Question

A national grocer’s magazine reports the typical shopper spends 9 minutes in line waiting to check out. A sample of 19 shoppers at the local Farmer Jack’s showed a mean of 8.2 minutes with a standard deviation of 2.8 minutes.

Is the waiting time at the local Farmer Jack’s less than that reported in the national magazine? Use the 0.010 significance level.

1. What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

2. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

3. What is your decision regarding H₀?

• Reject H₀

• Do not reject H₀

Answer 1

Solution:

Given: A national grocer’s magazine reports the typical shopper spends 9 minutes in line waiting to check out.

Sample size = n = 19

Sample mean = minutes

Sample Standard Deviation = s = 2.8 minutes

Level of significance =

We have to test the waiting time at the local Farmer Jack’s less than that reported in the national magazine.

that is test if mean .

Thus hypothesis are:

Vs

Part a) What is the decision rule?

Level of significance = and df = n - 1 = 19 - 1 = 18

Look in t table for one tail area = 0.01 and df = 18

t critical value = 2.552

Since this is left tailed test, t critical value = -2.552

Decision Rule:

Reject null hypothesis H0, if t test statistic value < t critical value = -2.552 , otherwise we fail to reject H0

Part b) Compute the value of the test statistic.

Part c) What is your decision regarding H₀?

Since t test statistic value = > t critical value = -2.552 , that is t test statistic value does not fall in rejection region, thus we failed to reject H0.

Thus correct answer is:

Do not reject H₀