1. **Problem statement:** We want to find the sample size needed to estimate the proportion of customers who prefer "express shipping" with 95% confidence and a margin of error (MOE) of 4%. 2. **Formula for sample size when estimating a proportion:** $$n = \left(\frac{Z_{\alpha/2}^2 \cdot p \cdot (1-p)}{E^2}\right)$$ where: - $Z_{\alpha/2}$ is the z-score for the confidence level (95% confidence means $Z_{0.025} = 1.96$), - $p$ is the estimated proportion, - $E$ is the margin of error. 3. **Case 1: No preliminary estimate available** When no estimate for $p$ is available, use $p=0.5$ to maximize the product $p(1-p)$ and thus the sample size. Calculate: $$n = \frac{(1.96)^2 \times 0.5 \times (1-0.5)}{(0.04)^2} = \frac{3.8416 \times 0.25}{0.0016} = \frac{0.9604}{0.0016} = 600.25$$ Round up: $$n = 601$$ 4. **Case 2: Previous survey showed 38% support ($p=0.38$)** Calculate: $$n = \frac{(1.96)^2 \times 0.38 \times (1-0.38)}{(0.04)^2} = \frac{3.8416 \times 0.38 \times 0.62}{0.0016} = \frac{3.8416 \times 0.2356}{0.0016} = \frac{0.9049}{0.0016} = 565.56$$ Round up: $$n = 566$$ 5. **Explanation of difference:** The sample size is larger when no preliminary estimate is available because using $p=0.5$ maximizes the variance $p(1-p)$, leading to a more conservative (larger) sample size to ensure the margin of error is met regardless of the true proportion. When a prior estimate is available, the sample size can be smaller because the variance is likely less than the maximum possible.