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# ch10 SOLUTION.doc - Chapter 10 Statistical Inferences About...

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Chapter 10: Statistical Inferences About Two Populations 1 Chapter 10 - Statistical Inferences about Two Populations SOLUTIONS TO PROBLEMS IN CHAPTER 10 10.1 Sample 1 x 1 = 51.3 1 2 = 52 n 1 = 32 Sample 2 x 2 = 53.2 2 2 = 60 n 2 = 32 a) H o : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 < 0 For one-tail test, = .10 z .10 = -1.28 z = 32 60 32 52 ) 0 ( ) 2 . 53 3 . 51 ( ) ( ) ( 2 2 2 1 2 1 2 1 2 1 n n x x = -1.02 Since the observed z = -1.02 > z c = -1.645, the decision is to fail to reject the null hypothesis . b) Critical value method: z c = 2 2 2 1 2 1 2 1 2 1 ) ( ) ( n n x x c -1.645 = 32 60 32 52 ) 0 ( ) ( 2 1 c x x ( 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 10.3 a) Sample 1 x 1 = 88.23 1 2 = 22.74 n 1 = 30 Sample 2 x 2 = 81.2 2 2 = 26.65 n 2 = 30 H o : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 0 For two-tail test, use /2 = .01 z .01 = + 2.33 z = 30 65 . 26 30 74 . 22 ) 0 ( ) 2 . 81 23 . 88 ( ) ( ) ( 2 2 2 1 2 1 2 1 2 1 n n x x = 5.48 Since the observed z = 5.48 > z .01 = 2.33, the decision is to reject the null hypothesis . 10.4 Computers/electronics Food/Beverage x 1 = 1.96 x 2 = 3.02 1 2 = 1.0188 2 2 = 0.9180 n 1 = 50 n 2 = 50 H o : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 0 For two-tail test, /2 = .005 z . 005 = ±2.575 z = 50 9180 . 0 50 0188 . 1 ) 0 ( ) 02 . 3 96 . 1 ( ) ( ) ( 2 2 2 1 2 1 2 1 2 1 n n x x = -5.39 Since the observed z = -5.39 < z c = -2.575, the decision is to reject the null hypothesis . 10.6 Managers n 1 = 35 x 1 = 1.84 1 = .38 Specialty n 2 = 41 x 2 = 1.99 2 = .51 Hypothesis Test: 1) H o : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 0 2) z = 2 2 2 1 2 1 2 1 2 1 ) ( ) ( n n x x 3) = .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 . 1 84 . 1 ( 2 2 = -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.

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Chapter 10: Statistical Inferences About Two Populations 2 10.7 1994 2001 x 1 = 190 x 2 = 198 1 = 18.50 2 = 15.60 n 1 = 51 n 2 = 47 = .01 H 0 : 1 - 2 = 0 H a : 1 - 2 < 0 For a one-tailed test, z .01 = -2.33 z = 47 ) 60 . 15 ( 51 ) 50 . 18 ( ) 0 ( ) 198 190 ( ) ( ) ( 2 2 2 2 2 1 2 1 2 1 2 1 n n x x = -2.32 Since the observed z = -2.32 > z .01 = -2.33, the decision is to fail to reject the null hypothesis . 10.9 Canon Pioneer x 1 = 5.8 x 2 = 5.0 1 = 1.7 2 = 1.4 n 1 = 36 n 2 = 45 H o : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 0 For two-tail test, /2 = .025 z .025 = ±1.96 z = 45 ) 4 . 1 ( 36 ) 7 . 1 ( ) 0 ( ) 0 . 5 8 . 5 ( ) ( ) ( 2 2 2 2 1 2 1 2 1 2 1 n n x x = 2.27 Since the observed z = 2.27 > z c = 1.96, the decision is to reject the null hypothesis . 10.10 A B x 1 = 8.05 x 2 = 7.26 1 = 1.36 2 = 1.06 n 1 = 50 n 2 = 38 H o : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 > 0 For one-tail test, = .10 z .10 = 1.28
Chapter 10: Statistical Inferences About Two Populations 3 z = 38 ) 06 . 1 ( 50 ) 36 . 1 ( ) 0 ( ) 26 . 7 05 . 8 ( ) ( ) ( 2 2 2 2 2 1 2 1 2 1 2 1 n n x x = 3.06 Since the observed z = 3.06 > z c = 1.28, the decision is to reject the null hypothesis .

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• Summer '18
• Keyur Popat
• Null hypothesis

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