Question

a magazine claims that the mean amount spent by a customer at burger stop is greater than the mean a customer at fry world. the results for samples of customer transactions for the two fast food restaurants below. at \\( \alpha = 0.05 \\), can you support the magazines claim? assume the population variances are equal, samples are random and independent, and the populations are normally distributed. complete parts (a) below.
\\( \

$$\begin{array}{c|c} \\text{burger stop} & \\text{fry world} \\\\ \\hline \\bar{x}_1 = \\$9.89 & \\bar{x}_2 = \\$9.26 \\\\ s_1 = \\$0.79 & s_2 = \\$0.68 \\\\ n_1 = 13 & n_2 = 11 \\end{array}$$

enter the critical value(s) below.
1.717
(type an integer or decimal rounded to three decimal places as needed. use a comma to separate
select the correct rejection region(s) below.
\\( \bigcirc \\) a. \\( t < -t_0 \\)
\\( \bigcirc \\) b. \\( t > t_0 \\)
\\( \bigcirc \\) c. \\( -t_0 < t < t_0 \\)
\\( \bigcirc \\) d. \\( t < -t_0, t > t_0 \\)

Explanation:

Step1: Identify Test Type

This is a two - sample t - test for means (since population variances are equal, samples are random, independent, and populations are normal). The claim is that $\mu_1>\mu_2$ (mean at Burger Stop > mean at Fry World), so it's a right - tailed test.

Step2: Determine Degrees of Freedom

The formula for degrees of freedom for two - sample t - test with equal variances is $df=n_1 + n_2-2$. Here, $n_1 = 13$ and $n_2=11$, so $df=13 + 11-2=22$.

Step3: Find Critical Value

For a right - tailed test with $\alpha = 0.05$ and $df = 22$, we look up the t - critical value in the t - distribution table. The critical value $t_0$ is such that $P(T>t_0)=\alpha = 0.05$. From t - table, $t_{0.05,22}\approx1.717$.

Step4: Determine Rejection Region

In a right - tailed t - test, the rejection region is where the test statistic $t>t_0$. So the correct rejection region is $t > t_0$, which corresponds to option B.

Answer:

Critical value: $1.717$; Rejection region: B. $t>t_0$