t .025, 22  –2.074 or t  t  / 2,  t .025, 22  2.074
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1

 n1 n 2



=
(76  71)  0
 (12  1)18 2  (12  1)16 2


12  12  2

 1 1 
  
 12 12 

= .72, p-value = .4796. There is not
enough evidence to infer that the population means differ.
g The value of the test statistic increases and the p-value decreases.
13.3
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
a Unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 200.4 (rounded to 200)
Rejection region: t  t  ,  t.05, 200  1.653
t
( x 1  x 2 )  (1   2 )
 s12 s 22 
 

 n1 n 2 


=
(412  405 )  0
 128 2 54 2 



 150 150 


= .62, p-value = .2689. There is not enough evidence to infer
that 1 is greater than  2 .
b Unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 223.1 (rounded to 223)
Rejection region: t  t ,  t .05, 223  1.645
t
( x 1  x 2 )  (1   2 )
 s12 s 22 
 

 n1 n 2 


=
(412  405 )  0
 31 2 16 2 



 150 150 


= 2.46, p-value = .0074. There is enough evidence to infer that
1 is greater than  2 .
270
c The value of the test statistic increases and the p-value decreases.
d Unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 25.6 (rounded to 26)
Rejection region: t  t ,  t.05, 26  1.706
t
( x 1  x 2 )  (1   2 )
 s12 s 22 
 

 n1 n 2 


=
(412  405 )  0
 128 2 54 2 



 20
20 

= .23, p-value = .4118. There is not enough evidence to infer
that 1 is greater than  2 .
e The value of the test statistic decreases and the p-value increases.
f Unequal-variances test statistic
Rejection region: t  t  ,  t.05, 200  1.653
t
( x 1  x 2 )  (1   2 )
 s12 s 22 
 

 n1 n 2 


=
(409  405 )  0
 128 2 54 2 



 150 150 


= .35, p-value = .3624. There is not enough evidence to infer
that 1 is greater than  2 .
g The value of the test statistic decreases and the p-value increases.
13.4 a Unequal-variances estimator

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 64.8 (rounded to 65, approximated by   70 )
 s2 s2 
( x1  x 2 )  t  / 2  1  2  = (63 – 60)  1.667
n

 1 n2 
 18 2 7 2 

 = 3  4.59

 50 45 


b Unequal-variances estimator

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 63.1 (rounded to 63, approximated by   60 )
 s2 s2 
( x1  x 2 )  t  / 2  1  2  = (63 – 60)  1.671
n

 1 n2 
 41 2 15 2 

 = 3  10.38

 50
45 

c The interval widens.
271
d Unequal-variances estimator
(s12 / n 1  s 22 / n 2 ) 2

(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 131 (approximated by   140 )
 s2 s2 
( x1  x 2 )  t  / 2  1  2  = (63 – 60)  1.656
n

 1 n2 
 18 2 7 2 

 = 3  3.22

 100 90 


e The interval narrows.
13.5 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
13.6 a In all cases the equal-variances t-test degrees of freedom is greater than the unequal-variances t-test
degrees of freedom.
13.7
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
Two-tail F test: F = 1.02, p-value = .9856; use equal-variances test statistic
Rejection region: t   t ,   t .10,10  1.372
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1 


 n1 n 2 
=
(361 .50  381 .83)  0
 (6  1)6767 .5  (6  1)6653 .4  1 1 

  
662

 6 6 
 .43, p-value = .3382. The manager
should choose to use cameras.
13.8
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
Two-tail F test: F = 1.02, p-value = .9823; use equal-variances test statistic
Rejection region: t   t ,   t .10,18  1.330
t
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
=
(5.10  7.30 )  0
 (10  1)5.88  (10  1)5.79  1 1 

  
10  10  2

 10 10 
 2.04 , p-value = .0283. There is
enough evidence to infer that there are fewer errors when the yellow ball is used.
272
H 0 : (1   2 ) = 0
13.9
H1 : (1   2 )
0
Two-tail F test: F = .11, p-value = .0106; use unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 8.59 (rounded to 9)
Rejection region: t   t  / 2,   t .05,9  1.833 or t  t  / 2,  t .05,9  1.833
t
( x 1  x 2 )  (1   2 )
 s12




 n1 n 2 


=
s 22
(18 .25  16 .50 )  0
 42 .21 367 .71 



8 
 8
 .24, p-value = .8130. There is not enough evidence to
infer that the two groups differ in the number of pictures printed.
13.10
H 0 : (1   2 ) = 0
H1 : (1   2 )
0
Two-tail F test: F = 1.04, p-value = .9873; use equal-variances test statistic
Rejection region: t   t  / 2,   t .05,13  1.771 or t  t  / 2,  t .05,13  1.771
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1 


n
n
2 
 1
=
(3,372  4,093 )  0
 (9  1)755 ,196  (6  1)725 ,778  1 1 

  
962

 9 6 
 1.59, p-value = .1368. There is
not enough evidence to infer a difference between the two types of vacation expenses.
13.11a
H 0 : (1   2 ) = 0
H1 : (1   2 ) > 0
Two-tail F test: F = 1.59, p-value = .3050; use equal-variances test statistic
Rejection region: t  t ,  t.05,38  1.684
t
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
=
(36 .93  31 .36 )  0
 (15  1)4.23 2  (25  1)3.35 2


15  25  2

 1
  1 
 15 25 

= 4.61, p-value = 0. There is
enough evidence to infer that Tastee is superior.
1
1 
 = (36.93 – 31.36)  2.021
b ( x1  x 2 )  t  / 2 s 2p  
n
n
2
 1
 (15  1)4.23 2  (25  1)3.35 2


15  25  2

 1
  1 
 15 25 

= 5.57  2.43; LCL = 3.14, UCL = 8.00
c The histograms are somewhat bell shaped. The weight gains may be normally distributed.
273
13.12 H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = .99, p-value = .9571; use equal-variances test statistic
Rejection region: t   t  / 2,   t .025, 238  1.960 or t  t  / 2,  t .025, 238  1.960
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1 


n
n
2 
 1
=
(10 .01  9.12 )  0
 (120  1)4.43 2  (120  1)4.45 2


120  120  2

 1
1 


 120 120 

 1.55, p-value = .1218.
There is not enough evidence to infer that oat bran is different from other cereals in terms of cholesterol
reduction?
13.13 a H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = .50, p-value = 0; use unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 449
Rejection region: t   t  / 2,   t .025, 449  1.960 or t  t  / 2,  t .025, 449  1.960
t
( x 1  x 2 )  (1   2 )
 s12 s 22 
 

 n1 n 2 


=
(58 .99  52 .96 )  0
 30 .77 2 43 .32 2


 250
250





= 1.79, p-value = .0734. There is not enough evidence
to conclude that a difference in mean listening times exist between the two populations.
 s2 s2 
b ( x 1  x 2 )  t  / 2  1  2  = (58.99 –52.96)  1.960
 n1 n 2 


 30 .77 2 43 .32 2 

 = 6.03  6.59;

 250
250 

LCL = –.56, UCL = 12.62
c The histograms are bell shaped.
13.14 a H 0 : (1   2 ) = 0
H1 : (1   2 ) > 0
Two-tail F test: F = 1.01, p-value = .9619; use equal-variances test statistic
Rejection region: t  t ,  t .05, 282  1.645
t
( x 1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



=
(59 .81  57 .40 )  0
 (125  1)7.02 2  (159  1)6.99 2


125  159  2

 1
1 


 125 159 

 2.88, p-value = .0021. There
is enough evidence to infer that the cruise ships are attracting younger customers.
274
1
1 
 = (59.81 – 57.40)
b ( x1  x 2 )  t  / 2 s 2p  
 n1 n 2 
 (125  1)7.02 2  (159  1)6.99 2
 2.576 
125  159  2

 1
1 

= 2.41  2.16;

 125 159 

LCL = .25, UCL = 4.57
13.15a
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = .98, p-value = .9254; use equal-variances test statistic
Rejection region: t   t  / 2,   t.025,198  1.972 or t  t  / 2,  t.025,198  1.972
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1 


n
n
2 
 1
=
(10 .23  9.66 )  0
 (100  1)2.87 2  (100  1)2.90 2


100  100  2

 1
1 


 100 100 

= 1.40, p-value = .1640.
There is not enough evidence to infer that the distance males and females drive differs.
1
1 
 = (10.23 – 9.66)  1.972
b ( x1  x 2 )  t  / 2 s 2p  
 n1 n 2 
 (100  1)2.87 2  (100  1)2.90 2


100  100  2

 1
1 


 100 100 

= .57  .80; LCL = –.23, UCL = 1.37
c The histograms are bell shaped.
13.16
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = .98, p-value = .9520; use equal-variances test statistic
Rejection region: t  t ,  t .05,58  1.671
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1 


 n1 n 2 
=
(115 .50  110 .20 )  0
 (30  1)21 .69 2  (30  1)21 .93 2


30  30  2

 1
  1 
 30 30 

 .94 , p-value = .1753. There is
not enough evidence to retain supplier A - switch to supplier B.
13.17
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = .92, p-value = .5000; use equal-variances test statistic
Rejection region: t   t  / 2,   t .025,594  1.960 or t  t  / 2,  t .025,594  1.960
275
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1 


n
n
2 
 1
=
(5.56  5.49 )  0
 (306  1)5.36 2  (290  1)5.58 2


306  290  2

 1
1 


 306 290 

 .16 , p-value = .8759. There
is no evidence of a difference in job tenures between men and women.
13.18a
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = 5.18, p-value = .0019; use unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 33.9 (rounded to 34)
Rejection region: t   t  / 2,   t .005,34  2.724 or t  t  / 2,  t .005,34  2.724
t
( x 1  x 2 )  (1   2 )
 s12




 n1 n 2 


=
s 22
(70 .42  56 .44 )  0
 20 .54 2 9.03 2 



 24
16 

 2.94 , p-value = .0060. 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 
b ( x1  x 2 )  t  / 2  1  2  = (70.42 –56.44)  2.030
n

 1 n2 
 20 .54 2 9.03 2 

 = 13.98  9.67;

 24

16


LCL = 4.31, UCL = 23.65
c The amount of time is required to be normally distributed.
d The histograms are somewhat bell shaped.
13.19a H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
Two-tail F test: F = 5.18, p-value = .0019; use unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 276.5 (rounded to 277)
Rejection region: t   t ,   t .01, 277  2.326
t
( x 1  x 2 )  (1   2 )
 s12




 n1 n 2 


s 22
=
(5.02  7.80 )  0
 1.39 2 3.09 2 



 200

200


= –11.60, p-value = 0. There is enough evidence to infer
that the amount of time wasted in unsuccessful firms exceeds that of successful firms.
276
 s2 s2 
b ( x1  x 2 )  t  / 2  1  2 
n

 1 n2 

2
2

1.39
3.09 
= (5.02 – 7.80)  1.960 
= –2.78  .47;


 200
200 
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.
13.20
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = .73, p-value = .0699; use equal-variances test statistic
Rejection region: t  t ,  t .05, 268  1.645
t
( x 1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



=
(.646  .601)  0
 (125  1).045 2  (145  1).053 2


125  145  2

 1
1 


 125 145 

= 7.54, 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.
13.21
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = 1.10, p-value = .6406; use equal-variances test statistic
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 )

s 2p 
1
1 


n
n
2 
 1
=
(.654  .662 )  0
 (95  1).048 2  (90  1).045 2


95  90  2

 1
  1 
 95 90 

= –1.17, p-value = .2444. There is
not enough evidence to infer that the type of discussion affects reaction times.
13.22
H 0 : (1   2 ) = 0
H1 : (1   2 ) > 0
Two-tail F test: F = .97, p-value = .9054; use equal-variances test statistic
Rejection region: t  t ,  t.05,143  1.656
t
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
=
(6.18  5.94 )  0
 (64  1)1.59 2  (81  1)1.61 2


64  81  2

 1
  1 
 64 81 

= .90, p-value = .1858. There is not
enough evidence to infer that people spend more time researching for a financial planner than they do for a
stock broker.
277
13.23a
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
Two-tail F test: F = .74, p-value = .0446; use unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 373
Rejection region: t   t ,   t.05,373  1.645
t
( x 1  x 2 )  (1   2 )
 s12




 n1 n 2 


=
s 22
(63 .71  66 .80 )  0
 5.90 2 6.85 2 



 173
202 

= –4.69, p-value = 0. There is enough evidence to infer
that students without textbooks outperform those with textbooks.
13.24
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = .85, p-value = .2494; use equal-variances test statistic
Rejection region: t   t  / 2,   t .025, 413  1.960 or t  t  / 2,  t .025, 413  1.960
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1 


n
n
2 
 1
=
(149 .85  154 .43)  0
 (213  1)21 .82 2  (202  1)23 .64 2


213  202  2

 1
1 


 213 202 

= –2.05, p-value = .0412.
There is enough evidence to conclude that there are differences in service times between the two chains.
13.25
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = .71, p-value = .1214; use equal-variances test statistic
Rejection region: t   t  / 2,   t .025,168  1.975 or t  t  / 2,  t .025,168  1.975
t
( x 1  x 2 )  (1   2 )

s 2p 
1
1

 n1 n 2



=
(53 .05  51 .67 )  0
 (79  1)3.06 2  (91  1)3.64 2


79  91  2

 1
  1 
 79 91 

= 2.65, p-value = .0087. There is
enough evidence to conclude that the two types of specialties differ in the time devoted to patients.
13.26 a H 0 : (1   2 ) = 0
H1 : (1   2 )  0
Two-tail F test: F = 1.51, p-value = .0402; use unequal-variances test statistic
278

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 190
Rejection region: t   t  / 2,   t.025,190  1.973 or t  t  / 2,  t.025,190  1.973
t
( x1  x 2 )  (1   2 )
 s12


 
n n 
2
 1
=
s 22
(130 .93  126 .14 )  0
 31 .99 2 26 .00 2


 100
100





= 1.16, p-value = .2467. There is not enough evidence to
infer that differences exist between the two types of customers.
13.27
H 0 : (1   2 ) = 0
H1 : (1   2 ) > 0
Two-tail F test: F = 1.07, p-value = .8792; use equal-variances test statistic
Rejection region: t  t ,  t.05,38  1.684
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 
 
 n1 n 2 
=
(73 .60  69 .20 )  0
 (20  1)15 .60 2  (20  1)15 .06 2


20  20  2

 1
  1 
 20 20 

= .91, p-value = .1849. There is
not enough evidence to infer that the new design tire lasts longer than the existing design tire.
13.28
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
Two-tail F test: F = 1.51, p-value = .0402; use unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 531
Rejection region: t   t ,   t.05,531  1.645
t
( x 1  x 2 )  (1   2 )
 s12 s 22 
 

 n1 n 2 


=
(237 .99  251 .99 )  0
 12 .24 2 14 .84 2 



 263
279 

= –12.01, p-value = 0. There is enough evidence to
infer that British golfers play golf in less time than do American golfers.
13.29
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
Two-tail F test: F = 1.64, p-value = .1815; use equal-variances test statistic
Rejection region: t   t ,   t .05,59  1.671
279
t
( x1  x 2 )  (1   2 )

s 2p 
1
1 
 
 n1 n 2 
=
(6,345 .5  6,358 .3)  0
 (28  1)71 .30 2  (33  1)55 .72 2


28  33  2

 1
  1 
 28 33 

= –.85, p-value = .1999. There is
not enough evidence to conclude that British courses are shorter than American courses.
H 0 : (1   2 ) = 0
13.30
H1 : (1   2 ) < 0
Two-tail F test: F = .35, p-value = .0061; use unequal-variances test statistic

(s12 / n 1  s 22 / n 2 ) 2
(s12 / n 1 ) 2 (s 22 / n 2 ) 2

n1 1
n 2 1
= 53
Rejection region: t   t ,   t.05,53  1.676
t
( x1  x 2 )  (1   2 )
 s12


 
n n 
2
 1
=
s 22
(7,137 .2  9,303 .7)  0
 195 .07 2 331 .89 2


 28
33

= –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.
13.31 The data are observational. Experimental data could be produced by randomly assigning babies to
either Tastee or the competitor’s product.
13.32 Assuming that the volunteers were randomly assigned to eat either oat bran or another grain cereal
the data are experimental.
13.33 The data are observational. It is not possible to conduct a controlled experiment.
13.34 a The data are observational because the students decided themselves which package each would use.
b The professor can randomly assign one of the two packages to each student.
c Better students tend to choose Program B and better students learn how to use computer software more
quickly.
13.35 a The data are observational.
b A powerful dictator can randomly order medical students to either general/family practice or pediatrics.
C It may be that physicians who choose to be general practitioners are more patient-oriented and as a result
such physicians spend more time with their patients.
13.36a Let students select the section they wish to attend and compare test results.
280
b Randomly assign students to either section and compare test results.
13.37 Randomly assign patients with the disease to receive either the new drug or a placebo.
13.38a 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.
13.39a 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.
13.40
H0 : D = 0
H1 :  D < 0
Rejection region: t   t ,   t .05,7  1.895
t
x D  D
sD / n D

 4.75  0
4.17 / 8
 3.22 , p-value = .0073. There is enough evidence to infer that the Brand A is
better than Brand B.
13.41
H0 : D = 0
H1 :  D < 0
Rejection region: t   t ,   t .05,7  1.895
t
xD  D

sD / nD
13.42
 .175  0
 2.20, p-value = .0320. There is enough evidence to infer that ABS is better.
.225 / 8
H0 : D = 0
H1 :  D > 0
Rejection region: t  t ,  t .05,6  1.943
t
xD  D
sD / nD

1.86  0
 1.98, p-value = .0473. There is enough evidence to infer that the camera
2.48 / 7
reduces the number of red-light runners.
281
13.43a
H0 : D = 0
H1 :  D  0
Rejection region: t   t ,   t.05,11  1.796
t
x D  D
 1.00  0

sD / nD
 –1.15, p-value = .1375. There is not enough evidence to infer that the new
3.02 / 12
fertilizer is better.
b xD  t / 2
sD
nD
=  1.00  2.201
3.02
 1.00  1.92 ; LCL = –2.92, UCL = .92
12
c The differences are required to be normally distributed
d No, the histogram is bimodal.
e The data are experimental.
f The experimental design should be independent samples.
13.44 a H 0 :  D = 0
H1 :  D  0
Rejection region: t  t ,  t.05,11  1.796
t
x D  D
3.08  0

sD / nD
 1.82, p-value = .0484. There is enough evidence to infer that companies with
5.88 / 12
exercise programs have lower medical expenses.
b xD  t / 2
sD
nD
= 3.08  2.201
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.
13.45
H0 : D = 0
H1 :  D  0
Rejection region: t  t ,  t.05,149  1.656
t
x D  D
sD / n D

12 .4  0
99 .1 / 150
 1.53, p-value = .0638. There is not enough evidence to infer that mortgage
payments have increases in the past 5 years.
13.46
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
282
t
x D  D
sD / n D

 1.16  0
2.22 / 50
 –3.70, p-value = .0006. There is enough evidence to infer that waiters and
waitresses earn different amounts in tips.
13.47 a x D  t  / 2
sD
nD
= 19.75  1.684
30 .63
 19 .75  8.16 ; LCL = 11.59, UCL = 27.91
40
H0 : D = 0
b
H1 :  D  0
Rejection region: t  t ,  t.05,39  1.684
t
x D  D
sD / n D

19 .75  0
30 .63 / 40
 4.08, p-value = .0001. 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.
13.48a
H0 : D = 0
H1 :  D  0
Rejection region: t  t ,  t.10,14  1.345
t
x D  D
57 .40  0

sD / nD
 16 .92, p-value = 0. There is enough evidence to conclude that heating costs
13 .14 / 15
for insulated homes is less than that for uninsulated homes.
b xD  t / 2
sD
nD
= 57 .40  2.145
13 .14
 57 .40  7.28 ; LCL = 50.12, UCL = 64.68
15
c Differences are required to be normally distributed.
13.49
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
x D  D
sD / n D

 42 .94  0
317 .16 / 45
 –.91, p-value = .3687. There is not enough evidence to infer men and
women spend different amounts on health care.
283
13.50
H0 : D = 0
H1 :  D  0
Rejection region: t   t ,   t.05,169  1.654
t
x D  D
sD / n D

 183 .35  0
1568 .94 / 170
 –1.52, p-value = .0647. There is not enough to infer stock holdings have
decreased.
13.51
H0 : D = 0
H1 :  D > 0
Rejection region: t  t ,  t.05,37  1.690
t
xD  D

sD / nD
.0422  0
 1.59, p-value = .0599. There is not enough evidence to conclude that ratios
.1634 / 38
are higher this year.
13.52
H0 : D = 0
H1 :  D > 0
Rejection region: t  t ,  t.05,54  1.676
t
xD  D

sD / nD
520 .85  0
 2.08, p-value = .0210. There is enough evidence to infer that company 1’s
1854 .92 / 55
calculated tax payable is higher than company 2’s.
13.53
H0 : D = 0
H1 :  D > 0
Rejection region: t  t ,  t.05,19  1.729
t
x D  D
sD / n D

4.55  0
7.22 / 20
 2.82, p-value = .0055. There is enough evidence to that the new design tire
lasts longer than the existing design.
13.54 The matched pairs experiment reduced the variation caused by different drivers.
13.55
H0 : D = 0
H1 :  D > 0
Rejection region: t  t ,  t.05, 24  1.711
284
t
x D  D
sD / n D

4587  0
22 ,851 / 25
 1.00, p-value = .1628. There is not enough evidence to infer that finance
majors attract higher salary offers than do marketing majors.
13.56 Salary offers and undergraduate GPA are not as strongly linked as are salary offers and MBA GPA.
13.57 a 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. There is not enough evidence to conclude that the population
variances differ.
b 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 = .2071. There is not enough evidence to conclude that the population
variances differ.
c The value of the test statistic is unchanged. The p-value increases.
 s2
13.58 a LCL =  12
s
 2
 s2
b LCL =  12
s
 2

1
 28  1

=  
= .366, UCL =
 F / 2, ,
19
4
.
03



1 2

1
 28  1

= 
= .649, UCL =
 F / 2, ,
 19  2.27

1 2
 s12

 s2
 2
 s12

 s2
 2

F / 2, , =  28 4.03 = 5.939
2 1

 19 


F / 2, , =  28 2.27 = 3.345
2 1

 19 

c The interval narrows.
13.59
H 0 : 12 /  22  1
H1 : 12 /  22  1
Rejection region: F  F / 2,1 ,2  F.025,9,10  3.78 or
F  F1 / 2,1 ,2  1 / F / 2,2 ,1  1 / F.025,10,9  1 / 3.96  .25
F = s12 / s 22 = .0000057/.0000114 =.50, p-value = .3179. There is not enough evidence to conclude that the
two machines differ in their consistency of fills.
285
13.60
H 0 : 12 /  22  1
H1 : 12 /  22  1
Rejection region: F  F / 2,1 , 2  F.025,8,5  6.76 or
F  F1 / 2,1 ,2  1 / F / 2,2 ,1  1 / F.025,5,8  1 / 4.82  .21
F = s12 / s 22 = 755,196/725,778 =1.04, p-value = .9873. The equal-variances t-test in Exercise 13.10 was
justified.
13.61
H 0 : 12 /  22  1
H1 : 12 /  22  1
Rejection region: F  F1,1 ,2  1 / F,2 ,1  1 / F.05,9,9  1 / 3.18  .314
F = s12 / s 22 = .1854/.1893 =.98, p-value = .4879. There is not enough evidence to infer that the second
method is more consistent than the first method.
13.62
H 0 : 12 /  22  1
H1 : 12 /  22  1
Rejection region: F  F / 2,1 , 2  F.025,10,10  3.72 or
F  F1 / 2,1 ,2  1 / F / 2,2 ,1  1 / F.025,10,10  1 / 3.72  .269
F = s12 / s 22 = 193.67/60.00 = 3.23, p-value = .0784. There is not enough evidence to infer that the variances
of the marks differ between the two sections.
13.63
H 0 : 12 /  22  1
H1 : 12 /  22  1
Rejection region: F  F,1 ,2  F.05,99,99  1.35
F = s12 / s 22 = 19.38/12.70 = 1.53, p-value = .0183. There is enough evidence to infer that limiting the
minimum and maximum speeds reduces the variation in speeds.
13.64
H 0 : 12 /  22  1
H1 : 12 /  22  1
Rejection region: F  F / 2,1 , 2  F.025,212,201  1.43 or
F  F1 / 2,1 ,2  1 / F / 2,2 ,1  1 / F.025,212,201  1 / 1.43  .70
286
F = s12 / s 22 = 476.1/558.9 =.85, p-value = .2494. There is not enough evidence to infer a difference between
variances; the equal-variance t test in Exercise 13.24 was justified.
13.65 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 = 41,309/19,850 = 2.08, p-value = .0003. There is enough evidence to conclude that the
variances differ.
13.66
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 = 1,023/676 =1.51, p-value = .0405. There is enough evidence to infer a difference between
variances; the unequal-variance t test in Exercise 13.26 was appropriate.
13.67
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. There is enough evidence to infer that portfolio 2 is riskier than
portfolio 1.
13.68
H 0 : 12 /  22  1
H1 : 12 /  22  1
Rejection region: F  F / 2,1 ,2  F.025,27,32  2.14 or
F  F1 / 2,1 ,2  1 / F / 2,2 ,1  1 / F.025,32,27  1 / 2.13  .47
F = s12 / s 22 = 38,051/110,151 =.35, p-value = .0061. There is enough evidence to infer a difference between
variances; the unequal-variance t test in Exercise 13.30 was the correct technique.
287
13.69
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. There is enough evidence to conclude that the variance in
service times differ.
13.70
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
a z
(p̂1  p̂ 2 )
=
 1
1 

p̂(1  p̂) 
 n1 n 2 
(.45  .40 )
1 
 1
.425 (1  .425 )


 100 100 
= .72, p-value = 2P(Z > .72) = 2(.5 – .2642)
= .4716.
b z
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2
=



(.45  .40 )
1 
 1
.425 (1  .425 )


400
400


= 1.43, p-value = 2P(Z > 1.43) = 2(.5 – .4236)
= .1528.
c The p-value decreases.
13.71
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
a z
(p̂1  p̂ 2 )
=
 1
1 

p̂(1  p̂) 
 n1 n 2 
(.60  .55)
1 
 1
.575 (1  .575 )


 225 225 
 1.07 , p-value = 2P(Z > 1.07) = 2(.5 – .3577)
= .2846
bz
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



=
(.95  .90 )
1 
 1
.925 (1  .925 )


225
225


 2.01, p-value = 2P(Z > 2.01) = 2(.5 – .4778)
= .0444.
c. The p-value decreases.
d z
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



=
(.10  .05)
1 
 1
.075 (1  .075 )


225
225


= .0444.
e. The p-value decreases.
288
 2.01, p-value = 2P(Z > 2.01) = 2(.5 – .4778)
13.72 a (p̂ 1  p̂ 2 )  z  / 2
p̂ 1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
= (.18–.22)  1.645
.18(1  .18) .22 (1  .22 )

100
100
= –.040  .0929
b (p̂1  p̂ 2 )  z  / 2
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
= (.48–.52)  1.645
.48(1  .48) .52 (1  .52 )

100
100
= –.040  .1162
c The interval widens.
13.73a
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z   z .05 = 1.645
z
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



=
(.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.
b
H 0 : (p1  p 2 ) = .05
H1 : (p1  p 2 )  .05
Rejection region: z  z   z .05 = 1.645
z
(p̂1  p̂ 2 )  (p1  p 2 )
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 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 (p̂1  p̂ 2 )  z  / 2
13.74
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
= (.56  .46 )  1.96
.56 (1  .56 ) .46 (1  .46 )
= .10  .045

1100
800
H 0 : (p1  p 2 ) = –.08
H1 : (p1  p 2 )  –.08
Rejection region: z  z  z.01 = –2.33
z
(p̂1  p̂ 2 )  (p1  p 2 )
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 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.
289
13.75
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z   z .05 = 1.645
(p̂1  p̂ 2 )
z
 1
1 

p̂(1  p̂) 
 n1 n 2 
=
(.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.
13.76 a H 0 : (p1  p 2 ) = 0
H1 : (p1  p 2 )  0
Rejection region: z  z   z .05 = –1.645
(p̂1  p̂ 2 )
z
 1
1 

p̂(1  p̂) 
n
n
2 
 1
=
(.093  .115 )
1 
 1
.104 (1  .104 )


 6281 6281 
= –4.04, p-value = 0. There is enough evidence
to infer that Plavix is effective.
13.77 a H 0 : (p1  p 2 ) = 0
H1 : (p1  p 2 )  0
Rejection region: z  z   z .01 = –2.33
p̂1 
104
189
104  189
 .0095 p̂ 2 
 .0172 p̂ 
 .01335
11,000
11,000
22,000
(p̂1  p̂ 2 )
z
 1
1
p̂(1  p̂) 
 n1 n 2



=
(.0095  .0172 )
 1
1 
.01335 (1  .01335 )


11
,
000
11
,
000 

= –4.98, p-value = 0. There is enough
evidence to infer that aspirin is effective in reducing the incidence of heart attacks.
13.78
H 0 : ( p1  p 2 ) = 0
H1 : ( p1  p 2 )  0
Rejection region: z  z   z .05  1.645
p̂1 
1,084
997
1,084  997
 .0985 p̂ 2 
 .0906 p̂ 
 .0946
11,000
11,000
22,000
290
z
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
n
n
2 
 1
(.0985  .0906 )
=
 1
1 
.0946 (1  .0946 )


11
,
000
11
,
000 

 2.00 , p-value = P(Z > 2.00) = .5 – .4772
= .0228. There is enough evidence to infer that aspirin leads to more cataracts.
H 0 : ( p1  p 2 ) = 0
13.79
H1 : (p1  p 2 )  0
Rejection region: z  z   z .01 = –2.33
p̂1 
z
75
132
75  132
 .0289 p̂ 2 
 .0509 p̂ 
 .0399
2,594
2,594
2,594  2,594
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



(.0289  .0509 )
=
 1
1 
.0399 (1  .0399 )


2
,
594
2
,
594


 4.04 , p-value = 0.
There is enough evidence to infer that Letrozole works.
H 0 : ( p1  p 2 ) = 0
13.80
H1 : (p1  p 2 )  0
Rejection region: z  z   z .05  1.645
p̂1 
z
88
105
88  105
 .2228 p̂ 2 
 .2586 p̂ 
 .2409
395
406
395  406
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
(.2228  .2586 )
=
1 
 1
.2409 (1  .2409 )


 395 406 
 1.19, p-value = P(Z < –1.19) = .5 – .3830
= .1170. There is not enough evidence to infer that exercise training reduces mortality.
13.81
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
a z
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
=
(.9057  .8878
1 
 1
.8975 (1  .8975 )


 350 294 
 .75, p-value = 2P(Z > .75) = 2(.5 – .2734)
= .4532.
There is not enough evidence to infer that the two populations of car owners differ in their satisfaction
levels.
291
H 0 : ( p1  p 2 ) = 0
13.82
H 1 : ( p1  p 2 ) > 0
Rejection region: z  z   z .10 = 1.28
z
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
=
(.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 (p̂1  p̂ 2 )  z  / 2
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
= (.2632–0741)  1.645
13.83a
.2632 (1  .2632 ) .0741 (1  .0741 )
=.1891  .1223; LCL = .0668, UCL = .3114

38
162
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z   z .10 = 1.28
z
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



=
(.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 (p̂1  p̂ 2 )  z  / 2
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
= (.62–.52)  1.645
.62 (1  .62 ) .52 (1  .52 )
=.10  .0543;

400
500
LCL = .0457 and UCL = .1543; the change in the public’s opinion is estimated to lie between 4.57 and
15.43%
13.84
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z   z .05 = 1.645
z
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



=
(.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.
292
H 0 : (p1  p 2 ) = –.02
13.85
H1 : (p1  p 2 )  –.02
Rejection region: z  z   z .05 = –1.645
(p̂1  p̂ 2 )  (p1  p 2 )
z
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 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.
H 0 : ( p1  p 2 ) = 0
13.86
H 1 : ( p1  p 2 )  0
Rejection region: z  z   z .05 = 1.645
z
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
=
(.1385  .0905 )
1 
 1
.1035 (1  .1035 )


231
619


z = 2.04, p-value = P(Z > 2.04) = .5 – .4793
= .0207.
There is enough evidence to conclude that health conscious adults are more likely to buy Special X.
13.87a
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
Rejection region: z  z   z .05 = 1.645
p̂ 1 
z
68
29
68  29
 .4172 p̂ 2 
 .2685 p̂ 
 .3579
163
108
163  108
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
n
n
2 
 1
=
(.4172  .2685 )
1 
 1
.3579 (1  .3579 )


 163 108 
 2.50, p-value = P(Z > 2.50) = .5 – .4938
= .0062.
There is enough evidence to conclude that members of segment 1 are more likely to use the service than
members of segment 4.
b
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
p̂ 1 
20
10
20  10
 .3704 p̂ 2 
 .4348 p̂ 
 .3896
54
23
54  23
293
z
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
n
n
2 
 1
=
(.3704  .4348
1 
 1
.3896 (1  .3896 )  
 54 23 
 .53, p-value = 2P(Z < –.53) = 2(.5 – .2019)
= .5962.
There is not enough evidence to infer that retired persons and spouses that work in the home differ in their
use of services such as Quik Lube.
13.88
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
z
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
n
n
2 
 1
=
(.0995  .1297 )
1 
 1
.1132 (1  .1132 )


 382 316 
= –1.25, p-value = 2P(Z < –1.25)
= 2(.5 – .3944) = .2112.
There is not enough evidence to infer differences between the two sources.
13.89
H 0 : (1   2 ) = 0
H1 : (1   2 ) > 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
C
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.
13.90 Gross sales must increase by 50/.20 = $250 to pay for ads.
H 0 : (1   2 ) = 250
H1 : (1   2 ) > 250
294
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
C
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.
13.91
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
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.
13.92
H0 : D = 0
H1 :  D  0
295
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
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
C
Drug
Placebo
18.43
22.03
30.39
66.37
100
100
0.69
0
99
-6.09
0.0000
1.6604
0.0000
1.9842
t = –6.09, p-value = 0. There is enough evidence to infer that the new drug is effective.
13.93
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
C
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.
13.94
H 0 : ( p1  p 2 ) = 0
H1 : (p1  p 2 )  0
296
1
2
3
4
5
6
7
8
9
10
11
A
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
B
C
D
Last Year This Year
0.6758
0.7539
327
382
0
-2.30
0.0106
1.6449
0.0212
1.96
z = –2.30, p-value = .0106. There is enough evidence to infer an increase in seatbelt use.
13.95a
H 0 : (1   2 ) = 0
H1 : (1   2 ) < 0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
C
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.
13.96a
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
297
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
C
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.
H 0 : (1   2 ) = 0
b
H1 : (1   2 ) < 0
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
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
Low
33.10
278.69
21.00
0.00
26.00
-2.87
0.0040
1.3150
0.0080
1.7056
High
56.84
1047.81
19
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.
13.97a
H0 : D = 0
H1 :  D  0
298
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
t-Test: Paired Two Sample for Means
C
Female
55.68
105.64
25
0.96
0
24
-1.13
0.1355
1.3178
0.2710
1.7109
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
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.
13.98
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
1
2
3
4
5
6
7
8
9
10
11
A
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
B
C
D
This Year 3 Years Ago
0.4351
0.3558
393
385
0
2.26
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.
13.99
H 0 : (1   2 ) = 25
H1 : (1   2 ) > 25
299
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
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
C
A nondefectives A 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.
13.100
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
The totals in columns A through D are 5788, 265, 5154, and 332, respectively.
A
B
C
D
1 z-Test of the Difference Between Two Proportions (Case 1)
2
3
Sample 1 Sample 2 z Stat
4 Sample proportion 0.045800 0.064400 P(Z<=z) one-tail
5 Sample size
5788
5154
z Critical one-tail
6 Alpha
0.05
P(Z<=z) two-tail
7
z Critical two-tail
E
-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.
13.101
H0 : D = 0
H1 :  D  0
300
Dry Cleaner
A
B
C
1 t-Test: Paired Two Sample for Means
2
3
Dry C Before Dry C After
4 Mean
168.00
165.50
5 Variance
351.38
321.96
6 Observations
14
14
7 Pearson Correlation
0.86
8 Hypothesized Mean Difference
0
9 df
13
10 t Stat
0.96
11 P(T<=t) one-tail
0.1780
12 t Critical one-tail
1.7709
13 P(T<=t) two-tail
0.3559
14 t Critical two-tail
2.1604
t = .96, p-value = .1780. There is not enough evidence to conclude that the dry cleaner sales have
decreased.
Doughnut shop
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
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
B
C
Donut Before
308.14
809.67
14
0.86
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.
301
Convenience store
A
B
C
1 t-Test: Paired Two Sample for Means
2
3
Convenience Before Convenience After
4 Mean
374.64
348.14
5 Variance
2270.40
2941.82
6 Observations
14
14
7 Pearson Correlation
0.97
8 Hypothesized Mean Difference
0
9 df
13
10 t Stat
7.34
11 P(T<=t) one-tail
0.0000
12 t Critical one-tail
1.7709
13 P(T<=t) two-tail
0.0000
14 t Critical two-tail
2.1604
t = 7.34, p-value = 0. There is enough evidence to conclude that the convenience store sales have
decreased.
13.102a H 0 : (1   2 ) = 0
H1 : (1   2 )  0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
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
C
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
302
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
t-Test: Two-Sample Assuming Equal Variances
C
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
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
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.
13.103a H 0 : (p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
A
1 z-Test: Two Proportions
2
3
4 Sample Proportions
5 Observations
6 Hypothesized Difference
7 z Stat
8 P(Z<=z) one tail
9 z Critical one-tail
10 P(Z<=z) two-tail
11 z Critical two-tail
B
C
D
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.
13.104 a H 0 : (1   2 ) = 0
H1 : (1   2 )  0
303
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
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
Exercise
13.52
5.76
25
0
42
4.14
0.0001
2.4185
0.0002
2.6981
Drug
9.92
13.16
25
t = 4.14, p-value = .0001. There is enough evidence that exercise is more effective than medication in
reducing hypertension.
b
1
2
3
4
5
6
7
8
A
B
C
D
t-Estimate of the Difference Between Two Means (Unequal-Variances)
Mean
Variance
Sample size
Degrees of freedom
Confidence level
Sample 1 Sample 2 Confidence Interval Estimate
13.52
9.92
3.60

5.76
13.16 Lower confidence limit
25
25
Upper confidence limit
41.63
0.95
LCL = 1.84, UCL = 5.36
c The histograms are bell shaped.
13.105
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
1
2
3
4
5
6
7
8
9
10
11
12
13
E
A
B
C
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
304
F
1.76
1.84
5.36
t = –1.14, p-value = .1288. There is not enough evidence to conclude that people who exercise moderately
more frequently lose weight faster
13.106
H0 : D = 0
H1 :  D  0
A
B
C
1 t-Test: Paired Two Sample for Means
2
3
Group 1 Group 2
4 Mean
7.53
8.57
5 Variance
29.77
43.37
6 Observations
50
50
7 Pearson Correlation
0.89
8 Hypothesized Mean Difference
0
9 df
49
10 t Stat
-2.40
11 P(T<=t) one-tail
0.0100
12 t Critical one-tail
1.6766
13 P(T<=t) two-tail
0.0201
14 t Critical two-tail
2.0096
t = –2.40, p-value = .0100. There is enough evidence to conclude that people who exercise moderately
more frequently lose weight faster
13.107
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
1
2
3
4
5
6
7
8
9
10
11
A
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
B
C
D
Special K Other
0.635
0.53
200
200
0
2.13
0.0166
1.6449
0.0332
1.9600
z = 2.13, p-value = .0166. There is enough evidence to conclude that Special K buyers like the ad more
than non-buyers.
13.108
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
305
1
2
3
4
5
6
7
8
9
10
11
A
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
B
C
D
Special K Other
0.575
0.515
200
200
0
1.20
0.1141
1.6449
0.2282
1.9600
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.
13.109
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
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
C
Small space Large space
1245.7
1915.8
23812
65566
25
25
0
39
-11.21
0.0000
1.6849
0.0000
2.0227
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.
13.110
H0 : (1  2 ) = 0
H1 : (1   2 )  0
306
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
t-Test: Two-Sample Assuming Unequal Variances
C
Computer No Computer
69,933
48,246
63,359,040 101,588,525
89
61
0
109
14.07
0.0000
1.2894
0.0000
1.6590
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
t = 14.07, p-value = 0. There is enough evidence to conclude that single-person businesses that use a PC
earn more.
13.111
H 0 : ( p1  p 2 ) = 0
H 1 : ( p1  p 2 )  0
1
2
3
4
5
6
7
8
9
10
11
A
z-Test: Two Proportions
B
C
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
0.948
250
0
1.26
0.1037
1.2816
0.2074
1.6449
D
Older
0.920
250
z = 1.26, p-value = .1037. There is not enough evidence to conclude that the new company is better.
13.112
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
307
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
t-Test: Two-Sample Assuming Equal Variances
C
Supplement Placebo
19.02
21.85
41.34
25.49
48
48
33.41
0
94
-2.40
0.0092
1.6612
0.0183
1.9855
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
t = –2.40, p-value = .0092. There is enough evidence to infer that taking vitamin and mineral supplements
daily increases the body's immune system?
13.113
H 0 : ( p1  p 2 ) = 0
H1 : (p1  p 2 )  0
A
1 z-Test: Two Proportions
2
3
4 Sample Proportions
5 Observations
6 Hypothesized Difference
7 z Stat
8 P(Z<=z) one tail
9 z Critical one-tail
10 P(Z<=z) two-tail
11 z Critical two-tail
B
C
D
$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.
13.114
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
308
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
C
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
Female
7.27
2.57
103
2.71
0
198
1.08
0.1410
1.6526
0.2820
1.9720
Male
7.02
2.85
97
t = 1.08, p-value = .2820. There is not enough evidence to conclude that female and male high school
students differ in the amount of time spent at part-time jobs.
13.115
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
A
B
C
1 t-Test: Two-Sample Assuming Unequal Variances
2
3
City
Suburbs
4 Mean
2.20
2.00
5 Variance
0.388
0.737
6 Observations
47
53
7 Hypothesized Mean Difference
0
8 df
94
9 t Stat
1.35
10 P(T<=t) one-tail
0.0907
11 t Critical one-tail
1.6612
12 P(T<=t) two-tail
0.1813
13 t Critical two-tail
1.9855
t = 1.35, p-value = .0907. There is not enough evidence to infer that city households discard more
newspaper than do suburban households.
13.116
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
309
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
t-Test: Two-Sample Assuming Unequal Variances
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
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
t = 2.28, p-value = .0115.There is enough evidence to infer that teenagers see more movies than do twenty
to thirty year olds.
13.117
H 0 : ( p1  p 2 ) = 0
H1 : (p1  p 2 )  0
1
2
3
4
5
6
7
8
9
10
11
A
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
B
C
D
No HS
HS
0.127
0.358
63
257
0
-3.54
0.0002
1.6449
0.0004
1.96
z = –3.54, p-value = .0002. There is enough evidence to conclude that Californians who did not complete
high school are less likely to take a course in the university’s evening program.
13.118
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
310
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
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 Groups 2-4
11.58
10.60
9.28
21.41
269
981
0
644
4.15
0.0000
1.6472
0.0000
1.9637
t = 4.15, p-value = 0. There is enough evidence to conclude that on average the market segment concerned
about eating healthy food (group 1) outspends the other market segments.
13.119
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
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
Sale-CDs Sale-fax
59.04
65.57
425.4
849.7
122
144
0
256
-2.13
0.0171
1.6508
0.0341
1.9693
t = –2.13, p-value = .0171. There is enough evidence to conclude that those who buy the fax/copier
outspend those who buy the package of CD-ROMS.
13.120
H 0 : (1   2 ) = 0
H1 : (1   2 )  0
311
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
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
C
No
Yes
91,467
97,836
461,917,705 401,930,840
466
55
455,676,297
0
519
-2.09
0.0184
1.6478
0.0369
1.9645
t = –2.09, p-value = .0184. There is enough evidence to infer that professors aged 55 to 64 who plan to
retire early have higher salaries than those who don’t plan to retire early.
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
ACT 241 0 versus 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
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
C
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.
312
ACT 241 0 versus 2
A
B
C
1 t-Test: Two-Sample Assuming Unequal Variances
2
3
ACT 241-0 ACT 241-2
4 Mean
73.49
80.26
5 Variance
45.62
26.72
6 Observations
296
54
7 Hypothesized Mean Difference
0
8 df
90
9 t Stat
-8.40
10 P(T<=t) one-tail
0.0000
11 t Critical one-tail
1.6620
12 P(T<=t) two-tail
0.0000
13 t Critical two-tail
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.
ACT 242 0 versus 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
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
C
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.
313
ACT 242 0 versus 2
A
B
C
1 t-Test: Two-Sample Assuming Equal Variances
2
3
ACT242-0 ACT 242-2
4 Mean
76.89
78.92
5 Variance
15.09
14.12
6 Observations
210
50
7 Pooled Variance
14.90
8 Hypothesized Mean Difference
0
9 df
258
10 t Stat
-3.35
11 P(T<=t) one-tail
0.0005
12 t Critical one-tail
1.6508
13 P(T<=t) two-tail
0.0009
14 t Critical two-tail
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.
Case 13.2 a
H 0 : ( p1  p 2 ) = 0
H1 : (p1  p 2 )  0
1
2
3
4
5
6
7
8
9
10
11
A
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
B
C
D
No Degree Degree
0.470
0.623
417
183
0
-3.45
0.0003
1.6449
0.0006
1.96
z = –3.45, p-value = .0003. There is enough evidence to conclude that Canadians with a university degree
are more likely to take vitamin pills than are less-educated Canadians.
314
b
H 0 : ( p1  p 2 ) = 0
H1 : (p1  p 2 )  0
A
1 z-Test: Two Proportions
2
3
4 Sample Proportions
5 Observations
6 Hypothesized Difference
7 z Stat
8 P(Z<=z) one tail
9 z Critical one-tail
10 P(Z<=z) two-tail
11 z Critical two-tail
B
C
D
<$35,000 >$70,000
0.468
0.651
252
152
0
-3.57
0.0002
1.6449
0.0004
1.96
z = –3.57, p-value = .0002. There is enough evidence to conclude that Canadians with household incomes
that exceed $70,000 are more likely to take vitamin pills than Canadians with household incomes that are
less than $35,000.
315
316