Ken Black QA 5th chapter 10 Solution

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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 two population means using the z statistic.2. Test hypotheses and establish confidence intervals about the difference in two

population means using the t statistic.3. Test hypotheses and construct confidence intervals about the difference in two

related populations.4. Test hypotheses and construct confidence intervals about the difference in two

population proportions.5. Test hypotheses and construct confidence intervals about two population

variances.

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.

In this chapter, the approach as to whether to use a z statistic or a t statistic for analyzing the differences in two sample means is the same as that used in chapters 8and 9. When the population variances are known, the z statistic can be used. However, if the population variances are unknown and sample variances are being used, then thet test is the appropriate statistic for the analysis. It is always an assumption underlyingthe use of the t statistic that the populations are normally distributed. If sample sizes aresmall and the population variances are known, the z statistic can be used if the

populations are normally distributed.




In conducting a t test for the difference of two means from independent populations, there are two different formulas given in the chapter. One version of thistest uses a "pooled" estimate of the population variance and assumes that the populationvariances are equal. The other version does not assume equal population variances and is

simpler to compute. In doing hand calculations, it is generally easier to use the “pooled”variance formula because the degrees of freedom formula for the unequal varianceformula is quite complex. However, it is good to expose students to both formulas sincecomputer software packages often give you the option of using the “pooled” that assumesequal population variances or the formula for unequal variances.

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 the

potential difference in the outcome of the two techniques, it is sometimes valuable toanalyze some related measures data with both techniques and demonstrate that the results

and 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 for two 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 Means

using 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

TestConfidence Intervals

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

PopulationsConfidence Intervals

10.4 Statistical Inferences About Two Population Proportions, p1- p2

Hypothesis TestingConfidence IntervalsUsing the Computer to Analyze the Difference in Two Proportions

10.5 Testing Hypotheses About Two Population VariancesUsing the Computer to Test Hypotheses about Two Population Variances

KEY TERMS

Dependent Samples Independent Samples F Distribution Matched-Pairs Test F Value Related Measures




SOLUTIONS TO PROBLEMS IN CHAPTER 10

10.1 Sample 1 Sample 2 x

1 = 51.3 x

2 = 53.2 s12 = 52 s2

2 = 60 n1 = 31 n2 = 32

a) H o: µ1 - µ2 = 0Ha: µ1 - µ2 < 0

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

z =3260

3152

)0()2.533.51()()(

2

22

1

21

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ

= -1.01

Since the observed z = -1.01 > z c = -1.28, the decision is to fail to rejectthe null hypothesis .

b) Critical value method:

z c =

2

2

2

1

2

1

2121 )()(

nn

x xc

σ σ

µ µ

+

−−−

-1.28 =3260

3152

)0()( 21

+

−− c x x

( x 1 - x 2)c = -2.41

c) The area for z = -1.01 using Table A.5 is .3438.The p-value is .5000 - .3438 = .1562




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

22

1

21

21 )(nn

z x xσ σ

+±−

(70.4) – 68.7) + 1.64531

1.63276.5 22

+

1.7 ± 2.46

-.76 < µ1 - µ2 < 4.16

10.3 a) Sample 1 Sample 2

x 1 = 88.23 x 2 = 81.2 σ 1

2 = 22.74 σ 22 = 26.65

n1 = 30 n2 = 30

Ho: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0

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

z =30

65.2630

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

22

1

21

21 )(nn

z x xσ σ

+±−

(88.23 – 81.2) + 2.3330

65.2630

74.22 +

7.03 + 2.99

4.04 < 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 1

2 = 1.0188 σ 22 = .9180

n1 = 50 n2 = 50

Ho: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0

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

z =509180.0

500188.1

)0()02.396.1()()(

2

22

1

21

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ

= -5.39

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




10.5 A Bn1 = 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

22

1

21

21 )(nn

z x xσ σ

+±−

(5.3 – 6.5) + 1.9637

36.24099.1

+

-1.2 ± .66 -1.86 < µ < -.54

The results indicate that we are 95% confident that, on average, Plumber B does between 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 = 1.99σ 1 = .38 σ 2 = .51

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

2

22

1

21

21 )(nn

z x xσ σ

+±−

(1.84 - 1.99) ± 2.334151.

3538. 22

+

-.15 ± .2384

-.3884 < µ1 - µ2 < .0884

Point Estimate = -.15

Hypothesis Test:




1) H o: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0

2) z =

2

22

1

21

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 rates of the two groups.

10.7 1996 2006 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 = 0Ha: µ 1 - µ 2 < 0For a one-tailed test, z .01 = -2.33

z =47

)60.15(51

)50.18(

)0()198190()()(22

2

22

1

21

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 = 2.36σ 1

2 = .03 σ 22 = .015

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

2

22

1

21

21 )(nn

z x xσ σ

+±−

(2.64-2.36) ± 2.57531015.

3103.

+

.28 ± .10 .18 < 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 = 1.4 n1 = 36 n2 = 45

Ho: µ1 - µ2 = 0Ha: 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 > z c = 1.96, the decision is to reject the nullhypothesis .




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

22

1

21

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ

= 3.06


10.11 H o: 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 = 26.42

s12 = 12.4 s22 = 15.8

For one-tail test, α = .01 Critical t .01,17 = -2.567

t =

2121

22

212

1

2121

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

=111

81

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) H o: µ1 - µ2 = 0 α =.10Ha: µ1 - µ2 ≠ 0 df = 20 + 20 - 2 = 38

Sample 1 Sample 2

n1 = 20 n2 = 20 x 1 = 118 x 2 = 113 s1 = 23.9 s2 = 21.6

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

t =

2121

22

212

1

2121

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

=

t =201

201

22020)19()6.21()19()9.23()0()113118(

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

22

212

121

112

)1()1()(

nnnn

n sn st x x +

−+

−+−±− =

(118 – 113) + 1.697201

201

22020)19()6.21()19()9.23( 22

+−+

+

5 + 12.224

-7.224 < µ 1 - µ 2 < 17.224




10.13 H o: µ1 - µ2 = 0 α = .05Ha: 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 = 40.49 s1 = 2.357 s2 = 2.355

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

t =

2121

22

212

1

2121

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

=

t =101

101

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 H o: µ1 - µ2 = 0 α =.01Ha: µ1 - µ2 ≠ 0 df = 18 + 18 - 2 = 34

Sample 1 Sample 2 n1 = 18 n2 = 18 x 1 = 5.333 x 2 = 9.444 s1

2 = 12 s22 = 2.026

For two-tail test, α /2 = .005 Critical t .005,34 = ±2.75 (used df=30)

t =

2121

22

212

1

2121

11

2

)1()1(

)()(

nnnn

n sn s

x x

+−+

−+−

−−− µ µ

=

t =181

181

2181817)026.2()17(12

)0()444.9333.5(

+−+

+

−−

= -4.66

Since the observed t = -4.66 < t .005,34 = -2.75, reject the null hypothesis.




b) For 98% confidence, t .01, 30 = 2.457

2121

22

212

121

112

)1()1()(

nnnn

n sn st x x +

−+

−+−±− =

(5.333 – 9.444) + 2.457181

181

21818)17)(026.2()17)(12(

+−+

+

-4.111 + 2.1689

-6.2799 < µ 1 - µ 2 < -1.9421

10.15 Peoria Evansville n1 = 21 n2 = 26

1 x = 116,900 2 x = 114,000 s1 = 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

22

212

121

112

)1()1()(

nnnn

n sn st x x +

−+

−+−±− =

(116,900 – 114,000) + 1.684261

211

22621)25()1750()20()2300( 22

+−+

+=

2,900 + 994.62

1905.38 < µ 1 - µ 2 < 3894.62




10.16 H o: µ1 - µ2 = 0 α = .10Ha: µ1 - µ2 ≠ 0 df = 12 + 12 - 2 = 22

Co-op Internsn1 = 12 n2 = 12 x 1 = $15.645 x 2 = $15.439

s1 = $1.093 s2 = $0.958

For two-tail test, α /2 = .05Critical t .05,22 = ± 1.717

t =

2121

22

212

1

2121

11

2

)1()1(

)()(

nnnn

n sn s

x x

+

−+

−+−

−−− µ µ

=

t =121

121

21212)11()958.0()11()093.1(

)0()439.15645.15(22

+−+

+

−−

= 0.49

Since the observed t = 0.49 < t .05,22 = 1.717, the decision is to fail reject the nullhypothesis .

90% Confidence Interval: t .05,22 = ± 1.717

2121

22

212

121

112

)1()1()(

nnnn

n sn st x x +

−+

−+−±− =

(15.645 – 15.439) + 1.717121

121

21212)11()958.0()11()093.1( 22

+−+

+=

0.206 + 0.7204

-0.5144 < µ 1 - µ 2 < 0.9264




10.17 Let Boston be group 1

1) H o: µ1 - µ2 = 0Ha: µ1 - µ2 > 0

2) t =

2121

22

212

1

2121

112

)1()1()()(

nnnnn sn s

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 observedvalue of 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 = 44 s1 = 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 nullhypothesis.

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

10.18 nm = 22 nno = 20 x m = 112 x no = 122

sm = 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

22

212

121

112

)1()1()(

nnnn

n sn st x x +

−+

−+−±− =

(112 – 122) + 2.423201

221

22022)19()12()21()11( 22

+−+

+




-10 ± 8.60

-$18.60 < µ1 - µ2 < -$1.40

Point Estimate = -$10

10.19 H o: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0

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

Toronto Mexico Cityn1 = 11 n2 = 11 x 1 = $67,381.82 x 2 = $63,481.82

s1 = $2,067.28 s2 = $1,594.25

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

t =

2121

22

212

1

2121

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

=

t =111

111

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 H o: µ1 - µ2 = 0Ha: µ1 - µ2 > 0

df = n1 + n2 - 2 = 9 + 10 - 2 = 17

Men Womenn1 = 9 n2 = 10 x 1 = $110.92 x 2 = $75.48 s1 = $28.79 s2 = $30.51

This is a one-tail test, α = .01 Critical t .01,17 = 2.567




t =

2121

22

212

1

2121

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

=

t =

( . . ) ( )

( . ) ( ) ( . ) ( )

1 1 09 2 7 54 8 0

2 87 9 8 3 05 1 99 1 0 2

19

11 0

2 2

− −

+

+ −+

= 2.60

Since the observed t = 2.60 > t .01,17 = 2.567, the decision is to Reject the nullhypothesis .

10.21 H o: 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 H o: D = 0Ha: D ≠ 0

Before After d 107 102 5

99 98 1110 100 10113 108 5

96 89 798 101 -3

100 99 1102 102 0107 105 2

109 110 -1104 102 299 96 3

101 100 1

n = 13 d = 2.5385 sd=3.4789 α = .05df = 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 = .025df = n - 1 = 9 - 1 = 8

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

t =n

st d d ±

t = -3 + (1.86) 9

6347.5 = -3 ± 3.49

-6.49 < D < 0.49

10.25 City Cost Resale d




Atlanta 20427 25163 -4736Boston 27255 24625 2630Des Moines 22115 12600 9515Kansas City 23256 24588 -1332

Louisville 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 H o: 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

716.10778.1 −−=−

n s

Dd d = -3.11

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

10.27 Before After d 255 197 58230 225 5290 215 75242 215 27300 240 60

250 235 15215 190 25230 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 H 0: D = 0Ha: 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

5071.3 −

=−

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 H o: D = 0Ha: 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 =159.1

085.2 −−=−

n

s Dd

d = -5.81


10.31 a) Sample 1 Sample 2




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

368175

ˆ1

11

==

n x

p = .476405182

ˆ2

22

==

n x

p = .449

773357

405368182175

21

21 =+

+=

+

+=

nn x x

p = .462

Ho: p1 - p2 = 0Ha: p1 - p2 ≠ 0

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

+

−−=

+⋅

−−−=

4051

3681)538)(.462(.

)0()449.476(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z

= 0.75

Since the observed z = 0.75 < z c = 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

+

−−=

+⋅

−−−=

5581

6491

)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(

nq p

nq 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(

nq p

nq 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

430275ˆ

1

11

==

n x

p = .64399275

ˆ2

22

==

n x

p = .69

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

2

22

1

1121

ˆˆˆˆ)ˆˆ(

nq p

nq 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

15001050

ˆ1

11

==

n x

p = .7015001100ˆ

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(

nq p

nq p

z p p +±−

(.70 - .733) ± 1.281500

)267)(.733(.1500

)30)(.70(.+ = -.033 ± .02

-.053 < p1 - p2 < -.013

10.33 H 0: pm - pw = 0Ha: 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

+

−−=

+⋅

−−−=

4811

3741

)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

1121

ˆˆˆˆ)ˆˆ(n

q pn

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 = .56 n1 = 56 n2 = 89

8956)56(.89)48(.56ˆˆ

21

2211

++=

++=

nn pn pn

p = .529

Ho: p1 - p2 = 0Ha: p1 - p2 ≠ 0

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

+

−−=

+⋅

−−−=

891

561

)471)(.529(.

)0()56.48(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z

= -0.94

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




10.36 A Bn1 = 35 n2 = 35

x1 = 5 x2 = 7

355ˆ

1

11

==

n x

p = .14357

ˆ2

22

==

n x

p = .20


2

22

1

1121

ˆˆˆˆ)ˆˆ(

nq p

nq p

z p p +±−

(.14 - .20) ± 2.3335

)80)(.20(.

35

)86)(.14(.+ = -.06 ± .21

-.27 < p1 - p2 < .15

10.37 H 0: p1 – p2 = 0Ha: p1 – p2 ≠ 0

α = .10 p̂ 1 = .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

+

−−=

+⋅

−−−=

9151

7801

)9262)(.0738(.

)0()06.09(.

11

)()ˆˆ(

1

2121

nnq p

p p p p Z

= 2.35





10.38 n1 = 850 n2 = 910 p̂ 1 = .60 p̂ 2 = .52

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

2

22

1

1121

ˆˆˆˆ)ˆˆ(

nq p

nq p

z p p +±−

(.60 - .52) + 1.96910

)48)(.52(.850

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

.034 < p1 – p2 < .126

10.39 H 0: σ 12 = σ 2

2 α = .01 n1 = 10 s12 = 562

Ha: σ 12 < σ 2

2 n2 = 12 s22 = 1013

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

Table F .01,10,9 = 5.26

F =562

10132

1

22 =

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 H 0: σ 12 = σ 2

2 α = .05 n1 = 5 s1 = 4.68Ha: σ 1

2 ≠ σ 22 n2 = 19 s2 = 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

22

21

)78.2(

)68.4(=

s

s= 2.83





10.41 City 1 City 2

3.43 3.333.40 3.423.39 3.393.32 3.303.39 3.463.38 3.393.34 3.363.38 3.443.38 3.373.28 3.38

n1 = 10 df 1 = 9 n2 = 10 df 2 = 9 s1

2 = .0018989 s22 = .0023378

H0: σ 12 = σ 2

2 α = .10 α /2 = .05Ha: σ 1

2 ≠ σ 22

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.

22

21 =

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) H 0: σ 12 = σ 2

2 Ha: σ 1

2 ≠ σ 22

2) F = 22

21

s s

3) α = .01

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

The critical table F values are: F .005,12,10 = 5.66 F .995,10,12 = .177

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 of number of days between Houston and Chicago.

10.43 H 0: σ 12 = σ 2

2 α = .05 n1 = 12 s1 = 7.52Ha: σ 1

2 > σ 22 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 = 2.60

F = 2

2

22

21

)08.6()52.7(

= s

s

= 1.53





10.44 H 0: 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 The critical table F values are: F .005,12,14 = 4.43 F .995,14,12 = .226

F =3.675.91

22

21 =

s

s= 1.36

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

10.45 H o: µ1 - µ2 = 0Ha: µ1 - µ2 ≠ 0

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

Sample 1 Sample 2 x 1 = 138.4 x 2 = 142.5σ 1 = 6.71 σ 2 = 8.92

n1 = 48 n2 = 39

z =39

)92.8(48

)71.6(

)0()5.1424.138()()(2

2

22

1

21

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.




10.46 Sample 1 Sample 2 x 1 = 34.9 x 2 = 27.6σ 1

2 = 2.97 σ 22 = 3.50

n1 = 34 n2 = 31

For 98% Confidence Level, z .01 = 2.33

2

22

1

21

21 )(n

s

n

s z x x +±−

(34.9 – 27.6) + 2.333150.3

3497.2

+ = 7.3 + 1.04

6.26 < µ 1 - µ 2 < 8.34

10.47 H o: µ1 - µ2 = 0Ha: µ1 - µ2 > 0

Sample 1 Sample 2 x 1= 2.06 x 2 = 1.93

s12 = .176 s2

2 = .143 n1 = 12 n2 = 15 α = .05

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

22

212

1

2121

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

t =151

121

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.




10.48 Sample 1 Sample 2 x 1 = 74.6 x 2 = 70.9 s1

2 = 10.5 s22 = 11.4

n1 = 18 n2 = 19

For 95% confidence, α /2 = .025.Using df = 18 + 19 - 2 = 35, t 30,.025 = 2.042

2121

22

212

121

112

)1()1()(

nnnn

n sn st x x +

−+

−+−±−

(74.6 – 70.9) + 2.042191

181

21918)18)(4.11()17)(5.10(

+−+

+

3.7 + 2.22

1.48 < µ 1 - µ 2 < 5.92

10.49 H o: 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 =2101.1

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 - 6

5 39 44 - 56 27 29 - 27 35 32 38 46 54 - 89 41 47 - 6

d = -4.44 sd = 5.703 df = 8For 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 H o: p1 - p2 = 0 α = .05 α /2 = .025Ha: 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

896783421345

21

21

+

+=

+

+=

nn x x

p = .4562

783345ˆ

1

11

==

n x

p = .4406896421

ˆ2

22

==

n x

p = .4699

+

−−=

+⋅

−−−=

8961

7831

)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 fail toreject the null hypothesis . There is no significant difference.




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


2

22

1

1121

ˆˆˆˆ)ˆˆ(

nq p

nq p

z p p +±−

(.71 - .67) + 2.575378

)33)(.67(.409

)29)(.71(.+ = .04 ± .085

-.045 < p1 - p2 < .125

10.53 H 0: σ 12 = σ 2

2 α = .05 n1 = 8 s12 = 46

Ha: σ 12 ≠ σ 2

2 n2 = 10 s22 = 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 =3746

22

21

=

s

s= 1.24

Since the observed F = 1.24 is less than F .025,7,9 =4.20 and greater than F .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 s t = $22,000 sw = $15,500 nt = 27 nw = 29

df = 27 + 29 - 2 = 54

For a 95% Confidence Level, α /2 = .025 and t .025,50 = 2.009 (used df=50)

2121

22

212

121

112

)1()1()(

nnnn

n sn st x x +

−+

−+−±−

(75,000 – 45,000) + 2.009291

271

22927)28()500,15()26()000,22( 22

+−+

+

30,000 ± 10,160.11

19,839.89 < µ1 - µ2 < 40,160.11

10.55 Morning Afternoon d 43 41 251 49 237 44 -7

24 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 Marketing Accountants n1 = 400 n2 = 450 x1 = 220 x2 = 216

Ho: p1 - p2 = 0

Ha: p1 - p2 > 0 α = .01

The critical table z value is: z .01 = 2.33

1ˆ p =

400220

= .55 2ˆ p =

450216

= .48

450400216220

21

21

+

+=

+

+=

nn x x

p = .513

+

−−=

+⋅

−−−=

4501

4001

)487)(.513(.

)0()48.55(.

11

)()ˆˆ(

1

2121

nnq p

p p p p z = 2.04

Since the observed z = 2.04 is less than z .01 = 2.33, the decision is to fail to rejectthe null hypothesis . There is no significant difference between marketingmanagers and accountants in the proportion who keep track of obligations “intheir head”.

10.57 Accounting Data Entryn1 = 16 n2 = 14

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

H0: σ 12 = σ 2

2

Ha: σ 12 ≠ σ 2

2 α = .05 and α /2 = .025

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 =500,102,1000,440,1

22

21 =

s

s= 1.31

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




10.58 Men Women

n1 = 60 n2 = 41 x 1 = 631 x 2 = 848σ 1 = 100 σ 2 = 100


2

22

1

21

21 )(n

s

n

s z x x +±−

(631 – 848) + 1.9641

100

60

100 22

+ = -217 ± 39.7

-256.7 < µ1 - µ2 < -177.3

10.59 H o: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 ≠ 0 df = 20 + 24 - 2 = 42

Detroit Charlotte n1 = 20 n2 = 24 x 1 = 17.53 x 2 = 14.89 s1 = 3.2 s2 = 2.7

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

t =

2121

22

212

1

2121

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

t =241

201

42)23()7.2()19()2.3(

)0()89.1453.17(22

++

−−

= 2.97


10.60 With Fertilizer Without Fertilizer




x 1 = 38.4 x 2 = 23.1 σ 1 = 9.8 σ 2 = 7.4 n1 = 35 n2 = 35

Ho: µ1 - µ2 = 0

Ha: 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()()(22

2

22

1

21

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ

= 7.37


10.61 Specialty Discountn1 = 350 n2 = 500 p̂ 1 = .75 p̂ 2 = .52


2

22

1

1121

ˆˆˆˆ)ˆˆ( n

q pn

q p z p p +±−

(.75 - .52) + 1.645500

)48)(.52(.350

)25)(.75(.+ = .23 ± .053

.177 < p1 - p2 < .283




10.62 H 0: σ 12 = σ 2

2 α = .01 n1 = 8 n2 = 7Ha: σ 1

2 ≠ σ 22 s1

2 = 72,909 s22 = 129,569

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

F =909,72569,129

22

21 =

s

s= 1.78

Since F = 1.78 < 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.

10.63 Name Brand Store Brand d 54 49 555 50 559 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




10.64 H o: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 < 0 df = 23 + 19 - 2 = 40

Wisconsin Tennesseen1 = 23 n2 = 19 x 1 = 69.652 x 2 = 71.7368 s1

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

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

t =191

231

40)18)(6491.4()22)(9644.9(

)0()7368.71652.69(

++

−−

= -2.44


10.65 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 = 0 α = .05Ha: D > 0

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

t =5

263.508.15 −

=−

n

s

Dd

d = 6.71





10.66 H o: p1 - p2 = 0 α = .05Ha: p1 - p2 ≠ 0

Machine 1 Machine 2 x1 = 38 x2 = 21

n1 = 191 n2 = 202

19138ˆ

1

11 ==

n x

p = .19920221

ˆ2

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

+

−−=

+⋅

−−−=

2021

1911

)85)(.15(.)0()104.199(.

11)()ˆˆ(

1

2121

nnq p

p p p p z = 2.64


10.67 Construction Telephone Repair

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

338297ˆ

1

11

==

n x

p = .879281192ˆ

2

22

==

n x

p = .683


2

22

1

1121

ˆˆˆˆ)ˆˆ(

nq p

nq p

z p p +±−

(.879 - .683) + 1.645281

)317)(.683(.338

)121)(.879(.+ = .196 ± .054

.142 < p1 - p2 < .250




10.68 Aerospace Automobile n1 = 33 n2 = 35 x 1 = 12.4 x 2 = 4.6 σ 1 = 2.9 σ 2 = 1.8


2

22

1

21

21 )(nn

z x xσ σ

+±−

(12.4 – 4.6) + 2.57535

)8.1(33

)9.2( 22

+ = 7.8 ± 1.52

6.28 < µ1 - µ2 < 9.32

10.69 Discount Specialty x 1 = $47.20 x 2 = $27.40

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

Ho: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 ≠ 0

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

z =40

)82.9(60

)45.12(

)0()40.2720.47()()(22

2

22

1

21

2121

+

−−=

+

−−−

nn

x x

σ σ

µ µ

= 8.86





10.70 Before After d 12 8 4

7 3 410 8 216 9 7

8 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

8708.100.4 −

=−

n

s

Dd

d = 4.78


10.71 H o: µ1 - µ2 = 0 α = .01Ha: µ1 - µ2 ≠ 0 df = 10 + 6 - 2 = 14

A B___ n1 = 10 n2 = 6

x 1 = 18.3 x 2 = 9.667 s1

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

112

)1()1(

)()(

nnnnn sn s

x x

+−+

−+−

−−− µ µ

t =61

101

14)5)(467.7()9)(122.17(

)0()667.93.18(

++

−−

= 4.52





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

Ho: µ1 - µ2 = 0

Ha: µ1 - µ2 ≠ 0n1 = 19 n2 = 23 x 1 = 130.4 x 2 = 128.4 s1 = 12.9 s2 = 13.9 98% C.I. and α /2 = .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.

10.73 H 0: D = 0Ha: 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 thetreatment, 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 and as a result, the researcher rejects the nullhypothesis. This is underscored by a p-value of .0081 which is less than α = .05.The study concludes that there is a significantly higher likelihood of “blowing the

whistle” after the treatment.

10.74 The point estimates from the sample data indicate that in the northern city themarket share is .31078 and in the southern city the market share is .27013. The

point estimate for the difference in the two proportions of market share are .04065. Since the 99% confidence interval ranges from -.03936 to +.12067 andzero is in the interval, any hypothesis testing decision based on this interval wouldresult in failure to reject the null hypothesis. Alpha is .01 with a two-tailed test.

This is underscored by an observed 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 .




10.75 A test of differences of the variances of the populations of the two machines is being 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-eight pipes weremeasured for sample two. The observed F = 2.0575 is significant at α = .05 for aone-tailed test since the associated p-value is .034787. The variance of pipelengths for machine 1 is significantly greater than the variance of pipe lengths for machine 2.

Ken Black QA 5th chapter 10 Solution

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Transcript of Ken Black QA 5th chapter 10 Solution