Question

A local franchise of a national chain of day-old pastry stores recorded the number of customers that came in to the store for five weeks this fall. These data are recorded in the attached file. The national franchisor states that the percentage of customers is distributed as follows:   Sunday – 20%, Monday – 8%, Tuesday – 6%, Wednesday – 7%, Thursday – 12%, Friday – 22%, and Saturday – 25%. The manager doesn’t trust the franchisor and wants to show his store has different percentages. Use a χ² Goodness-of-Fit test to determine if the manager has evidence that his store is varies from the national percentages.

		Customers per day
		First five weeks after September
	Sun	53
	Mon	22
	Tue	18
	Wed	24
	Thu	36
	Fri	51
	Sat	83
	Sun	69
	Mon	14
	Tue	19
	Wed	24
	Thu	37
	Fri	57
	Sat	68
	Sun	60
	Mon	23
	Tue	16
	Wed	22
	Thu	34
	Fri	60
	Sat	72
	Sun	68
	Mon	22
	Tue	17
	Wed	25
	Thu	39
	Fri	64
	Sat	83
	Sun	84
	Mon	25
	Tue	15
	Wed	19
	Thu	36
	Fri	60
	Sat	81
		1500

Answer 1

SOLUTION

The Hypothesis:

H0: The sample has a distribution that agrees with the national percentage.

Ha: The sample has a distribution that is different from the national percentage.

The Test Statistic:

Each Expected value = (% / 100) * 1500

	Observed	Expected %	Expected	(O-E)2	(O-E)2/E
Sunday	334	20	300	1156	3.85333
Monday	106	8	120	196	1.63333
Tuesday	85	6	90	25	0.27778
Wednesday	114	7	105	81	0.77143
Thursday	182	12	180	4	0.02222
Friday	292	22	330	1444	4.37576
Saturday	387	25	375	144	0.38400
Total	1500.00	100.00	1500.00		11.318

$\chi^2$ test = 11.318

The Critical Value: The critical value at $\alpha$ = 0.05, df= n - 1 = 6

$\chi^2$critical = 11.07

The p value: The p value: The p value at $\chi^2$test = 11.318, df = 6; P value = 0.0454

The Decision Rule:

The Critical Value Method: If $\chi^2$test is >$\chi^2$ critical, then Reject H0.

The p - value Method: If p value is < $\alpha$ , Then Reject H0.

The Decision:  

The Critical Value Method: Since $\chi^2$test (11.318) is > $\chi^2$critical (11.07), We Reject H0.

The p - value Method: Since p value (0.0454) is < $\alpha$ (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that sample has a distribution that is different from the national percentage.