Welch’s t-test is a nonparametric univariate test that tests for a significant difference between the mean of two unrelated groups. It is an adaptation of the Student’s t-test and is more reliable when the two samples have unequal variances and/or unequal sample sizes.
The Welch’s t-test is defined as:
- H0 (null hypothesis): the two population means are equal
- H1 (alternative hypothesis): the two population means are not equal
The choice to use Welch’s t-test over the independent samples t-test depends upon the violation of the assumption of homogeneity of variances. Welch’s t-test is more robust than the independent samples t-test against the heterogeneity of variances.
In this article, we will explain how to perform a Welch’s t-test in R, including setting up your data, testing assumptions, performing the test, and interpreting the results.
Preparing Your Dataset
Your data should be organized in a way where one column represents the dependent variable (the variable you’re interested in comparing across groups), and another column represents the grouping variable.
Let’s say you’re studying the effect of two different teaching methods on student’s scores. You randomly assign 50 students to two groups: method A and method B. After the study period, you give the students a test and record their scores. Here is an example of how you might set up this data in R:
# Set seed for reproducibility set.seed(123) # Create the dataset data <- data.frame( group = rep(c('A', 'B'), each = 25), score = c(rnorm(25, mean = 75, sd = 10), rnorm(25, mean = 80, sd = 15)) )
Before performing the Welch’s t-test, there are several assumptions that we need to check:
- Independence: The cases should be independent. In our case, because we’re randomly assigning students to each group, we can assume independence.
- Normality: Each group’s scores should follow a normal distribution. We can visually check this using a histogram or a Q-Q plot, or statistically with a Shapiro-Wilk test. Here’s how you might do this in R:
# Check distribution of scores in group A hist(data$score[data$group == 'A']) # Check distribution of scores in group B hist(data$score[data$group == 'B'])
3. Homogeneity of variances: The variances of the two groups should be equal. However, the Welch’s t-test is more robust against this assumption, and it is often used when this assumption is violated. We can check this assumption with Levene’s test:
library(car) # Perform Levene's test for equality of variances leveneTest(score ~ group, data = data)
Performing Welch’s t-Test
After checking the assumptions, you can perform the Welch’s t-test using the
t.test() function in R. The syntax for performing a Welch’s t-test is as follows:
t.test(dependent_variable ~ grouping_variable, data = your_data)
t.test() performs a Welch’s t-test (not assuming equal variances) when you pass in a formula. So, you don’t need to specify
var.equal = FALSE (which is what you would do to explicitly perform a Welch’s t-test).
Here’s how you can perform a Welch’s t-test with our dataset:
# Perform Welch's t-test t.test(score ~ group, data = data)
Interpreting the Results
After running the t-test, R provides an output that includes the t-value, degrees of freedom, p-value, confidence interval, and the mean of each group. Here’s an example of what the output might look like:
Welch Two Sample t-test data: score by group t = -2.2681, df = 42.822, p-value = 0.02848 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -9.3024609 -0.4898898 sample estimates: mean in group A mean in group B 74.19422 79.09039
Here’s how to interpret this output:
t: The t-value is the calculated difference represented in units of standard error. The greater the magnitude of T (either positive or negative), the greater the evidence against the null hypothesis. In this case, t is -2.2681.
df: This is the degrees of freedom, which is the number of independent pieces of information that went into calculating the estimate. In this case, df is 42.822.
p-value: The p-value is the probability of obtaining results as extreme as the observed results of a statistical hypothesis test, assuming that the null hypothesis is correct. A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis. In this case, the p-value is 0.02848, which is less than 0.05, so we reject the null hypothesis.
alternative hypothesis: This is the alternative hypothesis you specified (or the default). In this case, the alternative hypothesis was that the true difference in means is not equal to 0.
95 percent confidence interval: This is a range of values, derived from the sample, that is likely to contain the population mean difference. In this case, the 95% confidence interval is between -9.302 and -0.489.
sample estimates: These are the sample means of each group. In this case, the mean score for group A is 74.19, and the mean score for group B is 79.09.
In conclusion, we reject the null hypothesis that the means of the two groups are equal, and conclude that there is a significant difference in means between the two teaching methods.
Welch’s t-test is a powerful statistical test for comparing the means of two independent groups, especially when the assumptions of the independent samples t-test are violated. As always, the results of a t-test should be interpreted within the context of the research question and study design.