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:

 

 H0μ1 = μ2 = μ3

 

 HA:  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 F2,11 > 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.

 

 

 

 

 

 

 

 

 

 

 

 

 

copyright 2004 Elisabeth Knowlton