Question

Question
In a survey of 75 students, 18 said they would benefit from summer school.

Given a 95% level of confidence with a z-score of 1.96, what is the margin of error?

(1 point)
Responses

24%
24%

10%
10%

0.096%
0.096%

47%

Answer 1

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:

1. Calculate \( p(1-p) \):
\[
p(1-p) = 0.24 \times (1 - 0.24) = 0.24 \times 0.76 = 0.1824
\]

2. Calculate the standard error:
\[
\text{Standard Error} = \sqrt{\frac{0.1824}{75}} \approx \sqrt{0.002432} \approx 0.04932
\]

3. 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%**.