Question

12. There are four McBurger restaurants in the Columbus, Georgia, area. The numbers of burgers sold at the respective restaurants for each of the last 6 weeks are shown below. At the 0.05 significance level, is there a difference in the mean number sold among the four restaurants when the factor of week is considered?

	Restaurant
Week	Metro	Interstate	University	River
1	124	160	320	190
2	234	220	340	230
3	430	290	290	240
4	105	245	310	170
5	240	205	280	180
6	310	260	270	205

Complete the ANOVA table. (Round your SS, MS to 2 decimal places, and F to 3 decimal places. Round df to nearest whole number.)

Answer 1

The summary statistics obtained from the given data are as below.

	Metro	Interstate	University	River
Total	1443	1380	1810	1215
n	6	6	6	6
Mean	240.50	230.00	301.67	202.50
Sum Of Squares	72715.50	10350.00	3483.33	3887.50

The Hypothesis:

H0: There is no difference between the mean sales in the four restaurant.

Ha: The mean sales of at least one restaurant is different from the others..

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The ANOVA table is as below.

Source	SS	DF	Mean Square	F	Fcv	p
Between	31533.23	3	10511.08	2.325	3.10	0.1056
Within/Error	90436.33	20	4521.82
Total	121969.57	23

The p value is calculated for F = 2.325 for df1 = 3 and df2 = 20

The Fcritical is calculated at = 0.05 for df1 = 3 and df2 = 20

The Decision Rule:

If Ftest is > F critical, Then Reject H0.

Also if p-value is < , Then reject H0.

The Decision:

Since Ftest (2.33 is > F critical (3.10), We Fail to Reject H0.

Also since p-value (0.1056) is > (0.05), We Fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 95% level of significance to conclude that the mean sales of at least one restaurant is different from the others.

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Calculations For the ANOVA Table:

Overall Mean = [240.5 + 230 + 301.67 + 202.5] / 4 = 243.67

SS treatment = SUM [n* ( - overall mean)²] = 6 * (240.5 - 243.67)² + 6 * (230 - 243.67)² + 6 * (301.67 - 243.67)² + 6 * (202.5 - 243.67)² = 31533.23

df1 = k - 1 = 4 - 1 = 3

MSTR = SS treatment/df1 = 31533.23 / 3 = 10511.08

SSerror = SUM (Sum of Squares) = 72715 + 10350 + 3483.33 + 3887.5 = 90436.33

df2 = N - k = 24 - 4 = 20

Therefore MS error = SSerror/df2 = 90436.33 / 20 = 4521.82

F = MSTR/MSE = 31533.23 / 4521.82 = 2.325

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