Customer Distribution by Weekday: A drop-in auto repair shop staffs the same num

Question

Customer Distribution by Weekday: A drop-in auto repair shop staffs the same number of mechanics on every weekday (weekends are not counted here). One of the mechanics thinks this is a bad idea because he suspects the number of customers is not evenly distributed across these days. For a sample of 289 customers, the counts by weekday are given in the table.

Number of Customers by Day (n = 289)

The Test: Test the claim that the number of customers is not evenly distributed across the five weekdays. Test this claim at the 0.05 significance level.

(a) What is the null hypothesis for this test in terms of the probabilities of the outcomes?

H0: At least one of the probabilities doesn't equal 1/5.

H0:  pmon = ptue = pwed = pthur = pfri = 1/5.   

H0: None of the probabilities are equal to 1/5.

H0:  pmon = 0.51, ptue = 0.68, pwed = 0.57, pthur = 0.67, pfri = 0.46.

(b) What is the value of the test statistic? Round to 3 decimal places unless your software automatically rounds to 2 decimal places.

2 =

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places unless your software automatically rounds to 3 decimal places.
P-value =  

(d) What is the conclusion regarding the null hypothesis?

reject H0

fail to reject H0    

(e) Choose the appropriate concluding statement.

We have proven that the number of customers is evenly distributed across the five weekdays.

The data supports the claim that the number of customers is not evenly distributed across the five weekdays.    

There is not enough data to support the claim that the number of customers is not evenly distributed across the five weekdays.

Monday Tuesday Wednesday Thursday Friday Count   51     68     57     67     46  

Explanation / Answer

a) Since we are testing if the number of customers is not evenly distributed across the five weekdays, our null hypothesis is H0:  pmon = ptue = pwed = pthur = pfri = 1/5 Also, the alternate hypothesis is Ha:  Probabilities vary over weekdays b) We will use the chi-square test and find the test statistic chi square (2) Using 1/5 = 0.2 as the estimated proportion, we calculate the estimated count for each week day which is 0.2 * 289 Mon Tue Wed Thurs Fri Observed Count (O) 51 68 57 67 46 Estimated Count (E ) 57.8 57.8 57.8 57.8 57.8 (O - E)2 46.24 104 0.64 84.64 139.2 (O - E)2 / E 0.8 1.8 0.011 1.464 2.409 2 = [(O - E)2 / E] 2 = 6.48             Value of the test statistic c) Finding p-value of the test statistic 2 = 6.48 degrees of freedom = number of probabilities - 1 = 5 - 1 = 4 p-value for 2 = 6.48 and 4 degrees of freedom is 0.1661 p-value = 0.1661 d) We consider level of significance as 5% Since p-value 0.1661 > = 0.05, we do not reject the null hypothesis Conclusion : Fail to reject Ho e) Appropriate concluding statement is There is not enough data to support the claim that the number of customers is not evenly distributed across the five weekdays.

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