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DS212_HW9_Solutions

# DS212_HW9_Solutions - DS 212 Business Statistics Soorapanth...

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Unformatted text preview: DS 212 Business Statistics Soorapanth Homework #9: Chapter 9 Statistical Inference: Hypothesis Testing for Single Populations 9.2 H o : µ = 7.48 H a : µ < 7.48 = 6.91 n = 24 σ = 1.21 α =.01 For one-tail, α = .01 z c = -2.33 z = = -2.31 observed z = -2.31 > z c = -2.33 Fail to reject the null hypothesis 9.3 a) H o : µ = 1,200 H a : µ > 1,200 = 1,215 n = 113 σ = 100 α = .10 For one-tail, α = .10 z c = 1.28 z = = 1.59 observed z = 1.59 > z c = 1.28 Reject the null hypothesis b) Probability > observed z = 1.59 is .5000 - .4441 = .0559 (the p-value) which is less than α = .10. Reject the null hypothesis . c) Critical mean value: 1 DS 212 Business Statistics Soorapanth z c = 1.28 = c = 1,200 + 12.04 Since the observed = 1,215 is greater than the critical = 1212.04, the decision is to reject the null hypothesis . 9.5 H : μ = \$424.20 H a : μ ≠ \$424.20 = \$432.69 n = 54 σ = \$33.90 α = .05 2-tailed test, α /2 = .025 z .025 = + 1.96 z = = 1.84 Since the observed z = 1.84 < z .025 = 1.96, the decision is to fail to reject the null hypothesis . 9.6 H : μ = \$62,600 H a : μ < \$62,600 = \$58,974 n = 18 σ = \$7,810 α = .01 1-tailed test, α = .01 z .01 = -2.33 z = = - 1.97 2 DS 212 Business Statistics Soorapanth Since the observed z = -1.97 > z .01 = -2.33, the decision is to fail to reject the null hypothesis ....
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DS212_HW9_Solutions - DS 212 Business Statistics Soorapanth...

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