Question

2. See problem 1. Corporate sends 20 "mystery shoppers" to a certain location to see if the location's overall speed is consistent with company standards. There will be a problem if the location consistently takes longer than expected to process orders, so the company wishes to test the null hypothesis

?0:? = 4.5

versus the alternate

?1:? > 4.5

a. To test the null, the company sends 20 mystery shoppers and computes the mean time ? needed to process those 20 orders. If ? = .05 is used, find the rejection region of the test; that is what sample mean would convince corporate that the location's order processing is slower than expected?

b. See part a. In the sample of 20 mystery shoppers, it is found that ? = 4.9 minutes. Based on part a, decide whether or not to reject the null, and then state what should be concluded.

c. See part b. Run a Z-test to answer the same question. Find the p-value of the test, decide whether to reject the null hypothesis, and then state what should be concluded.

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PROBLEM 1 WITH ANSWERS:

1.

A take-out chain restaurant believes that the time needed to process any given customer order (place the order, process payment, give the customer the food) should be normally distributed with mean 4.5 minutes and standard deviation 0.75 minutes.

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a. If the restaurant sampled 20 orders and computed the mean time ? to process those 20 orders, find  the mean and standard deviation of the random variable ?.

MEAN = 4.5, SD = 0.1677

b. If the restaurant sampled 20 orders and computed the total time ∑ ? needed to process those 20 hours, find the mean and standard deviation of the random variable ∑ ?.

MEAN = 90, SD = 3.3541

Answer 1

(a) The rejection region of the test is:

Reject ?0 if t > 2.09.

The sample mean is 4.9 which convinces corporate that the location's order processing is slower than expected.

(b) The test statistic, t = (x - µ)/s/√n

t = (4.9 - 4.5)/0.75/√20 = 2.385

The p-value is 0.0138.

Since the p-value (0.0138) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the location's order processing is slower than expected.

(c) The test statistic, z = (x - µ)/σ/√n

z = (4.9 - 4.5)/0.75/√20 = 2.39

The p-value is 0.0085.

Since the p-value (0.0085) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the location's order processing is slower than expected.

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