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)

	Monday	Tuesday	Wednesday	Thursday	Friday
Count	53	68	55	65	48

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: 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 = 0.53, p_tue = 0.68, p_wed = 0.55, p_thur = 0.65, p_fri = 0.48.

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.

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

using minitab>stat>tables>chi square >goodness of fit

we have

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: count

Using category names in day

Test Contribution
Category Observed Proportion Expected to Chi-Sq
monday 53 0.2 57.8 0.39862
tuesday 68 0.2 57.8 1.80000
wednesday 55 0.2 57.8 0.13564
thursday 65 0.2 57.8 0.89689
friday 48 0.2 57.8 1.66159

N DF Chi-Sq P-Value
289 4 4.89273 0.298
ans a ) the null hypothesis is given by

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

Ans b ) the value of the test statistic χ² = 4.89

ans c )   the P-value of the test statistic = 0.298

ans d ) since p value is greater than 0.05

so fail to reject H₀    

Ans e ) We have proven that the number of customers is evenly distributed across the five weekdays.