To determine the sample size based on a margin of error, we typically use the margin of error formula for proportions:

$\text{Margin of Error} = z \times \sqrt{\frac{p(1 - p)}{n}}$

where:

• $z$ is the z-score for the confidence level (commonly 1.96 for a 95% confidence level),
• $p$ is the estimated proportion (if unknown, 0.5 is often used to maximize the margin of error),
• $n$ is the sample size.

Given:

• Margin of Error = $1.4\% = 0.014$
• Assuming a 95% confidence level, $z \approx 1.96$
• Assuming $p = 0.5$

Let's solve for $n$:

1. Rearrange the formula to isolate $n$: $n = \frac{z^2 \times p(1 - p)}{(\text{Margin of Error})^2}$

2. Substitute the values: $n = \frac{(1.96)^2 \times 0.5 \times (1 - 0.5)}{(0.014)^2}$

I'll calculate this now.The survey would require approximately 4,900 people to achieve a margin of error of ±1.4% at a 95% confidence level.

Would you like more details on the calculations or have any other questions?

Here are some related questions:

1. How does changing the confidence level affect the sample size required?
2. What happens to the margin of error if we survey fewer people?
3. Can we calculate a margin of error for a different confidence level, such as 90%?
4. How does the value of $p$ affect the sample size calculation?
5. What methods can reduce the margin of error without increasing sample size?

Tip: Using a smaller $p$ when proportions are known can reduce the sample size required, as it adjusts the calculation to real distribution.