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	26	22	19	24	19

Copy Data

Step 1 of 10:

State the null and alternative hypotheses.

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 hypotheses 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 0.025 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 0.025 level of significance.

Step 10 of 10:

State the conclusion of the hypothesis test at the 0.025 level of significance.

Answer 1

Step 1 of 10:

Null hypothesis: Ho: number of service calls are uniformly distributed over days of the week

Alternate hypothesis: Ha: number of service calls are not uniformly distributed over days of the week

step 2: expected proportion pi =0.20

Step3 of 10:

Ho: pMon=ptue=pwed =pthu=pFri=0.20

Ha: at least pi is different from 0.20

Step 4: expected value =Np =110*0.2 =22

Step 5: expected value =Np =110*0.2 =22

step 6:

applying chi square goodness of fit test:

	relative	observed	Expected		Chi square
Category	frequency(p)	Oi	Ei=total*p		R2i=(Oi-Ei)2/Ei
1	0.2000	26	22.00		0.727
2	0.2000	22	22.00		0.000
3	0.2000	19	22.00		0.409
4	0.2000	24	22.00		0.182
5	0.2000	19	22.00		0.409
total	1.00	110	110		1.73
test statistic X2=		1.727

Step 7 of 10:

degree of freedom =categories-1=

4

Step 8 of 10:

for 0.025 level and 4 df :crtiical value X2 =

11.143

from excel: chiinv(0.025,4)

step 9:

Fail to reject the null (since test statistic < critical value)

step 10)

we do not have sufficient evidence to conclude that,,,,,,,,,,,,,