InferenceScenarios_2013

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Comments: If the question had asked whether the average reimbursement is different from \$100, then the test would have been two-sided; H a : μ ≠ 100, P-value = 2x0.059= 0.12; with the same conclusion – no evidence that the average reimbursement is different from \$100. A 95% CI is 93.70 ± 7.96 or (85.74,101.66). Note that 100 is inside the CI, confirming that the null hypothesis is supported. b) Is there evidence that the percentage of incorrect reimbursements in the population is greater than 10%? SOLUTION: One-sample z-test of a single proportion. This is also a one-sided alternative hypothesis. H o : p = 0.10 H a : p > 0.10 One-sample z-test of a proportion: z = (0.16 – 0.10)/√[(0.10)(0.90)/75] = 1.73 P-value = Pr(z > 1.73) = 0.0418 (from Excel, NORMDIST) Since the P-value is less than 0.05 (but just barely), we conclude that there is weak evidence to say that the percentage of incorrect reimbursements is greater than 10%. Comments: If the question had asked whether the percentage of incorrect reimbursements is different from 10%, then the test would have been two-sided; H a : p ≠ 0.10, P-value = 2x0.0418 = 0.084; with the conclusion that there is not evidence that the percentage of incorrect reimbursements is different from 10%. This illustrates the importance of choosing your alternative hypothesis first, AND the importance of how you word the conclusion. In this case, the evidence against the null hypothesis is very weak. 2

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Scenario 3. Gender Equity in Hiring The federal Ministry of Labour plans to submit a report to parliament that will support a new employment law which aims to ensure women workers to receive fair treatment in the job market. To support its argument in the report, they commission an independent research institution to conduct a study about the job market. In particular, the research institution plans a survey with the objective of determining if women are, on the average, less likely to succeed in job applications than their male counterparts. The institution selects 100 male and 100 female job applicants (i.e., 200 in total) and records the number of successes in job applications for each of the groups. The samples yield success rates for men and women or 76% and 60%, respectively. Do the data provide evidence that women are less likely to succeed in job application than men with similar backgrounds? Carry out the appropriate hypothesis test. SOLUTION: Two-sample z-test of two proportions. H 0 : p m – p w = 0 H a : p m – p w > 0 Pooled = (76+60)/(100+100) = 0.68 Test statistic: z = (0.76–0.60) / √[0.68)(1-0.68)(1/100+1/100)] = 2.425 P-value = Pr(Z > 2.425) = 0.0076 Conclusion: The men’s success rate is statistically significantly higher than the women's success rate. *** 95% confidence interval for the difference in the probabilities of success of male and female job applicants: (0.76–0.60) ± 1.96√[(0.76)(1-0.76)/100 + (0.60)(1-0.60)/100] = 0.16 ± 0.127 or (0.033 , 0.287) The CI does not include 0 so the difference in proportions is statistically significant 3
Scenario 4. Market Research A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores. Marketing managers are prone to look at the point estimate only and ignore sampling error (i.e., the margin of error in a confidence

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