To find the margin of error (M.E.) for estimating a population proportion, we use the formula:

$M.E. = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

where:

• $z^*$ is the z-score corresponding to the desired confidence level,
• $\hat{p}$ is the sample proportion,
• $n$ is the sample size.

Step 1: Identify the Given Values

• Sample size ($n$) = 373
• Sample proportion ($\hat{p}$) = 61.1% = 0.611
• Confidence level = 90%

Step 2: Find the Critical Value ($z^*$) for a 90% Confidence Level

For a 90% confidence level, the critical value $z^*$ is approximately 1.645.

Step 3: Substitute Values into the Formula

$M.E. = 1.645 \cdot \sqrt{\frac{0.611 \cdot (1 - 0.611)}{373}}$

Step 4: Calculate the Margin of Error

1. Calculate $0.611 \cdot (1 - 0.611) = 0.611 \cdot 0.389 = 0.237879$.
2. Divide by the sample size: $\frac{0.237879}{373} = 0.00063785$.
3. Take the square root: $\sqrt{0.00063785} = 0.02525$.
4. Multiply by the critical value: $1.645 \cdot 0.02525 = 0.04153$.

Step 5: Round the Final Answer

The margin of error, rounded to one decimal place, is:

$M.E. \approx 4.2\%$

Final Answer:

$\text{M.E.} = 4.2\%$

Would you like more details or have any questions?

Here are some related questions to consider:

1. How would the margin of error change with a 95% confidence level?
2. What happens to the margin of error if the sample size increases?
3. How would a lower sample proportion affect the margin of error?
4. Why is it important to calculate the margin of error when estimating a population parameter?
5. How does margin of error relate to the concept of statistical significance?

Tip: Higher confidence levels increase the critical value, thereby increasing the margin of error.