Question

An advertising executive claims that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold. A random survey of 9 Visa Gold cardholders resulted in a mean household income of $51,270 with a standard deviation of $11,300. A random survey of 17 MasterCard Gold cardholders resulted in a mean household income of $45,060 with a standard deviation of $8200. Is there enough evidence to support the executive's claim? Let μ1 be the true mean household income for Visa Gold cardholders and μ2 be the true mean household income for MasterCard Gold cardholders. Use a significance level of α=0.01 for the test. Assume that the population variances are not equal and that the two populations are normally distributed.

Step 1 of 4: State the null and alternative hypotheses for the test.

Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round your answer to three decimal places.

Step 4 of 4: State the test's conclusion.

Answer 1

1)

H0 : mu1 = mu2
Ha: mu1 not = mu2

2)

t = (x1-x2)/sqrt(s1^2/n1+s2^2/n2)
= ( 51270 - 45060)/sqrt(11300^2/9 + 8200^2/17)
= 1.458

3)

This is two tailed test, for α = 0.01 and df = 8
Critical value of t are -3.355 and 3.355.
Hence reject H0 if t < -3.355 or t > 3.355

4)

Fail to reject the null hypothesis
There is not sufficient evidence to support the claim that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold.