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)

Answer 1

To calculate the margin of error (E) for a proportion at a given confidence level, you can use the formula:

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

Where:
- \( z \) is the z-score corresponding to the confidence level (1.96 for 95% confidence),
- \( p \) is the sample proportion (the number of students who said they would benefit from summer school divided by the total number of students surveyed),
- \( n \) is the sample size.

Given:
- Number of students who would benefit from summer school = 18
- Total number of students surveyed = 75

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

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

Now, we can calculate \( p(1 - p) \):

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

Next, we can calculate the standard error \( \sqrt{\frac{p(1 - p)}{n}} \):

\[
\sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.1824}{75}} = \sqrt{0.002432} \approx 0.04932
\]

Finally, we can calculate the margin of error \( E \):

\[
E = z \times \sqrt{\frac{p(1 - p)}{n}} = 1.96 \times 0.04932 \approx 0.0967
\]

Rounding to four decimal places, the margin of error is approximately:

\[
E \approx 0.0967
\]

Thus, the margin of error is **approximately 0.097**.