A consumer advocacy group feels that Walmart provides a less safe shopping envir

Question

A consumer advocacy group feels that Walmart provides a less safe shopping environment than Target. To try to prove their point, they randomly selected 24 Walmart stores and found the annual number of crime reports filed for these stores. Each of these Walmart stores was then paired with a Target store within a 10 mile radius, and the annual crime reports for the associated Target stores were also recorded. The data is shown below. Test to to see if the mean number of crime incidents at Walmart stores is significantly larger than the mean number of crime incidents at Target stores. Please note the data shown below is not real, but if you would like to see the real example, check out walmartcrimereport.com. Unfortunately, the real data was collected in a way that was not fair to Walmart. Assume normality. Assignment 8q5 data a) Let uw represent the mean incidents associated with Walmart, and let uT represent the mean incidents associated with Target. What are the proper hypotheses? O Ho: ww AT versus Ha: Auw AT Ho: puw T versus Ha: uw AT O Ho: HW AIT versus Ha: Auw AT O Ho: Atw AIT versus Ha: Auw AuT b) What is the test statistic? Give your answer to four decimal places. c) Using a 0.05 level of significance, what is the critical point for this test? Give your answer to four decimal places. d) What is the appropriate conclusion? O Conclude that uw is greater than umr because the test statistic is larger than the critical point. O Fail to reject the claim that uw is equal to pT because the test statistic is larger than the critical point. O Reject the claim that uw is greater than AT because the test statistic is larger than the critical point. O Fail to reject the claim that uw is equal to pT because the test statistic is less than the critical point.

Explanation / Answer

Part a)

Answer:

Ho: µw = µT versus Ha: µw > µT       [First option]

Part b)

We go to Data then Data Analysis. There we select t-test: Two samples Assuming Equal Variances. We get the following output here.

t-Test: Two-Sample Assuming Equal Variances

Walmart

Target

Mean

762.875

759.125

Variance

39917.77

45299.33

Observations

24

24

Pooled Variance

42608.55

Hypothesized Mean Difference

0

df

46

t Stat

0.0629

P(T<=t) one-tail

0.475047

t Critical one-tail

1.6787

P(T<=t) two-tail

0.950093

t Critical two-tail

2.012896

Test Statistics:

Answer:

t = 0.0629

Part c)

For 46 degrees of freedom at 5% level of significance we get the one tailed critical t from t table as 1.6787.

Answer:

Critical value = 1.6787

Part d)

The calculated test statistics is less than the critical value (0.0629<1.6787); we fail to reject the null hypothesis.

Answer:

Fail to reject the claim that µw is equal to µT because the test statistic is less than the critical point.

t-Test: Two-Sample Assuming Equal Variances

Walmart

Target

Mean

762.875

759.125

Variance

39917.77

45299.33

Observations

24

24

Pooled Variance

42608.55

Hypothesized Mean Difference

0

df

46

t Stat

0.0629

P(T<=t) one-tail

0.475047

t Critical one-tail

1.6787

P(T<=t) two-tail

0.950093

t Critical two-tail

2.012896

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