To calculate the required sample size $n$ for estimating the proportion of voters with a specified margin of error, we can use the formula for the sample size in proportion estimation:

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

Where:

• $Z$ is the Z-value corresponding to the desired confidence level.
• $p$ is the estimated proportion from the pilot study.
• $E$ is the margin of error.

Given:

• Confidence level = 95%, so $Z \approx 1.96$ (from Z-tables).
• Estimated proportion $\hat{p} = 0.57$.
• Margin of error $E = 0.015$.

Now, substitute these values into the formula:

1. Calculate $p(1 - p)$: $p(1 - p) = 0.57 \times (1 - 0.57) = 0.57 \times 0.43 = 0.2451$

2. Substitute into the sample size formula: $n = \frac{(1.96^2) \cdot 0.2451}{(0.015^2)}$

   $n = \frac{(3.8416) \cdot 0.2451}{0.000225}$

   $n = \frac{0.9429}{0.000225} \approx 4186.67$

3. Round up to the nearest whole integer: $n = 4187$

Thus, the required sample size is 4187.

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