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- 1. Chapter 10: Statistical Inferences About Two Populations 1 Chapter 10 Statistical Inferences about Two Populations LEARNING OBJECTIVES The general focus of Chapter 10 is on testing hypotheses and constructing confidence intervals 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 when the differences are normally distributed. 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 when the two populations are normally distributed. CHAPTER TEACHING STRATEGY The major emphasis of chapter 10 is on analyzing data from two samples. The student should be ready to deal with this topic given that he/she has tested hypotheses and computed 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 shown initially in Chapter 7. In analyzing the differences in two sample means where the population variances are unknown, if it can be assumed that the populations are normally distributed, a t test for independent samples can be used. There are two different
- 2. Chapter 10: Statistical Inferences About Two Populations 2 formulas given in the chapter to conduct this t test. One version uses a "pooled" estimate of the population variance and assumes that the population variances are equal. The other 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 that the student be able to recognize when two samples are related and when they are independent. 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 to analyze some related measures data with both techniques and demonstrate that the results and conclusions are usually quite different. You can have your students work problems like this using both techniques to help them understand the differences between the two tests (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 begin to understand that the F values have two different degrees of freedom. The F distribution tables are upper tailed only. For this reason, formula 10.14 is given in the chapter to be used 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 Testing Confidence Intervals Using 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 Testing Using 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 Populations Hypothesis Testing Using the Computer to Make Statistical Inferences about Two Related Populations Confidence Intervals
- 3. Chapter 10: Statistical Inferences About Two Populations 3 10.4 Statistical Inferences About Two Population Proportions, p1 – p2 Hypothesis Testing Confidence Intervals Using the Computer to Analyze the Difference in Two Proportions 10.5 Testing Hypotheses About Two Population Variances Using 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 σ1 2 = 52 σ2 2 = 60 n1 = 32 n2 = 32 a) Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 < 0 For one-tail test, α = .10 z.10 = -1.28 z = 32 60 32 52 )0()2.533.51()()( 2 2 2 1 2 1 2121 + −− = + −−− nn xx σσ µµ = -1.02 Since the observed z = -1.02 > zc = -1.645, the decision is to fail to reject the null hypothesis.
- 4. Chapter 10: Statistical Inferences About Two Populations 4 b) Critical value method: zc = 2 2 2 1 2 1 2121 )()( nn xx c σσ µµ + −−− -1.645 = 32 60 32 52 )0()( 21 + −− cxx ( x 1 - x 2)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 σ1 = 5.76 σ2 = 6.1 For a 90% C.I., z.05 = 1.645 2 2 2 1 2 1 21 )( nn zxx σσ +±− (70.4) – 68.7) + 1.645 31 1.6 32 76.5 22 + 1.7 ± 2.465 -.76 < µ1 - µ2 < 4.16
- 5. Chapter 10: Statistical Inferences About Two Populations 5 10.3 a) Sample 1 Sample 2 x 1 = 88.23 x 2 = 81.2 σ1 2 = 22.74 σ2 2 = 26.65 n1 = 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 2 2 1 2 1 2121 + −− = + −−− nn xx σσ µµ = 5.48 Since the observed z = 5.48 > z.01 = 2.33, the decision is to reject the null hypothesis. b) 2 2 2 1 2 1 21 )( nn zxx σσ +±− (88.23 – 81.2) + 2.33 30 65.26 30 74.22 + 7.03 + 2.99 4.04 < µµµµ < 10.02 This supports the decision made in a) to reject the null hypothesis because zero is not in the interval. 10.4 Computers/electronics Food/Beverage x 1 = 1.96 x 2 = 3.02 σ1 2 = 1.0188 σ2 2 = 0.9180 n1 = 50 n2 = 50 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 ≠ 0 For two-tail test, α/2 = .005 z.005 = ±2.575
- 6. Chapter 10: Statistical Inferences About Two Populations 6 z = 50 9180.0 50 0188.1 )0()02.396.1()()( 2 2 2 1 2 1 2121 + −− = + −−− nn xx σσ µµ = -5.39 Since the observed z = -5.39 < zc = -2.575, the decision is to reject the null hypothesis. 10.5 A B n1 = 40 n2 = 37 x 1 = 5.3 x 2 = 6.5 σ1 2 = 1.99 σ2 2 = 2.36 For a 95% C.I., z.025 = 1.96 2 2 2 1 2 1 21 )( nn zxx σσ +±− (5.3 – 6.5) + 1.96 37 36.2 40 99.1 + -1.2 ± .66 -1.86 < µµµµ < -.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 lie in this interval, we are confident that there is a difference between Plumber A and Plumber B. 10.6 Managers Specialty n1 = 35 n2 = 41 x 1 = 1.84 x 2 = 1.99 σ1 = .38 σ2 = .51 for a 98% C.I., z.01 = 2.33 2 2 2 1 2 1 21 )( nn zxx σσ +±−
- 7. Chapter 10: Statistical Inferences About Two Populations 7 (1.84 - 1.99) ± 2.33 41 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 xx σσ µµ + −−− 3) α = .02 4) For a two-tailed test, z.01 = + 2.33. If the observed z value is greater than 2.33 or 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 null hypothesis. 8) There is no significant difference in the hourly rates of the two groups.
- 8. Chapter 10: Statistical Inferences About Two Populations 8 10.7 1994 2001 x 1 = 190 x 2 = 198 σ1 = 18.50 σ2 = 15.60 n1 = 51 n2 = 47 α = .01 H0: µ1 - µ2 = 0 Ha: µ1 - µ2 < 0 For 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 xx σσ µµ = -2.32 Since the observed z = -2.32 > z.01 = -2.33, the decision is to fail to reject the null hypothesis. 10.8 Seattle Atlanta n1 = 31 n2 = 31 x 1 = 2.64 x 2 = 2.36 σ1 2 = .03 σ2 2 = .015 For a 99% C.I., z.005 = 2.575 2 2 2 1 2 1 21 )( nn zxx σσ +±− (2.64-2.36) ± 2.575 31 015. 31 03. + .28 ± .10 .18 < µµµµ < .38 Between $ .18 and $ .38 difference with Seattle being more expensive.
- 9. Chapter 10: Statistical Inferences About Two Populations 9 10.9 Canon Pioneer x 1 = 5.8 x 2 = 5.0 σ1 = 1.7 σ2 = 1.4 n1 = 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 2 2 1 2 1 2121 + −− = + −−− nn xx σσ µµ = 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 σ1 = 1.36 σ2 = 1.06 n1 = 50 n2 = 38 Ho: µ1 - µ2 = 0 Ha: 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 xx σσ µµ = 3.06 Since the observed z = 3.06 > zc = 1.28, the decision is to reject the null hypothesis.
- 10. Chapter 10: Statistical Inferences About Two Populations 10 10.11 Ho: µ1 - µ2 = 0 α = .01 Ha: µ1 - µ2 < 0 df = 8 + 11 - 2 = 17 Sample 1 Sample 2 n1 = 8 n2 = 11 x 1 = 24.56 x 2 = 26.42 s1 2 = 12.4 s2 2 = 15.8 For one-tail test, α = .01 Critical t.01,19 = -2.567 t = 2121 2 2 21 2 1 2121 11 2 )1()1( )()( nnnn nsns xx + −+ −+− −−− µµ = 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 the null hypothesis. 10.12 a) Ho: µ1 - µ2 = 0 α =.10 Ha: 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 2 2 21 2 1 2121 11 2 )1()1( )()( nnnn nsns xx + −+ −+− −−− µµ = 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 reject the null hypothesis.
- 11. Chapter 10: Statistical Inferences About Two Populations 11 b) 2121 2 2 21 2 1 21 11 2 )1()1( )( nnnn nsns txx + −+ −+− ±− = (118 – 113) + 1.697 20 1 20 1 22020 )19()6.21()19()9.23( 22 + −+ + 5 + 12.224 -7.224 < µµµµ1 - µµµµ2 < 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 = 40.49 s1 = 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 xx + −+ −+− −−− µµ = 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 the null 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 = 9.444 s1 2 = 12 s2 2 = 2.026
- 12. Chapter 10: Statistical Inferences About Two Populations 12 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 xx + −+ −+− −−− µµ = 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 1 21 11 2 )1()1( )( nnnn nsns txx + −+ −+− ±− = (5.333 – 9.444) + 2.457 18 1 18 1 21818 )17)(026.2()17)(12( + −+ + -4.111 + 2.1689 -6.2799 < µµµµ1 - µµµµ2 < -1.9421 10.15 Peoria Evansville n1 = 21 n2 = 26 1x = 86,900 2x = 84,000 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 2 2 21 2 1 21 11 2 )1()1( )( nnnn nsns txx + −+ −+− ±− = (86,900 – 84,000) + 1.684 26 1 21 1 22621 )25()1750()20()2300( 22 + −+ + = 2,900 + 994.62 1905.38 < µµµµ1 - µµµµ2 < 3894.62
- 13. Chapter 10: Statistical Inferences About Two Populations 13 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 = $84,000 s1 = $2,300 s2 = $1,750 For two-tail test, α/2 = .025 Critical t.025,45 = ± 2.021 (used df=40) t = 2121 2 2 21 2 1 2121 11 2 )1()1( )()( nnnn nsns xx + −+ −+− −−− µµ = 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 = 0 Ha: µ1 - µ2 > 0 2) t = 2121 2 2 21 2 1 2121 11 2 )1()1( )()( nnnn nsns xx + −+ −+− −−− µµ 3) α = .01 4) For a one-tailed test and df = 8 + 9 - 2 = 15, t.01,15 = 2.602. If the observed value 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
- 14. Chapter 10: Statistical Inferences About Two Populations 14 6) t = 9 1 8 1 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 = 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 2 2 21 2 1 21 11 2 )1()1( )( nnnn nsns txx + −+ −+− ±− = (112 – 122) + 2.423 20 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 = $63,481.82 s1 = $2,067.28 s2 = $1,594.25 For a two-tail test, α/2 = .005 Critical t.005,20 = ±2.845
- 15. Chapter 10: Statistical Inferences About Two Populations 15 t = 2121 2 2 21 2 1 2121 11 2 )1()1( )()( nnnn nsns xx + −+ −+− −−− µµ = 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 null hypothesis. 10.20 Toronto Mexico City n1 = 11 n2 = 11 x 1 = $67,381.82 x 2 = $63,481.82 s1 = $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 1 21 11 2 )1()1( )( nnnn nsns txx + −+ −+− ±− = ($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
- 16. Chapter 10: Statistical Inferences About Two Populations 16 10.21 Ho: D = 0 Ha: D > 0 Sample 1 Sample 2 d 38 22 16 27 28 -1 30 21 9 41 38 3 36 38 -2 38 26 12 33 19 14 35 31 4 44 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 null hypothesis. 10.22 Ho: D = 0 Ha: D ≠ 0 Before After d 107 102 5 99 98 1 110 100 10 113 108 5 96 89 7 98 101 -3 100 99 1 102 102 0 107 105 2 109 110 -1 104 102 2 99 96 3 101 100 1
- 17. Chapter 10: Statistical Inferences About Two Populations 17 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 Since the observed t = 2.63 > t.025,12 = 2.179, the decision is to reject the null hypothesis. 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 s td d ± 40.56 ± (2.518) 22 58.26 40.56 ± 14.27 26.29 < D < 54.83 10.24 Before After d 32 40 -8 28 25 3 35 36 -1 32 32 0 26 29 -3 25 31 -6 37 39 -2 16 30 -14 35 31 4
- 18. Chapter 10: Statistical Inferences About Two Populations 18 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 s td 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 -4736 Boston 27255 24625 2630 Des Moines 22115 12600 9515 Kansas City 23256 24588 -1332 Louisville 21887 19267 2620 Portland 24255 20150 4105 Raleigh-Durham 19852 22500 -2648 Reno 23624 16667 6957 Ridgewood 25885 26875 - 990 San Francisco 28999 35333 -6334 Tulsa 20836 16292 4544 d = 1302.82 sd = 4938.22 n = 11, df = 10 α = .01 α/2 = .005 t.005,10= 3.169 n s td d ± = 1302.82 + 3.169 11 22.4938 = 1302.82 + 4718.42 -3415.6 < D < 6021.2
- 19. Chapter 10: Statistical Inferences About Two Populations 19 10.26 Ho: D = 0 Ha: D < 0 Before After d 2 4 -2 4 5 -1 1 3 -2 3 3 0 4 3 1 2 5 -3 2 6 -4 3 4 -1 1 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 58 230 225 5 290 215 75 242 215 27 300 240 60 250 235 15 215 190 25 230 240 -10 225 200 25 219 203 16 236 223 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
- 20. Chapter 10: Statistical Inferences About Two Populations 20 n s td 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 Since the observed t = 3.86 > t.01,26 = 2.479, the decision is to reject the null hypothesis. 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 s td d ± 75 + 1.725 21 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
- 21. Chapter 10: Statistical Inferences About Two Populations 21 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 null hypothesis. 10.31 a) Sample 1 Sample 2 n1 = 368 n2 = 405 x1 = 175 x2 = 182 368 175 ˆ 1 1 1 == n x p = .476 405 182 ˆ 2 2 2 == n x p = .449 773 357 405368 182175 21 21 = + + = + + = nn xx 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 nn qp pppp z = 0.75 Since the observed z = 0.75 < zc = 1.96, the decision is to fail to reject the null hypothesis. b) Sample 1 Sample 2 pˆ 1 = .38 pˆ 2 = .25 n1 = 649 n2 = 558 558649 )25(.558)38(.649ˆˆ 21 2211 + + = + + = nn pnpn p = .32
- 22. Chapter 10: Statistical Inferences About Two Populations 22 Ho: p1 - p2 = 0 Ha: 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 nn qp pppp z = 4.83 Since the observed z = 4.83 > zc = 1.28, the decision is to reject the null hypothesis. 10.32 a) n1 = 85 n2 = 90 pˆ 1 = .75 pˆ 2 = .67 For a 90% Confidence Level, z.05 = 1.645 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.75 - .67) ± 1.645 90 )33)(.67(. 85 )25)(.75(. + = .08 ± .11 -.03 < p1 - p2 < .19 b) n1 = 1100 n2 = 1300 pˆ 1 = .19 pˆ 2 = .17 For a 95% Confidence Level, α/2 = .025 and z.025 = 1.96 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.19 - .17) + 1.96 1300 )83)(.17(. 1100 )81)(.19(. + = .02 ± .03 -.01 < p1 - p2 < .05
- 23. Chapter 10: Statistical Inferences About Two Populations 23 c) n1 = 430 n2 = 399 x1 = 275 x2 = 275 430 275 ˆ 1 1 1 == n x p = .64 399 275 ˆ 2 2 2 == n x p = .69 For an 85% Confidence Level, α/2 = .075 and z.075 = 1.44 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.64 - .69) + 1.44 399 )31)(.69(. 430 )36)(.64(. + = -.05 ± .047 -.097 < p1 - p2 < -.003 d) n1 = 1500 n2 = 1500 x1 = 1050 x2 = 1100 1500 1050 ˆ 1 1 1 == n x p = .70 1500 1100 ˆ 2 2 2 == n x p = .733 For an 80% Confidence Level, α/2 = .10 and z.10 = 1.28 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.70 - .733) ± 1.28 1500 )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 pˆ m = .59 pˆ w = .70 For a one-tailed test and α = .05, z.05 = -1.645 481374 )70(.481)59(.374ˆˆ + + = + + = wm wwmm nn pnpn p = .652
- 24. Chapter 10: Statistical Inferences About Two Populations 24 + −− = +⋅ −−− = 481 1 374 1 )348)(.652(. )0()70.59(. 11 )()ˆˆ( 1 2121 nn qp pppp z = -3.35 Since the observed z = -3.35 < z.05 = -1.645, the decision is to reject the null hypothesis. 10.34 n1 = 210 n2 = 176 1 ˆp = .24 2 ˆp = .35 For a 90% Confidence Level, α/2 = .05 and z.05 = + 1.645 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.24 - .35) + 1.645 176 )65)(.35(. 210 )76)(.24(. + = -.11 + .0765 -.1865 < p1 – p2 < -.0335 10.35 Computer Firms Banks pˆ 1 = .48 pˆ 2 = .56 n1 = 56 n2 = 89 8956 )56(.89)48(.56ˆˆ 21 2211 + + = + + = nn pnpn 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 nn qp pppp z = -0.94 Since the observed z = -0.94 > zc = -1.28, the decision is to fail to reject the null hypothesis.
- 25. Chapter 10: Statistical Inferences About Two Populations 25 10.36 A B n1 = 35 n2 = 35 x1 = 5 x2 = 7 35 5 ˆ 1 1 1 == n x p = .14 35 7 ˆ 2 2 2 == n x p = .20 For a 98% Confidence Level, α/2 = .01 and z.01 = 2.33 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.14 - .20) ± 2.33 35 )80)(.20(. 35 )86)(.14(. + = -.06 ± .21 -.27 < p1 - p2 < .15 10.37 H0: p1 – p2 = 0 Ha: p1 – p2 ≠ 0 α = .10 pˆ 1 = .09 pˆ 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 pnpn p = .0738 + −− = +⋅ −−− = 915 1 780 1 )9262)(.0738(. )0()06.09(. 11 )()ˆˆ( 1 2121 nn qp pppp Z = 2.35 Since the observed z = 2.35 > z.05 = 1.645, the decision is to reject the null hypothesis.
- 26. Chapter 10: Statistical Inferences About Two Populations 26 10.38 n1 = 850 n2 = 910 pˆ 1 = .60 pˆ 2 = .52 For a 95% Confidence Level, α/2 = .025 and z.025 = + 1.96 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.60 - .52) + 1.96 910 )48)(.52(. 850 )40)(.60(. + = .08 + .046 .034 < p1 – p2 < .126 10.39 H0: σ1 2 = σ2 2 α = .01 n1 = 10 s1 2 = 562 Ha: σ1 2 < σ2 2 n2 = 12 s2 2 = 1013 dfnum = 12 - 1 = 11 dfdenom = 10 - 1 = 9 Table F.01,10,9 = 5.26 F = 562 1013 2 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 the null hypothesis. 10.40 H0: σ1 2 = σ2 2 α = .05 n1 = 5 S1 = 4.68 Ha: σ1 2 ≠ σ2 2 n2 = 19 S2 = 2.78 dfnum = 5 - 1 = 4 dfdenom = 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 Since the observed F = 2.83 < F.025,4,18 = 3.61, the decision is to fail to reject the null hypothesis.
- 27. Chapter 10: Statistical Inferences About Two Populations 27 10.41 City 1 City 2 1.18 1.08 1.15 1.17 1.14 1.14 1.07 1.05 1.14 1.21 1.13 1.14 1.09 1.11 1.13 1.19 1.13 1.12 1.03 1.13 n1 = 10 df1 = 9 n2 = 10 df2 = 9 s1 2 = .0018989 s2 2 = .0023378 H0: σ1 2 = σ2 2 α = .10 α/2 = .05 Ha: σ1 2 ≠ σ2 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 0.314 and less than the upper tail critical value of 3.18, the decision is to fail to reject the null hypothesis. 10.42 Let Houston = group 1 and Chicago = group 2 1) H0: σ1 2 = σ2 2 Ha: σ1 2 ≠ σ2 2 2) F = 2 2 2 1 s s 3) α = .01 4) df1 = 12 df2 = 10 This is a two-tailed test The critical table F values are: F.005,12,10 = 5.66 F.995,10,12 = .177
- 28. Chapter 10: Statistical Inferences About Two Populations 28 If the observed value is greater than 5.66 or less than .177, the decision will be to reject the null hypothesis. 5) s1 2 = 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 H0: σ1 2 = σ2 2 α = .05 n1 = 12 s1 = 7.52 Ha: σ1 2 > σ2 2 n2 = 15 s2 = 6.08 dfnum = 12 - 1 = 11 dfdenom = 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 Since the observed F = 1.53 < F.05,10,14 = 2.60, the decision is to fail to reject the null hypothesis. 10.44 H0: σ1 2 = σ2 2 α = .01 n1 = 15 s1 2 = 91.5 Ha: σ1 2 ≠ σ2 2 n2 = 15 s2 2 = 67.3 dfnum = 15 - 1 = 14 dfdenom = 15 - 1 = 14 The critical table F values are: F.005,12,14 = 4.43 F.995,14,12 = .226 F = 3.67 5.91 2 2 2 1 = s s = 1.36 Since the observed F = 1.36 < F.005,12,14 = 4.43 and > F.995,14,12 = .226, the decision is to fail to reject the null hypothesis.
- 29. Chapter 10: Statistical Inferences About Two Populations 29 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 1x = 138.4 2x = 142.5 σ1 = 6.71 σ2 = 8.92 n1 = 48 n2 = 39 z = 39 )92.8( 48 )71.6( )0()5.1424.138()()( 2 2 2 2 1 2 1 2121 + −− = + −−− nn xx σσ µµ = -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 in the means of the two populations. 10.46 Sample 1 Sample 2 1x = 34.9 2x = 27.6 σ1 2 = 2.97 σ2 2 = 3.50 n1 = 34 n2 = 31 For 98% Confidence Level, z.01 = 2.33 2 2 2 1 2 1 21 )( nn zxx σσ +±− (34.9 – 27.6) + 2.33 31 50.3 34 97.2 + = 7.3 + 1.04 6.26 < µµµµ1 - µµµµ2 < 8.34
- 30. Chapter 10: Statistical Inferences About Two Populations 30 10.47 Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 >0 Sample 1 Sample 2 1x = 2.06 2x = 1.93 s1 2 = .176 s2 2 = .143 n1 = 12 n2 = 15 This is a one-tailed test with df = 12 + 15 - 2 = 25. The critical value is t.05,25 = 1.708. If the observed value is greater than 1.708, the decision will be to reject the null hypothesis. t = 2121 2 2 21 2 1 2121 11 2 )1()1( )()( nnnn nsns xx + −+ −+− −−− µµ 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, the decision is to fail to reject the null hypothesis. The mean for population one is not 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 s2 2 = 11.4 n1 = 18 n2 = 19 For 95% confidence, α/2 = .025. Using df = 18 + 19 - 2 = 35, t35,.025 = 2.042 2121 2 2 21 2 1 21 11 2 )1()1( )( nnnn nsns txx + −+ −+− ±− (74.6 – 70.9) + 2.042 20 1 20 1 22020 )19()6.21()19()9.23( 22 + −+ + 3.7 + 2.22 1.48 < µµµµ1 - µµµµ2 < 5.92
- 31. Chapter 10: Statistical Inferences About Two Populations 31 10.49 Ho: D = 0 α = .01 Ha: 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 decision will 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, the decision is to reject the null hypothesis. The population difference is less than zero. 10.50 Respondent Before After d 1 47 63 -16 2 33 35 - 2 3 38 36 2 4 50 56 - 6 5 39 44 - 5 6 27 29 - 2 7 35 32 3 8 46 54 - 8 9 41 47 - 6 d = -4.44 sd = 5.703 df = 8 For a 99% Confidence Level, α/2 = .005 and t8,.005 = 3.355 n s td 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 decision will be to reject the null hypothesis.
- 32. Chapter 10: Statistical Inferences About Two Populations 32 Sample 1 Sample 2 x1 = 345 x2 = 421 n1 = 783 n2 = 896 896783 421345 21 21 + + = + + = nn xx p = .4562 783 345 ˆ 1 1 1 == n x p = .4406 896 421 ˆ 2 2 2 == n x p = .4699 + −− = +⋅ −−− = 896 1 783 1 )5438)(.4562(. )0()4699.4406(. 11 )()ˆˆ( 1 2121 nn qp pppp z = -1.20 Since the observed value of z = -1.20 is greater than -1.96, the decision is to fail to reject the null hypothesis. There is no significant difference in the population proportions. 10.52 Sample 1 Sample 2 n1 = 409 n2 = 378 pˆ 1 = .71 pˆ 2 = .67 For a 99% Confidence Level, α/2 = .005 and z.005 = 2.575 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.71 - .67) + 2.575 378 )33)(.67(. 409 )29)(.71(. + = .04 ± .085 -.045 < p1 - p2 < .125 10.53 H0: σ1 2 = σ2 2 α = .05 n1 = 8 s1 2 = 46 Ha: σ1 2 ≠ σ2 2 n2 = 10 S2 2 = 37 dfnum = 8 - 1 = 7 dfdenom = 10 - 1 = 9 The critical F values are: F.025,7,9 = 4.20 F.975,9,7 = .238
- 33. Chapter 10: Statistical Inferences About Two Populations 33 If the observed value of F is greater than 4.20 or less than .238, then the decision will be to reject the null hypothesis. F = 37 46 2 2 2 1 = 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 no significant 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 = $15,500 nt = 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 1 21 11 2 )1()1( )( nnnn nsns txx + −+ −+− ±− (75,000 – 45,000) + 2.021 29 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 2 51 49 2 37 44 -7 24 32 -8 47 46 1 44 42 2 50 47 3 55 51 4 46 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
- 34. Chapter 10: Statistical Inferences About Two Populations 34 n s td 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 1 ˆp = .447 2 ˆp = .487 14501300 )1450)(487(.)1300)(447(.ˆˆ 21 2211 + + = + + = nn pnpn p = .468 + −− = +⋅ −−− = 1450 1 1300 1 )532)(.468(. )0()487.447(. 11 )()ˆˆ( 1 2121 nn qp pppp z = -2.10 Since the observed z = -3.73 is less than z.05 = -1.645, the decision is to reject the null hypothesis. 1997 has a significantly higher proportion. 10.57 Accounting Data Entry n1 = 16 n2 = 14 x 1 = 26,400 x 2 = 25,800 s1 = 1,200 s2 = 1,050 H0: σ1 2 = σ2 2 Ha: σ1 2 ≠ σ2 2 dfnum = 16 – 1 = 15 dfdenom = 14 – 1 = 13 The critical F values are: F.025,15,13 = 3.05 F.975,15,13 = 0.33
- 35. Chapter 10: Statistical Inferences About Two Populations 35 F = 500,102,1 000,440,1 2 2 2 1 = 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 H0: σ1 2 = σ2 2 α = .01 n1 = 8 n2 = 7 Ha: σ1 2 ≠ σ2 2 S1 2 = 72,909 S2 2 = 129,569 dfnum = 6 dfdenom = 7 The critical F values are: F.005,6,7 = 9.16 F.995,7,6 = .11 F = 909,72 569,129 2 2 2 1 = 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 to reject 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 σ1 = 100 σ2 = 100 For a 95% Confidence Level, α/2 = .025 and z.025 = 1.96 2 2 2 1 2 1 21 )( nn zxx σσ +±− (631 – 848) + 1.96 41 100 60 100 22 + = -217 ± 39.7 -256.7 < µ1 - µ2 < -177.3
- 36. Chapter 10: Statistical Inferences About Two Populations 36 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 = 14.89 s1 = 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 xx + −+ −+− −−− µµ t = 24 1 20 1 42 )23()7.2()19()2.3( )0()89.1453.17( 22 + + −− = 2.97 Since the observed t = 2.97 > t.005,42 = 2.704, the decision is to reject the null hypothesis. 10.61 With Fertilizer Without Fertilizer x 1 = 38.4 x 2 = 23.1 σ1 = 9.8 σ2 = 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()()( 2 2 2 2 1 2 1 2121 + −− = + −−− nn xx σσ µµ = 7.37 Since the observed z = 7.37 > z.01 = 2.33, the decision is to reject the null hypothesis.
- 37. Chapter 10: Statistical Inferences About Two Populations 37 10.62 Specialty Discount n1 = 350 n2 = 500 pˆ 1 = .75 pˆ 2 = .52 For a 90% Confidence Level, α/2 = .05 and z.05 = 1.645 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.75 - .52) + 1.645 500 )48)(.52(. 350 )25)(.75(. + = .23 ± .053 .177 < p1 - p2 < .283 10.63 H0: σ1 2 = σ2 2 α = .05 n1 = 27 s1 = 22,000 Ha: σ1 2 ≠ σ2 2 n2 = 29 s2 = 15,500 dfnum = 27 - 1 = 26 dfdenom = 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, the decision is to fail to reject the null hypothesis.
- 38. Chapter 10: Statistical Inferences About Two Populations 38 10.64 Name Brand Store Brand d 54 49 5 55 50 5 59 52 7 53 51 2 54 50 4 61 56 5 51 47 4 53 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 s td d ± 4.5 + 1.895 8 414.1 = 4.5 ± .947 3.553 < D < 5.447 10.65 Ho: µ1 - µ2 = 0 α = .01 Ha: µ1 - µ2 < 0 df = 23 + 19 - 2 = 40 Wisconsin Tennessee n1 = 23 n2 = 19 x 1 = 69.652 x 2 = 71.7368 s1 2 = 9.9644 s2 2 = 4.6491 For one-tail test, α = .01 and the critical t.01,40 = -2.423 t = 2121 2 2 21 2 1 2121 11 2 )1()1( )()( nnnn nsns xx + −+ −+− −−− µµ t = 19 1 23 1 40 )18)(6491.4()22)(9644.9( )0()7368.71652.69( + + −− = -2.44
- 39. Chapter 10: Statistical Inferences About Two Populations 39 Since the observed t = -2.44 < t.01,40 = -2.423, the decision is to reject the null hypothesis. 10.66 Wednesday Friday d 71 53 18 56 47 9 75 52 23 68 55 13 74 58 16 n = 5 d = 15.8 sd = 5.263 df = 5 - 1 = 4 Ho: D = 0 α = .05 Ha: 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 Since the observed t = 6.71 > t.05,4 = 2.132, the decision is to reject the null hypothesis. 10.67 Ho: P1 - P2 = 0 α = .05 Ha: P1 - P2 ≠ 0 Machine 1 Machine 2 x1 = 38 x2 = 21 n1 = 191 n2 = 202 191 38 ˆ 1 1 1 == n x p = .199 202 21 ˆ 2 2 2 == n x p = .104 202191 )202)(104(.)191)(199(.ˆˆ 21 2211 + + = + + = nn pnpn p = .15 For two-tail, α/2 = .025 and the critical z values are: z.025 = ±1.96
- 40. Chapter 10: Statistical Inferences About Two Populations 40 + −− = +⋅ −−− = 202 1 191 1 )85)(.15(. )0()104.199(. 11 )()ˆˆ( 1 2121 nn qp pppp z = 2.64 Since the observed z = 2.64 > zc = 1.96, the decision is to reject the null hypothesis. 10.68 Construction Telephone Repair n1 = 338 n2 = 281 x1 = 297 x2 = 192 338 297 ˆ 1 1 1 == n x p = .879 281 192 ˆ 2 2 2 == n x p = .683 For a 90% Confidence Level, α/2 = .05 and z.05 = 1.645 2 22 1 11 21 ˆˆˆˆ )ˆˆ( n qp n qp zpp +±− (.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 σ1 = 2.9 σ2 = 1.8 For a 99% Confidence Level, α/2 = .005 and z.005 = 2.575 2 2 2 1 2 1 21 )( nn zxx σσ +±− (12.4 – 4.6) + 2.575 35 )8.1( 33 )9.2( 22 + = 7.8 ± 1.52 6.28 < µ1 - µ2 < 9.32
- 41. Chapter 10: Statistical Inferences About Two Populations 41 10.70 Discount Specialty x 1 = $47.20 x 2 = $27.40 σ1 = $12.45 σ2 = $9.82 n1 = 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 xx σσ µµ = 8.86 Since the observed z = 8.86 > zc = 2.575, the decision is to reject the null hypothesis. 10.71 Before After d 12 8 4 7 3 4 10 8 2 16 9 7 8 5 3 n = 5 d = 4.0 sd = 1.8708 df = 5 - 1 = 4 Ho: D = 0 α = .01 Ha: 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 Since the observed t = 4.78 > t.01,4 = 3.747, the decision is to reject the null hypothesis.
- 42. Chapter 10: Statistical Inferences About Two Populations 42 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 = 9.667 s1 2 = 17.122 s2 2 = 7.467 For two-tail test, α/2 = .005 and the critical t.005,14 = ±2.977 t = 2121 2 2 21 2 1 2121 11 2 )1()1( )()( nnnn nsns xx + −+ −+− −−− µµ t = 6 1 10 1 14 )5)(467.7()9)(122.17( )0()667.93.18( + + −− = 4.52 Since the observed t = 4.52 > t.005,14 = 2.977, the decision is to reject the null hypothesis. 10.73 A t test was used to test to determine if Hong Kong has significantly different rates than Bombay. Let group 1 be Hong Kong. Ho: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0 n1 = 19 n2 = 23 x1 = 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 not enough evidence in these data to declare that there is a difference in the average rental 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 involved in the study. Before the treatment, the sample mean was 4.357 and after the
- 43. Chapter 10: Statistical Inferences About Two Populations 43 treatment, the mean was 5.214. The higher number after the treatment indicates that subjects were more likely to “blow the whistle” after having been through the treatment. 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 study concludes 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 the market share is .3108 and in the southern city the market share is .2701. The point 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 the interval, any hypothesis testing decision based on this interval would result in failure 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 is being computed. The hypotheses are: H0: σ1 2 = σ2 2 Ha: σ1 2 ≠ σ2 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 for a two-tailed test since the associated p-value is .0758. There is no significant difference in the variance of pipe lengths for pipes produced by machine A versus machine B.

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