To determine the number of voters that should be sampled for a 95% confidence interval, we use the following formula for margin of error (MOE) in a proportion:

$MOE = z \times \sqrt{\frac{p(1 - p)}{n}}$

Where:

• $MOE$ is the margin of error (1.52% or 0.0152),
• $z$ is the z-score corresponding to the desired confidence level (for a 95% confidence level, $z = 1.96$),
• $p$ is the sample proportion (0.51 for 51%),
• $n$ is the sample size (what we're solving for).

Step 1: Rearrange the formula to solve for $n$:

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

Step 2: Plug in the known values:

$n = \frac{(1.96)^2 \times (0.51)(0.49)}{(0.0152)^2}$

Step 3: Perform the calculations:

$n = \frac{3.8416 \times 0.2499}{0.00023104}$ $n = \frac{0.96000584}{0.00023104}$ $n \approx 4154.44$

Step 4: Round up to the nearest whole number:

$n = 4155$

Thus, 4155 voters should be sampled to achieve a 95% confidence interval with a margin of error of 1.52%.

Would you like more details or have any questions?

Here are 5 related questions for deeper understanding:

1. How does changing the confidence level affect the sample size calculation?
2. What happens to the sample size if the margin of error is decreased?
3. Why is the proportion used as 0.51 and how does this impact the result?
4. What is the significance of rounding up in sample size calculations?
5. How is the z-score determined for different confidence levels?

Tip: Increasing the sample size decreases the margin of error, improving the precision of the estimate.