EXCEL

                                   One-Way ANOVA

Perform a One-Way ANOVA Analysis:

Assumptions:

1.  The populations are normally distributed.

2.  The populations have equal standard deviations.

3.  The samples are select independently.

The following are the travel times in minutes of ambulances for 3 different routes.

A

B

C

1

Route 1

Route 2

Route 3

2

2.1

1.8

2.4

3

2.2

2

2.1

4

1.9

2.1

2

5

1.8

2.2

2

6

2

1.9

Enter the data; place the travel times for Route 1 in column A, Route 2 in column B  

and Route 3 in column C.

Compare the travel times of the three different routes:

Perform a ANOVA test at the 5% significance level to determine if there exists a   

significant difference between the travel times of the different routes.

Perform One-Way ANOVA Test with EXCEL:

1.  Enter the data into columns A, B and C.

2.  Select Tools and Data Analysis.

3.  Select ANOVA: Single Factor and click OK.

4.  Input the following in the dialog box:

    a.  Input Range:  A1:C6.

    b.  Click on Labels (since you included the labels in your input range).

    c.  Alpha: 0.05.

    d.  Check Output Range and specify the range, if you would like the output on the same   

        spreadsheet. Output Range is A10.  Choose a New Work Sheet Ply if you like a

        new sheet.

    e.  Click OK and you will see the output below.

Anova: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Route 1

5

10

2

0.025

Route 2

4

8.1

2.025

0.029167

Route 3

5

10.4

2.08

0.037

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

0.016642857

2

0.00832143

0.272834

0.766223

3.982308

Within Groups

0.3355

11

0.0305

Total

0.352142857

13

Interpret Your Results:

1. State Hypothesis:

 H₀:  μ₁ = μ₂ = μ₃

 H_A:  At least one of the mean travel times are not equal.

2.  Level of Significance:  α = 0.05.

3. Determine Test Statistic:  F = .272834

4. Decision Rule:  Critical F-Value:  Reject Ho if F_2,11 > 3.982.

Degrees of freedom in the numerator is k - 1 = 3 - 1 = 2

Degrees of freedom in the denominator is n - k = 14 - 3 = 11

5. Make a Decision:  Do not reject Ho since F= .2728 does fall into the rejection region.

6. Conclusion:  There is not enough evidence at the 5% significance level to conclude 

that there is a significant difference in the travel times of the ambulances on the three

different routes.