Question

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 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 55 68 59 67 40

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?

H0:  pmon = ptue = pwed = pthur = pfri = 1/5.H0: At least one of the probabilities doesn't equal 1/5.     H0: None of the probabilities are equal to 1/5.H0:  pmon = 0.55, ptue = 0.68, pwed = 0.59, pthur = 0.67, pfri = 0.40.

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

χ2

=

(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.

a)

H0:  pmon = ptue = pwed = pthur = pfri = 1/5.H0: At least one of the probabilities doesn't equal 1/5

b)

applying chi square test:

 relative observed Expected Chi square category frequency Oi Ei=total*p R2i=(Oi-Ei)2/Ei mon 0.200 55.000 57.800 0.136 tue 0.200 68.000 57.800 1.800 wed 0.200 59.000 57.800 0.025 thu 0.200 67.000 57.800 1.464 fri 0.200 40.000 57.800 5.482 total 1.000 289 289 8.907

χ2 =8.91

c)

p value =0.0635 (please try 0.0634 if this comes wrong)

d)

fail to reject H0

e)

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