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 national 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 CardNational Credit CardSample Size:3250Mean Monthly Purchases:$45.67$39.87Standard Deviation:$10.90$12.47Suppose that the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level α of 0.05, what is the critical value assuming the population standard deviations are not known but that the populations are normally distributed with equal variances?A. t = 3.49B. z = 1.645C. z = 1.96D. t = 1.9901

Critical value is t = 1.9901

Step-by-step explanation:

We are given the results of the sampling :

                                     In-House Credit Card                National Credit Card                    Sample Size :                              32                                                    50

Mean Monthly Purchases :      $45.67                                            $39.87

Standard Deviation :                $10.90                                             $12.47

Also, the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level α of 0.05.

Firstly, we will specify our null and alternate hypothesis;

Let = Mean Monthly Purchases of In-House Credit Card

      = Mean Monthly purchases of National Credit Card

So, Null Hypothesis, : = 0  {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}

Alternate Hypothesis, : 0  {means that there is statistical difference in the mean monthly purchases by customers using the two types of credit cards}

The test statistics that will be used here is Two-sample t-test statistics;

              T.S. =  ~

where, = Sample mean Purchases of In-House Credit Card = $45.67

            = Sample mean Purchases of National Credit Card = $39.87

            = pooled variance

            = In-house credit card sample = 32

            = National credit card sample = 50

So, degree of freedom of t-value here is (32 + 50 - 2) = 80

Now, at 0.05 significance level, t table gives critical value of t = 1.9901 at 80 degree of freedom.

Therefore, the critical value assuming the population standard deviations are not known but that the populations are normally distributed with equal variances is t = 1.9901.

175

step-by-step explanation:

2 inches x 10 miles = 200 miles - 3 inches = 175