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 12 visa gold cardholders resulted in a mean household income of $73,540 with a standard deviation of $9500. a random survey of 8 mastercard gold cardholders resulted in a mean household income of $63,360 with a standard deviation of $8500. is there enough evidence to support the executives claim? let $mu_1$ be the true mean household income for visa gold cardholders and $mu_2$ be the true mean household income for mastercard gold cardholders. use a significance level of $alpha = 0.05$ for the test. assume that the population variances are not equal and that the two populations are normally distributed. step 4 of 4: state the tests conclusion. answer 1 point reject null hypothesis fail to reject null hypothesis

Explanation:

Step1: Identify degrees of freedom

The degrees of freedom for two - sample t - test with unequal variances is calculated using the formula $df=\frac{(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2})^{2}}{\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}}$. Here, $n_1 = 12$, $s_1=9500$, $n_2 = 8$, $s_2 = 8500$. First, calculate $\frac{s_1^{2}}{n_1}=\frac{9500^{2}}{12}\approx7504166.67$ and $\frac{s_2^{2}}{n_2}=\frac{8500^{2}}{8}\approx9031250$. Then $(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2})^{2}=(7504166.67 + 9031250)^{2}\approx(16535416.67)^{2}\approx2.734\times10^{14}$. Also, $\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}=\frac{7504166.67^{2}}{11}\approx5.11\times10^{12}$ and $\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}=\frac{9031250^{2}}{7}\approx1.16\times10^{13}$. So, $df=\frac{2.734\times10^{14}}{5.11\times10^{12}+1.16\times10^{13}}\approx16.7\approx16$ (rounded down).

Step2: Determine critical values

For a two - tailed test with $\alpha = 0.05$ and $df = 16$, the critical values are $t_{\alpha/2,df}=t_{0.025,16}=\pm 2.120$.

Step3: Calculate the test statistic

The test statistic for two - sample t - test with unequal variances is $t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}$, where $\bar{x}_1 = 73540$, $\bar{x}_2=63360$, $\frac{s_1^{2}}{n_1}\approx7504166.67$, $\frac{s_2^{2}}{n_2}\approx9031250$. So, $t=\frac{73540 - 63360}{\sqrt{7504166.67+9031250}}=\frac{10180}{\sqrt{16535416.67}}\approx\frac{10180}{4066.38}\approx2.50$.

Step4: Make a decision

Since the calculated $t$ - value ($2.50$) is greater than the critical value $t_{0.025,16}=2.120$, we reject the null hypothesis.

Answer:

Reject Null Hypothesis