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Jackson's Wait Time Trina Appleton, Jackson's Food Marts marketing director, read from a national grocery magazine...

Jackson's Wait Time

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 2.5 minutes. She took a random sample of 20 shoppers at the Jackson's Food Mart on Overland Road. The average wait for the sample shoppers was 2.8 minutes with a standard deviation of 1.5 minutes. She asked you if the average wait time at the Overland Jackson's was really less than the national average stated in the magazine or was it just sampling error. Run a hypothesis test on the sample with a 0.05 significance level and write a memo the Ms. Appleton stating your conclusion. (Hint: This is a lower-tailed, t-test.)

Do your work in Excel and use the six-step process.

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Answer #1

To test 2.5 against H1 : μ < 2.5

Here

sample mean ar{x} = 2.8

sample standard deviation s = 1.5

and sample size n20

The test statistic can be written as

t=vili- which under H0 follows a t distribution with n-1 df.

We reject H0 at 5% level of significance if P-value < 0.05

Now,

The value of the test statistic tots = 0.894427

P-value P(tn-1 < tobs) = P(t19 < 0.894427) = 0.808858

Since P-value > 0.05, so we fail to reject H0 at 5% level of significance and we can conclude that the average wait time at the Overland Jackson's was definitely not less than the national average stated in the magazine.

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