Chapter 13

13.1a Equal-variances estimator

+±− α21

2

p2/21 n

1

n

1

st)xx(

= (524 – 469)

±

2.009

+

−+ −+−

25

1

25

1

22525

131)125(129)125( 22

= 55

±

73.87

b Equal-variances estimator

+±− α21

2

p2/21 n

1

n

1

st)xx(

= (524 – 469)

±

2.009

+

−+ −+−

25

1

25

1

22525

260)125(255)125( 22

= 55

±

146.33

c The interval widens.

d Equal-variances estimator

+±− α21

2

p2/21 n

1

n

1

st)xx(

= (524 – 469)

±

1.972

+

−+ −+−

100

1

100

1

2100100

131)1100(129)1100( 22

= 55

±

36.26

e The interval narrows.

13.2

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

≠

0

a Equal-variances test statistic

Rejection region:

=−<

να

,2/

tt

=− 22,025.

t

–2.074 or

=>

να ,2/

tt

=

22,025.

t

2.074

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

12

1

12

1

21212

16)112(18)112(

0)7174(

22

= .43, p-value = .6703. There is not enough

evidence to infer that the population means differ.

b Equal-variances test statistic

Rejection region:

=−<

να

,2/

tt

=− 22,025.

t

–2.074 or

=>

να ,2/

tt

=

22,025.

t

2.074

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

12

1

12

1

21212

198)112(210)112(

0)7174(

22

= .04, p-value = .9716. There is not enough

evidence to infer that the population means differ.

c The value of the test statistic decreases and the p-value increases.

25

d Equal-variances test statistic

Rejection region:

=−<

να

,2/

tt

=− 298,025.

t

–1.960 or

=>

να ,2/

tt

=

298,025.

t

1.960

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

150

1

150

1

2150150

16)1150(18)1150(

0)7174(

22

= 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:

=−<

να

,2/

tt

=− 22,025.

t

–2.074 or

=> να ,2/

tt

=

22,025.

t

2.074

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

12

1

12

1

21212

16)112(18)112(

0)7176(

22

= .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 a Unequal-variances estimator

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 64.8 (rounded to 65, approximated by

70=ν

)

+±− α2

2

2

1

2

1

2/21 n

s

n

s

t)xx(

= (63 – 60)

±

1.667

+45

7

50

18 22

= 3

±

4.59

b Unequal-variances estimator

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 63.1 (rounded to 63, approximated by

60=ν

)

+±− α2

2

2

1

2

1

2/21 n

s

n

s

t)xx(

= (63 – 60)

±

1.671

+45

15

50

41 22

= 3

±

10.38

c The interval widens.

d Unequal-variances estimator

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 131 (approximated by

140=ν

)

+±− α2

2

2

1

2

1

2/21 n

s

n

s

t)xx(

= (63 – 60)

±

1.656

+90

7

100

18 22

= 3

±

3.22

26

e The interval narrows.

13.4

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

>

0

a Unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 200.4 (rounded to 200)

Rejection region:

653.1ttt 200,05., ==> να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

150

54

150

128

0)405412(

22

= .62, p-value = .2689. There is not enough evidence to infer that

1

µ

is greater than

2

µ

.

b Unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 223.1 (rounded to 223)

Rejection region:

645.1ttt 223,05., ≈=> να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

150

16

150

31

0)405412(

22

= 2.46, p-value = .0074. There is enough evidence to infer that

1

µ

is

greater than

2

µ

.

c The value of the test statistic increases and the p-value decreases.

d Unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 25.6 (rounded to 26)

Rejection region:

706.1ttt

26,05.,

==>

να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

20

54

20

128

0)405412(

22

= .23, p-value = .4118. There is not enough evidence to infer that

1

µ

is greater than

2

µ

.

e The value of the test statistic decreases and the p-value increases.

27

f Unequal-variances test statistic

Rejection region:

653.1ttt 200,05., ==> να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

150

54

150

128

0)405409(

22

= .35, p-value = .3624. There is not enough evidence to infer that

1

µ

is greater than

2

µ

.

g The value of the test statistic decreases and the p-value increases.

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 210 µ−µ

= 0

)(:H 211 µ−µ

< 0

Two-tail F test: F = 1.02, p-value = .9856; use equal-variances test statistic

Rejection region:

=−<

να

,

tt

372.1t 10,10. −=−

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

,43.

6

1

6

1

266

4.6653)16(5.6767)16(

0)83.38150.361( −=

+

−+ −+−

−−

p-value = .3382. The manager should

choose to use cameras.

13.8

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

< 0

Two-tail F test: F = 1.02, p-value = .9823; use equal-variances test statistic

Rejection region:

=−<

να

,

tt

330.1t 18,10. −=−

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

,04.2

10

1

10

1

21010

79.5)110(88.5)110(

0)30.710.5( −=

+

−+ −+−

−−

p-value = .0283. There is enough

evidence to infer that there are fewer errors when the yellow ball is used.

28

13.9

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

≠

0

Two-tail F test: F = .11, p-value = .0106; use unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 8.59 (rounded to 9)

Rejection region:

=−<

να

,2/

tt

833.1t 9,05. −=−

or

=>

να ,2/

tt

833.1t 9,05. =

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

,24.

8

71.367

8

21.42

0)50.1625.18( =

+

−−

p-value = .8123. There is not enough evidence to infer that

the two groups differ in the number of pictures printed.

13.10

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

≠

0

Two-tail F test: F = 1.04, p-value = .9873; use equal-variances test statistic

Rejection region:

=−<

να

,2/

tt

771.1t 13,05. −=−

or

=>

να ,2/

tt

771.1t 13,05. =

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

,59.1

6

1

9

1

269

778,725)16(196,755)19(

0)093,4372,3( −=

+

−+ −+−

−−

p-value = .1368. There is not

enough evidence to infer a difference between the two types of vacation expenses.

13.11a

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

> 0

Two-tail F test: F = 1.59, p-value = .3050; use equal-variances test statistic

Rejection region:

684.1ttt

38,05.,

≈=>

να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

25

1

15

1

22515

35.3)125(23.4)115(

0)36.3193.36(

22

= 4.61, p-value = 0. There is enough

evidence to infer that Tastee is superior.

b

+±− α21

2

p2/21 n

1

n

1

st)xx(

= (36.93 – 31.36)

±

2.021

+

−+ −+−

25

1

15

1

22515

35.3)125(23.4)115( 22

= 5.57

±

2.43; LCL = 3.14, UCL = 8.00

c The histograms are somewhat bell shaped. The weight gains may be normally distributed.

29

13.12

)(:H 210 µ−µ

= 0

≠µ−µ )(:H 211

0

Two-tail F test: F = .99, p-value = .9571; use equal-variances test statistic

Rejection region:

=−<

να

,2/

tt

960.1t 238,025. −=−

or

=>

να ,2/

tt

960.1t 238,025. =

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

,55.1

120

1

120

1

2120120

45.4)1120(43.4)1120(

0)12.901.10(

22 =

+

−+ −+−

−−

p-value = .1204. There is not

enough evidence to infer that oat bran is different from other cereals in terms of cholesterol reduction?

13.13 a

)(:H 210 µ−µ

= 0

≠µ−µ )(:H 211

0

Two-tail F test: F = .50, p-value = 0; use unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 449

Rejection region:

960.1ttt

449,025.,2/

−≈−=−<

να

or

960.1ttt

449,025.,2/

≈=>

να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

250

32.43

250

77.30

0)96.5299.58(

22

= 1.79, p-value = .0734. There is not enough evidence to

conclude that a difference in mean listening times exist between the two populations.

b

+±− α2

2

2

1

2

1

2/21 n

s

n

s

t)xx(

= (58.99 –52.96)

±

1.960

+250

32.43

250

77.30 22

= 6.03

±

6.59;

LCL = –.56, UCL = 12.62

c The histograms are bell shaped.

13.14 a

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

> 0

Two-tail F test: F = 1.01, p-value = .9619; use equal-variances test statistic

Rejection region:

=>

να

,

tt

645.1t 282,05. ≈

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

,88.2

159

1

125

1

2159125

99.6)1159(02.7)1125(

0)40.5781.59(

22 =

+

−+ −+−

−−

p-value = .0021. There is

enough evidence to infer that the cruise ships are attracting younger customers.

30

b

+±− α21

2

p2/21 n

1

n

1

st)xx(

= (59.81 – 57.40)

±

2.576

+

−+ −+−

159

1

125

1

2159125

99.6)1159(02.7)1125( 22

= 2.41

±

2.16;

LCL = .25, UCL = 4.57

13.15a

)(:H 210 µ−µ

= 0

≠µ−µ )(:H 211

0

Two-tail F test: F = .98, p-value = .9254; use equal-variances test statistic

Rejection region:

972.1ttt 198,025.,2/ −≈−=−< να

or

972.1ttt 198,025.,2/ ≈=> να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

100

1

100

1

2100100

90.2)1100(87.2)1100(

0)66.923.10(

22

= 1.40, p-value = .1640. There is not

enough evidence to infer that the distance males and females drive differs.

b

+±− α21

2

p2/21 n

1

n

1

st)xx(

= (10.23 – 9.66)

±

1.972

+

−+ −+−

100

1

100

1

2100100

90.2)1100(87.2)1100( 22

= .57

±

.80; LCL = –.23, UCL = 1.37

c The histograms are bell shaped.

13.16

)(:H 210 µ−µ

= 0

>µ−µ )(:H 211

0

Two-tail F test: F = .98, p-value = .9520; use equal-variances test statistic

Rejection region:

671.1ttt 58,05., ≈=> να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

,94.

30

1

30

1

23030

93.21)130(69.21)130(

0)20.11050.115(

22 =

+

−+ −+−

−−

p-value = .1753. There is not

enough evidence to retain supplier A - switch to supplier B.

13.17

)(:H 210 µ−µ

= 0

≠µ−µ )(:H 211

0

Two-tail F test: F = .92, p-value = .5000; use equal-variances test statistic

Rejection region:

960.1ttt 594,025.,2/ −≈−=−< να

or

960.1ttt 594,025.,2/ ≈=> να

31

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

,16.

290

1

306

1

2290306

58.5)1290(36.5)1306(

0)49.556.5(

22 =

+

−+ −+−

−−

p-value = .8759. There is no

evidence of a difference in job tenures between men and women.

13.18a

)(:H 210 µ−µ

= 0

≠µ−µ )(:H 211

0

Two-tail F test: F = 5.18, p-value = .0019; use unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 33.9 (rounded to 34)

Rejection region:

724.2ttt 34,005.,2/ −≈−=−< να

or

724.2ttt 34,005.,2/ ≈=> να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

,94.2

16

03.9

24

54.20

0)44.5642.70(

22 =

+

−−

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.

b

+±− α2

2

2

1

2

1

2/21 n

s

n

s

t)xx(

= (70.42 –56.44)

±

2.030

+16

03.9

24

54.20 22

= 13.98

±

9.67;

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 210 µ−µ

= 0

)(:H 211 µ−µ

< 0

Two-tail F test: F = 5.18, p-value = .0019; use unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 276.5 (rounded to 277)

Rejection region:

326.2ttt 277,01., −≈−=−< να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

200

09.3

200

39.1

0)80.702.5(

22

= –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.

32

b

+±− α2

2

2

1

2

1

2/21 n

s

n

s

t)xx(

= (5.02 – 7.80)

±

1.960

+200

09.3

200

39.1 22

= –2.78

±

.47;

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 210 µ−µ

= 0

>µ−µ )(:H 211

0

Two-tail F test: F = .73, p-value = .0699; use equal-variances test statistic

Rejection region:

645.1ttt 268,05., ≈=> να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

145

1

125

1

2145125

053).1145(045).1125(

0)601.646(.

22

= 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 210 µ−µ

= 0

≠µ−µ )(:H 211

0

Two-tail F test: F = 1.10, p-value = .6406; use equal-variances test statistic

Rejection region:

973.1ttt 183,025.,2/ −≈−=−< να

or

973.1ttt 183,025.,2/ ≈=> να

+

µ−µ−−

=

21

2

2121

11

)()(

nn

s

xx

t

p

=

+

−+ −+−

−−

90

1

95

1

29095

045).190(048).195(

0)662.654(.

22

= –1.17, p-value = .2444. There is not

enough evidence to infer that the type of discussion affects reaction times.

13.22

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

> 0

Two-tail F test: F = .97, p-value = .9054; use equal-variances test statistic

Rejection region:

656.1ttt 143,05., ≈=> να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

81

1

64

1

28164

61.1)181(59.1)164(

0)94.518.6(

22

= .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.

13.23

)(:H 210 µ−µ

= 0

33

)(:H 211 µ−µ

< 0

Two-tail F test: F = .74, p-value = .0446; use unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 373

Rejection region:

645.1ttt 373,05., −≈−=−< να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

202

85.6

173

90.5

0)80.6671.63(

22

= –4.69, p-value = 0. There is enough evidence to infer that

students without textbooks outperform those with textbooks.

13.24

)(:H 210 µ−µ

= 0

≠µ−µ )(:H 211

0

Two-tail F test: F = .85, p-value = .2494; use equal-variances test statistic

Rejection region:

960.1ttt 413,025.,2/ −≈−=−< να

or

960.1ttt 413,025.,2/ ≈=> να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

202

1

213

1

2202213

64.23)1202(82.21)1213(

0)43.15485.149(

22

= –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 210 µ−µ

= 0

≠µ−µ )(:H 211

0

Two-tail F test: F = .71, p-value = .1214; use equal-variances test statistic

Rejection region:

975.1ttt 168,025.,2/ −≈−=−< να

or

975.1ttt 168,025.,2/ ≈=> να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

91

1

79

1

29179

64.3)191(06.3)179(

0)67.5105.53(

22

= 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

)(: 210 µ−µH

= 0

≠µ−µ )(: 211

H

0

Two-tail F test: F = 1.51, p-value = .0402; use unequal-variances test statistic

34

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 190

Rejection region:

973.1ttt 190,025.,2/ −≈−=−< να

or

973.1ttt 190,025.,2/ ≈=> να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

100

00.26

100

99.31

0)14.12693.130(

22

= 1.16, p-value = .2467. There is not enough evidence to infer

that differences exist between the two types of customers.

13.27

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

> 0

Two-tail F test: F = 1.07, p-value = .8792; use equal-variances test statistic

Rejection region:

684.1ttt

38,05.,

≈=>

να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

20

1

20

1

22020

06.15)120(60.15)120(

0)20.6960.73(

22

= .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

0)(:H 210 =µ−µ

)(:H 211 µ−µ

< 0

Two-tail F test: F = .95, p-value = .8252; use equal-variances test statistic

Rejection region:

653.1ttt 178,05., ≈=> να

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

90

1

90

1

22020

755,10)190(506,10)190(

0)563,63245,60

22

= −2.09

t =

09.2−

, p-value = .0189. There is enough evidence to conclude that commission salespeople outperform fixed-

salary salespersons

13.29

0)(:H 210 =µ−µ

)(:H 211 µ−µ

≠

0

Two-tail F test: F = .88, p-value = .4709; use equal-variances test statistic

Rejection region:

645.1ttt 429,025.,2/ ≈=> να

and

645.1ttt 429,025.,2/ −≈−=− να

35

+

µ−µ−−

=

21

2

p

2121

n

1

n

1

s

)()xx(

t

=

+

−+ −+−

−−

338

1

93

1

233893

69.52)1338(45.49)193(

0)86.66197.633

22

= −4.58

t = −4.58, p-value = 0. There is enough evidence to conclude there is a difference in scores between those who have

and those who do not have accidents in a three-year period.

13.30

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

> 0

Two-tail F test: F = .41, p-value = 0; use unequal-variances test statistic

1n

)n/s(

1n

)n/s(

)n/sn/s(

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1

−

+

−

+

=ν

= 222

Rejection region:

645.1ttt 222,05., −≈−=> να

+

µ−µ−−

=

2

2

2

1

2

1

2121

n

s

n

s

)()xx(

t

=

+

−−

130

42.4

130

84.2

0)27.1120.14(

22

= 6.28, p-value = 0. There is enough evidence to conclude that

bottles of wine with metal caps are perceived to be cheaper.

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.

13.36a Let students select the section they wish to attend and compare test results.

b Randomly assign students to either section and compare test results.

36

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

D0 :H µ

= 0

D1 :H µ

< 0

Rejection region:

895.1ttt 7,05., −=−=−< να

,22.3

8/17.4

075.4

n/s

x

t

DD

D

D−=

−−

=

µ−

=

p-value = .0073. There is enough evidence to infer that the Brand A is better

than Brand B.

13.41

D0 :H µ

= 0

D1 :H µ

< 0

Rejection region:

895.1ttt 7,05., −=−=−< να

,20.2

8/225.

0175.

n/s

x

t

DD

D

D−=

−−

=

µ−

=

p-value = .0320. There is enough evidence to infer that ABS is better.

13.42

D0 :H µ

= 0

D1 :H µ

> 0

Rejection region:

943.1ttt 6,05., ==> να

,98.1

7/48.2

086.1

n/s

x

t

DD

D

D=

−

=

µ−

=

p-value = .0473. There is enough evidence to infer that the camera reduces the

number of red-light runners.

13.43a

D0 :H µ

= 0

<µD1 :H

0

37

Rejection region:

796.1ttt 11,05., −=−=−< να

=

−−

=

µ−

=12/02.3

000.1

n/s

x

t

DD

D

D

–1.15, p-value = .1375. There is not enough evidence to infer that the new fertilizer is

better.

b

D

D

2/D n

s

tx α

±

=

92.100.1

12

02.3

201.200.1 ±−=±−

; LCL = –2.92, UCL = .92

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

D0 :H µ

= 0

>µD1 :H

0

Rejection region:

796.1ttt 11,05., ==> να

=

−

=

µ−

=12/88.5

008.3

n/s

x

t

DD

D

D

1.82, p-value = .0484. There is enough evidence to infer that companies with exercise

programs have lower medical expenses.

b

D

D

2/D n

s

tx α

±

=

74.308.3

12

88.5

201.208.3 ±=±

; LCL = –.66, UCL = 6.82

c Yes because medical expenses will vary by the month of the year.

13.45

D0 :H µ

= 0

>µD1 :H

0

Rejection region:

656.1ttt 149,05., ≈=> να

=

−

=

µ−

=150/1.99

04.12

n/s

x

t

DD

D

D

1.53, p-value = .0638. There is not enough evidence to infer that mortgage payments

have increases in the past 5 years.

13.46

D0 :H µ

= 0

≠µD1 :H

0

Rejection region:

009.2ttt 49,025.,2/ −≈−=−< να

or

009.2ttt 49,025.,2/ ≈=> να

38

=

−−

=

µ−

=50/22.2

016.1

n/s

x

t

DD

D

D

–3.70, p-value = .0006. There is enough evidence to infer that waiters and waitresses

earn different amounts in tips.

13.47 a

D

D

2/D n

s

tx α

±

=

16.875.19

40

63.30

684.175.19 ±=±

; LCL = 11.59, UCL = 27.91

b

D0 :H µ

= 0

>µD1 :H

0

Rejection region:

684.1ttt

39,05.,

≈=>

να

=

−

=

µ−

=40/63.30

075.19

n/s

x

t

DD

D

D

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

D0 :H µ

= 0

>µD1 :H

0

Rejection region:

345.1ttt

14,10.,

==>

να

,92.16

15/14.13

040.57

n/s

x

t

DD

D

D=

−

=

µ−

=

p-value = 0. There is enough evidence to conclude that heating costs for

insulated homes is less than that for uninsulated homes.

b

D

D

2/D n

s

tx α

±

=

28.740.57

15

14.13

145.240.57 ±=±

; LCL = 50.12, UCL = 64.68

c Differences are required to be normally distributed.

13.49

D0 :H µ

= 0

≠µD1 :H

0

Rejection region:

014.2ttt 44,025.,2/ −≈−=−< να

or

014.2ttt 44,025.,2/ ≈=> να

=

−−

=

µ−

=45/16.317

094.42

n/s

x

t

DD

D

D

–.91, p-value = .3687. There is not enough evidence to infer men and women spend

different amounts on health care.

39

13.50

D0 :H µ

= 0

<µD1 :H

0

Rejection region:

654.1ttt 169,05., −≈−=−< να

=

−−

=

µ−

=170/94.1568

035.183

n/s

x

t

DD

D

D

–1.52, p-value = .0647. There is not enough to infer stock holdings have

decreased.

13.51

D0 :H µ

= 0

D1 :H µ

> 0

Rejection region:

690.1ttt

37,05.,

≈=>

να

=

−

=

µ−

=38/1634.

00422.

n/s

x

t

DD

D

D

1.59, p-value = .0599. There is not enough evidence to conclude that ratios are higher

this year.

13.52

D0 :H µ

= 0

D1 :H µ

> 0

Rejection region:

676.1ttt

54,05.,

≈=>

να

=

−

=

µ−

=55/92.1854

085.520

n/s

x

t

DD

D

D

2.08, p-value = .0210. There is enough evidence to infer that company 1’s

calculated tax payable is higher than company 2’s.

13.53

D0 :H µ

= 0

D1 :H µ

> 0

Rejection region:

729.1ttt

19,05.,

==>

να

=

−

=

µ−

=20/22.7

055.4

n/s

x

t

DD

D

D

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

D0 :H µ

= 0

D1 :H µ

> 0

Rejection region:

711.1ttt

24,05.,

==>

να

40

=

−

=

µ−

=25/851,22

04587

n/s

x

t

DD

D

D

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

D0 :H µ

= 0

<µD1 :H

0

Rejection region:

684.1ttt

41,05.,

−≈−=−<

να

=

−−

=

µ−

=42/95.1

010.

n/s

x

t

DD

D

D

–.33, p-value = .3704. There is not enough evidence to infer that for companies where

an offspring takes the helm there is a decrease in operating income.

b

D0 :H µ

= 0

D1 :H µ

> 0

Rejection region:

660.1ttt 97,05., ==> να

=

−

=

µ−

=98/83.2

024.1

n/s

x

t

DD

D

D

4.34, p-value = 0. There is enough evidence to infer that when an outsider becomes

CEO the operating income increases.

13.58 a

1/:H 2

2

2

10 =σσ

1/:H 2

2

2

11 ≠σσ

Rejection region:

88.1FFF 29,29,05.,,2/ 21 ≈=> ννα

or

53.88.1/1F/1F/1FF 29,29,05.,,2/,,2/1 1221 =≈==< νναννα−

F =

2

2

2

1s/s

= 350/700 =.50, p-value = .0669. There is enough evidence to conclude that the population variances

differ.

b Rejection region:

98.2FFF 14,14,025.,,2/ 21 ==> ννα

or

34.98.2/1F/1F/1FF 14,14,025.,,2/,,2/1 1221 ====< νναννα−

F =

2

2

2

1s/s

= 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 and in this exercise the conclusion changed as well..

13.59 a LCL =

21 ,,2/

2

2

2

1

F

1

s

s

ννα

=

03.4

1

19

28

= .366, UCL =

12 ,,2/

2

2

2

1F

s

sννα

=

03.4

19

28

= 5.939

41

b LCL =

21 ,,2/

2

2

2

1

F

1

s

s

ννα

=

27.2

1

19

28

= .649, UCL =

12 ,,2/

2

2

2

1F

s

sννα

=

27.2

19

28

= 3.345

c The interval narrows.

13.60

1/:H 2

2

2

10 =σσ

1/:H 2

2

2

11 ≠σσ

Rejection region:

78.3FFF 10,9,025.,,2/ 21 ==> ννα

or

25.96.3/1F/1F/1FF 9,10,025.,,2/,,2/1 1221 ====< νναννα−

F =

2

2

2

1s/s

= .0000057/.0000114 =.50, p-value = .3179. There is not enough evidence to conclude that the two

machines differ in their consistency of fills.

13.61

1/:H 2

2

2

10 =σσ

1/:H 2

2

2

11 <σσ

Rejection region:

314.18.3/1F/1F/1FF 9,9,05.,,,,1 1221 ====< νναννα−

F =

2

2

2

1s/s

= .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

1/:H 2

2

2

10 =σσ

1/:H 2

2

2

11 ≠σσ

Rejection region:

72.3FFF 10,10,025.,,2/ 21 ==> ννα

or

269.72.3/1F/1F/1FF 10,10,025.,,2/,,2/1 1221 ====< νναννα−

F =

2

2

2

1s/s

= 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

1/:H 2

2

2

10 =σσ

1/:H 2

2

2

11 >σσ

Rejection region:

39.1FFF 99,99,05.,, 21 ≈=> ννα

F =

2

2

2

1s/s

= 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

1/:H 2

2

2

10 =σσ

1/:H 2

2

2

11 ≠σσ

Rejection region:

48.1FFF 99,99,025.,,2/ 21 ≈=> ννα

or

68.48.1/1F/1F/1FF 99,99,025.,,2/,,2/1 1221 =≈==< νναννα−

42

F =

2

2

2

1s/s

= 41,309/19,850 = 2.08, p-value = .0003. There is enough evidence to conclude that the variances differ.

13.65

2

2

2

10 /:H σσ

= 1

2

2

2

11 /:H σσ

< 1

Rejection region:

63.60.1/1F/1F/1FF 51,51,05.,,,,1 1221 =≈==< νναννα−

F =

2

2

2

1/ss

= .0261/.0875 = .298, p-value = 0. There is enough evidence to infer that portfolio 2 is riskier than

portfolio 1.

13.66

2

2

2

10 /:H σσ

= 1

≠σσ 2

2

2

11 /:H

1

Rejection region:

39.1FFF 99,99,05.,,2/ 21 ≈=> ννα

or

72.39.1/1F/1F/1FF 99,99,05.,,,,1 1221 =≈==< νναννα−

F =

2

2

2

1/ss

= 3.35/10.95 = .31, p-value = 0. There is enough evidence to conclude that the variance in service times

differ.

13.67

)pp(:H 210 −

= 0

≠− )pp(:H 211

0

a

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

100

1

100

1

)425.1(425.

)40.45(.

= .72, p-value = 2P(Z > .72) = 2(1 – .7642)

= .4716.

b

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

400

1

400

1

)425.1(425.

)40.45(.

= 1.43, p-value = 2P(Z > 1.43) = 2(1 – .9236)

= .1528.

c The p-value decreases.

13.68

)pp(:H 210 −

= 0

≠− )pp(:H 211

0

a

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,07.1

225

1

225

1

)575.1(575.

)55.60(. =

+−

−

p-value = 2P(Z > 1.07) = 2(1 – .8577)

= .2846

43

b

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,01.2

225

1

225

1

)925.1(925.

)90.95(. =

+−

−

p-value = 2P(Z > 2.01) = 2(1 – .9778)

= .0444.

c. The p-value decreases.

d

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,01.2

225

1

225

1

)075.1(075.

)05.10(. =

+−

−

p-value = 2P(Z > 2.01) = 2(1 – .94778)

= .0444.

e. The p-value decreases.

13.69 a

2

22

1

11

2/21 n

)p

ˆ

1(p

ˆ

n

)p

ˆ

1(p

ˆ

z)p

ˆ

p

ˆ

(−

+

−

±− α

= (.18–.22)

100

)22.1(22.

100

)18.1(18.

645.1 −

+

−

±

= –.040

±

.0929

b

2

22

1

11

2/21 n

)p

ˆ

1(p

ˆ

n

)p

ˆ

1(p

ˆ

z)p

ˆ

p

ˆ

(−

+

−

±− α

= (.48–.52)

100

)52.1(52.

100

)48.1(48.

645.1 −

+

−

±

= –.040

±

.1162

c The interval widens.

13.70

)pp(:H 210 −

= 0

>− )pp(:H 211

0

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,70.1

178

1

229

1

)177.1(177.

)140.205(. =

+−

−

p-value = P(Z > 1.70) = 1 – .9554 = .0446.

There is enough evidence to conclude that those who paid the regular price are more likely to buy an extended

warranty.

13.71

)pp(:H 210 −

= 0

0)pp(:H 211 ≠−

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,69.1

209

1

83

1

)209.1(209.

)234.145(. −=

+−

−

p-value = 2P(Z < −1.69) = 2(.0455)

= .0910.

There is not enough evidence to conclude that new and old accounts are different with respect to overdue accounts.

44

13.72

)pp(:H 210 −

= 0

>− )pp(:H 211

0

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,74.1

804

1

562

1

)0132.1(0132.

)0087.0196(. =

+−

−

p-value = P(Z > 1.74) = 1 – .9591 = .0409.

There is enough evidence to conclude that those who score under 600 are more likely to default than those who

score 60 or more.

13.73a

)pp(:H 210 −

= 0

>− )pp(:H 211

0

Rejection region:

05.

zzz => α

= 1.645

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

800

1

1100

1

)518.1(518.

)46.56(.

= 4.31, p-value = 0. There is enough evidence to infer that

the leader’s popularity has decreased.

b

)pp(:H 210 −

= .05

>− )pp(:H 211

.05

Rejection region:

05.

zzz => α

= 1.645

2

22

1

11

2121

n

)p

ˆ

1(p

ˆ

n

)p

ˆ

1(p

ˆ

)pp()p

ˆ

p

ˆ

(

z−

+

−−−−

=

=

800

)46.1(46.

1100

)56.1(56.

05.)46.56(.

−

+

−−−

= 2.16, p-value = P(Z > 2.16) = 1 – .9846

= .0154.

There is enough evidence to infer that the leader’s popularity has decreased by more than 5%.

c

2

22

1

11

2/21

n

)p

ˆ

1(p

ˆ

n

)p

ˆ

1(p

ˆ

z)p

ˆ

p

ˆ

(−

+

−

±−

α

=

800

)46.1(46.

1100

)56.1(56.

96.1)46.56(. −

+

−

±−

= .10

±

.045

13.74

)pp(:H 210 −

= –.08

<− )pp(:H 211

–.08

Rejection region:

01.

zzz −=−< α

= –2.33

2

22

1

11

2121

n

)p

ˆ

1(p

ˆ

n

)p

ˆ

1(p

ˆ

)pp()p

ˆ

p

ˆ

(

z−

+

−−−−

=

=

300

)28.1(28.

300

)11.1(11.

)08.()28.11(.

−

+

−−−−

= –2.85, p-value =P(Z < –2.85) = 1 – .9978

= .0022.

There is enough evidence to conclude that management should adopt process 1.

45

13.75

)pp(:H 210 −

= 0

>− )pp(:H 211

0

Rejection region:

05.

zzz => α

= 1.645

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

1109

1

1604

1

)068.1(068.

)064.071(.

= .71, p-value = P(Z > .71) = 1 – .7611

= .2389.

There is not enough evidence to infer that the claim is false.

13.76 a

)pp(:H 210 −

= 0

<− )pp(:H 211

0

Rejection region:

05.

zzz −=−< α

= –1.645

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

6281

1

6281

1

)104.1(104.

)115.093(.

= –4.04, p-value = 0. There is enough evidence to infer

that Plavix is effective.

13.77 a

)pp(:H 210 −

= 0

<− )pp(:H 211

0

Rejection region:

01.

zzz −=−< α

= –2.33

0095.

000,11

104

p

ˆ1==

0172.

000,11

189

p

ˆ2==

01335.

000,22

189104

p

ˆ=

+

=

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

000,11

1

000,11

1

)01335.1(01335.

)0172.0095(.

= –4.98, p-value = 0. There is enough evidence

to infer that aspirin is effective in reducing the incidence of heart attacks.

13.78

)pp(:H 210 −

= 0

>− )pp(:H 211

0

Rejection region:

645.1zzz 05. ==> α

0985.

000,11

084,1

p

ˆ1==

0906.

000,11

997

p

ˆ2==

0946.

000,22

997084,1

p

ˆ=

+

=

46

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,00.2

000,11

1

000,11

1

)0946.1(0946.

)0906.0985(. =

+−

−

p-value = P(Z > 2.00) = 1 – .9772 = .0228.

There is enough evidence to infer that aspirin leads to more cataracts.

13.79

)pp(:H 210 −

= 0

<− )pp(:H 211

0

Rejection region:

01.

zzz −=−< α

= –2.33

0289.

594,2

75

p

ˆ1==

0509.

594,2

132

p

ˆ2==

0399.

594,2594,2

13275

p

ˆ=

+

+

=

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,04.4

594,2

1

594,2

1

)0399.1(0399.

)0509.0289(. −=

+−

−

p-value = 0.

There is enough evidence to infer that Letrozole works.

13.80

)pp(:H 210 −

= 0

<− )pp(:H 211

0

Rejection region:

645.1zzz 05. −=−=−< α

2228.

395

88

p

ˆ1==

2586.

406

105

p

ˆ2==

2409.

406395

10588

p

ˆ=

+

+

=

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,19.1

406

1

395

1

)2409.1(2409.

)2586.2228(. −=

+−

−

p-value = P(Z < –1.19) = .1170. There is not

enough evidence to infer that exercise training reduces mortality.

13.81

)pp(:H 210 −

= 0

≠− )pp(:H 211

0

a

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,75.

294

1

350

1

)8975.1(8975.

8878.9057(. =

+−

−

p-value = 2P(Z > .75) = 2(1 – .7734)

= .4532.

There is not enough evidence to infer that the two populations of car owners differ in their satisfaction levels.

13.82

)pp(:H 210 −

= 0

47

)pp(:H 211 −

> 0

Rejection region:

10.

zzz => α

= 1.28

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

162

1

38

1

)11.1(11.

)0741.2632(.

= 3.35, p-value = 0. There is enough evidence to conclude that

smokers have a higher incidence of heart diseases than nonsmokers.

b

2

22

1

11

2/21 n

)p

ˆ

1(p

ˆ

n

)p

ˆ

1(p

ˆ

z)p

ˆ

p

ˆ

(−

+

−

±− α

= (.2632–0741)

162

)0741.1(0741.

38

)2632.1(2632.

645.1 −

+

−

±

=.1891

±

.1223; LCL = .0668, UCL = .3114

13.83a

)pp(:H 210 −

= 0

<− )pp(:H 211

0

Rejection region:

10.

zzz −=−< α

= −1.28

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,90.

500

1

400

1

)842.1(842.

)852.830(. −=

+−

−

p-value = P(Z < −.90) = .1841. There is not enough

evidence to infer that there has been an increase in belief in the greenhouse effect.

b

2

22

1

11

2/21 n

)p

ˆ

1(p

ˆ

n

)p

ˆ

1(p

ˆ

z)p

ˆ

p

ˆ

(−

+

−

±− α

= (.830–.852)

500

)852.1(852.

400

)830.1(830.

645.1 −

+

−

±

= −.022

±

.040;

LCL = −.062 and UCL = .018.

13.84

)pp(:H 210 −

= 0

>− )pp(:H 211

0

Rejection region:

05.

zzz => α

= 1.645

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

462

1

477

1

)3504.1(3504.

)3420.3585(.

= .53, p-value = P(Z > .53) = 1 – .7019

= .2981.

There is not enough evidence to infer that the use of illicit drugs in the United States has increased in the past

decade.

13.85

)pp(:H 210 −

= –.02

48

<− )pp(:H 211

–.02

Rejection region:

05.

zzz −=−< α

= –1.645

2

22

1

11

2121

n

)p

ˆ

1(p

ˆ

n

)p

ˆ

1(p

ˆ

)pp()p

ˆ

p

ˆ

(

z−

+

−−−−

=

=

200

)11.1(11.

200

)055.1(055.

)02.()11.055(.

−

+

−−−−

= –1.28, p-value = P(Z < –1.28) = .1003. There is not

enough evidence to choose machine A.

13.86

)pp(:H 210 −

= 0

>− )pp(:H 211

0

Rejection region:

05.

zzz => α

= 1.645

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

619

1

231

1

)1035.1(1035.

)0905.1385(.

z = 2.04, p-value = P(Z > 2.04) = 1 – .9793

= .0207.

There is enough evidence to conclude that health conscious adults are more likely to buy Special X.

13.87a

)pp(:H 210 −

= 0

>− )pp(:H 211

0

Rejection region:

05.

zzz => α

= 1.645

4172.

163

68

p

ˆ1==

2685.

108

29

p

ˆ2==

3579.

108163

2968

p

ˆ=

+

+

=

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,50.2

108

1

163

1

)3579.1(3579.

)2685.4172(. =

+−

−

p-value = P(Z > 2.50) = 1 – .9938

= .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

)pp(:H 210 −

= 0

≠− )pp(:H 211

0

Rejection region:

05.2/ zzz −=−< α

= –1.96 or

05.2/ zzz => α

= 1.96

3704.

54

20

p

ˆ1==

4348.

23

10

p

ˆ2==

3896.

2354

1020

p

ˆ=

+

+

=

49

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

,53.

23

1

54

1

)3896.1(3896.

4348.3704(. −=

+−

−

p-value = 2P(Z < –.53) = 2(.2981)

= .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

)pp(:H 210 −

= 0

≠− )pp(:H 211

0

Rejection region:

025.2/ zzz −=−< α

= –1.96 or

025.2/ zzz => α

= 1.96

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

=

+−

−

316

1

382

1

)1132.1(1132.

)1297.0995(.

= –1.25, p-value = 2P(Z < –1.25) = 2(.1056) =

.2112.

There is not enough evidence to infer differences between the two sources.

13.89

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

> 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

During Before

Mean 5746.07 5372.13

Variance 167289 194772

Observations 15 24

Pooled Variance 184373

Hypothesized Mean Difference 0

df 37

t Stat 2.65

P(T<=t) one-tail 0.0059

t Critical one-tail 1.6871

P(T<=t) two-tail 0.0119

t Critical two-tail 2.0262

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 210 µ−µ

= 250

)(:H 211 µ−µ

> 250

50

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Two-Sample Assuming Equal Variances

During Before

Mean 5746.07 5372.13

Variance 167289 194772

Observations 15 24

Pooled Variance 184373

Hypothesized Mean Difference 250

df 37

t Stat 0.88

P(T<=t) one-tail 0.1931

t Critical one-tail 1.6871

P(T<=t) two-tail 0.3862

t Critical two-tail 2.0262

t = .88, p-value = .1931. There is not enough evidence to conclude that the ads are profitable.

13.91

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

< 0

1

2

3

4

5

6

7

8

9

10

11

12

13

A B C

t-Test: Two-Sample Assuming Unequal Variances

Group 1 Group 2

Mean 4.94 9.48

Variance 11.19 20.29

Observations 68 193

Hypothesized Mean Difference 0

df 158

t Stat -8.73

P(T<=t) one-tail 0.0000

t Critical one-tail 1.6546

P(T<=t) two-tail 0.0000

t Critical two-tail 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

D0 :H µ

= 0

<µD1 :H

0

51

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Paired Two Sample for Means

Drug Placebo

Mean 18.43 22.03

Variance 30.39 66.37

Observations 100 100

Pearson Correlation 0.69

Hypothesized Mean Difference 0

df 99

t Stat -6.09

P(T<=t) one-tail 0.0000

t Critical one-tail 1.6604

P(T<=t) two-tail 0.0000

t Critical two-tail 1.9842

t = –6.09, p-value = 0. There is enough evidence to infer that the new drug is effective.

13.93

)(:H 210 µ−µ

= 0

≠µ−µ )(:H 211

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

Vendor Delivered

Mean 19.50 20.03

Variance 14.35 14.97

Observations 205 155

Pooled Variance 14.62

Hypothesized Mean Difference 0

df 358

t Stat -1.29

P(T<=t) one-tail 0.0996

t Critical one-tail 1.6491

P(T<=t) two-tail 0.1993

t Critical two-tail 1.9666

t = –1.29, p-value = .1993. There is not enough evidence of a difference in reading time between the two groups.

13.94

)pp(:H 210 −

= 0

<− )pp(:H 211

0

52

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

Last Year This Year

Sample Proportions 0.6758 0.7539

Observations 327 382

Hypothesized Difference 0

z Stat -2.30

P(Z<=z) one tail 0.0106

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.0212

z Critical two-tail 1.96

z = –2.30, p-value = .0106. There is enough evidence to infer an increase in seatbelt use.

13.95a

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

< 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

5 Years This Year

Mean 32.42 33.72

Variance 36.92 45.52

Observations 200 200

Pooled Variance 41.22

Hypothesized Mean Difference 0

df 398

t Stat -2.02

P(T<=t) one-tail 0.0218

t Critical one-tail 1.2837

P(T<=t) two-tail 0.0436

t Critical two-tail 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 210 µ−µ

= 0

≠µ−µ )(:H 211

0

53

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Two-Sample Assuming Equal Variances

Male Female

Mean 39.75 49.00

Variance 803.88 733.16

Observations 20 20

Pooled Variance 768.52

Hypothesized Mean Difference 0

df 38

t Stat -1.06

P(T<=t) one-tail 0.1490

t Critical one-tail 1.3042

P(T<=t) two-tail 0.2980

t Critical two-tail 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 210 µ−µ

= 0

)(:H 211 µ−µ

< 0

1

2

3

4

5

6

7

8

9

10

11

12

13

A B C

t-Test: Two-Sample Assuming Unequal Variances

Low High

Mean 33.10 56.84

Variance 278.69 1047.81

Observations 21.00 19

Hypothesized Mean Difference 0.00

df 26.00

t Stat -2.87

P(T<=t) one-tail 0.0040

t Critical one-tail 1.3150

P(T<=t) two-tail 0.0080

t Critical two-tail 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.

13.97a

D0 :H µ

= 0

≠µD1 :H

0

54

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Paired Two Sample for Means

Female Male

Mean 55.68 56.40

Variance 105.64 116.75

Observations 25 25

Pearson Correlation 0.96

Hypothesized Mean Difference 0

df 24

t Stat -1.13

P(T<=t) one-tail 0.1355

t Critical one-tail 1.3178

P(T<=t) two-tail 0.2710

t Critical two-tail 1.7109

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

)pp(:H 210 −

= 0

>− )pp(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

This Year 3 Years Ago

Sample Proportions 0.4351 0.3558

Observations 393 385

Hypothesized Difference 0

z Stat 2.26

P(Z<=z) one tail 0.0119

z Critical one-tail 1.2816

P(Z<=z) two-tail 0.0238

z Critical two-tail 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 210 µ−µ

= 25

)(:H 211 µ−µ

> 25

55

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Two-Sample Assuming Equal Variances

A nondefectives A nondefectives

Mean 230.13 200.92

Variance 79.51 59.04

Observations 24 24

Pooled Variance 69.27

Hypothesized Mean Difference 25

df 46

t Stat 1.75

P(T<=t) one-tail 0.0433

t Critical one-tail 1.6787

P(T<=t) two-tail 0.0865

t Critical two-tail 2.0129

t = 1.75, p-value = .0433. There is enough evidence to conclude that machine A should be purchased.

13.100

)pp(:H 210 −

= 0

≠− )pp(:H 211

0

The totals in columns A through D are 5788, 265, 5154, and 332, respectively.

1

2

3

4

5

6

7

A B C D E

z-Test of the Difference Between Two Proportions (Case 1)

Sample 1 Sample 2 z Stat -4.28

Sample proportion 0.045800 0.064400 P(Z<=z) one-tail 0.0000

Sample size 5788 5154 z Critical one-tail 1.6449

Alpha 0.05 P(Z<=z) two-tail 0.0000

z Critical two-tail 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

D0 :H µ

= 0

>µD1 :H

0

56

Dry Cleaner

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Paired Two Sample for Means

Dry C Before Dry C After

Mean 168.00 165.50

Variance 351.38 321.96

Observations 14 14

Pearson Correlation 0.86

Hypothesized Mean Difference 0

df 13

t Stat 0.96

P(T<=t) one-tail 0.1780

t Critical one-tail 1.7709

P(T<=t) two-tail 0.3559

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 B C

t-Test: Paired Two Sample for Means

Donut Before Donut After

Mean 308.14 295.29

Variance 809.67 812.07

Observations 14 14

Pearson Correlation 0.86

Hypothesized Mean Difference 0

df 13

t Stat 3.24

P(T<=t) one-tail 0.0032

t Critical one-tail 1.7709

P(T<=t) two-tail 0.0065

t Critical two-tail 2.1604

t = 3.24, p-value = .0032. There is enough evidence to conclude that the doughnut shop sales have decreased.

57

Convenience store

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Paired Two Sample for Means

Convenience Before Convenience After

Mean 374.64 348.14

Variance 2270.40 2941.82

Observations 14 14

Pearson Correlation 0.97

Hypothesized Mean Difference 0

df 13

t Stat 7.34

P(T<=t) one-tail 0.0000

t Critical one-tail 1.7709

P(T<=t) two-tail 0.0000

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 210 µ−µ

= 0

<µ−µ )(:H 211

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

20-year-old 40-year-old

Mean 125.74 129.93

Variance 31.90 31.95

Observations 26 24

Pooled Variance 31.92

Hypothesized Mean Difference 0

df 48

t Stat -2.62

P(T<=t) one-tail 0.0059

t Critical one-tail 1.2994

P(T<=t) two-tail 0.0119

t Critical two-tail 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 210 µ−µ

= 0

<µ−µ )(:H 211

0

58

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Two-Sample Assuming Equal Variances

20-year-old 40-year-old

Mean 134.02 141.11

Variance 36.15 39.47

Observations 26 24

Pooled Variance 37.74

Hypothesized Mean Difference 0

df 48

t Stat -4.08

P(T<=t) one-tail 0.0001

t Critical one-tail 1.2994

P(T<=t) two-tail 0.0002

t Critical two-tail 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.

13.103a

)pp(:H 210 −

= 0

>− )pp(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

Depressed Not Depressed

Sample Proportions 0.2879 0.2004

Observations 132 1058

Hypothesized Difference 0

z Stat 2.33

P(Z<=z) one tail 0.0100

z Critical one-tail 2.3263

P(Z<=z) two-tail 0.0200

z Critical two-tail 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 210 µ−µ

= 0

>µ−µ )(:H 211

0

59

1

2

3

4

5

6

7

8

9

10

11

12

13

A B C

t-Test: Two-Sample Assuming Unequal Variances

Exercise Drug

Mean 13.52 9.92

Variance 5.76 13.16

Observations 25 25

Hypothesized Mean Difference 0

df 42

t Stat 4.14

P(T<=t) one-tail 0.0001

t Critical one-tail 2.4185

P(T<=t) two-tail 0.0002

t Critical two-tail 2.6981

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 E F

t-Estimate of the Difference Between Two Means (Unequal-Variances)

Sample 1 Sample 2 Confidence Interval Estimate

Mean 13.52 9.92 3.60 1.76

Variance 5.76 13.16 Lower confidence limit 1.84

Sample size 25 25 Upper confidence limit 5.36

Degrees of freedom 41.63

Confidence level 0.95

±

LCL = 1.84, UCL = 5.36

c The histograms are bell shaped.

13.105

)(:H 210 µ−µ

= 0

<µ−µ )(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

12

13

A B C

t-Test: Two-Sample Assuming Unequal Variances

Group 1 Group 2

Mean 7.46 8.46

Variance 25.06 12.98

Observations 50 50

Hypothesized Mean Difference 0

df 89

t Stat -1.14

P(T<=t) one-tail 0.1288

t Critical one-tail 1.6622

P(T<=t) two-tail 0.2575

t Critical two-tail 1.9870

60

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

D0 :H µ

= 0

<µD1 :H

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Paired Two Sample for Means

Group 1 Group 2

Mean 7.53 8.57

Variance 29.77 43.37

Observations 50 50

Pearson Correlation 0.89

Hypothesized Mean Difference 0

df 49

t Stat -2.40

P(T<=t) one-tail 0.0100

t Critical one-tail 1.6766

P(T<=t) two-tail 0.0201

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

)pp(:H 210 −

= 0

>− )pp(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

Special K Other

Sample Proportions 0.635 0.53

Observations 200 200

Hypothesized Difference 0

z Stat 2.13

P(Z<=z) one tail 0.0166

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.0332

z Critical two-tail 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

)pp(:H 210 −

= 0

>− )pp(:H 211

0

61

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

Special K Other

Sample Proportions 0.575 0.515

Observations 200 200

Hypothesized Difference 0

z Stat 1.20

P(Z<=z) one tail 0.1141

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.2282

z Critical two-tail 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 210 µ−µ

= 0

<µ−µ )(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

12

13

A B C

t-Test: Two-Sample Assuming Unequal Variances

Small space Large space

Mean 1245.7 1915.8

Variance 23812 65566

Observations 25 25

Hypothesized Mean Difference 0

df 39

t Stat -11.21

P(T<=t) one-tail 0.0000

t Critical one-tail 1.6849

P(T<=t) two-tail 0.0000

t Critical two-tail 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

)(:H 210 µ−µ

= 0

>µ−µ )(:H 211

0

62

1

2

3

4

5

6

7

8

9

10

11

12

13

A B C

t-Test: Two-Sample Assuming Unequal Variances

Computer No Computer

Mean 69,933 48,246

Variance 63,359,040 101,588,525

Observations 89 61

Hypothesized Mean Difference 0

df 109

t Stat 14.07

P(T<=t) one-tail 0.0000

t Critical one-tail 1.2894

P(T<=t) two-tail 0.0000

t Critical two-tail 1.6590

t = 14.07, p-value = 0. There is enough evidence to conclude that single-person businesses that use a PC earn more.

13.111

)pp(:H 210 −

= 0

>− )pp(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

New Older

Sample Proportions 0.948 0.920

Observations 250 250

Hypothesized Difference 0

z Stat 1.26

P(Z<=z) one tail 0.1037

z Critical one-tail 1.2816

P(Z<=z) two-tail 0.2074

z Critical two-tail 1.6449

z = 1.26, p-value = .1037. There is not enough evidence to conclude that the new company is better.

13.112

)(:H 210 µ−µ

= 0

)(:H 211 µ−µ

<

0

63

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Two-Sample Assuming Equal Variances

Supplement Placebo

Mean 19.02 21.85

Variance 41.34 25.49

Observations 48 48

Pooled Variance 33.41

Hypothesized Mean Difference 0

df 94

t Stat -2.40

P(T<=t) one-tail 0.0092

t Critical one-tail 1.6612

P(T<=t) two-tail 0.0183

t Critical two-tail 1.9855

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

)pp(:H 210 −

= 0

<− )pp(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

$100-Limit $3000-Limit

Sample Proportions 0.5234 0.5551

Observations 491 490

Hypothesized Difference 0

z Stat -1.00

P(Z<=z) one tail 0.1598

z Critical one-tail 1.2816

P(Z<=z) two-tail 0.3196

z Critical two-tail 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 210 µ−µ

= 0

≠µ−µ )(:H 211

0

64

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Two-Sample Assuming Equal Variances

Female Male

Mean 7.27 7.02

Variance 2.57 2.85

Observations 103 97

Pooled Variance 2.71

Hypothesized Mean Difference 0

df 198

t Stat 1.08

P(T<=t) one-tail 0.1410

t Critical one-tail 1.6526

P(T<=t) two-tail 0.2820

t Critical two-tail 1.9720

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 210 µ−µ

= 0

>µ−µ )(:H 211

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

City Suburb

Mean 2.42 1.97

Variance 1.08 0.77

Observations 70 78

Pooled Variance 0.92

Hypothesized Mean Difference 0

df 146

t Stat 2.85

P(T<=t) one-tail 0.0025

t Critical one-tail 1.6554

P(T<=t) two-tail 0.0050

t Critical two-tail 1.9763

t = 2.85, p-value = .0025. There is enough evidence to infer that city households discard more newspaper than do

suburban households.

13.116

)(:H 210 µ−µ

= 0

>µ−µ )(:H 211

0

65

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

Mean 18.18 14.30

Variance 357.32 130.79

Observations 176 154

Hypothesized Mean Difference 0

df 293

t Stat 2.28

P(T<=t) one-tail 0.0115

t Critical one-tail 1.6501

P(T<=t) two-tail 0.0230

t Critical two-tail 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.

13.117

)pp(:H 210 −

= 0

<− )pp(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

No HS HS

Sample Proportions 0.127 0.358

Observations 63 257

Hypothesized Difference 0

z Stat -3.54

P(Z<=z) one tail 0.0002

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.0004

z Critical two-tail 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 210 µ−µ

= 0

>µ−µ )(:H 211

0

66

1

2

3

4

5

6

7

8

9

10

11

12

13

A B C

t-Test: Two-Sample Assuming Unequal Variances

Group 1 Groups 2-4

Mean 11.58 10.60

Variance 9.28 21.41

Observations 269 981

Hypothesized Mean Difference 0

df 644

t Stat 4.15

P(T<=t) one-tail 0.0000

t Critical one-tail 1.6472

P(T<=t) two-tail 0.0000

t Critical two-tail 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 210 µ−µ

= 0

<µ−µ )(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

12

13

A B C

t-Test: Two-Sample Assuming Unequal Variances

Sale-CDs Sale-fax

Mean 59.04 65.57

Variance 425.4 849.7

Observations 122 144

Hypothesized Mean Difference 0

df 256

t Stat -2.13

P(T<=t) one-tail 0.0171

t Critical one-tail 1.6508

P(T<=t) two-tail 0.0341

t Critical two-tail 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 210 µ−µ

= 0

<µ−µ )(:H 211

0

67

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Two-Sample Assuming Equal Variances

No Yes

Mean 91,467 97,836

Variance 461,917,705 401,930,840

Observations 466 55

Pooled Variance 455,676,297

Hypothesized Mean Difference 0

df 519

t Stat -2.09

P(T<=t) one-tail 0.0184

t Critical one-tail 1.6478

P(T<=t) two-tail 0.0369

t Critical two-tail 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.

13.121

)pp(:H 210 −

= 0

≠− )pp(:H 211

0

The data must first be unstacked. Success = 2

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

Female Male

Sample Proportions 0.5945 0.6059

Observations 762 746

Hypothesized Difference 0

z Stat -0.45

P(Z<=z) one tail 0.3256

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.6512

z Critical two-tail 1.96

z = −.45, p-value = .6512. There is not enough evidence to conclude that men and women differ in their choices of

Christmas trees.

Case 13.1To test whether gender bias exists, we test to determine whether the proportions of loans denied is greater

for women (

1

p

) than for men (

2

p

).

0pp:H 210 =−

0pp:H 211 >−

68

1

2

3

4

5

6

7

A B C D E

z-Test of the Difference Between Two Proportions (Case 1)

Sample 1 Sample 2 z Stat 0.98

Sample proportion 0.1217 0.0933 P(Z<=z) one-tail 0.1633

Sample size 115 1050 z Critical one-tail 1.6449

Alpha 0.05 P(Z<=z) two-tail 0.3266

z Critical two-tail 1.9600

z = .98, p-value = .1633. There is not enough evidence to infer that loan requests by women are denied more

frequently than loan requests by men.

Interest rates (points above prime)

)(:H 210 µ−µ

= 0

>µ−µ )(:H 211

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

W Rate M Rate

Mean 1.55 1.28

Variance 0.41 0.45

Observations 101 952

Pooled Variance 0.44

Hypothesized Mean Difference 0

df 1051

t Stat 3.85

P(T<=t) one-tail 6.28E-05

t Critical one-tail 1.6463

P(T<=t) two-tail 0.0001

t Critical two-tail 1.9622

t = 3.85, p-value = 0. There is enough evidence to infer that the mean rate of women’s loans is greater than the mean

rate for men’s loans.

Case 13.2 Arrival times t-tests of

D

µ

D0 :H µ

= 1

D1 :H µ

> 1

DD

DD

n/s

x

tµ−

=

69

Cambridge:

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Paired Two Sample for Means

C Ambulance C Fire

Mean 12.86 10.64

Variance 9.71 3.39

Observations 280 280

Pearson Correlation 0.28

Hypothesized Mean Difference 1

df 279

t Stat 6.50

P(T<=t) one-tail 0.0000

t Critical one-tail 1.6503

P(T<=t) two-tail 0.0000

t Critical two-tail 1.9685

t = 6.50, p-value = 0. There is sufficient evidence to infer that in Cambridge fire trucks arrive at the scene more than

one minute sooner than ambulances.

Kitchener:

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Paired Two Sample for Means

K Ambulance K Fire

Mean 9.79 8.55

Variance 10.73 3.26

Observations 506 506

Pearson Correlation 0.55

Hypothesized Mean Difference 1

df 505

t Stat 1.98

P(T<=t) one-tail 0.0243

t Critical one-tail 1.6479

P(T<=t) two-tail 0.0487

t Critical two-tail 1.9647

t = 1.98, p-value = .0243. There is sufficient evidence to infer that in Kitchener fire trucks arrive at the scene more

than one minute sooner than ambulances.

70

Waterloo:

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Paired Two Sample for Means

W Ambulance W Fire

Mean 12.86 12.17

Variance 11.01 19.00

Observations 150 150

Pearson Correlation 0.80

Hypothesized Mean Difference 1

df 149

t Stat -1.46

P(T<=t) one-tail 0.0733

t Critical one-tail 1.6551

P(T<=t) two-tail 0.1466

t Critical two-tail 1.9760

t = –1.46, p-value = 1– .0733 = .9267. There is no evidence to infer that in Waterloo fire trucks arrive at the scene

more than one minute sooner than ambulances.

Frequency of arrivals in less than 8 minutes: z-tests of

21

pp −

(case 1) (Data were recoded so that 2 = yes, 1 = no)

0)pp(:H 210 =−

)pp(:H 211 −

< 0

+−

−

=

21

21

n

1

n

1

)p

ˆ

1(p

ˆ

)p

ˆ

p

ˆ

(

z

Cambridge

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

C Ambulance C Fire

Sample Proportions 0.0536 0.0857

Observations 280 280

Hypothesized Difference 0

z Stat -1.49

P(Z<=z) one tail 0.0676

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.1352

z Critical two-tail 1.9600

z = –1.49, p-value = .0676. There is not enough evidence to infer that in Cambridge fire trucks arrive at the scene in

less than 8 minutes more frequently than do ambulances.

Kitchener:

71

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

K Ambulance K Fire

Sample Proportions 0.2866 0.3992

Observations 506 506

Hypothesized Difference 0

z Stat -3.77

P(Z<=z) one tail 0.0001

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.0002

z Critical two-tail 1.9600

z = –3.77, p-value = .0001. There is enough evidence to infer that in Kitchener fire trucks arrive at the scene in less

than 8 minutes more frequently than do ambulances.

Waterloo:

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

W Ambulance W Fire

Sample Proportions 0.1000 0.1667

Observations 150 150

Hypothesized Difference 0

z Stat -1.70

P(Z<=z) one tail 0.0447

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.0894

z Critical two-tail 1.9600

z = –1.70, p-value = .0447. There is enough evidence to infer that in Waterloo fire trucks arrive at the scene in less

than 8 minutes more frequently than ambulances.

Case 13.3 Length of ownership:

)(:H 210 µ−µ

= 0

≠µ−µ )(:H 211

0

72

1

2

3

4

5

6

7

8

9

10

11

12

13

14

A B C

t-Test: Two-Sample Assuming Equal Variances

Would buy Would not buy

Mean 5.98 7.11

Variance 8.88 14.87

Observations 43 187

Pooled Variance 13.77

Hypothesized Mean Difference 0

df 228

t Stat -1.81

P(T<=t) one-tail 0.0359

t Critical one-tail 1.65

P(T<=t) two-tail 0.0717

t Critical two-tail 1.97

t = −1.81, p-value = .0717. There is some but not enough evidence to infer that the years of ownership differ

between those who would and those who would not buy a generator.

Compare homeowners who do not consider the resale value of their home versus homeowners who do

The data were sorted. Success = buy

0)pp(:H 210 =−

<− )pp(:H 211

0

1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

Do not consider resale Consider resale

Sample Proportions 0.1262 0.2362

Observations 103 127

Hypothesized Difference 0

z Stat -2.13

P(Z<=z) one tail 0.0167

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.0334

z Critical two-tail 1.96

z = −2.13, p-value = .0167. There is enough evidence to infer that the proportion of those who would buy a

generator is greater for homeowners who consider resale value than homeowners who do not consider resale value.

County

0)pp(:H 210 =−

≠− )pp(:H 211

0

The data were sorted. Success = buy

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1

2

3

4

5

6

7

8

9

10

11

A B C D

z-Test: Two Proportions

Miami-Dade Other

Sample Proportions 0.1791 0.1979

Observations 134 96

Hypothesized Difference 0

z Stat -0.36

P(Z<=z) one tail 0.3591

z Critical one-tail 1.6449

P(Z<=z) two-tail 0.7182

z Critical two-tail 1.96

z = −.36, p-value = .7182. There is not enough evidence to conclude that the proportion of those who would buy a

generator differ between Miami-Dade county and the other counties.

d. Advertising should focus on all counties with an emphasis on newer developments. Advertising should stress how

a generator would increase the resale value of a home.

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