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 0.01 significance level.

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

H0:  pmon = 0.43, ptue = 0.48, pwed = 0.57, pthur = 0.75, pfri = 0.66.

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.

(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   43     48     57     75     66  

Explanation / Answer

Ans:

a)

H0:Number of customers is evenly distributed across the five weekdays

or

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

Ha:Number of customers is not evenly distributed across the five weekdays

b)

Expected count=289/5=57.8

Test statistic:

Chi square score=11.744

c)df=5-1=4

p-value=CHIDIST(11.744,4)=0.0194

d)

As,p-value>0.01,we fail to reject 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.

Observed(O) Expected(E) (O-E)^2/E Monday 43 57.8 3.790 Tuesday 48 57.8 1.662 Wednesday 57 57.8 0.011 Thursday 75 57.8 5.118 Friday 66 57.8 1.163 Total 289 289 11.744

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