1. **State the problem:** We have two independent simple random samples (SRS): - Sample A: size $n_A=100$, proportion of successes $p_A=0.8$ - Sample B: size $n_B=400$, proportion of successes $p_B=0.5$ We want to find the mean and standard deviation of the sampling distribution of the difference in sample proportions $\hat{p}_A - \hat{p}_B$. 2. **Formula for the mean of the difference in sample proportions:** $$\mu_{\hat{p}_A - \hat{p}_B} = p_A - p_B$$ This is because the expected value of the difference of independent sample proportions equals the difference of their population proportions. 3. **Formula for the standard deviation of the difference in sample proportions:** $$\sigma_{\hat{p}_A - \hat{p}_B} = \sqrt{\frac{p_A(1-p_A)}{n_A} + \frac{p_B(1-p_B)}{n_B}}$$ This comes from the variance sum rule for independent variables. 4. **Calculate the mean:** $$\mu_{\hat{p}_A - \hat{p}_B} = 0.8 - 0.5 = 0.3$$ 5. **Calculate the standard deviation:** $$\sigma_{\hat{p}_A - \hat{p}_B} = \sqrt{\frac{0.8(1-0.8)}{100} + \frac{0.5(1-0.5)}{400}} = \sqrt{\frac{0.8 \times 0.2}{100} + \frac{0.5 \times 0.5}{400}}$$ $$= \sqrt{\frac{0.16}{100} + \frac{0.25}{400}} = \sqrt{0.0016 + 0.000625} = \sqrt{0.002225}$$ $$= 0.0472$$ (rounded to 3 decimal places) **Final answers:** - Mean = $0.3$ - Standard deviation = $0.047$