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Customer Distribution by Weekday: A drop-in auto repair shop staffs the same number of mechanics on...

Customer Distribution by Weekday: A drop-in auto repair shop staffs the same number of mechanics on every weekday (weekends a

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) | Monday Tuesday Wednesday Thursday Friday Count 51 68 55 48 67 The Test: Test the claim that the number of customers is not evenly distributed across the five weekdays. Test this claim at the 0.01 significance level. (a) What is the null hypothesis for this test in terms of the probabilities of the outcomes? Ho: At least one of the probabilities doesn't equal 1/5. Ho: Pmon = 0.51, Ptue = 0.68, pwed = 0.55, Pthur = 0.67, Pfri = 0.48. Ho: None of the probabilities are equal to 1/5. O Ho: Pmon = Ptue = Pwed = pthur = Pfri = 1/5. (b) What is the value of the test statistic? Round to 3 decimal places unless your software automatically rounds to 2 decimal places. x2 = (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 Ho fail to reject Ho (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.
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Answer #1

The statistical software output for this problem is :

Chi-Square goodness-of-fit results: Observed: Oi Expected: All cells in equal proportion N DF Chi-Square P-value 289 4 5.8615

Ho: Pmon = Ptue = Pwed = pthur = Peri = 1/5.

Test statistics \chi^2 = 5.86

P-value = 0.2097

fail to reject Ho

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