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%

10%
10%

47%
47%

24%

Answer 1

To calculate the margin of error (ME) for the proportion of students who would benefit from summer school, you can use the formula:

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

Where:
- \( z \) is the z-score for the desired confidence level (1.96 for 95% confidence)
- \( p \) is the proportion of students who said they would benefit from summer school
- \( n \) is the total number of students surveyed

First, we need to find \( p \):
- The number of students who would benefit from summer school = 18
- Total number of students surveyed = 75

So,
\[
p = \frac{18}{75} = 0.24
\]

Next, we calculate \( 1 - p \):
\[
1 - p = 1 - 0.24 = 0.76
\]

Now, substitute the values into the margin of error formula:
\[
ME = 1.96 \times \sqrt{\frac{0.24 \times 0.76}{75}}
\]

Calculating the fraction first:
\[
0.24 \times 0.76 = 0.1824
\]
Then:
\[
\frac{0.1824}{75} = 0.002432
\]

Taking the square root:
\[
\sqrt{0.002432} \approx 0.0493
\]

Now, multiply by the z-score:
\[
ME = 1.96 \times 0.0493 \approx 0.0966
\]

To express the margin of error in percentage form, multiply by 100:
\[
ME \approx 0.0966 \times 100 \approx 9.66\%
\]

Rounding this to whole numbers gives approximately **10%**.

So the answer is:

**10%**