Let’s consider a simple example to illustrate how a t-test can be used to compare the means of two groups.

**Example: Exam Scores of Two Study Groups**

Suppose we have two groups of students, Group A and Group B. Both groups recently took a math exam, and we want to determine if there is a significant difference in their average scores. Group A received extra tutoring sessions before the exam, while Group B did not receive any additional support.

Here are the exam scores of the two groups:

Group A: 78, 85, 90, 82, 88

Group B: 72, 79, 68, 75, 80

**Step 1: Calculate the means of both groups.**

Mean of Group A = (78 + 85 + 90 + 82 + 88) / 5 = 84.6

Mean of Group B = (72 + 79 + 68 + 75 + 80) / 5 = 74.8

**Step 2: Perform the t-test.**

Since we have two independent groups and want to compare their means, we’ll use the independent samples t-test.

The t-test will calculate the t-value based on the means, standard deviations, and sample sizes of the two groups.

Let’s assume that after performing the calculations, we get a t-value of 2.73.

**Step 3: Determine the critical t-value.**

We need to look up the critical t-value from the t-distribution table for the chosen significance level (e.g., 0.05) and degrees of freedom (which is *nA*+*nB*−2 for independent samples t-test).

Suppose the critical t-value for a significance level of 0.05 and 8 degrees of freedom is 2.306.

**Step 4: Compare the t-value and critical t-value.**

** Since our calculated t-value (2.73) is greater than the critical t-value (2.306), we can conclude that there is a significant difference in the average math exam scores between Group A and Group B. **The extra tutoring sessions seem to have had a positive impact on the performance of students in Group A compared to Group B.

Remember, the t-test helps us determine if the observed difference in means is significant or just due to random variability. In this example, it suggests that the tutoring sessions likely contributed to the difference in exam scores between the two groups.