Chapter 11.6, Problem 12E

To determine

(a)

To Find:

All the values for the ANOVA table.

Answer to Problem 12E

Solution:

Completed ANOVA table is-

	SS	df	MS	F	P-value	F crit
Treatments (T)	1.24218	2	0.62109	2.2949	0.12949	3.55456
Error (E)	4.87151	18	0.27064
Total	6.1137	20

Explanation of Solution

Given:

A regional manager wants to know if there is a difference between the mean amounts of time that customers wait in line at the drive-through window for the three stores in her region. She samples the wait limes at each store. Her data are given in the following table. Use an ANOVA test to determine if there is a difference between the mean wait times for the three stores, at the 0.05 level of significance.

Drive-Through Wait Times (in Minutes)
Store 1	Store 2	Store 3
2.34	2.87	1.32
1.23	1.94	1.45
1.89	2.36	1.78
2.31	1.85	2.01
3.02	1.75	2.45
1.95	2.82	1.92
2.45	3.32	1.83

Enter the data into Microsoft Excel, including the column headings into columns A, B, and C. Under the Data tab, choose Data Analysis. In the Data Analysis menu, choose ANOVA : Single Factor. Fill out the ANOVA table and then click ok, the ANOVA table will be displayed along with a summary table, which contains the descriptive statistics for each sample.

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Store 1	7	15.19	2.17	0.30997
Store 2	7	16.91	2.41571	0.36336
Store 3	7	12.76	1.82286	0.13859
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	1.24218	2	0.62109	2.2949	0.12949	3.55456
Within Groups	4.87151	18	0.27064
Total	6.1137	20

In the ANOVA table, Between Groups corresponds to Treatments (T) and Within Groups corresponds to Error (E).

Therefore, the completed ANOVA table is-

	SS	df	MS	F	P-value	F crit
Treatments (T)	1.24218	2	0.62109	2.2949	0.12949	3.55456
Error (E)	4.87151	18	0.27064
Total	6.1137	20

To determine

(b)

The conclusion and interpret the decision.

Answer to Problem 12E

Solution:

The null hypothesis is accepted and it is concluded that there is no sufficient evidence, at the 0.05 level of significance, to support the claim that there is a difference between the mean wait times for the three stores.

Explanation of Solution

Given:

A regional manager wants to know if there is a difference between the mean amounts of time that customers wait in line at the drive-through window for the three stores in her region. She samples the wait limes at each store. Her data are given in the following table. Use an ANOVA test to determine if there is a difference between the mean wait times for the three stores, at the 0.05 level of significance.

Drive-Through Wait Times (in Minutes)
Store 1	Store 2	Store 3
2.34	2.87	1.32
1.23	1.94	1.45
1.89	2.36	1.78
2.31	1.85	2.01
3.02	1.75	2.45
1.95	2.82	1.92
2.45	3.32	1.83

Calculation:

Rejection Region for ANOVA tests

Reject the null Hypothesis, ${\text{H}}_{\text{0}}$, if :

${\text{F > F}}_{\text{α}}$

Or $\text{P - value < α}$.

Let store 1, store 2, and store 3 be the population 1, 2 and 3 respectively and let ${\text{μ}}_{\text{1}}{\text{, μ}}_{\text{2}}{\text{, and μ}}_{\text{3}}$ be the mean amount of time for population 1, 2, and 3 respectively.

Let the null and alternative hypothesis for this test is

$H_0:{\mu }_1={\mu }_2={\mu }_3$

H₁ : At least two means are different from each other

Level of significance is 0.05, thus $\text{α = 0}\text{.05}$. From the ANOVA table P-value is 0.12949. Since $\text{0}\text{.12949 > 0}\text{.05}$, null hypothesis is accepted. Thus, the regional manager can conclude that there is no sufficient evidence, at the 0.05 level of significance, to support the claim that there is a difference between the mean wait times for the three stores.