To find the margin of error for the survey at the 95% confidence level, we can use the formula for the margin of error (ME) in a proportion:

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

Where:

• $z$ is the z-score corresponding to the desired confidence level (for 95% confidence, $z \approx 1.96$),
• $p$ is the sample proportion,
• $n$ is the sample size.

Given:

• Sample size $n = 50$,
• Number of favorable responses $x = 20$,
• Proportion $p = \frac{20}{50} = 0.4$.

Let's calculate the margin of error:

$\text{ME} = 1.96 \times \sqrt{\frac{0.4 \times (1 - 0.4)}{50}}$

$\text{ME} = 1.96 \times \sqrt{\frac{0.4 \times 0.6}{50}}$

$\text{ME} = 1.96 \times \sqrt{\frac{0.24}{50}}$

$\text{ME} = 1.96 \times \sqrt{0.0048}$

$\text{ME} = 1.96 \times 0.0693$

$\text{ME} \approx 0.135$

So, the margin of error for this survey at the 95% confidence level is approximately 0.135.

Would you like to go through the steps in more detail or ask about related concepts? Here are some questions that might be useful:

1. What is the significance of the 95% confidence level?
2. How does the sample size affect the margin of error?
3. What would happen to the margin of error if the sample proportion $p$ were closer to 0 or 1?
4. How is the z-score determined for different confidence levels?
5. What would the margin of error be for a different sample size?

Tip: The margin of error decreases as the sample size increases, which means larger samples provide more precise estimates.