Question

Can someone please show me how to do these type problems using Excell? Please?

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	57	67	42

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: At least one of the probabilities doesn't equal 1/5.      H₀: p_mon = p_tue = p_wed = p_thur = p_fri = 1/5. H₀: p_mon = 0.55, p_tue = 0.68, p_wed = 0.57, p_thur = 0.67, p_fri = 0.42.

(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

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 7.7301038 0.102

Hence,

a) H₀: p_mon = p_tue = p_wed = p_thur = p_fri = 1/5

b) Test statistic = 7.73

c) P - value = 0.102

d) Fail to reject Ho

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