To determine the sample size $n$ required to achieve a specified margin of error for a proportion, we can use the formula:

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

where:

• $Z$ is the Z-score corresponding to the desired confidence level,
• $p$ is the estimated proportion,
• $E$ is the margin of error.

For a 95% confidence level, the Z-score $Z$ is approximately 1.96.

Given:

• $p = 0.57$
• $E = 0.015$

Now, we can substitute these values into the formula:

1. Calculate $p(1 - p)$:

$p(1 - p) = 0.57 \times (1 - 0.57) = 0.57 \times 0.43 \approx 0.2451$

2. Now, plug the values into the sample size formula:

$n = \frac{(1.96^2) \cdot 0.2451}{(0.015^2)}$

Calculating $1.96^2$:

$1.96^2 \approx 3.8416$

So,

$n = \frac{3.8416 \cdot 0.2451}{0.000225} \approx \frac{0.94173}{0.000225} \approx 4185.73$

3. Finally, round up to the nearest whole integer:

$n \approx 4186$

Thus, the required sample size is 4186 voters.

Would you like to delve into any specific details or have any questions about this calculation? Here are some related questions you might find interesting:

1. How do different confidence levels affect the required sample size?
2. What is the significance of the margin of error in survey research?
3. How can we estimate the proportion if we don't have a pilot study?
4. What assumptions are made when using this formula for sample size determination?
5. How would the sample size change if the estimated proportion were closer to 0.5?

Tip: Always consider conducting a pilot study to obtain a more accurate estimate of $p$ if possible, as it significantly influences the sample size calculation.