Question

a magazine claims that the mean amount spent by a customer at burger stop is greater than th a customer at fry world. the results for samples of customer transactions for the two fast food below. at α = 0.05, can you support the magazine’s claim? assume the population variances ar samples are random and independent, and the populations are normally distributed. complete below.

burger stop | fry world
\\(\bar{x}_1 = \\$9.89\\) | \\(\bar{x}_2 = \\$9.26\\)
\\(s_1 = \\$0.79\\) | \\(s_2 = \\$0.68\\)
\\(n_1 = 13\\) | \\(n_2 = 11\\)

the null hypothesis, \\(h_0\\), is \\(\mu_1 \leq \mu_2\\). the alternative hypothesis, \\(h_a\\), is \\(\mu_1 > \mu_2\\).

which hypothesis is the claim?
\\(\circ\\) the null hypothesis, \\(h_0\\)
\\(\circ\\) the alternative hypothesis, \\(h_a\\)

(b) find the critical value(s) and identify the rejection region(s).

enter the critical value(s) below.
\\(\square\\)
(type an integer or decimal rounded to three decimal places as needed. use a comma to sep

Explanation:

Step1: Determine Degrees of Freedom

For two - sample t - test (since population variances are unknown, we use t - test), the degrees of freedom formula for unequal variances (or we can also use the formula \(df=\frac{(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}})^{2}}{\frac{(s_{1}^{2}/n_{1})^{2}}{n_{1}-1}+\frac{(s_{2}^{2}/n_{2})^{2}}{n_{2}-1}}\)) or we can use the conservative approach. But here, let's use the formula for degrees of freedom:

\(n_1 = 13\), \(n_2=11\), \(s_1 = 0.79\), \(s_2=0.68\)

First, calculate \(\frac{s_{1}^{2}}{n_{1}}=\frac{0.79^{2}}{13}=\frac{0.6241}{13}\approx0.048\)

\(\frac{s_{2}^{2}}{n_{2}}=\frac{0.68^{2}}{11}=\frac{0.4624}{11}\approx0.042\)

Then, \((\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}})^{2}=(0.048 + 0.042)^{2}=0.09^{2}=0.0081\)

\(\frac{(s_{1}^{2}/n_{1})^{2}}{n_{1}-1}=\frac{0.048^{2}}{12}=\frac{0.002304}{12}=0.000192\)

\(\frac{(s_{2}^{2}/n_{2})^{2}}{n_{2}-1}=\frac{0.042^{2}}{10}=\frac{0.001764}{10}=0.0001764\)

\(df=\frac{0.0081}{0.000192 + 0.0001764}=\frac{0.0081}{0.0003684}\approx21.99\approx22\) (we can also use the smaller of \(n_1 - 1\) and \(n_2 - 1\), which is \(10\), but the formula - based approach gives around 22)

Step2: Find Critical Value

We have a one - tailed test (since \(H_a:\mu_1>\mu_2\)) with \(\alpha = 0.05\) and \(df\approx22\) (using the t - distribution table or calculator).

Looking up in the t - table, for a one - tailed test with \(\alpha = 0.05\) and \(df = 22\), the critical value \(t_{\alpha,df}\) is approximately \(1.717\)

Answer:

\(1.717\)