1. **Problem Statement:** We are given the test scores of 12 students before and after intensive coaching. We want to determine if the coaching led to a significant improvement in scores using hypothesis testing. 2. **Formulating Hypotheses:** - Null hypothesis ($H_0$): The coaching has no effect, i.e., mean difference in scores before and after coaching is zero ($\mu_d = 0$). - Alternative hypothesis ($H_1$): The coaching improves scores, i.e., mean difference is greater than zero ($\mu_d > 0$). 3. **Data and Differences:** Calculate the difference $d_i = \text{After} - \text{Before}$ for each student: | Student | Before | After | Difference $d_i$ | |---------|--------|-------|-----------------| | 1 | 50 | 60 | 10 | | 2 | 42 | 40 | -2 | | 3 | 51 | 61 | 10 | | 4 | 26 | 35 | 9 | | 5 | 35 | 30 | -5 | | 6 | 42 | 52 | 10 | | 7 | 60 | 68 | 8 | | 8 | 41 | 51 | 10 | | 9 | 70 | 84 | 14 | | 10 | 55 | 63 | 8 | | 11 | 62 | 72 | 10 | | 12 | 38 | 50 | 12 | 4. **Calculate Mean and Standard Deviation of Differences:** - Mean difference: $$\bar{d} = \frac{10 - 2 + 10 + 9 - 5 + 10 + 8 + 10 + 14 + 8 + 10 + 12}{12} = \frac{94}{12} \approx 7.83$$ - Calculate each squared deviation $(d_i - \bar{d})^2$ and sum: | $d_i$ | $d_i - \bar{d}$ | $(d_i - \bar{d})^2$ | |-------|-----------------|---------------------| | 10 | 2.17 | 4.71 | | -2 | -9.83 | 96.61 | | 10 | 2.17 | 4.71 | | 9 | 1.17 | 1.37 | | -5 | -12.83 | 164.61 | | 10 | 2.17 | 4.71 | | 8 | 0.17 | 0.03 | | 10 | 2.17 | 4.71 | | 14 | 6.17 | 38.07 | | 8 | 0.17 | 0.03 | | 10 | 2.17 | 4.71 | | 12 | 4.17 | 17.39 | Sum of squared deviations $= 336.65$ - Sample variance: $$s_d^2 = \frac{336.65}{12 - 1} = \frac{336.65}{11} \approx 30.60$$ - Sample standard deviation: $$s_d = \sqrt{30.60} \approx 5.53$$ 5. **Calculate t-statistic:** $$t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{7.83}{5.53 / \sqrt{12}} = \frac{7.83}{5.53 / 3.464} = \frac{7.83}{1.596} \approx 4.91$$ 6. **Degrees of Freedom and Critical Value:** - Degrees of freedom $df = n - 1 = 11$ - Given table value of $t$ at 5% significance level for $df=11$ is 2.201 7. **Decision:** - Since calculated $t = 4.91 > 2.201$, we reject the null hypothesis. 8. **Conclusion:** There is significant evidence at 5% level to conclude that coaching improved the students' test scores.