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 17 visa gold cardholders resulted in a mean household income of $54,800 with a standard deviation of $9500. a random survey of 11 mastercard gold cardholders resulted in a mean household income of $47,390 with a standard deviation of $10,900. 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 $alpha = 0.02$ 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 t - test statistic

The formula for the t - test statistic when population variances are not equal is $t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}$. Here, we are testing the claim that $\mu_1
eq\mu_2$, so under the null hypothesis $H_0:\mu_1 - \mu_2 = 0$.

Step2: Substitute the given values

We have $\bar{x}_1 = 54800$, $s_1=9500$, $n_1 = 17$, $\bar{x}_2=47390$, $s_2 = 10900$, $n_2=11$.
\[

$$\begin{align*} t&=\frac{(54800 - 47390)-0}{\sqrt{\frac{9500^{2}}{17}+\frac{10900^{2}}{11}}}\\ &=\frac{7410}{\sqrt{\frac{90250000}{17}+\frac{118810000}{11}}}\\ &=\frac{7410}{\sqrt{5308823.529+ 10800909.091}}\\ &=\frac{7410}{\sqrt{16109732.62}}\\ &=\frac{7410}{4013.693}\\ &\approx1.846 \end{align*}$$

\]

Answer:

$1.846$