Question

A telephone company claims that the service calls which they receive are equally distributed among the five working days of the week. A survey of 110 randomly selected service calls was conducted. Is there enough evidence to refute the telephone company's claim that the number of service calls does not change from day-to-day?

Days of the Week	Mon	Tue	Wed	Thu	Fri
Number of Calls	24	26	17	20	23

Copy Data

Step 2 of 10:

What does the null hypothesis indicate about the proportions of service calls received each day?

Step 3 of 10:

State the null and alternative hypothesis in terms of the expected proportions for each category.

Step 4 of 10:

Find the expected value for the number of service calls received on Monday. Round your answer to two decimal places.

Step 5 of 10:

Find the expected value for the number of service calls received on Tuesday. Round your answer to two decimal places.

Step 6 of 10:

Find the value of the test statistic. Round your answer to three decimal places.

Step 7 of 10:

Find the degrees of freedom associated with the test statistic for this problem.

Step 8 of 10:

Find the critical value of the test at the 0.010.01 level of significance. Round your answer to three decimal places.

Step 9 of 10:

Make the decision to reject or fail to reject the null hypothesis at the .01 level of significance.

Step 10 of 10:

State the conclusion of the hypothesis test at the .01 level of significance.

Answer 1

As we are testing here whether the number of service calls does not change from day-to-day, therefore the null and the alternative hypothesis here are given as:

H0: The number of service calls does not change from day-to-day

Ha: The number of service calls do change from day to day.

Total Frequency = 24 + 26 + 17 + 20 + 23 = 110

Therefore the expected number of calls received on monday is computed as: 110/5 = 22
Therefore 22 is the expected number of calls received on monday.

Therefore the expected number of calls received on tuesday is computed as: 110/5 = 22
Therefore 22 is the expected number of calls received on tuesday.

The chi square test statistic value here is computed as:

Therefore 2.2727 is the test statistic value here.

The degrees of freedom here is computed as:
= Number of categories - 1 = 4

Tehrefore 4 is the number of categories here.

For 0.01 level of significance, we have from chi square distribution tables here:

Therefore 13.2767 is the required critical value here.

As the test statistic value is lower than the critical value here, therefore it lies in the non rejection region. Therefore we cannot reject the null hypothesis here.

Therefore we dont have sufficient evidence here to reject the claim that the number of service calls does not change from day-to-day