To determine the required sample size $n$ for a 95% confidence interval with a margin of error of 0.03, we use the formula for the margin of error in estimating a population proportion:

$E = Z_{\alpha/2} \times \sqrt{\frac{p^*(1 - p^*)}{n}}$

Where:

• $E$ is the margin of error (0.03 in this case),
• $Z_{\alpha/2}$ is the critical value for a 95% confidence level (which is 1.96),
• $p^*$ is the estimated population proportion (0.25),
• $n$ is the sample size.

Rearranging the formula to solve for $n$:

$n = \frac{(Z_{\alpha/2})^2 \times p^* \times (1 - p^*)}{E^2}$

Substituting the known values:

$n = \frac{(1.96)^2 \times 0.25 \times (1 - 0.25)}{(0.03)^2}$ $n = \frac{(1.96)^2 \times 0.25 \times 0.75}{0.0009}$ $n = \frac{3.8416 \times 0.25 \times 0.75}{0.0009}$ $n = \frac{0.7203}{0.0009} \approx 800.34$

Rounding up to the next whole number, the required sample size is 801.

Would you like further details on this calculation or help with another problem?

Here are some related questions:

1. How is the critical value $Z_{\alpha/2}$ determined for different confidence levels?
2. How does increasing the confidence level affect the sample size?
3. What happens if the margin of error is reduced to 0.01?
4. What if the population proportion estimate $p^*$ is not known?
5. How does changing the population size affect sample size calculations?

Tip: Always round the sample size up, as a fractional sample size does not meet the required confidence and margin of error conditions.