Question:
Problem 9.62:The American Water Works Association reports that, on average, men use between 10 and 15 gallons of water daily to shave when they leave the water running. Suppose the data below are the numbers of gallons of water used in a day to shave by 12 randomly selected men and the data come from a normal distribution of data. Use these data and a 5% level of significance to test to determine whether the population variance for such water usage is 2.5 gallons.

Given:

Manual Solutions:

Step 1:
This is a two -tailed test. The null and alternative hypothesis are:

Ho:
Ha:

Step 2:
The test statistic is:

Step 3:
Alpha of .05, since test is two-tailed divide it by 2, alpha is = .025

Step 4:

The degrees of freedom (df): 12 - 1 = 11. As follows,there are two critical chi-square values:

The decision is to reject the null hypothesis if the observed value of the test statistic is less than 3.81575 or greater than 21.9200.

Step 5:

Step 6:

The sample variance is:

The calculated chi-square value is:

Step 7:

This calculated chi-square value is in the rejection region, because: X2.975, 11 = 3.81575 < X2calculated = 45.89 > X2 .025, 11 = 21.9200, the decision is to reject the null hypothesis. The graph below shows the chi-square distribution with the critical values, the rejection region, the non-rejection region, the value of alpha, and the calculated chi-square value:



Step 8:
Based on this result, the variance of water usage of men is beyond 2.5 gallons.

Back to: Finals Main Page