To determine the required sample size for a confidence interval, you can use the following formula for estimating proportions:

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

Where:

• $Z$ is the Z-score for a 95% confidence level (which is 1.96),
• $p$ is the estimated proportion of the population, which is 36% or 0.36,
• $E$ is the margin of error, which is 1% or 0.01.

Let's calculate the sample size:

$n = \left( \frac{1.96^2 \cdot 0.36 \cdot (1-0.36)}{0.01^2} \right)$

I'll compute that for you.The required sample size for the poll to achieve a 95% confidence interval with a 1 percentage point margin of error is approximately 8,852 voters.

Would you like more details or have any further questions?

Here are 5 related questions to expand on this concept:

1. How does the margin of error affect the required sample size?
2. What is the significance of the 95% confidence level in polls?
3. How would the sample size change with a different margin of error?
4. Why is it necessary to round up the sample size in such calculations?
5. How does the estimated proportion (36% in this case) influence the sample size?

Tip: A larger sample size reduces the margin of error, increasing the accuracy of poll results.