ch09 SOLUTION.doc - Chapter 9 Statistical Inference...

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Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 1 Chapter 9 Statistical Inference: Hypothesis Testing for Single Populations SOLUTIONS TO PROBLEMS IN CHAPTER 9 9.1 a) H o : µ = 25 H a : µ 25 x = 28.1 n = 57 = 8.46 = .01 For two-tail, /2 = .005 z c = 2.575 z = 57 46 . 8 25 1 . 28 n x = 2.77 observed z = 2.77 > z c = 2.575 Reject the null hypothesis b) from Table A.5, inside area between z = 0 and z = 2.77 is .4972 p -value = .5000 - .4972 = .0028 Since the p -value of .0028 is less than 2 = .005, the decision is to: Reject the null hypothesis c) critical mean values: z c = n s x c ± 2.575 = 57 46 . 8 25 c x x c = 25 ± 2.885 x c = 27.885 (upper value) x c = 22.115 (lower value) 9.2 H o : µ = 7.48 H a : µ < 7.48 x = 6.91 n = 24 = 1.21 =.01 For one-tail, = .01 z c = -2.33 z = 24 21 . 1 48 . 7 91 . 6 n x = -2.31 observed z = -2.31 > z c = -2.33 Fail to reject the null hypothesis 9.3 a) H o : µ = 1,200
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Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 2 H a : µ > 1,200 x = 1,215 n = 113 = 100 = .10 For one-tail, = .10 z c = 1.28 z = 113 100 200 , 1 215 , 1 n x = 1.59 observed z = 1.59 > z c = 1.28 Reject the null hypothesis b) Probability > observed z = 1.59 is .0559 which is less than = .10. Reject the null hypothesis . c) Critical mean value: z c = n s x c 1.28 = 113 100 200 , 1 c x x c = 1,200 + 12.04 Since calculated x = 1,215 which is greater than the critical x = 1212.04, reject the null hypothesis. 9.4 H o : µ = 82 H a : µ < 82 x = 78.125 n = 32 = 9.184 = .01 z .01 = -2.33 z = 32 184 . 9 82 125 . 78 n x = -2.39 Since observed z = -2.39 < z .01 = -2.33 Reject the null hypothesis
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Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 3 Statistically, we can conclude that urban air soot is significantly lower. From a business and community point-of-view, assuming that the sample result is representative of how the air actually is now, is a reduction of suspended particles from 82 to 78.125 really an important reduction in air pollution? Certainly it marks an important first step and perhaps a significant start. Whether or not it would really make a difference in the quality of life for people in the city of St. Louis remains to be seen. Most likely, politicians and city chamber of commerce folks would jump on such results as indications of improvement in city conditions. 9.5 H 0 : = $424.20 H a : $424.20 x = $432.69 n = 54 = $33.90 = .05 2-tailed test, /2 = .025 z .025 = + 1.96 z = 54 90 . 33 20 . 424 69 . 432 n x = 1.84 Since the observed z = 1.85 < z .025 = 1.96, the decision is to fail to reject the null hypothesis. 9.6 H 0 : = $62,600 H a : < $62,600 x = $58,974 n = 18 = $7,810 = .01 1-tailed test, = .01 z .01 = -2.33 z = 18 810 , 7 600 , 62 974 , 58 n x = - 1.97 Since the observed z = -1.97 > z .01 = -2.33, the decision is to fail to reject the null hypothesis.
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Chapter 9: Statistical Inference: Hypothesis Testing for Single Populations 4 9.7 H 0 : = 5 H a : 5 x = 5.0611 n = 42 = 0.2803 = .10 2-tailed test, /2 = .05 z .05 = + 1.645 z = 42 2803 . 0 5 0611 . 5 n x = 1.41 Since the observed z = 1.41 < z .05 = 1.645, the decision is to fail to reject the null hypothesis.
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  • Summer '18
  • Keyur Popat
  • Null hypothesis, Statistical hypothesis testing, Type I and type II errors

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