Question

A fire insurance company thought that the mean distance from a home to the nearest fire department in a suburb of Chicago was at least 5.9 miles. It set its fire insurance rates accordingly. Members of the community set out to show that the mean distance was less than 5.9 miles. This, they thought, would convince the insurance company to lower its rates. They randomly indentified 60 homes and measured the distance to the nearest fire department from each. The resulting sample mean was 5.3. If σ = 2.2 miles, does the sample show sufficient evidence to support the community's claim at the α = .05 level of significance?

(a) Find z. (Give your answer correct to two decimal places.)

(ii) Find the p-value. (Give your answer correct to four decimal places.)

(b) State the appropriate conclusion. Reject the null hypothesis, there is not significant evidence that the mean distance is less than 5.9 miles. Reject the null hypothesis, there is significant evidence that the mean distance is less than 5.9 miles. Fail to reject the null hypothesis, there is significant evidence that the mean distance is less than 5.9 miles. Fail to reject the null hypothesis, there is not significant evidence that the mean distance is less than 5.9 miles.

You may need to use the appropriate table in Appendix B to answer this question.

Answer 1

Here Z test and Left tail test will be acceptable as n=60 which is greater than 30, to support the assumption that the distribution is normal and random then,

The Hypotheses are:

Rejection Region:

Reject Ho if Z_obs<-Z_0.05= -1.645

a) Test Statistic:

Z is calculated as

b) P-value associated with Z score calculated using Excel, Calculator or Z table as

P value =0.0179

c) Conclusion:

Since the Z(obs) <Z(0.05) and P value is less than 0.05( Level of significance) hence Reject the null hypothesis, there is significant evidence that the mean distance is less than 5.9 miles.

Z table used to find P value shown below