To determine the minimum sample size needed to estimate a population proportion with a specified confidence level and margin of error, we can use the formula for sample size in proportion estimation:

$n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right)$

Where:

• $n$ = required sample size
• $Z$ = Z-score corresponding to the desired confidence level
• $p$ = estimated proportion (if unknown, use $p = 0.5$ for maximum variability)
• $E$ = margin of error

Step 1: Identify the values

• Confidence Level: 90% (Z-score ≈ 1.645)
• Margin of Error (E): 0.05
• Estimated Proportion (p): If unknown, we use 0.5.

Step 2: Substitute into the formula

Using $p = 0.5$:

$n = \left( \frac{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.05)^2} \right)$

Calculating:

$n = \left( \frac{2.706025 \cdot 0.5 \cdot 0.5}{0.0025} \right) = \left( \frac{0.67650625}{0.0025} \right) = 270.6025$

Step 3: Round up

Since we cannot survey a fraction of a person, we round up to the next whole number:

$n = 271$

Conclusion

The minimum number of respondents needed to be 90% confident that the population proportion is estimated within 0.05 is 271.

Would you like more details or have any questions? Here are some related questions you might consider:

1. What is the significance of choosing $p = 0.5$ for maximum variability?
2. How does the confidence level affect the required sample size?
3. What are the potential consequences of using a sample size that is too small?
4. Can this sample size calculation be applied to other populations or studies?
5. What are some methods to ensure a representative sample in surveys?

Tip: Always check if the sample is randomly selected to minimize bias in your results.