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.
Solution:
Step 1: Null Hypothesis: The average (mean) annual income was equal to $45,000.
H_0: μ=45,0000
Step2: Alternate Hypothesis: The average (mean) annual was less than $50,000.
H_a: μ 45 , a z-test for the mean will be used to test the given hypothesis.
As for the alternative hypothesis, which is Ha:μ0.45 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 1.645.
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
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 Hypothesis: The average (mean) annual was less than $50,000.
H_a: μ 45 , a z-test for the mean will be used to test the given hypothesis.
As for the alternative hypothesis, which is Ha:μ0.45 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 1.645.
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