# Essay on Aj Davis Department Store Part B

892 Words4 Pages
AJ Davis Department Store Part B AJ Davis Department Store Introduction The following information will show whether or not the manager’s speculations are correct. He wants to know the following information: Is the average mean greater than \$45,000? Does the true population proportion of customers who live in an urban area exceed 45%? Is the average number of years lived in the current home less than 8 years? Is the credit balance for suburban customers more than \$3200? Hypothesis testing and confidence intervals for situations A-D are calculated. A. THE AVERAGE (MEAN) ANNUAL INCOME WAS LESS THAN \$45,000. Solution: Step 1: Null Hypothesis: The average (mean) annual income was equal to \$45,000. H_0: μ=45,0000 Step2: Alternate…show more content…
Solution: Step 1: Null Hypothesis: The average (mean) number of years lived in the current home is equal to 8 years. H_0: μ=8 Step 2: Alternate Hypothesis: The average (mean) number of years lived in the current home is less than 8 years. H_a: μ 50 requires that the z-test for mean be used to test the given hypothesis. The alternative hypothesis is Ha:μ 3200 Step 3: Test Statistic: z= Following the provided information, the Significance Level is α=0.05. The alternative hypothesis is Ha: μ>3200; therefore, the given test is a one-tailed (upper-tailed) z-test. Step 4: Critical Value and Rejection Region: The critical value for significance level α=0.05 for an upper-tailed z-test is given as 1.645. Rejection Region: Reject H_0,if z-statistic>1.645. Step 5: Assumptions: The sample size in this speculation is greater than 30, therefore, The Central Limit Theorem (CLT) will apply, and no assumptions need to be made. Step 6: Calculation of test statistic: One-Sample Z: Credit Balance (\$) Test of mu = 3200 vs > 3200 The assumed standard deviation = 742.365 95% Lower Variable N Mean StDev SE Mean Bound Z P Credit Balance (\$) 15 4675 742 192 4360 1.96 0.025 Step 7: Interpretation: According to the above results from MINITAB, the p-value of 0.038 is smaller than the significance level of 0.05; consequently, the null hypothesis will be rejected. There is sufficient evidence to support the