A Brief Demonstration of 1-Way Anova Using Minitab

Example:

Scientists conducted an experiment to test the effects of 5 different diets in turkeys. They randomly assigned 6 turkeys to each of the 5 diet groups and fed them for a fixed period of time:

Group / Weight gained (pounds)
Control diet / 4.1, 3.3, 3.1, 4.2, 3.6, 4.4
Control diet + level 1 of additive A / 5.2, 4.8, 4.5, 6.8, 5.5, 6.2
Control diet + level 2 of additive A / 6.3, 6.5, 7.2, 7.4, 7.8, 6.7
Control diet + level 1 of additive B / 6.5, 6.8, 7.3, 7.5, 6.9, 7.0
Control diet + level 2 of additive B / 9.5, 9.6, 9.2, 9.1, 9.8, 9.1

Step 1: Key in the data.

Step 2: Conduct 1-way Anova.

Minitab: Stat > Anova > One-way

Click the button “Graphs…” for assumption checking:

Click the button “Comparisons…” for multiple comparisons:

Then click OK to close all windows.

Step 3: Interpret outputs.

Assumption checking:

The assumption of equal variance seems invalid. You need to conduct a test to verify this assumption. You can also conduct a test for the normality assumption.

· Test for equal variance.

Ho: variances of residuals are equal vs. Ha: they are not equal

Minitab: Stat> Anova > Test for Equal variances

Output:

Test for Equal Variances: weight versus group

Bartlett's Test (normal distribution)

Test statistic = 6.52, p-value = 0.164

Levene's Test (any continuous distribution)

Test statistic = 2.23, p-value = 0.095

· Test for normality:

Ho: residuals follow a normal distribution vs. Ha: they do not follow a normal distribution

Output:

Since the assumptions are all passed, we can use Anova methods to analyze data.

Output:

One-way ANOVA: weight versus group

Source DF SS MS F P

group 4 103.038 25.760 81.67 0.000

Error 25 7.885 0.315

Total 29 110.923

S = 0.5616 R-Sq = 92.89% R-Sq(adj) = 91.75%

Since p-value is almost 0, we reject the null hypothesis that there is no difference among groups. We therefore do multiple comparisons to analyze further. See lecture notes for details how to interpret the outputs.