To calculate the margin of error for the survey results, we can use the formula:
\[ \text{Margin of Error} = z \times \sqrt{\frac{p(1-p)}{n}} \]
where:
- \( z \) is the z-score for the desired confidence level,
- \( p \) is the sample proportion,
- \( n \) is the sample size.
In this case:
- The z-score for a 95% confidence level is 1.96.
- The sample size (\( n \)) is 75.
- The number of students who would benefit from summer school is 18, so the sample proportion (\( p \)) is \( \frac{18}{75} = 0.24 \).
Now we can calculate the margin of error:
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Calculate \( p(1-p) \): \[ p(1-p) = 0.24 \times (1 - 0.24) = 0.24 \times 0.76 = 0.1824 \]
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Calculate the standard error: \[ \text{Standard Error} = \sqrt{\frac{0.1824}{75}} \approx \sqrt{0.002432} \approx 0.04932 \]
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Finally, calculate the margin of error using the z-score: \[ \text{Margin of Error} = 1.96 \times 0.04932 \approx 0.0967 \]
To express the margin of error as a percentage, multiply by 100: \[ \text{Margin of Error} \approx 0.0967 \times 100 \approx 9.67% \]
Thus, rounding appropriately, the margin of error is approximately 10%.
The correct answer is 10%.