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 CardSample Size: 86 113Mean 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 of .05, what is the value of the test statistic assuming the standard deviations are known?z = 11.91z = 2.86t = 3.49z = 3.49

Answer 1

Answer: 3.49

Explanation:

When standard deviation is known then the test statistic for difference of two population means (independent population) is given by :-

`z=\frac{\overline{X}-\overline{Y}}{\sqrt{(\sigma_1^2)/(n_1)+(\sigma_2^2)/(n_2)}}`

Given :
`n_1=86\ \ , \ n_2=113`

`\overline{X}=45.67\ \ , \ \overline{Y}=39.87`

`\sigma_1=10.90\ \ ,\ \sigma_2=12.47`

Then , the value of the test statistic will be :-

`z=\frac{45.67-39.87}{\sqrt{((10.9)^2)/(86)+((12.47)^2)/(113)}}\\\\=(5.8)/(√(2.757625787))=3.4926923081\approx3.49`

Hence, the value of the test statistic = 3.49