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 8 Visa Gold cardholders resulted in a mean household income of $78,320 with a standard deviation of $11,100. A random survey of 12 MasterCard Gold cardholders resulted in a mean household income of $68,070 with a standard deviation of $10,700. 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=0.05α=0.05 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 H0H0. Round your answer to three decimal places.

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

....reject null hypothesis...fail to reject null hypothesis

Answer 1

1) H₀:

   H₁:

2) The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                            = (78320 - 68070)/sqrt((11100)^2/8 + (10700)^2/12)

                            = 2.052

3) DF = (s1^2/n1 + s2^2/n2)^2/((s1^2/n1)^2/(n1 - 1) + (s2^2/n2)^2/(n2 - 1))

     = ((11100)^2/8 + (10700)^2/12)^2/(((11100)^2/8)^2/7 + ((10700)^2/12)^2/11)

     = 14

At 0.05 significance level, the critical values are +/- t_0.025,14 = +/- 2.145

Reject H₀, if t < -2.145 or t > 2.145

4) Since the test statistic value is not greater than the positive critical value(2.052 < 2.145), so we fail to reject the null hypothesis.