EXCEL 

OneWay ANOVA 











Perform a OneWay 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 OneWay 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 
Pvalue 
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: 


H_{0}: μ_{1} = μ_{2 }= μ_{3} 


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 FValue: 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. 













© copyright 2004 Elisabeth Knowlton