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% 0.096% 0.096% 47% 47% 10%

Answer 1

To calculate the margin of error for a proportion, we can use the formula:

\[
\text{Margin of Error} = z \times \sqrt{\frac{p(1-p)}{n}}
\]

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

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

\[
p = \frac{\text{number of students who benefit from summer school}}{\text{total number of students}} = \frac{18}{75} = 0.24
\]

Next, we can substitute the values into the margin of error formula:

\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.24(1-0.24)}{75}}
\]

Calculating \( 1 - p \):

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

Now we calculate the expression inside the square root:

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

Now take the square root:

\[
\sqrt{0.002432} \approx 0.0493
\]

Now multiply by the z-score:

\[
\text{Margin of Error} = 1.96 \times 0.0493 \approx 0.0967
\]

So, rounding off to three decimal places, the margin of error is approximately 0.097.

Among the given options, the correct one corresponding to the margin of error is **0.096**.