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Chapter 10: Statistical Inferences About Two Populations 1

Chapter 10Statistical Inferences about Two Populations

LEARNING OBJECTIVES

The general focus of Chapter 10 is on testing hypotheses and constructing confidenceintervals about parameters from two populations, thereby enabling you to

1. Test hypotheses and construct confidence intervals about the difference in twopopulation means using the z statistic.

2. Test hypotheses and establish confidence intervals about the difference in twopopulation means using the t statistic.

3. Test hypotheses and construct confidence intervals about the difference in tworelated populations when the differences are normally distributed.

4. Test hypotheses and construct confidence intervals about the difference in twopopulation proportions.

5. Test hypotheses and construct confidence intervals about two populationvariances when the two populations are normally distributed.

CHAPTER TEACHING STRATEGY

The major emphasis of chapter 10 is on analyzing data from two samples. Thestudent should be ready to deal with this topic given that he/she has tested hypotheses andcomputed confidence intervals in previous chapters on single sample data.

The z test for analyzing the differences in two sample means is presented here.Conceptually, this is not radically different than the z test for a single sample mean showninitially in Chapter 7. In analyzing the differences in two sample means where thepopulation variances are unknown, if it can be assumed that the populations are normallydistributed, a t test for independent samples can be used. There are two different




formulas given in the chapter to conduct this t test. One version uses a "pooled" estimateof the population variance and assumes that the population variances are equal. Theother version does not assume equal population variances and is simpler to compute.

However, the degrees of freedom formula for this version is quite complex.

A t test is also included for related (non independent) samples. It is important thatthe student be able to recognize when two samples are related and when they areindependent. The first portion of section 10.3 addresses this issue. To underscore thepotential difference in the outcome of the two techniques, it is sometimes valuable toanalyze some related measures data with both techniques and demonstrate that the resultsand conclusions are usually quite different. You can have your students work problemslike this using both techniques to help them understand the differences between the twotests (independent and dependent t tests) and the different outcomes they will obtain.

A z test of proportions for two samples is presented here along with an F test fortwo population variances. This is a good place to introduce the student to the F distribution in preparation for analysis of variance in Chapter 11. The student will beginto understand that the F values have two different degrees of freedom. The F distributiontables are upper tailed only. For this reason, formula 10.14 is given in the chapter to beused to compute lower tailed F values for two-tailed tests.

CHAPTER OUTLINE

10.1 Hypothesis Testing and Confidence Intervals about the Difference in Two Meansusing the z Statistic (population variances known)

Hypothesis TestingConfidence IntervalsUsing the Computer to Test Hypotheses about the Difference in Two

Population Means Using the z Test

10.2 Hypothesis Testing and Confidence Intervals about the Difference in Two Means:Independent Samples and Population Variances Unknown

Hypothesis TestingUsing the Computer to Test Hypotheses and Construct Confidence

Intervals about the Difference in Two Population Means Using the t Test

Confidence Intervals

10.3 Statistical Inferences For Two Related PopulationsHypothesis TestingUsing the Computer to Make Statistical Inferences about Two Related

PopulationsConfidence Intervals






b) Critical value method:

zc =

2

2

2

1

2

1

2121 )()(

nn

x xc

σ σ

µ µ

+

−−−

-1.645 =

32

60

32

52

)0()( 21

+

−− c x x

( x1 -

x2)c = -3.08

c) The area for z = -1.02 using Table A.5 is .3461.The p-value is .5000 - .3461 = .1539

10.2 Sample 1 Sample 2

n1 = 32 n2 = 31

x 1 = 70.4 x 2 = 68.7

σ 1 = 5.76 σ 2 = 6.1

For a 90% C.I., z.05 = 1.645

2

2

2

1

2

121 )(

nn z x x σ σ

+±−

(70.4) – 68.7) + 1.64531

1.6

32

76.5 22

+

1.7 ± 2.465

-.76 < µ1 - µ2 < 4.16




10.3 a) Sample 1 Sample 2

x 1 = 88.23 x 2 = 81.2

σ 12 = 22.74 σ 2

2 = 26.65n1 = 30 n2 = 30

Ho: µ1 - µ2 = 0

Ha: µ1 - µ2 ≠ 0

For two-tail test, use α /2 = .01 z.01 = + 2.33

z =

30

65.26

30

74.22

)0()2.8123.88()()(

2

22

1

21

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ = 5.48

Since the observed z = 5.48 > z.01 = 2.33, the decision is to reject the nullhypothesis.

b)2

2

2

1

2

121 )(

nn z x x σ σ

+±−

(88.23 – 81.2) + 2.3330

65.26

30

74.22+

7.03 + 2.99

4.04 < µ µµ µ < 10.02

This supports the decision made in a) to reject the null hypothesis becausezero is not in the interval.

10.4 Computers/electronics Food/Beverage

x 1 = 1.96 x 2 = 3.02

σ 12 = 1.0188 σ 2

2 = 0.9180n1 = 50 n2 = 50

Ho: µ1 - µ2 = 0

Ha: µ1 - µ2 ≠ 0

For two-tail test, α /2 = .005 z.005 = ±2.575




z =

509180.0

500188.1

)0()02.396.1()()(

2

2

2

1

2

1

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ = -5.39

Since the observed z = -5.39 < zc = -2.575, the decision is to reject the nullhypothesis.

10.5 A B

n1 = 40 n2 = 37

x 1 = 5.3 x 2 = 6.5σ 1

2 = 1.99 σ 22 = 2.36

For a 95% C.I., z.025 = 1.96

2

2

2

1

2

121 )(

nn z x x σ σ

+±−

(5.3 – 6.5) + 1.9637

36.2

40

99.1+

-1.2 ± .66 -1.86 < µ µµ µ < -.54

The results indicate that we are 95% confident that, on average, Plumber B doesbetween 0.54 and 1.86 more jobs per day than Plumber A. Since zero does not liein this interval, we are confident that there is a difference between Plumber A andPlumber B.

10.6 Managers Specialty

n1 = 35 n2 = 41 x 1 = 1.84 x 2 = 1.99

σ 1 = .38 σ 2 = .51

for a 98% C.I., z.01 = 2.33

2

2

2

1

2

121 )(

nn z x x σ σ

+±−




(1.84 - 1.99) ± 2.3341

51.

35

38. 22

+

-.15 ± .2384

-.3884 < µ1 - µ2 < .0884

Point Estimate = -.15

Hypothesis Test:

1) Ho: µ1 - µ2 = 0

Ha: µ1 - µ2 ≠ 0

2) z =

2

2

2

1

2

1

2121 )()(

nn

x x

σ σ

µ µ

+

−−−

3) α = .02

4) For a two-tailed test, z.01 = + 2.33. If the observed z value is greater than 2.33or less than -2.33, then the decision will be to reject the null hypothesis.

5) Data given above

6) z =

41

)51(.

35

)38(.

)0()99.184.1(22

+

−− = -1.47

7) Since z = -1.47 > z.01 = -2.33, the decision is to fail to reject the nullhypothesis.

8) There is no significant difference in the hourly ratesof the two groups.




10.7 1994 2001

x 1 = 190 x 2 = 198

σ 1 = 18.50 σ 2 = 15.60

n1 = 51 n2 = 47 α = .01

H0: µ 1 - µ 2 = 0

Ha: µ 1 - µ 2 < 0For a one-tailed test, z.01 = -2.33

z =

47

)60.15(

51

)50.18(

)0()198190()()(

22

2

2

2

1

2

1

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ = -2.32

Since the observed z = -2.32 > z.01 = -2.33, the decision is to fail to reject the nullhypothesis.

10.8 Seattle Atlanta

n1 = 31 n2 = 31

x 1 = 2.64 x 2 = 2.36

σ 12 = .03 σ 2

2 = .015

For a 99% C.I., z.005 = 2.575

2

2

2

1

2

121 )(

nn z x x σ σ

+±−

(2.64-2.36) ± 2.57531

015.

31

03.+

.28 ± .10 .18 < µ µµ µ < .38

Between $ .18 and $ .38 difference with Seattle being more expensive.




10.9 Canon Pioneer

x 1 = 5.8 x 2 = 5.0

σ 1 = 1.7 σ 2 = 1.4n1 = 36 n2 = 45

Ho: µ1 - µ2 = 0

Ha: µ1 - µ2 ≠ 0

For two-tail test, α /2 = .025 z.025 = ±1.96

z =

45

)4.1(

36

)7.1(

)0()0.58.5()()(

2

2

22

1

21

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ = 2.27

Since the observed z = 2.27 > zc = 1.96, the decision is to reject the null hypothesis.

10.10 A B

x 1 = 8.05 x 2 = 7.26

σ 1 = 1.36 σ 2 = 1.06

n1 = 50 n2 = 38

Ho: µ1 - µ2 = 0Ha: µ1 - µ2 > 0

For one-tail test, α = .10 z.10 = 1.28

z =

38

)06.1(

50

)36.1(

)0()26.705.8()()(

22

2

2

2

1

2

1

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ = 3.06

Since the observed z = 3.06 > zc = 1.28, the decision is to reject the nullhypothesis.




10.11 Ho: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 < 0 df = 8 + 11 - 2 = 17

Sample 1 Sample 2

n1 = 8 n2 = 11

x 1 = 24.56 x 2 = 26.42s1

2 = 12.4 s22 = 15.8

For one-tail test, α = .01 Critical t .01,19 = -2.567

t =

2121

22

212

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ =

11

1

8

1

2118

)10(8.15)7(4.12

)0()42.2656.24(

+−+

+

−− = -1.05

Since the observed t = -1.05 > t .01,19 = -2.567, the decision is to fail to reject thenull hypothesis.

10.12 a) Ho: µ1 - µ2 = 0 α =.10

Ha: µ1 - µ2 ≠ 0 df = 20 + 20 - 2 = 38

Sample 1 Sample 2n1 = 20 n2 = 20

x 1 = 118 x 2 = 113s1 = 23.9 s2 = 21.6

For two-tail test, α /2 = .05 Critical t .05,38 = 1.697 (used df=30)

t =

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+

−+

−+−

−−− µ µ =

t =

20

1

20

1

22020

)19()6.21()19()9.23(

)0()42.2656.24(22

+−+

+

−− = 0.69

Since the observed t = 0.69 < t .05,38 = 1.697, the decision is to fail to rejectthe null hypothesis.




b)2121

2

2

21

2

121

11

2

)1()1()(

nnnn

nsnst x x +

−+

−+−±− =

(118 – 113) + 1.69720

1

20

1

22020

)19()6.21()19()9.23( 22

+−+

+

5 + 12.224

-7.224 < µ µµ µ 1 - µ µµ µ 2 < 17.224

10.13 Ho: µ1 - µ2 = 0 α = .05

Ha: µ1 - µ2 > 0 df = n1 + n2 - 2 = 10 + 10 - 2 = 18

Sample 1 Sample 2

n1 = 10 n2 = 10

x 1 = 45.38 x 2 = 40.49s1 = 2.357 s2 = 2.355

For one-tail test, α = .05 Critical t .05,18 = 1.734

t =

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ

=

t =

10

1

10

1

21010

)9()355.2()9()357.2(

)0()49.4038.45(22

+−+

+

−− = 4.64

Since the observed t = 4.64 > t .05,18 = 1.734, the decision is to reject thenull hypothesis.

10.14 Ho: µ1 - µ2 = 0 α =.01

Ha: µ1 - µ2 ≠ 0 df = 18 + 18 - 2 = 34

Sample 1 Sample 2

n1 = 18 n2 = 18

x 1 = 5.333 x 2 = 9.444s1

2 = 12 s22 = 2.026




For two-tail test, α /2 = .005 Critical t .005,34 = ±2.457 (used df=30)

t =

2121

2

2

21

2

1

2121

11

2

)1()1()()(

nnnn

nsns x x

+−+

−+−

−−− µ µ =

t =

18

1

18

1

21818

17)026.2()17(12

)0()444.9333.5(

+−+

+

−− = -4.66

Since the observed t = -4.66 < t .005,34 = -2.457

Reject the null hypothesis

b)2121

2

2

21

2

121

11

2

)1()1()(

nnnn

nsnst x x +

−+

−+−±− =

(5.333 – 9.444) + 2.45718

1

18

1

21818

)17)(026.2()17)(12(+

−+

+

-4.111 + 2.1689

-6.2799 < µ µµ µ 1 - µ µµ µ 2 < -1.9421

10.15 Peoria Evansville

n1 = 21 n2 = 26

1 x = 86,900 2 x = 84,000s1 = 2,300 s2 = 1,750 df = 21 + 26 – 2

90% level of confidence, α /2 = .05 t .05,45 = 1.684 (used df = 40)

2121

2

2

21

2

121

11

2

)1()1()(

nnnn

nsnst x x +

−+

−+−±− =

(86,900 – 84,000) + 1.68426

1

21

1

22621

)25()1750()20()2300( 22

+−+

+ =

2,900 + 994.62

1905.38 < µ µµ µ 1 - µ µµ µ 2 < 3894.62




10.16 Ho: µ1 - µ2 = 0 α = .05

Ha: µ1 - µ2 ≠ 0!= 0 df = 21 + 26 - 2 = 45

Peoria Evansville

n1 = 21 n2 = 26

x 1 = $86,900 x 2 = $84,000s1 = $2,300 s2 = $1,750

For two-tail test, α /2 = .025Critical t .025,45 = ± 2.021 (used df=40)

t =

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ =

t =

26

1

21

1

22621

)25()750,1()20()300,2(

)0()000,84900,86(22

+−+

+

−− = 4.91

Since the observed t = 4.91 > t .025,45 = 2.021, the decision is to reject the null

hypothesis.

10.17 Let Boston be group 1

1) Ho: µ1 - µ2 = 0Ha: µ1 - µ2 > 0

2) t =

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ

3) α = .01

4) For a one-tailed test and df = 8 + 9 - 2 = 15, t .01,15 = 2.602. If the observed valueof t is greater than 2.602, the decision is to reject the null hypothesis.

5) Boston Dallas

n1 = 8 n2 = 9

x 1 = 47 x 2 = 44s1 = 3 s2 = 3




6) t =

91

81

15)3(8)3(7

)0()4447(22

++

−− = 2.06

7) Since t = 2.06 < t .01,15 = 2.602, the decision is to fail to reject the null hypothesis.

8) There is no significant difference in rental rates between Boston and Dallas.

10.18 nm = 22 nno = 20

x m = 112 x no = 122sm = 11 sno = 12

df = nm + nno - 2 = 22 + 20 - 2 = 40

For a 98% Confidence Interval, α /2 = .01 and t .01,40 = 2.423

2121

2

2

21

2

121

11

2

)1()1()(

nnnn

nsnst x x +

−+

−+−±− =

(112 – 122) + 2.42320

1

22

1

22022

)19()12()21()11( 22

+−+

+

-10 ± 8.63

-$18.63 < µ1 - µ2 < -$1.37

Point Estimate = -$10

10.19 Ho: µ1 - µ2 = 0

Ha: µ1 - µ2 ≠ 0

df = n1 + n2 - 2 = 11 + 11 - 2 = 20

Toronto Mexico City

n1 = 11 n2 = 11

x 1 = $67,381.82 x 2 = $63,481.82s1 = $2,067.28 s2 = $1,594.25

For a two-tail test, α /2 = .005 Critical t .005,20 = ±2.845




t =

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ =

t =

11

1

11

1

21111

)10()25.594,1()10()28.067,2(

)0()82.481,6382.381,67(22

+−+

+

−− = 4.95

Since the observed t = 4.95 > t .005,20 = 2.845, the decision is to Reject the nullhypothesis.

10.20 Toronto Mexico City

n1 = 11 n2 = 11

x 1 = $67,381.82 x 2 = $63,481.82s1 = $2,067.28 s2 = $1,594.25

df = n1 + n2 - 2 = 11 + 11 - 2 = 20

For a 95% Level of Confidence, α /2 = .025 and t .025,20 = 2.086

2121

2

2

21

2

121 11

2)1()1()(

nnnnnsnst x x +

−+−+−±− =

($67,381.82 - $63,481.82) ± (2.086) 11

1

11

1

21111

)10()25.594,1()10()28.067,2( 22

+−+

+

3,900 ± 1,641.9

2,258.1 < µ1 - µ2 < 5,541.9




10.21 Ho: D = 0Ha: D > 0

Sample 1 Sample 2 d 38 22 1627 28 -130 21 941 38 336 38 -238 26 1233 19 1435 31 444 35 9

n = 9 d =7.11 sd=6.45 α = .01

df = n - 1 = 9 - 1 = 8

For one-tail test and α = .01, the critical t .01,8 = ±2.896

t =

9

45.6

011.7 −=

−

n

s

Dd

d

= 3.31

Since the observed t = 3.31 > t .01,8 = 2.896, the decision is to reject the nullhypothesis.

10.22 Ho: D = 0

Ha: D ≠ 0

Before After d 107 102 5

99 98 1110 100 10

113 108 596 89 798 101 -3

100 99 1102 102 0107 105 2109 110 -1104 102 2

99 96 3101 100 1




n = 13 d = 2.5385 sd=3.4789 α = .05

df = n - 1 = 13 - 1 = 12

For a two-tail test and α /2 = .025 Critical t .025,12 = ±2.179

t =

13

4789.3

05385.2 −=

−

n

s

Dd

d

= 2.63


10.23 n = 22 d = 40.56 sd = 26.58

For a 98% Level of Confidence, α /2 = .01, and df = n - 1 = 22 - 1 = 21

t .01,21 = 2.518

n

st d d

±

40.56 ± (2.518)22

58.26

40.56 ± 14.27

26.29 < D < 54.83

10.24 Before After d 32 40 -828 25 335 36 -132 32 026 29 -325 31 -637 39 -216 30 -1435 31 4




n = 9 d = -3 sd = 5.6347 α = .025

df = n - 1 = 9 - 1 = 8

For 90% level of confidence and α /2 = .025, t .05,8 = 1.86

t =n

st d d

±

t = -3 + (1.86) 9

6347.5 = -3 ± 3.49

-0.49 < D < 6.49

10.25 City Cost Resale d

Atlanta 20427 25163 -4736Boston 27255 24625 2630Des Moines 22115 12600 9515Kansas City 23256 24588 -1332Louisville 21887 19267 2620Portland 24255 20150 4105Raleigh-Durham 19852 22500 -2648Reno 23624 16667 6957Ridgewood 25885 26875 - 990San Francisco 28999 35333 -6334Tulsa 20836 16292 4544

d = 1302.82 sd = 4938.22 n = 11, df = 10

α = .01 α /2 = .005 t .005,10= 3.169

n

st d d

± = 1302.82 + 3.169

11

22.4938 = 1302.82 + 4718.42

-3415.6 < D < 6021.2




10.26 Ho: D = 0Ha: D < 0

Before After d 2 4 -24 5 -11 3 -23 3 04 3 12 5 -32 6 -43 4 -11 5 -4

n = 9 d =-1.778 sd=1.716 α = .05 df = n - 1 = 9 - 1 = 8

For a one-tail test and α = .05, the critical t .05,8 = -1.86

t =

9

716.1

0778.1 −−=

−

n

s

Dd

d

= -3.11

Since the observed t = -3.11 < t .05,8 = -1.86, the decision is to reject the null

hypothesis.

10.27 Before After d 255 197 58230 225 5290 215 75242 215 27300 240 60250 235 15215 190 25

230 240 -10225 200 25219 203 16236 223 13

n = 11 d = 28.09 sd=25.813 df = n - 1 = 11 - 1 = 10

For a 98% level of confidence and α /2=.01, t .01,10 = 2.764




n

st d d

±

28.09 ± (2.764) 11

813.25 = 28.09 ± 21.51

6.58 < D < 49.60

10.28 H0: D = 0

Ha: D > 0 n = 27 df = 27 – 1 = 26 d = 3.17 sd = 5

Since α = .01, the critical t .01,26 = 2.479

t =

27

5

071.3 −=

−

n

s

Dd

d

= 3.86


10.29 n = 21 d = 75 sd=30 df = 21 - 1 = 20

For a 90% confidence level, α /2=.05 and t .05,20 = 1.725

n

st d d

±

75 + 1.72521

30 = 75 ± 11.29

63.71 < D < 86.29

10.30 Ho: D = 0

Ha: D ≠ 0

n = 15 d = -2.85 sd = 1.9 α = .01 df = 15 - 1 = 14

For a two-tail test, α /2 = .005 and the critical t .005,14 = + 2.977




t =

15

9.1

085.2 −−=

−

n

s

Dd

d

= -5.81

Since the observed t = -5.81 < t .005,14 = -2.977, the decision is to reject the nullhypothesis.

10.31 a) Sample 1 Sample 2

n1 = 368 n2 = 405 x1 = 175 x2 = 182

368

175ˆ

1

11 ==

n

x p = .476

405

182ˆ

2

22 ==

n

x p = .449

773

357

405368

182175

21

21=

+

+=

+

+=

nn

x x p = .462

Ho: p1 - p2 = 0

Ha: p1 - p2 ≠ 0

For two-tail, α /2 = .025 and z.025 = ±1.96

+

−−=

+⋅

−−−=

405

1

368

1)538)(.462(.

)0()449.476(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z = 0.75

Since the observed z = 0.75 < zc = 1.96, the decision is to fail to reject the nullhypothesis.

b) Sample 1 Sample 2

p̂ 1 = .38 p̂ 2 = .25

n1 = 649 n2 = 558

558649

)25(.558)38(.649ˆˆ

21

2211

+

+=

+

+=

nn

pn pn p = .32




Ho: p1 - p2 = 0Ha: p1 - p2 > 0

For a one-tail test and α = .10, z.10 = 1.28

+

−−=

+⋅

−−−=

558

1

649

1)68)(.32(.

)0()25.38(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z = 4.83


10.32 a) n1 = 85 n2 = 90 p̂ 1 = .75 p̂ 2 = .67

For a 90% Confidence Level, z.05 = 1.645

2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.75 - .67) ± 1.64590

)33)(.67(.

85

)25)(.75(.+ = .08 ± .11

-.03 < p1 - p2 < .19

b) n1 = 1100 n2 = 1300 p̂ 1 = .19 p̂ 2 = .17

For a 95% Confidence Level, α /2 = .025 and z.025 = 1.96

2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.19 - .17) + 1.961300

)83)(.17(.

1100

)81)(.19(.+ = .02 ± .03

-.01 < p1 - p2 < .05




c) n1 = 430 n2 = 399 x1 = 275 x2 = 275

430

275ˆ

1

11 == n

x p = .64 399

275ˆ

2

22 == n

x p = .69

For an 85% Confidence Level, α /2 = .075 and z.075 = 1.44

2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.64 - .69) + 1.44399

)31)(.69(.

430

)36)(.64(.+ = -.05 ± .047

-.097 < p1 - p2 < -.003

d) n1 = 1500 n2 = 1500 x1 = 1050 x2 = 1100

1500

1050ˆ

1

11 ==

n

x p = .70

1500

1100ˆ

2

22 ==

n

x p = .733

For an 80% Confidence Level, α /2 = .10 and z.10 = 1.28

2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.70 - .733) ± 1.281500

)267)(.733(.

1500

)30)(.70(.+ = -.033 ± .02

-.053 < p1 - p2 < -.013

10.33 H0: pm - pw = 0

Ha: pm - pw < 0 nm = 374 nw = 481 p̂ m = .59 p̂ w = .70

For a one-tailed test and α = .05, z.05 = -1.645

481374

)70(.481)59(.374ˆˆ

+

+=

+

+=

wm

wwmm

nn

pn pn p = .652




+

−−=

+⋅

−−−=

481

1

374

1)348)(.652(.

)0()70.59(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z = -3.35

Since the observed z = -3.35 < z.05 = -1.645, the decision is to reject the nullhypothesis.

10.34 n1 = 210 n2 = 176 1ˆ p = .24 2

ˆ p = .35

For a 90% Confidence Level, α /2 = .05 and z.05 = + 1.645

2

22

1

11

21

ˆˆˆˆ

)ˆˆ( n

q p

n

q p

z p p +±−

(.24 - .35) + 1.645176

)65)(.35(.

210

)76)(.24(.+ = -.11 + .0765

-.1865 < p1 – p2 < -.0335

10.35 Computer Firms Banks

p̂ 1 = .48 p̂ 2 = .56n1 = 56 n2 = 89

8956

)56(.89)48(.56ˆˆ

21

2211

+

+=

+

+=

nn

pn pn p = .529

Ho: p1 - p2 = 0

Ha: p1 - p2 ≠ 0

For two-tail test, α /2 = .10 and zc = ±1.28

+

−−=

+⋅

−−−=

89

1

56

1)471)(.529(.

)0()56.48(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z = -0.94

Since the observed z = -0.94 > zc = -1.28, the decision is to fail to reject the nullhypothesis.




10.36 A B

n1 = 35 n2 = 35 x1 = 5 x2 = 7

35

5ˆ

1

11 ==

n

x p = .14

35

7ˆ

2

22 ==

n

x p = .20


2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.14 - .20) ± 2.3335

)80)(.20(.

35

)86)(.14(.+ = -.06 ± .21

-.27 < p1 - p2 < .15

10.37 H0: p1 – p2 = 0

Ha: p1 – p2 ≠ 0

α = .10 p̂1 = .09 p̂

2 = .06 n1 = 780 n2 = 915

For a two-tailed test, α /2 = .05 and z.05 = + 1.645

915780

)06(.915)09(.780ˆˆ

21

2211

+

+=

+

+=

nn

pn pn p = .0738

+

−−=

+⋅

−−−=

915

1

780

1)9262)(.0738(.

)0()06.09(.

11

)()ˆˆ(

1

2121

nnq p

p p p p Z = 2.35





10.38 n1 = 850 n2 = 910 p̂1

= .60 p̂ 2 = .52

For a 95% Confidence Level, α /2 = .025 and z.025 = + 1.96

2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.60 - .52) + 1.96910

)48)(.52(.

850

)40)(.60(.+ = .08 + .046

.034 < p1 – p2 < .126

10.39 H0: σ 12 = σ22 α = .01 n1 = 10 s12 = 562Ha: σ 1

2 < σ 22 n2 = 12 s2

2 = 1013

df num = 12 - 1 = 11 df denom = 10 - 1 = 9

Table F .01,10,9 = 5.26

F =562

10132

1

2

2=

s

s = 1.80

Since the observed F = 1.80 < F .01,10,9 = 5.26, the decision is to fail to reject thenull hypothesis.

10.40 H0: σ 12 = σ 2

2 α = .05 n1 = 5 S 1 = 4.68

Ha: σ 12 ≠ σ 2

2 n2 = 19 S 2 = 2.78

df num = 5 - 1 = 4 df denom = 19 - 1 = 18

The critical table F values are: F .025,4,18 = 3.61 F .95,18,4 = .277

F = 2

2

2

2

2

1

)78.2(

)68.4(

=s

s

= 2.83





10.41 City 1 City 2

1.18 1.081.15 1.171.14 1.141.07 1.051.14 1.211.13 1.141.09 1.111.13 1.191.13 1.121.03 1.13

n1 = 10 df 1 = 9 n2 = 10 df 2 = 9

s12 = .0018989 s2

2 = .0023378

H0: σ 12 = σ 2

2 α = .10 α /2 = .05

Ha: σ 12 ≠ σ 2

2

Upper tail critical F value = F .05,9,9 = 3.18

Lower tail critical F value = F .95,9,9 = 0.314

F =0023378.

0018989.2

2

2

1=

s

s = 0.81

Since the observed F = 0.81 is greater than the lower tail critical value of 0.314and less than the upper tail critical value of 3.18, the decision is to failto reject the null hypothesis.

10.42 Let Houston = group 1 and Chicago = group 2

1) H0: σ 12 = σ 2

2

Ha: σ 12 ≠ σ 22

2) F =2

2

2

1

s

s

3) α = .01

4) df 1 = 12 df 2 = 10 This is a two-tailed test





If the observed value is greater than 5.66 or less than .177, the decision will beto reject the null hypothesis.

5) s12 = 393.4 s2

2 = 702.7

6) F =7.702

4.393 = 0.56

7) Since F = 0.56 is greater than .177 and less than 5.66,the decision is to fail to reject the null hypothesis.

8) There is no significant difference in the variances ofnumber of days between Houston and Chicago.

10.43 H0: σ 12 = σ 2

2 α = .05 n1 = 12 s1 = 7.52

Ha: σ 12 > σ 2

2 n2 = 15 s2 = 6.08

df num = 12 - 1 = 11 df denom = 15 - 1 = 14

The critical table F value is F .05,10,14 = 5.26

F =2

2

2

2

2

1

)08.6(

)52.7(=

s

s = 1.53


10.44 H0: σ 12 = σ 2

2 α = .01 n1 = 15 s12 = 91.5

Ha: σ 12 ≠ σ 2

2 n2 = 15 s22 = 67.3

df num = 15 - 1 = 14 df denom = 15 - 1 = 14


F =3.675.912

2

2

1 =ss = 1.36

Since the observed F = 1.36 < F .005,12,14 = 4.43 and > F .995,14,12 = .226, the decisionis to fail to reject the null hypothesis.




10.45 Ho: µ1 - µ2 = 0

Ha: µ1 - µ2 ≠ 0

For α = .10 and a two-tailed test, α /2 = .05 and z.05 = + 1.645

Sample 1 Sample 2

1 x = 138.4 2 x = 142.5

σ 1 = 6.71 σ 2 = 8.92n1 = 48 n2 = 39

z =

39

)92.8(

48

)71.6(

)0()5.1424.138()()(

2

2

2

2

1

2

1

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ

= -2.38

Since the observed value of z = -2.38 is less than the critical value of z = -1.645,the decision is to reject the null hypothesis. There is a significant difference inthe means of the two populations.


1 x = 34.9 2 x = 27.6

σ 12 = 2.97 σ 2

2 = 3.50n1 = 34 n2 = 31

For 98% Confidence Level, z.01 = 2.33

2

2

2

1

2

121 )(

nn z x x σ σ

+±−

(34.9 – 27.6) + 2.33

31

50.3

34

97.2+ = 7.3 + 1.04

6.26 < µ µµ µ 1 - µ µµ µ 2 < 8.34




10.47 Ho: µ1 - µ2 = 0Ha: µ1 - µ2 > 0

Sample 1 Sample 2

1 x = 2.06 2 x = 1.93

s12 = .176 s2

2 = .143n1 = 12 n2 = 15

This is a one-tailed test with df = 12 + 15 - 2 = 25. The critical value ist .05,25 = 1.708. If the observed value is greater than 1.708, the decision will be toreject the null hypothesis.

t =

2121

2

2

21

2

1

2121

11

2

)1()1()()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ

t =

15

1

12

1

25

)14)(143(.)11)(176(.

)0()93.106.2(

++

−− = 0.85

Since the observed value of t = 0.85 is less than the critical value of t = 1.708, thedecision is to fail to reject the null hypothesis. The mean for population one isnot significantly greater than the mean for population two.


x 1 = 74.6 x 2 = 70.9s1

2 = 10.5 s22 = 11.4

n1 = 18 n2 = 19

For 95% confidence, α /2 = .025.Using df = 18 + 19 - 2 = 35, t 35,.025 = 2.042

2121

2

2

21

2

121

112

)1()1()(nnnn

nsnst x x +−+

−+−±−

(74.6 – 70.9) + 2.04220

1

20

1

22020

)19()6.21()19()9.23( 22

+−+

+

3.7 + 2.22

1.48 < µ µµ µ 1 - µ µµ µ 2 < 5.92




10.49 Ho: D = 0 α = .01Ha: D < 0

n = 21 df = 20 d = -1.16 sd = 1.01

The critical t .01,20 = -2.528. If the observed t is less than -2.528, then the decisionwill be to reject the null hypothesis.

t =

21

01.1

016.1 −−=

−

n

s

Dd

d

= -5.26

Since the observed value of t = -5.26 is less than the critical t value of -2.528, thedecision is to reject the null hypothesis. The population difference is lessthan zero.

10.50 Respondent Before After d

1 47 63 -162 33 35 - 23 38 36 24 50 56 - 65 39 44 - 5

6 27 29 - 27 35 32 38 46 54 - 89 41 47 - 6

d = -4.44 sd = 5.703 df = 8

For a 99% Confidence Level, α /2 = .005 and t 8,.005 = 3.355

n

st d d

± = -4.44 + 3.355

9

703.5 = -4.44 + 6.38

-10.82 < D < 1.94

10.51 Ho: p1 - p2 = 0 α = .05 α /2 = .025

Ha: p1 - p2 ≠ 0 z.025 = + 1.96

If the observed value of z is greater than 1.96 or less than -1.96, then the decisionwill be to reject the null hypothesis.




Sample 1 Sample 2

x1 = 345 x2 = 421n1 = 783 n2 = 896

896783

421345

21

21

+

+=

+

+=

nn

x x p = .4562

783

345ˆ

1

11 ==

n

x p = .4406

896

421ˆ

2

22 ==

n

x p = .4699

+

−−

=

+⋅

−−−

=

896

1

783

1)5438)(.4562(.

)0()4699.4406(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z = -1.20

Since the observed value of z = -1.20 is greater than -1.96, the decision is to failto reject the null hypothesis. There is no significant difference in thepopulationproportions.


n1 = 409 n2 = 378 p̂ 1 = .71 p̂ 2 = .67


2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.71 - .67) + 2.575378

)33)(.67(.

409

)29)(.71(.+ = .04 ± .085

-.045 < p1 - p2 < .125

10.53 H0: σ 12 = σ 2

2 α = .05 n1 = 8 s12 = 46

Ha: σ 12 ≠ σ 2

2 n2 = 10 S 22 = 37

df num = 8 - 1 = 7 df denom = 10 - 1 = 9The critical F values are: F .025,7,9 = 4.20 F .975,9,7 = .238




If the observed value of F is greater than 4.20 or less than .238, then the decisionwill be to reject the null hypothesis.

F =37

462

2

2

1=

s

s = 1.24

Since the observed F = 1.24 is less than F .025,7,9 =4.20 and greater thanF .975,9,7 = .238, the decision is to fail to reject the null hypothesis. There is nosignificant difference in the variances of the two populations.

10.54 Term Whole Life

x t = $75,000 x w = $45,000

st = $22,000 sw = $15,500nt = 27 nw = 29

df = 27 + 29 - 2 = 54

For a 95% Confidence Level, α /2 = .025 and t .025,40 = 2.021 (used df=40)

2121

2

2

21

2

121

11

2

)1()1()(

nnnn

nsnst x x +

−+

−+−±−

(75,000 – 45,000) + 2.02129

1

27

1

22927

)28()500,15()26()000,22( 22

+−+

+

30,000 ± 10,220.73

19,779.27 < µ1 - µ2 < 40,220.73

10.55 Morning Afternoon d 43 41 251 49 2

37 44 -724 32 -847 46 144 42 250 47 355 51 446 49 -3

n = 9 d = -0.444 sd =4.447 df = 9 - 1 = 8

For a 90% Confidence Level: α /2 = .05 and t .05,8 = 1.86




n

st d d

±

-0.444 + (1.86) 9

447.4 = -0.444 ± 2.757

-3.201 < D < 2.313

10.56 Let group 1 be 1990

Ho: p1 - p2 = 0

Ha: p1 - p2 < 0 α = .05

The critical table z value is: z.05 = -1.645

n1 = 1300 n2 = 1450 1ˆ p = .447 2

ˆ p = .487

14501300

)1450)(487(.)1300)(447(.ˆˆ

21

2211

+

+=

+

+=

nn

pn pn p = .468

+

−−=

+⋅

−−−=

1450

1

1300

1)532)(.468(.

)0()487.447(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z = -2.10

Since the observed z = -3.73 is less than z.05 = -1.645, the decision is to reject thenull hypothesis. 1997 has a significantly higher proportion.

10.57 Accounting Data Entry

n1 = 16 n2 = 14

x 1 = 26,400 x 2 = 25,800s1 = 1,200 s2 = 1,050

H0: σ 12 = σ2

2

Ha: σ 12 ≠ σ 2

2

df num = 16 – 1 = 15 df denom = 14 – 1 = 13

The critical F values are: F .025,15,13 = 3.05 F .975,15,13 = 0.33




F =500,102,1

000,440,12

2

2

1=

s

s = 1.31

Since the observed F = 1.31 is less than F .025,15,13 = 3.05 and greater thanF .975,15,13 = 0.33, the decision is to fail to reject the null hypothesis.

10.58 H0: σ 12 = σ 2

2 α = .01 n1 = 8 n2 = 7

Ha: σ 12 ≠ σ 2

2 S 12 = 72,909 S 2

2 = 129,569

df num = 6 df denom = 7

The critical F values are: F .005,6,7 = 9.16 F .995,7,6 = .11

F =909,72

569,1292

2

21

=s

s = 1.78

Since F = 1.95 < F .005,6,7 = 9.16 but also > F .995,7,6 = .11, the decision is to fail toreject the null hypothesis. There is no difference in the variances of the shifts.

10.59 Men Women

n1 = 60 n2 = 41 x 1 = 631 x 2 = 848

σ 1 = 100 σ 2 = 100


2

2

2

1

2

121 )(

nn z x x σ σ

+±−

(631 – 848) + 1.96 41

100

60

100 22

+ = -217 ± 39.7

-256.7 < µ1 - µ2 < -177.3




10.60 Ho: µ1 - µ2 = 0 α = .01

Ha: µ1 - µ2 ≠ 0 df = 20 + 24 - 2 = 42

Detroit Charlotte

n1 = 20 n2 = 24

x 1 = 17.53 x 2 = 14.89s1 = 3.2 s2 = 2.7

For two-tail test, α /2 = .005 and the critical t.005,42 = ±2.704 (used df=40)

t =

2121

2

2

21

2

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ

t =

24

1

20

1

42

)23()7.2()19()2.3(

)0()89.1453.17(22

++

−− = 2.97


10.61 With Fertilizer Without Fertilizer

x 1 = 38.4 x 2 = 23.1

σ 1 = 9.8 σ 2 = 7.4n1 = 35 n2 = 35

Ho: µ1 - µ2 = 0Ha: µ1 - µ2 > 0

For one-tail test, α = .01 and z.01 = 2.33

z =

35

)4.7(

35

)8.9(

)0()1.234.38()()(

2

2

2

2

1

2

1

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ = 7.37





10.62 Specialty Discount

n1 = 350 n2 = 500

p̂ 1 = .75 p̂ 2 = .52


2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.75 - .52) + 1.645500

)48)(.52(.

350

)25)(.75(.+ = .23 ± .053

.177 < p1 - p2 < .283

10.63 H0: σ 12 = σ 2

2 α = .05 n1 = 27 s1 = 22,000

Ha: σ 12 ≠ σ 2

2 n2 = 29 s2 = 15,500

df num = 27 - 1 = 26 df denom = 29 - 1 = 28

The critical F values are: F .025,24,28 = 2.17 F .975,28,24 = .46

F =2

2

2

2

2

1

500,15

000,22=

s

s = 2.01

Since the observed F = 2.01 < F.025,24,28 = 2.17 and > than F .975,28,24 = .46, thedecision is to fail to reject the null hypothesis.




10.64 Name Brand Store Brand d 54 49 555 50 5

59 52 753 51 254 50 461 56 551 47 453 49 4

n = 8 d = 4.5 sd=1.414 df = 8 - 1 = 7

For a 90% Confidence Level, α /2 = .05 and t .05,7 = 1.895

n

st d d

±

4.5 + 1.8958

414.1 = 4.5 ± .947

3.553 < D < 5.447

10.65 Ho: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 < 0 df = 23 + 19 - 2 = 40

Wisconsin Tennessee

n1 = 23 n2 = 19

x 1 = 69.652 x 2 = 71.7368s1

2 = 9.9644 s22 = 4.6491

For one-tail test, α = .01 and the critical t .01,40 = -2.423

t =

2121

22

212

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ

t =

19

1

23

1

40

)18)(6491.4()22)(9644.9(

)0()7368.71652.69(

++

−− = -2.44




Since the observed t = -2.44 < t .01,40 = -2.423, the decision is to reject the nullhypothesis.

10.66 Wednesday Friday d 71 53 1856 47 975 52 2368 55 1374 58 16

n = 5 d = 15.8 sd = 5.263 df = 5 - 1 = 4

Ho: D = 0α

= .05Ha: D > 0

For one-tail test, α = .05 and the critical t .05,4 = 2.132

t =

5

263.5

08.15 −=

−

n

s

Dd

d

= 6.71


10.67 Ho: P1 - P2 = 0 α = .05

Ha: P1 - P2 ≠ 0

Machine 1 Machine 2

x1 = 38 x2 = 21n1 = 191 n2 = 202

191

38ˆ

1

11 ==

n

x p = .199

202

21ˆ

2

22 ==

n

x p = .104

202191

)202)(104(.)191)(199(.ˆˆ

21

2211

+

+=

+

+=

nn

pn pn p = .15

For two-tail, α /2 = .025 and the critical z values are: z.025 = ±1.96




+

−−=

+⋅

−−−=

202

1

191

1)85)(.15(.

)0()104.199(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z = 2.64


10.68 Construction Telephone Repair

n1 = 338 n2 = 281 x1 = 297 x2 = 192

338

297ˆ

1

11 ==

n

x p = .879

281

192ˆ

2

22 ==

n

x p = .683


2

22

1

1121

ˆˆˆˆ)ˆˆ(

n

q p

n

q p z p p +±−

(.879 - .683) + 1.645 281

)317)(.683(.

338

)121)(.879(.

+ = .196 ± .054

.142 < p1 - p2 < .250

10.69 Aerospace Automobile

n1 = 33 n2 = 35

x 1 = 12.4 x 2 = 4.6

σ 1 = 2.9 σ 2 = 1.8


2

2

2

1

2

121 )(

nn z x x σ σ

+±−

(12.4 – 4.6) + 2.57535

)8.1(

33

)9.2( 22

+ = 7.8 ± 1.52

6.28 < µ1 - µ2 < 9.32




10.70 Discount Specialty

x 1 = $47.20 x 2 = $27.40

σ 1 = $12.45 σ 2 = $9.82n1 = 60 n2 = 40

Ho: µ1 - µ2 = 0 α = .01

Ha: µ1 - µ2 ≠ 0

For two-tail test, α /2 = .005 and zc = ±2.575

z =

40

)82.9(

60

)45.12(

)0()40.2720.47()()(

22

2

2

2

1

2

1

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ

= 8.86


10.71 Before After d

12 8 47 3 4

10 8 216 9 78 5 3

n = 5 d = 4.0 sd = 1.8708 df = 5 - 1 = 4

Ho: D = 0 α = .01Ha: D > 0

For one-tail test, α = .01 and the critical t .01,4 = 3.747

t =

5

8708.1

00.4 −=

−

n

s

Dd

d

= 4.78





10.72 Ho: µ1 - µ2 = 0 α = .01

Ha: µ1 - µ2 ≠ 0 df = 10 + 6 - 2 = 14

A B

n1 = 10 n2 = 6

x 1 = 18.3 x 2 = 9.667s1

2 = 17.122 s22 = 7.467

For two-tail test, α /2 = .005 and the critical t .005,14 = ±2.977

t =

2121

22

212

1

2121

11

2

)1()1(

)()(

nnnn

nsns

x x

+−+

−+−

−−− µ µ

t =

6

1

10

1

14

)5)(467.7()9)(122.17(

)0()667.93.18(

++

−− = 4.52


10.73 A t test was used to test to determine if Hong Kong has significantly differentrates than Bombay. Let group 1 be Hong Kong.

Ho: µ1 - µ2 = 0Ha: µ1 - µ 2 > 0

n1 = 19 n2 = 23 x 1 = 130.4 x 2 = 128.4

S1 = 12.9 S2 = 13.9 α = .01

t = 0.48 with a p-value of .634 which is not significant at of .05. There is notenough evidence in these data to declare that there is a difference in the averagerental rates of the two cities.

10.74 H0: D = 0

Ha: D ≠ 0

This is a related measures before and after study. Fourteen people were involvedin the study. Before the treatment, the sample mean was 4.357 and after the




treatment, the mean was 5.214. The higher number after the treatment indicatesthat subjects were more likely to “blow the whistle” after having been through thetreatment. The observed t value was –3.12 which was more extreme than two-

tailed table t value of + 2.16 causing the researcher to reject the null hypothesis.

This is underscored by a p-value of .0081 which is less than α = .05. The studyconcludes that there is a significantly higher likelihood of “blowing the whistle”after the treatment.

10.75 The point estimates from the sample data indicate that in the northern city themarket share is .3108 and in the southern city the market share is .2701. Thepoint estimate for the difference in the two proportions of market share are .0407.Since the 99% confidence interval ranges from -.0394 to +.1207 and zero is in theinterval, any hypothesis testing decision based on this interval would result infailure to reject the null hypothesis. Alpha is .01 with a two-tailed test. This is

underscored by a calculated z value of 1.31 which has an associated p-value of

.191 which, of course, is not significant for any of the usual values of α.

10.76 A test of differences of the variances of the populations of the two machines isbeing computed. The hypotheses are:

H0: σ 12 = σ 2

2

Ha: σ 12 ≠ σ 2

2

Twenty-six pipes were measured for sample one and twenty-six pipes were

measured for sample two. The observed F = 1.79 is not significant at α = .05 fora two-tailed test since the associated p-value is .0758. There is no significantdifference in the variance of pipe lengths for pipes produced by machine A versusmachine B.

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