How to Perform One-Way ANOVA in SAS

Nikki Kamouneh

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For a fuller treatment on what one-way ANOVA is, see our One-Way ANOVA tutorial. This tutorial will go over how to conduct one-way ANOVA using SAS.

The data we are using can be downloaded as a file from our GitHub repository.

We want to study the effectiveness of different treatments on anxiety. We collect a sample of 75 subjects in the following categories:

  • No treatment (\(n_1\) = 27).
  • Biofeedback (\(n_2\) = 24).
  • Cognitive-behavioral Treatment (\(n_3\) = 24).

The dependent variable is anxiety levels. The null hypothesis is that all three means are equal.

  • \(H_0\) : \(\mu_1 = \mu_2 = \mu_3\)
  • \(H_A\) : At least two \(\mu_i\)’s are different

If even two of the means are significantly different, we reject the null hypothesis.

One-Way ANOVA in SAS

We can visualize the data with bar graphs to check for normality and equality of variances across groups. SAS will print this automatically when we run our ANOVA.

We can conduct our ANOVA and view the results below:

proc anova data=anova;  
class treatment;  
model anxiety = treatment; 
means treatment / tukey;

The class statement tells SAS that treatment is a categorical variable. The means statement will provide all pairwise comparisons, with p-values adjusted using Tukey’s method.

This will give us the following figure:

Boxplot of anxiety by each treatment type

Normality looks good. The equality of variances assumption appears suspect (this can be tested). We will also get the following output:

Class level information

ANOVA Results

Tukey Results

The first two tables give the class level variable (treatment) and its possible levels (1, 2 or 3), and the number of observations used (\(n=75\)).

The next three tables give the results of the ANOVA. We find the \(F\) statistic to be 8.75, which we compare to an \(F\) distribution with \(df_1\) = 2 and \(df_2\) = 72. The cut-off for significance given these \(df\) is 3.124, so we have a significant result (\(p<.001\)). We can reject the null hypothesis that all \(\mu_i\)’s are equal.

Although we see that some of the means are different, the \(F\)-test doesn’t tell us which means are different. We therefore carry out post hoc contrasts consisting of all pairwise comparisons. The last table give us the results based on the Tukey-adjusted p-values. We find that group 1 and group 3 are significantly differently.

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