A major retail clothing store is interested in estimating the difference in mean

Question

A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the store's in-house credit card versus using a Visa, Mastercard, or one of the other major credit cards. To do this, it has randomly selected a sample of customers who have made one or more purchases with each of the types of credit cards. The following represents the results of the sampling:

   In-House Credit Card    National Credit Card
Sample Size:                                         86   113
Mean Monthly Purchases: $45.67      $39.87
Standard Deviation:      $10.90     $12.47

Based on these sample data, what is the lower limit for the 95 percent confidence interval estimate for the difference between population means?

A) About $5.28

B) Approximately $4.85

C) Approximately $2.54

D) Approximately $3.41

Explanation / Answer

Given a=1-0.95= 0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower limit is

(xbar1-xbar2) - Z*sqrt(s1^2/n1+s2^2/n2)

=(45.67-39.87)-1.96*sqrt(10.9^2/86+12.47^2/113)

=2.545204

Answer: C) Approximately $2.54

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