892 WordsMay 9, 20134 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

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