To determine the required sample size for a proportion with a specified margin of error and confidence level, we can use the following formula:

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

Where:

• $n$ is the sample size,
• $Z_{\alpha/2}$ is the z-value corresponding to the confidence level (for 99% confidence, $Z_{\alpha/2} = 2.576$),
• $p$ is the estimated proportion of voters willing to pay additional taxes (which we can estimate from the survey as $p = \frac{50}{177}$),
• $E$ is the desired margin of error (given as 0.03).

Step 1: Estimate the proportion $p$

$p = \frac{50}{177} \approx 0.2825$

Step 2: Calculate $1 - p$

$1 - p = 1 - 0.2825 = 0.7175$

Step 3: Plug the values into the formula

$n = \left( \frac{2.576^2 \cdot 0.2825 \cdot 0.7175}{0.03^2} \right)$

Step 4: Compute

$n = \left( \frac{6.635 \cdot 0.2825 \cdot 0.7175}{0.0009} \right)$ $n = \left( \frac{1.3519}{0.0009} \right)$ $n \approx 1502.11$

Thus, the required sample size is approximately 1503 (since the sample size must be a whole number, we round up).

Conclusion:

To ensure a 99% confidence interval with a margin of error of 0.03, a sample size of approximately 1503 voters is needed.

Do you have any questions or want more details?

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Here are five related questions for further practice:

1. How would the required sample size change if the margin of error was reduced to 0.02?
2. What is the impact on sample size if the confidence level is reduced to 95%?
3. How can we determine the confidence interval from the data gathered in the original survey of 177 voters?
4. What assumptions are we making about the underlying population when using this method to calculate the sample size?
5. How would the sample size change if the estimated proportion $p$ were higher, say 0.5?

Tip: In sample size calculations, if the initial proportion $p$ is unknown, using $p = 0.5$ provides the maximum possible sample size, ensuring a conservative estimate.