To find the margin of error (E) in this problem, we use the formula for margin of error in estimating proportions:

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

Where:

• $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level (for 90% confidence, $Z_{\alpha/2}$ is approximately 1.645),
• $p$ is the proportion of successes (given as 40% or 0.40),
• $n$ is the sample size (given as 8,000).

Now, plugging in the values:

$E = 1.645 \times \sqrt{\frac{0.40(1-0.40)}{8000}}$ $E = 1.645 \times \sqrt{\frac{0.40 \times 0.60}{8000}}$ $E = 1.645 \times \sqrt{\frac{0.24}{8000}}$ $E = 1.645 \times \sqrt{0.00003}$ $E = 1.645 \times 0.005477$ $E \approx 0.009$

So, the margin of error (E) is approximately 0.009, or 0.9%.

Would you like further explanation or have any questions on this? Here's a related question for deeper understanding:

1. How would the margin of error change if the sample size were doubled?
2. How does the Z-score change with a different confidence level (e.g., 95%)?
3. What would happen to the margin of error if the proportion of successes were 50% instead of 40%?
4. How do you interpret the margin of error in the context of this survey?
5. Can you explain why the margin of error decreases as the sample size increases?

Tip: Larger sample sizes lead to smaller margins of error, which increases the accuracy of your estimate!