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:  pmon = ptue = pwed = pthur = pfri = 1/5.H0: None of the probabilities are equal to 1/5.    H0:  pmon = 0.55, ptue = 0.68, pwed = 0.57, pthur = 0.65, pfri = 0.44.H0: At least one of the probabilities doesn't equal 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.

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 H0fail 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   55     68     57     65     44  

Explanation / Answer

(a) The Test: Below are the null and alternate hypothesis

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

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

(b) Test statistics is given by

2 = [ (Oi - Ei)2 / Ei ]

2 = 6.1384

(c) p-value

p-value = P(2 > 6.1384) = 0.1890

(d) As p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis

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

Expected Count(Ei) Actual Count(Oi) (Ei-Oi)^2 (Ei-Oi)^2/Ei Monday 57.8 55 7.84 0.1356 Tuesday 57.8 68 104.04 1.8000 Wednesday 57.8 57 0.64 0.0111 Thusday 57.8 65 51.84 0.8969 Friday 57.8 44 190.44 3.2948 6.1384

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