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# Aj Davis Department Store Part B

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.

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Solution: Step 1: Null Hypothesis: The average (mean) annual income was equal to \$45,000. H_0: ? =45,0000 Step2: Alternate Hypothesis: The average (mean) annual was less than \$50,000. H_a: ? 45 , a z-test for the mean will be used to test the given hypothesis. As for the alternative hypothesis, which is Ha:? 0. 45 and the given test is a one-tailed (upper-tailed) z-test. Step 4: Critical Value and Rejection Region: The critical value for significance level is ? =0. 05. The upper tail z-test is 1. 45. Rejection Region: Reject H_0,if z-statistic>1. 645. Step 5: Assumptions: The sample size in this experiment is n 0. 4 95% Lower Sample X N Sample p Bound Z-Value P-Value 1 21 50 0. 420000 0. 305190 0. 29 0. 386 Step 7: Interpretation: According to the calculations, the p-value is 0. 386. This value is larger than the significance level of 0. 05. Therefore, we will not reject the null hypothesis. 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%.

Based on the results provided by MINITAB below, the lower 95% confidence limit is 0. 28. Since 0. 42 is greater than the 95% lower confidence limit, hence, we cannot support the claim that the true population proportion of customers who live in an urban area is greater than 45%. Confidence Interval: Test and CI for One Proportion Sample X N Sample p 95% CI 1 21 50 0. 420000 (0. 283195, 0. 556805) ? C. THE AVERAGE (MEAN) NUMBER OF YEARS LIVED IN THE CURRENT HOME IS LESS THAN 8 YEARS. Solution: Step 1: Null Hypothesis: The average (mean) number of years lived in the current home is equal to 8 years.

H_0: ? =8 Step 2: Alternate Hypothesis: The average (mean) number of years lived in the current home is less than 8 years. H_a: ? 50 requires that the z-test for mean be used to test the given hypothesis. The alternative hypothesis is Ha:? 3200 Step 3: Test Statistic: z= Following the provided information, the Significance Level is ? =0. 05. The alternative hypothesis is Ha: ? >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 ? =0. 5 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 claim that the average (mean) credit balance for suburban customers is more than \$3200. Based on the results from MINITAB, which are provided below, the significance level is 0. 05; therefore, there is sufficient evidence to support the claim that the average (mean) credit balance for suburban customers is more than \$3200.

MINITAB reveals that the 95% lower confidence limit is 4469; therefore, 3200 is smaller than the 95% lower confidence limit, which means that the claim can be supported regarding the average (mean) credit balance for suburban customers is more than \$3200. Confidence Interval One-Sample Z The assumed standard deviation = 742. 365 N Mean SE Mean 95% CI 50 4675 105 (4469, 4881) Conclusion After performing hypothesis testing and confidence intervals on each speculation, we can conclude that from the sample of 50 customers from AJ DAVIS department store, the average (mean) annual income is less than \$50,000.

We can also conclude that the number of customers that live in urban areas are less than or equal to 40%. The average number of years their customers have lived in their current home is more than 13. Also, the average credit balance for suburban customers is more than \$4300. The manager was correct about the speculations of the average income being less than \$50,000 and a customers’ credit balance being more than \$4,300. For AJ Davis department store to continue a great relationship with their customers, it wise and imperative that they continue to perform analyses of this magnitude.

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