Questions: 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 57,230 with a standard deviation of 9900. A random survey of 18 MasterCard Gold cardholders resulted in a mean household income of 48,920 with a standard deviation of 8400. 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.

Solution Steps

The standard error is calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{9900^2}{8} + \frac{8400^2}{18}} \]

Calculating the variances:

\[ s_1^2 = 9900^2 = 98010000, \quad s_2^2 = 8400^2 = 70560000 \]

Now substituting the values:

\[ SE = \sqrt{\frac{98010000}{8} + \frac{70560000}{18}} = \sqrt{12251250 + 3920000} = \sqrt{16171250} \approx 4016.0 \]

The test statistic is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{57230 - 48920}{4016.0} \approx \frac{8310}{4016.0} \approx 2.067 \]

The degrees of freedom for Welch's t-test is calculated using:

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]

Calculating the components:

\[ \frac{s_1^2}{n_1} = \frac{98010000}{8} = 12251250, \quad \frac{s_2^2}{n_2} = \frac{70560000}{18} = 3920000 \]

Now substituting into the df formula:

\[ df = \frac{(12251250 + 3920000)^2}{\frac{(12251250)^2}{7} + \frac{(3920000)^2}{17}} = \frac{(16171250)^2}{\frac{149086056250000}{7} + \frac{15366400000000}{17}} \approx 18.0 \]

Using the t-distribution, the p-value is calculated as:

\[ P = 2(1 - T(|t|)) = 2(1 - T(2.067)) \]

Assuming \( T(2.067) \) gives a p-value of approximately \( 0.025 \):

\[ P \approx 2(1 - 0.975) = 0.05 \]

Since the p-value \( P \approx 0.05 \) is greater than the significance level \( \alpha = 0.01 \), we fail to reject the null hypothesis.

Final Answer