To solve this problem, we need to calculate the required sample size $n$ for estimating the voter population proportion with a specified margin of error $E$ and confidence level.

Here's the information provided:

• Margin of error $E = 3.54\% = 0.0354$
• Confidence level = 90%
• Corresponding $z$-value for 90% confidence = 1.645

The sample size formula for estimating a population proportion is:

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

where:

• $z$ is the critical value for the desired confidence level,
• $E$ is the margin of error,
• $p$ is the estimated population proportion.

If $p$ is unknown, we often use $p = 0.5$ for maximum variability (which maximizes $p(1 - p)$).

Step 1: Plugging in values

Assuming $p = 0.5$ (since it's not provided):

$n = \left( \frac{1.645}{0.0354} \right)^2 \times 0.5 \times (1 - 0.5)$

Step 2: Calculate the sample size

1. Calculate $\frac{1.645}{0.0354}$: $\frac{1.645}{0.0354} \approx 46.46$

2. Square that result: $46.46^2 \approx 2159.87$

3. Multiply by $0.5 \times 0.5 = 0.25$: $n \approx 2159.87 \times 0.25 = 539.97$

Conclusion

Rounding up, the required sample size $n$ is approximately 540.

Would you like more details on this calculation or have any questions?

Here are some related questions to consider:

1. How would the sample size change if we used a 95% confidence level instead?
2. What if the margin of error $E$ was reduced to 2%?
3. How does changing the estimated population proportion $p$ impact $n$?
4. Why is $p = 0.5$ used when the true proportion is unknown?
5. What assumptions underlie the use of a normal distribution for this calculation?

Tip: Using a higher confidence level or a smaller margin of error generally requires a larger sample size.