Paired T-Test (Before and After Comparison)
Conduct a paired t-test to determine whether a training program significantly improves employee productivity scores in a before-and-after study.
Problem Scenario
A company implements a new training program and measures the productivity scores of 8 employees before and after the training. Before: 45, 52, 48, 55, 40, 50, 47, 53. After: 50, 58, 52, 60, 46, 55, 51, 59. At alpha = 0.05, test whether the training significantly increased productivity scores.
Given Data
Requirements
- Calculate the differences (After - Before) for each pair
- Compute the mean and standard deviation of the differences and the t-statistic
- Determine the p-value and draw a conclusion
Solution
Step 1:
Calculate the differences d_i = After - Before for each employee: 5, 6, 4, 5, 6, 5, 4, 6.
Step 2:
Calculate the mean of differences: d-bar = (5 + 6 + 4 + 5 + 6 + 5 + 4 + 6) / 8 = 41 / 8 = 5.125.
Step 3:
Calculate the standard deviation of differences: Deviations from d-bar: -0.125, 0.875, -1.125, -0.125, 0.875, -0.125, -1.125, 0.875. Squared deviations: 0.01563, 0.76563, 1.26563, 0.01563, 0.76563, 0.01563, 1.26563, 0.76563. Sum = 4.875. s_d = sqrt(4.875 / 7) = sqrt(0.6964) = 0.8345.
Step 4:
Calculate the t-statistic: t = d-bar / (s_d / sqrt(n)) = 5.125 / (0.8345 / sqrt(8)) = 5.125 / (0.8345 / 2.8284) = 5.125 / 0.2951 = 17.37.
Step 5:
State hypotheses and find p-value. H_0: mu_d <= 0 (no improvement or decrease). H_a: mu_d > 0 (improvement). With t = 17.37 and df = n - 1 = 7, the p-value < 0.0001 (critical value t(0.05, 7) = 1.895). Since 17.37 >> 1.895, we reject H_0.
Step 6:
Conclusion: At the 0.05 significance level, there is overwhelming evidence that the training program significantly increased productivity scores. The average improvement was 5.125 points.
Final Answer
t = 17.37, df = 7, p-value < 0.0001. We reject H_0. The training program produced a statistically significant increase in productivity scores, with an average improvement of 5.125 points per employee.
Key Takeaways
- ✓A paired t-test is used when observations come in natural pairs (e.g., before/after on the same subject). It controls for individual differences by analyzing the within-pair differences.
- ✓The paired t-test is more powerful than the independent two-sample t-test when subjects vary widely, because it removes between-subject variability.
- ✓Always check whether the differences are approximately normally distributed, especially with small sample sizes. A histogram or normal probability plot of the differences can help.
Common Errors to Avoid
- ✗Using an independent two-sample t-test instead of a paired t-test when the data are matched. This ignores the pairing structure and can lead to incorrect conclusions.
- ✗Subtracting in the wrong direction and getting confused about the sign of the t-statistic. Define differences consistently (e.g., After - Before) and match the hypothesis direction.
- ✗Forgetting that the degrees of freedom for a paired t-test are n - 1 (number of pairs minus one), not 2n - 2.
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Common questions about this problem type
Use a paired t-test when each observation in one group can be naturally paired with an observation in the other group (same subject before/after, matched pairs, twins, etc.). Use an independent two-sample t-test when the two groups contain different, unrelated subjects.
For small samples with non-normal differences, use the Wilcoxon signed-rank test as a nonparametric alternative. For larger samples (n >= 30), the Central Limit Theorem ensures the sampling distribution of d-bar is approximately normal, making the paired t-test robust to moderate departures from normality.