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

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 [tex]\mu_1[/tex] = Mean Monthly Purchases of In-House Credit Card

[tex]\mu_2[/tex] = Mean Monthly purchases of National Credit Card

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1-\mu_2[/tex] = 0 {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}

Alternate Hypothesis, [tex]H_0[/tex] : [tex]\mu_1-\mu_2\neq[/tex] 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. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t_n___1+n_2-_2[/tex]

where, [tex]\bar X_1[/tex] = Sample mean Purchases of In-House Credit Card = $45.67

[tex]\bar X_2[/tex] = Sample mean Purchases of National Credit Card = $39.87

[tex]s_p[/tex] = pooled variance

[tex]n_1[/tex] = In-house credit card sample = 32

[tex]n_2[/tex] = 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.