Keller Math 533 Project Part B

1734 WordsSep 29, 20137 Pages
Math 533 Project Part B In regards to the dataset from AJ Department store, your manager has speculated the following: the average (mean) annual income is less than \$50,000, the true population proportion of customers who live in an urban area exceeds 40%, the average (mean) number of years lived in the current home is less than 13 years, the average (mean) credit balance for suburban customers is more than \$4300. Part 1. Using the sample data, perform the hypothesis test for each of the above situations in order to see if there is evidence to support your manager’s belief in each case a.-d. In each case use the Seven Elements of a Test of Hypothesis, in Section 6.2 of your text book with α =…show more content…
Therefore, we will reject 〖the null hypothesis,H〗_0,if z>1.645 Element 5: Assumptions Element 6:Interpretation of Results Since the P-value (0.386) is greater than the significance level (0.05), we fail to reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis. Element 7: Conclusion At a significance level of 0.05, there is no sufficient evidence to support the claim that the true population proportion of customers who live in an urban area is greater than 40%. The average (mean) number of years lived in the current home is less than 13 years Element 1. Null hypothesis (H0): The average (mean) number of years lived in the current home is greater than or equal to 13 years H_0: μ≥13 Element 2. Alternative (research) hypothesis (Ha): The average (mean) number of years lived in the current home is less than 13 years. H_a: μ 30 we will use a z-test for the mean to test the given hypothesis. As the alternative hypothesis is Ha:μ 30 we will use z-test for mean to test the given hypothesis. As the alternative hypothesis is Ha:μ>4300 , the given test is a one-tailed (upper-tailed) z-test. One-Sample Z: Credit Balance (\$) _1 Test of mu = 4300 vs > 4300 The assumed standard deviation = 770.339