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

Step-by-step explanation:

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

$z=\dfrac{\overline{X}-\overline{Y}}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}$

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 :-

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

Hence, the value of the test statistic = 3.49