To find the minimum sample size required to estimate the proportion of voters who will vote for Candidate X, we can use the formula for determining the sample size for a population proportion:

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

Where:

• $n$ is the required sample size.
• $Z$ is the Z-score corresponding to the desired confidence level.
• $p$ is the estimated population proportion.
• $E$ is the margin of error.

Step 1: Identify the known values.

• Confidence level is 95%, so the Z-score is approximately $1.96$ (based on the standard normal distribution).
• From the problem, the estimated population proportion $p \leq 0.40$. To ensure the maximum sample size, we use $p = 0.40$.
• The margin of error $E = 0.03$.

Step 2: Plug the values into the formula.

$n = \frac{(1.96)^2 \cdot 0.40 \cdot (1 - 0.40)}{(0.03)^2}$

$n = \frac{(1.96)^2 \cdot 0.40 \cdot 0.60}{0.0009}$

Step 3: Calculate the sample size.

$n = \frac{3.8416 \cdot 0.24}{0.0009}$

$n = \frac{0.922}{0.0009}$

$n \approx 1024.44$

Since the sample size must be a whole number, round up to the next whole number.

$n = 1025$

Conclusion:

The minimum sample size required to estimate the proportion of voters for Candidate X with a 95% confidence level and a margin of error of 0.03 is 1025.

Would you like further clarification or details on any part of this process?

Related Questions:

1. How would the sample size change if the confidence level were increased to 99%?
2. What happens to the sample size if the margin of error is reduced to 0.02?
3. How is the sample size affected if the population proportion is lower, say $p = 0.30$?
4. What is the formula to estimate the sample size when the population proportion is unknown?
5. How can you interpret the confidence interval once the sample data is collected?

Tip:

To minimize the required sample size, adjusting the margin of error or confidence level can significantly impact the result. Keep this in mind when designing surveys or polls.