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 2 of 4: compute the value of the t - test statistic. round your answer to three decimal places. answerhow to enter your answer (opens in new window) 1 point tables keypad keyboard shortcuts

Explanation:

Step1: Identify the formula for the t - test statistic

The formula for the t - test statistic when variances are unequal is $t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}$. Since the claim is about the difference in means and we assume $\mu_1-\mu_2 = 0$ under the null hypothesis, the formula simplifies to $t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}$.

Step2: Define the given values

We have $\bar{x}_1 = 73540$, $s_1 = 9500$, $n_1=12$, $\bar{x}_2 = 63360$, $s_2 = 8500$, $n_2 = 8$.

Step3: Substitute the values into the formula

First, calculate the denominator:
$\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}=\sqrt{\frac{9500^{2}}{12}+\frac{8500^{2}}{8}}$
$=\sqrt{\frac{90250000}{12}+\frac{72250000}{8}}$
$=\sqrt{7520833.333 + 9031250}$
$=\sqrt{16552083.333}\approx4068.426$.
Then, calculate the numerator: $\bar{x}_1-\bar{x}_2=73540 - 63360=10180$.
Finally, calculate the t - statistic: $t=\frac{10180}{4068.426}\approx2.502$.

Answer:

$2.502$