Question

2.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 66 59 63 46 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: None of the probabilities are equal to 1/5. H0: pmon = 0.55, ptue = 0.66, pwed = 0.59, pthur = 0.63, pfri = 0.46. H0: pmon = ptue = pwed = pthur = pfri = 1/5. 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. 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 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.

Answer 1

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 goodness of fit test:

		% *students			pearson
	relative	observed	Expected		Chi square
category	frequency(p)	Oi	Ei=total*p		R2i=(Oi-Ei)2/Ei
1	0.20	55.0	57.80		0.136
2	0.20	66.0	57.80		1.163
3	0.20	59.0	57.80		0.025
4	0.20	63.0	57.80		0.468
5	0.20	46.0	57.80		2.409
total	1.000	289	289		4.2007
test statistic X2 =		4.201

c)

p value =

0.3795

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.