1. **Problem Statement:** We want to find the sample size $n$ needed to estimate the percentage of adults gambling online with a 95% confidence level and a margin of error of 4 percentage points (0.04). 2. **Given Data:** Confidence level = 95% \( \Rightarrow z_{\alpha/2} = 1.96 \) Margin of error \(E = 0.04\) 3. **Formula for sample size when estimating a proportion:** \[ n = \left(\frac{z_{\alpha/2}}{E}\right)^2 p (1-p) \] where $p$ is the estimated proportion of adults gambling online. ### Part (a) 4. **Assuming nothing is known about $p$:** We use $p = 0.5$ to maximize the product $p(1-p)$, giving the largest sample size. 5. Calculate $$ n = \left(\frac{1.96}{0.04}\right)^2 \times 0.5 \times (1-0.5) = (49)^2 \times 0.25 = 2401 \times 0.25 = 600.25 $$ 6. Round up: \[ n = 601\] ### Part (b) 7. **Assuming $p=0.19$ (19% gamble online):** Calculate $$ n = \left(\frac{1.96}{0.04}\right)^2 \times 0.19 \times (1-0.19) = 2401 \times 0.19 \times 0.81 = 2401 \times 0.1539 = 369.63 $$ 8. Round up: \[ n = 370\] **Final answers:** - (a) $n = 601$ - (b) $n = 370$