1. **Problem Statement:** A doctor wants to test if a new medication reduces systolic blood pressure. We have paired data (before and after) for 10 patients. We will use a paired t-test (T-Test for correlated samples) at the 0.05 significance level. 2. **Formula and Explanation:** The paired t-test statistic is calculated as: $$t = \frac{\bar{d}}{s_d / \sqrt{n}}$$ where: - $\bar{d}$ is the mean of the differences (Before - After), - $s_d$ is the standard deviation of the differences, - $n$ is the number of pairs (patients). We test the null hypothesis $H_0$: medication does not reduce blood pressure (mean difference $\leq 0$) against the alternative $H_a$: medication reduces blood pressure (mean difference $> 0$). 3. **Calculate Differences:** Patient differences $d_i = \text{Before} - \text{After}$: 1: 150 - 140 = 10 2: 160 - 145 = 15 3: 155 - 150 = 5 4: 148 - 140 = 8 5: 170 - 160 = 10 6: 165 - 155 = 10 7: 158 - 150 = 8 8: 162 - 152 = 10 9: 155 - 145 = 10 10: 160 - 150 = 10 4. **Calculate Mean Difference $\bar{d}$:** $$\bar{d} = \frac{10 + 15 + 5 + 8 + 10 + 10 + 8 + 10 + 10 + 10}{10} = \frac{96}{10} = 9.6$$ 5. **Calculate Standard Deviation $s_d$:** First, find squared deviations: $$(10 - 9.6)^2 = 0.16$$ $$(15 - 9.6)^2 = 29.16$$ $$(5 - 9.6)^2 = 21.16$$ $$(8 - 9.6)^2 = 2.56$$ $$(10 - 9.6)^2 = 0.16$$ $$(10 - 9.6)^2 = 0.16$$ $$(8 - 9.6)^2 = 2.56$$ $$(10 - 9.6)^2 = 0.16$$ $$(10 - 9.6)^2 = 0.16$$ $$(10 - 9.6)^2 = 0.16$$ Sum of squared deviations = 0.16 + 29.16 + 21.16 + 2.56 + 0.16 + 0.16 + 2.56 + 0.16 + 0.16 + 0.16 = 56.4 Variance: $$s_d^2 = \frac{56.4}{10 - 1} = \frac{56.4}{9} = 6.267$$ Standard deviation: $$s_d = \sqrt{6.267} \approx 2.503$$ 6. **Calculate t-statistic:** $$t = \frac{9.6}{2.503 / \sqrt{10}} = \frac{9.6}{2.503 / 3.162} = \frac{9.6}{0.791} \approx 12.14$$ 7. **Degrees of Freedom:** $$df = n - 1 = 10 - 1 = 9$$ 8. **Critical t-value:** At $\alpha = 0.05$ for a one-tailed test and $df=9$, the critical t-value is approximately 1.833. 9. **Decision:** Since calculated $t = 12.14$ is much greater than 1.833, we reject the null hypothesis. 10. **Conclusion:** There is significant evidence at the 0.05 level that the medication reduces systolic blood pressure.