Welch’s t-Test Calculator

Understanding Welch's T-Test

Welch's T-Test is a statistical test that is often used to determine whether there is a significant difference between the means of two groups that may have unequal variances and/or sample sizes. Unlike the traditional Student's t-test, Welch's T-Test does not assume equal variances, making it more reliable for real-world data where variances can differ.

When to Use Welch's T-Test

This test is particularly useful in scenarios where the two sample groups have noticeably different variances. It provides a more accurate estimation of the t-score and p-value under these conditions.

Real-Life Example: Comparing Test Scores

Imagine a researcher is investigating whether a new teaching method improves students' math test scores. They gather test scores from two groups:

• Group 1: Students taught using the traditional method
• Group 2: Students taught using the new teaching method

The data collected shows:

• Mean score for Group 1 = 75, Standard Deviation = 10, Sample Size = 30
• Mean score for Group 2 = 80, Standard Deviation = 12, Sample Size = 28

Using Welch's T-Test, the researcher can test whether the new teaching method (Group 2) significantly improves scores compared to the traditional method (Group 1). Here are the calculation steps:

Step-by-Step Calculation

1. Calculate the t-score:

$$ t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$

With:

• $ \bar{X}_1 = 75 $, $ s_1 = 10 $, $ n_1 = 30 $
• $ \bar{X}_2 = 80 $, $ s_2 = 12 $, $ n_2 = 28 $

2. Substitute these values into the formula to find:

$$ t = \frac{75 - 80}{\sqrt{\frac{10^2}{30} + \frac{12^2}{28}}} = -2.29 $$

3. Calculate the degrees of freedom (approximated) using the simplified formula:

$$ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} $$

This yields an approximate degrees of freedom of 55.28.

4. Determine the p-value:

The p-value for a t-score of -2.29 with 55.28 degrees of freedom is approximately 0.0259. Since this p-value is less than the significance level of 0.05, we reject the null hypothesis, suggesting that the new teaching method significantly improves test scores.

Formula

The t-score for Welch's T-Test is calculated as:

$$ t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$

where:

• $ \bar{X}_1 $ and $ \bar{X}_2 $ are the means of the two groups,
• $ s_1 $ and $ s_2 $ are the standard deviations of the two groups,
• $ n_1 $ and $ n_2 $ are the sample sizes of the two groups.

Hypothesis Testing

Welch's T-Test can be conducted as a right-tailed, left-tailed, or two-tailed test depending on the hypothesis:

• Right-tailed: Tests if the mean of Group 1 is significantly greater than Group 2.
• Left-tailed: Tests if the mean of Group 1 is significantly less than Group 2.
• Two-tailed: Tests if there is any significant difference between the two means.

Interpretation

Once the t-score is calculated, it is compared to a critical value from the t-distribution based on the degrees of freedom and the chosen significance level. The p-value helps decide whether to reject the null hypothesis:

• Reject the null hypothesis if the p-value is less than the significance level (e.g., 0.05), indicating a significant difference.
• Fail to reject the null hypothesis if the p-value is greater than the significance level, indicating no significant difference.