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Welch’s ANOVA is an alternative to the typical when the is violated.

The following step-by-step example shows how to perform Welch’s ANOVA in R.

Step 1: Create the Data

To determine if three different studying techniques lead to different exam scores, a professor randomly assigns 10 students to use each technique (Technique A, B, or C) for one week and then makes each student take an exam of equal difficulty. 

The exam scores of the 30 students are shown below:

#create data frame
df <-data.frame(group = rep(c('A','B', 'C'), each=10),
                score = c(64, 66, 68, 75, 78, 94, 98, 79, 71, 80,
                          91, 92, 93, 85, 87, 84, 82, 88, 95, 96,
                          79, 78, 88, 94, 92, 85, 83, 85, 82, 81))

#view first six rows of data frame
head(df)

   group score
1      A    64
2      A    66
3      A    68
4      A    75
5      A    78
6      A    94

Step 2: Test for Equal Variances

Next, we can perform to determine if the variances between each group is equal.

If the of the test statistic is less than some significance level (like α = .05) then we can reject the null hypothesis and conclude that not all groups have the same variance.

To perform Bartlett’s test we can use the bartlett.test function in base R, which uses the following syntax:

bartlett.test(formula, data)

Here’s how to use this function in our example:

#perform Bartlett's test
bartlett.test(score ~ group, data = df)

	Bartlett test of homogeneity of variances

data:  score by group
Bartlett's K-squared = 8.1066, df = 2, p-value = 0.01737

The p-value (.01737) from Bartlett’s test is less than α = .05, which means we can reject the null hypothesis that each group has the same variance.

Thus, the assumption of equal variances is violated and we can proceed to perform Welch’s ANOVA.

Step 3: Perform Welch’s ANOVA

To perform Welch’s ANOVA in R, we can use the oneway.test() function from base R as follows:

#perform Welch's ANOVA
oneway.test(score ~ group, data = df, var.equal = FALSE)

	One-way analysis of means (not assuming equal variances)

data:  score and group
F = 5.3492, num df = 2.00, denom df = 16.83, p-value = 0.01591

The overall p-value (.01591) from the ANOVA table is less than α = .05, which means we can reject the null hypothesis that the exam scores are equal between the three studying techniques.

We can then perform a post-hoc test to determine which group means are different. Refer to the following tutorials to see how to perform various post-hoc tests in R:

Reference to determine which post-hoc test is best to use depending on your situation.

Additional Resources