Question

a magazine claims that the mean amount spent by a customer at burger stop is great a customer at fry world. the results for samples of customer transactions for the two below. at α = 0.05, can you support the magazines claim? assume the population vari samples are random and independent, and the populations are normally distributed. c 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\\)

\\(\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\\)

(c) find the standardized test statistic.
t = 2.072 (type an integer or decimal rounded to three decimal places as needed.)

(d) decide whether to reject or fail to reject the null hypothesis.
\\(\boxed{}\\) the null hypothesis.

Explanation:

Step1: Determine Degrees of Freedom

The formula for degrees of freedom (df) for two - sample t - test (when population variances are assumed equal) is \(df=n_1 + n_2-2\). Here, \(n_1 = 13\) and \(n_2=11\), so \(df=13 + 11-2=22\).

Step2: Find Critical Value \(t_0\)

For a one - tailed test with \(\alpha = 0.05\) and \(df = 22\), we look up the critical value \(t_0\) from the t - distribution table. The critical value \(t_{0.05,22}\approx1.717\).

Step3: Compare Test Statistic and Critical Value

The standardized test statistic \(t = 2.072\). We compare \(t\) with \(t_0\). Since \(2.072>1.717\) (i.e., \(t>t_0\)), we reject the null hypothesis.

Answer:

Reject