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

0.096%
0.096%

24%
24%

47%
47%

10%

Answer 1

To calculate the margin of error for the proportion of students who would benefit from summer school, you can use the formula for the margin of error (MOE) at a given confidence level:

\[
MOE = z \times \sqrt{\frac{p(1-p)}{n}}
\]

Where:
- \( z \) is the z-score corresponding to the desired confidence level (1.96 for 95% confidence),
- \( p \) is the sample proportion,
- \( n \) is the sample size.

First, calculate the sample proportion \( p \):

\[
p = \frac{x}{n} = \frac{18}{75} = 0.24
\]

Now, substitute \( p \) and \( n \) into the margin of error formula:

\[
MOE = 1.96 \times \sqrt{\frac{0.24(1-0.24)}{75}}
\]

Calculate \( 1-p \):

\[
1 - p = 1 - 0.24 = 0.76
\]

Now compute the standard error (SE):

\[
SE = \sqrt{\frac{0.24 \times 0.76}{75}}
\]
\[
= \sqrt{\frac{0.1824}{75}}
\]
\[
= \sqrt{0.002432}
\]
\[
\approx 0.04932
\]

Now, compute the margin of error:

\[
MOE = 1.96 \times 0.04932 \approx 0.0967
\]

To express this as a percentage, multiply by 100:

\[
MOE \approx 0.0967 \times 100 \approx 9.67\%
\]

Rounding this to the nearest whole number gives approximately **10%**.

Thus, the correct answer is **10%**.