Question

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

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(\\( \text{type an integer or decimal rounded to three decimal places as needed. use a comma to separate answers as needed.} \\))\
\
select the correct rejection region(s) below.\
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\\( \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 \\)\
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(c) find the standardized test statistic.\
\\( t = \square \\) (\\( \text{type an integer or decimal rounded to three decimal places as needed.} \\))

Explanation:

Step1: Identify the test type

This is a two - sample t - test for means with equal variances. The formula for the standardized test statistic (t - statistic) for two - sample t - test with equal variances is:

$$t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$

where \(s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}\) is the pooled standard deviation, \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(n_1\) and \(n_2\) are the sample sizes, \(s_1\) and \(s_2\) are the sample standard deviations, and \(\mu_1-\mu_2 = 0\) (under the null hypothesis, and the alternative hypothesis is \(\mu_1-\mu_2>0\) since the magazine claims that the mean at Burger Stop is greater than at Fry World).

Step2: Calculate the pooled standard deviation \(s_p\)

First, calculate \((n_1 - 1)s_1^2\) and \((n_2 - 1)s_2^2\):
\(n_1 = 13\), \(s_1=0.79\), so \((n_1 - 1)s_1^2=(13 - 1)\times(0.79)^2=12\times0.6241 = 7.4892\)
\(n_2 = 11\), \(s_2 = 0.68\), so \((n_2 - 1)s_2^2=(11 - 1)\times(0.68)^2=10\times0.4624 = 4.624\)
Then, \(n_1 + n_2-2=13 + 11-2=22\)
\(s_p=\sqrt{\frac{7.4892 + 4.624}{22}}=\sqrt{\frac{12.1132}{22}}=\sqrt{0.5506}\approx0.742\)

Step3: Calculate the t - statistic

\(\bar{x}_1 = 9.89\), \(\bar{x}_2=9.26\), so \(\bar{x}_1-\bar{x}_2=9.89 - 9.26 = 0.63\)
\(\frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{13}+\frac{1}{11}=\frac{11 + 13}{13\times11}=\frac{24}{143}\approx0.1678\)
\(s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=0.742\times\sqrt{0.1678}\approx0.742\times0.4096\approx0.304\)
Then, \(t=\frac{0.63-0}{0.304}\approx2.072\)

Answer:

\(t\approx\boxed{2.072}\)