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.

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

yes the data was not prsented

step-by-step explanation: