Let's solve the problem step by step: --- ### **Given:** - Sample size (\( n \)) = 400 - Number of viewers who would watch the series (\( x \)) = 266 --- ## **a. Estimate the proportion of people in the population who would watch the new series.** The sample proportion (\( \hat{p} \)) is calculated as: \[ \hat{p} = \frac{x}{n} \] \[ \hat{p} = \frac{266}{400} = 0.665 \] **Population proportion (rounded to 3 decimal places):** \[ \boxed{0.665} \] --- ## **b. Develop a 99% confidence interval for the proportion** ### **Step 1: Find the standard error (SE)** \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] \[ SE = \sqrt{\frac{0.665 \times (1-0.665)}{400}} = \sqrt{\frac{0.665 \times 0.335}{400}} = \sqrt{\frac{0.222775}{400}} = \sqrt{0.0005569375} = 0.0236 \] ### **Step 2: Find the Z-value for 99% confidence** - For 99% confidence, \( Z = 2.576 \) ### **Step 3: Calculate the margin of error (E)** \[ E = Z \times SE = 2.576 \times 0.0236 = 0.0608 \] ### **Step 4: Calculate the confidence interval** \[ \text{Lower limit} = \hat{p} - E = 0.665 - 0.0608 = 0.604 \] \[ \text{Upper limit} = \hat{p} + E = 0.665 + 0.0608 = 0.726 \] **Confidence interval (rounded to 3 decimal places):** \[ \boxed{0.604} \text{ and } \boxed{0.726} \] --- ## **Final Answers:** ### a. Population proportion: \[ \boxed{0.665} \] ### b. 99% confidence interval: \[ \boxed{0.604} \text{ and } \boxed{0.726} \]