1. **State the problem:** We want to test if the calcium supplement reduces blood pressure more than the placebo on average. 2. **Hypotheses:** - Null hypothesis $H_0$: $\mu_1 - \mu_2 = 0$ (no difference in means) - Alternative hypothesis $H_a$: $\mu_1 - \mu_2 > 0$ (calcium reduces blood pressure more) 3. **Data:** Group 1 (calcium): $7, -4, 18, 17, -3, -5, 1, 10, 11, -2$ Group 2 (placebo): $-1, 12, -1, -3, 3, -5, 5, 2, -11, -1, -3$ 4. **Calculate sample means and standard deviations:** $$\bar{x}_1 = \frac{7 + (-4) + 18 + 17 + (-3) + (-5) + 1 + 10 + 11 + (-2)}{10} = \frac{50}{10} = 5$$ $$\bar{x}_2 = \frac{-1 + 12 + (-1) + (-3) + 3 + (-5) + 5 + 2 + (-11) + (-1) + (-3)}{11} = \frac{-3}{11} \approx -0.273$$ Calculate sample standard deviations $s_1$ and $s_2$ (using formula $s = \sqrt{\frac{1}{n-1} \sum (x_i - \bar{x})^2}$): $s_1 \approx 8.68$, $s_2 \approx 5.99$ 5. **Calculate the test statistic $t$:** $$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} = \frac{5 - (-0.273)}{\sqrt{\frac{8.68^2}{10} + \frac{5.99^2}{11}}} = \frac{5.273}{\sqrt{7.53 + 3.26}} = \frac{5.273}{\sqrt{10.79}} = \frac{5.273}{3.29} \approx 1.60$$ 6. **Degrees of freedom (Welch's approximation):** $$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{(7.53 + 3.26)^2}{\frac{7.53^2}{9} + \frac{3.26^2}{10}} = \frac{10.79^2}{6.30 + 1.06} = \frac{116.4}{7.36} \approx 15.8$$ 7. **Find the p-value:** Using $t=1.60$ and $df \approx 15.8$, the one-tailed p-value is approximately $0.0644$. 8. **Interpretation:** Since $p = 0.0644 > 0.05$, we do not reject the null hypothesis at the 5% significance level. There is not strong enough evidence to conclude calcium reduces blood pressure more than placebo. **Final answer:** $$\boxed{p \approx 0.0644}$$