Trina Appleton, Jackson\'s Food Marts marketing director, read from a national g

Question

Trina Appleton, Jackson's Food Marts marketing director, read from a national grocery magazine that the average wait time at a convenience food mart is 2 1/2 minutes. She took a random sample of 25 shoppers at the Jackson's Food Mart on Overland Road. The average wait for the sample shoppers was 2.2 minutes with a standard deviation of 1.5 minutes. She asked you if the average wait time at the Overland Jackson's was less than the national average stated in the magazine. Run a hypothesis test on the sample with a 0.05 significance level and write a memo the Ms. Appleton stating your conclusion. Do your work in Excel and use the six-step process.

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: > 2.50
Alternative hypothesis: < 2.50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.30
DF = n - 1

D.F = 24
t = (x - ) / SE

t = - 1

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of - 1.0.

Thus the P-value in this analysis is 0.1587.

Interpret results. Since the P-value (0.1587) is greater than the significance level (0.05), we cannot reject the null hypothesis.

We do not have sufficient evidence in the favor of the claim that the average wait time at the Overland Jackson's was less than the national average stated in the magazine.

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