To find the margin of error $E$ for the given statistics and confidence level, we can use the formula for the margin of error for a proportion:

$E = Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Where:

• $Z_{\alpha/2}$ is the z-score corresponding to the confidence level (for 95% confidence, $Z_{\alpha/2} = 1.96$),
• $\hat{p}$ is the sample proportion,
• $n$ is the sample size.

Step 1: Calculate the sample proportion

The sample proportion $\hat{p}$ is calculated as:

$\hat{p} = \frac{x}{n} = \frac{1872}{2888} \approx 0.648$

Step 2: Apply the formula for the margin of error

Now, plug the values into the formula for the margin of error:

$E = 1.96 \cdot \sqrt{\frac{0.648 \cdot (1 - 0.648)}{2888}}$

Let me calculate that.The margin of error $E$ is approximately $0.0174$.

Summary:

For a 95% confidence level, the margin of error is about 0.0174 (or 1.74%).

Would you like further details or have any questions?

Here are 5 related questions to consider:

1. How does the margin of error change with different confidence levels?
2. What is the relationship between sample size and the margin of error?
3. How do we interpret the confidence interval for the population proportion?
4. What happens to the margin of error if the sample proportion is closer to 0.5?
5. How is the z-score for different confidence levels determined?

Tip: Increasing the sample size decreases the margin of error, making your estimate more precise.