1. **State the problem:** We have two groups in a clinical trial: the treatment group with $n_1=194,791$ children and the placebo group with $n_2=195,762$ children. We want to find the sample proportions $\hat{p}_1$, $\hat{q}_1$, $\hat{p}_2$, $\hat{q}_2$, and the combined proportions $\bar{p}$ and $\bar{q}$. 2. **Recall the formulas:** - Sample proportion of success: $\hat{p} = \frac{x}{n}$ where $x$ is the number of successes (children who developed the disease). - Sample proportion of failure: $\hat{q} = 1 - \hat{p}$. - Combined proportion of success: $\bar{p} = \frac{x_1 + x_2}{n_1 + n_2}$. - Combined proportion of failure: $\bar{q} = 1 - \bar{p}$. 3. **Calculate $\hat{p}_1$ and $\hat{q}_1$ for the treatment group:** - Number of successes $x_1 = 40$. - $\hat{p}_1 = \frac{40}{194,791} \approx 0.00020528$. - $\hat{q}_1 = 1 - 0.00020528 = 0.99979472$. 4. **Calculate $\hat{p}_2$ and $\hat{q}_2$ for the placebo group:** - Number of successes $x_2 = 150$. - $\hat{p}_2 = \frac{150}{195,762} \approx 0.00076625$. - $\hat{q}_2 = 1 - 0.00076625 = 0.99923375$. 5. **Calculate combined proportions $\bar{p}$ and $\bar{q}$:** - Total successes $x_1 + x_2 = 40 + 150 = 190$. - Total sample size $n_1 + n_2 = 194,791 + 195,762 = 390,553$. - $\bar{p} = \frac{190}{390,553} \approx 0.00048657$. - $\bar{q} = 1 - 0.00048657 = 0.99951343$. **Final answers:** - $n_1 = 194,791$ - $\hat{p}_1 = 0.00020528$ - $\hat{q}_1 = 0.99979472$ - $n_2 = 195,762$ - $\hat{p}_2 = 0.00076625$ - $\hat{q}_2 = 0.99923375$ - $\bar{p} = 0.00048657$ - $\bar{q} = 0.99951343$