To calculate the margin of error for a proportion in a poll, we can use the formula:
\[ \text{Margin of Error} = z \times \sqrt{\frac{p(1-p)}{n}} \]
where:
- \( p \) is the sample proportion (in this case, 39% or 0.39),
- \( n \) is the sample size (350 students), and
- \( z \) is the z-score for the desired confidence level (typically 1.96 for a 95% confidence level).
- Calculate \( p(1-p) \):
\[ p(1-p) = 0.39(1-0.39) = 0.39 \times 0.61 = 0.2379 \]
- Calculate the standard error:
\[ \text{Standard Error} = \sqrt{\frac{0.2379}{350}} \approx \sqrt{0.000678} \approx 0.0260 \]
- Now multiply by the z-score (assuming 95% confidence, z ≈ 1.96):
\[ \text{Margin of Error} \approx 1.96 \times 0.0260 \approx 0.0510 \text{ or } 5.1% \]
Since the options are presented with slight rounding, we can round this to about 5.3%.
Thus, the margin of error for this poll is 5.3%.