To calculate the margin of error (ME) for the proportion of students who said they would benefit from summer school, we can use the formula for the margin of error for a proportion:
\[ ME = z \times \sqrt{\frac{p(1-p)}{n}} \]
where:
- \( z \) is the z-score (1.96 for a 95% confidence level),
- \( p \) is the sample proportion,
- \( n \) is the sample size.
First, we need to find the sample proportion \( p \):
\[ p = \frac{18}{75} = 0.24 \]
Now we can plug in the values into the margin of error formula:
- Calculate \( p(1 - p) \):
\[ p(1 - p) = 0.24 \times (1 - 0.24) = 0.24 \times 0.76 = 0.1824 \]
- Calculate \( \frac{p(1-p)}{n} \):
\[ \frac{p(1-p)}{n} = \frac{0.1824}{75} \approx 0.002432 \]
- Take the square root:
\[ \sqrt{\frac{p(1-p)}{n}} \approx \sqrt{0.002432} \approx 0.04932 \]
- Multiply by the z-score:
\[ ME = 1.96 \times 0.04932 \approx 0.0967 \]
To convert this to a percentage, we multiply by 100:
\[ ME \approx 0.0967 \times 100 \approx 9.67% \approx 10% \]
So, the margin of error is approximately 10%.
The correct response is 10%.