Question

According to a recent government survey, the daily one-way commuting distance of U.S. workers averages 13 miles with a standard deviation of 13 miles. An investigator wishes to determine whether the national average describes the mean commute distance for all workers in the Chicago area. Commute distances are obtained for a random sample of 168 workers from this area, and the mean distance is found to be 10.5 miles. Test the null hypothesis at the.05 level of significance.

a. state the null hpothesis

b. state the alternative hypothesis

c. is this a one or two tail test? why?

d. will you conduct a t-test or z-test? why?

e. what is (are) the critical value (s)?

f. What is the decision rule?

g. calculate the statistic

h. what is your decision rule regarding null hypothesis and why?

i. interpretation of the results

Answer 1

Let X be the daily one-way commuting distance of U.S. workers(in miles).

Given:

n=168

At 5% level of significance we want to test the claim that the national average describes the mean commute distance for all workers in the Chicago area. That is the mean commute distance for all workers in the Chicago area is 13 miles.

a) The  null hypothesis is

b) Alternative hypothesis is:

c) This is a two tail test. Since the alternative hypothesis does not state whether the mean is greater than of less than the given value.

d) We will used z test, since the population standard deviation is known and the sample size is greater than 30.

e) The critical value at 5% level of significance is:

f) We reject the null hypothesis if

g) The test statistic is:

Therefore,

h) Since , we reject the null hypothesis at 5% level of significance.

i) Therefore we have sufficient evidence to say that the national average does not describes the mean commute distance for all workers in the Chicago area.