To solve this problem, we need to find the sample size $n$ required to achieve a margin of error $E$ of 0.1 with a 95% confidence level, and a population proportion $p^* = 0.25$.

Formula for sample size:

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

Where:

• $Z_{\alpha/2}$ is the Z-score corresponding to the 95% confidence level.
• $p^* = 0.25$ is the estimated population proportion.
• $E = 0.1$ is the margin of error.

Step-by-step solution:

1. Z-score for 95% confidence: From standard normal distribution tables, the Z-score for 95% confidence is approximately $Z_{\alpha/2} = 1.96$.

2. Plug values into the formula: $n = \left( \frac{1.96 \cdot \sqrt{0.25(1 - 0.25)}}{0.1} \right)^2$ Simplifying inside the square root: $n = \left( \frac{1.96 \cdot \sqrt{0.25 \cdot 0.75}}{0.1} \right)^2$ $n = \left( \frac{1.96 \cdot \sqrt{0.1875}}{0.1} \right)^2$

Let's compute this value to find the sample size.The required sample size is $n = 73$, rounded up to the nearest whole number.

Would you like any further details or explanations? Here are a few related questions to deepen your understanding:

1. How is the Z-score determined for different confidence levels?
2. Why is the margin of error important in determining the sample size?
3. What happens to the sample size if the margin of error is decreased?
4. Can the sample size formula be applied when the population proportion $p^*$ is unknown?
5. How does the confidence level affect the sample size calculation?

Tip: A smaller margin of error or a higher confidence level always requires a larger sample size to maintain precision in estimates.