1. **Problem Statement:** We have marks of 10 employees before and after training. We want to determine if the training has significantly improved their performance using a paired t-test. 2. **Formula for paired t-test:** $$ t = \frac{\bar{d}}{s_d / \sqrt{n}} $$ where $\bar{d}$ is the mean of the differences, $s_d$ is the standard deviation of the differences, and $n$ is the number of pairs. 3. **Calculate differences (After - Before):** $$ d = [84-80, 70-76, 96-92, 80-60, 70-70, 52-56, 84-74, 72-56, 72-70, 50-56] = [4, -6, 4, 20, 0, -4, 10, 16, 2, -6] $$ 4. **Calculate mean of differences:** $$ \bar{d} = \frac{4 + (-6) + 4 + 20 + 0 + (-4) + 10 + 16 + 2 + (-6)}{10} = \frac{40}{10} = 4 $$ 5. **Calculate standard deviation of differences:** First, find squared deviations: $$ (4-4)^2=0, (-6-4)^2=100, (4-4)^2=0, (20-4)^2=256, (0-4)^2=16, (-4-4)^2=64, (10-4)^2=36, (16-4)^2=144, (2-4)^2=4, (-6-4)^2=100 $$ Sum of squared deviations: $$ 0 + 100 + 0 + 256 + 16 + 64 + 36 + 144 + 4 + 100 = 720 $$ Variance: $$ s_d^2 = \frac{720}{10-1} = \frac{720}{9} = 80 $$ Standard deviation: $$ s_d = \sqrt{80} \approx 8.944 $$ 6. **Calculate t-statistic:** $$ t = \frac{4}{8.944 / \sqrt{10}} = \frac{4}{8.944 / 3.162} = \frac{4}{2.828} \approx 1.414 $$ 7. **Degrees of freedom:** $$ df = n - 1 = 9 $$ 8. **Conclusion:** At $\alpha = 0.05$ significance level, the critical t-value for two-tailed test with 9 df is approximately 2.262. Since calculated $t = 1.414 < 2.262$, we fail to reject the null hypothesis. **Therefore, there is not enough evidence to conclude that the training significantly improved employee performance.**