In: Math
Consider the results from a completely randomized design showing commuting times in three states. Use an appropriate Excel ANOVA tool, to test for any significant differences in commuting times between the three states. Use α = 0.05.
| Illinois | Ohio | Texas | 
| 26.8 | 27.5 | 10.1 | 
| 17.6 | 28.9 | 18.8 | 
| 27 | 19.1 | 31.4 | 
| 20 | 36.9 | 44.2 | 
| 50.7 | 40.8 | 24.6 | 
| 24.4 | 9.5 | 29.5 | 
| 36.8 | 37.4 | 38.1 | 
| 42.2 | 38.9 | 30.3 | 
| 26.3 | 46.2 | 11.7 | 
| 14 | 35.8 | 35.8 | 
| 28.5 | 20.7 | 22.4 | 
| 36.9 | 37.8 | 17 | 
| 25.6 | 49.7 | 15.4 | 
| 25.9 | 44.3 | 15.4 | 
| 29.5 | 12.1 | 6.8 | 
| 29.7 | 43.7 | 14.8 | 
| 30.5 | 35.9 | 59.3 | 
| 20 | 30.2 | 5.3 | 
| 23.2 | 8.5 | 0.6 | 
| 20.7 | 34.6 | 20.7 | 
| 6.2 | 37.9 | 18.6 | 
| 44.2 | 50.9 | 24.9 | 
| 28.2 | 24.2 | 9.3 | 
| 28.8 | 39.1 | 11.9 | 
| 16.6 | 20.4 | 19.6 | 
| 20.2 | 12.4 | 31 | 
| 13.1 | 28 | 25.9 | 
| 16.9 | 28.4 | 52.6 | 
| 32.4 | 19.4 | 38.3 | 
| 19.6 | 42.5 | 34 | 
| 12.8 | 27.2 | 24.9 | 
| 30.2 | 22.6 | 32.1 | 
| 65.1 | 50.8 | 43 | 
| 25.5 | 34.1 | 31.1 | 
| 17.5 | 27.1 | 16.8 | 
| 11.1 | 38.9 | 34.1 | 
| 48.8 | 28.7 | 40.4 | 
| 38.9 | 54.2 | 29.4 | 
| 23.1 | 30.6 | 9.8 | 
| 21.6 | 15.9 | 19.5 | 
| 22.3 | 15.1 | 9.6 | 
| 27.3 | 30.1 | 21.6 | 
| 30.7 | 32.2 | 26.5 | 
Testing 
 against 
 at least one mean
is different
Using Excel, (Data -> Data Analysis -> ANOVA : single factor), we get the following output -
Now,

The value of F statistic F = 3.57436
and p-value = 0.03091
Since p-value < 0.05, so we reject the null hypothesis at 5% level of significance and we can conclude that there is significant difference in commuting times between the three states.