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 era 10% 47%, 24%, 0.096%

Answer 1

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

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

where:
- \( z \) is the z-score (1.96 for a 95% confidence level),
- \( p \) is the sample proportion,
- \( n \) is the sample size.

First, let's calculate the sample proportion \( p \):

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

Now we can plug the values into the margin of error formula:

1. Calculate \( 1 - p \):

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

2. Now substitute \( p \), \( 1 - p \), \( n \), and \( z \) into the formula:

\[
\text{ME} = 1.96 \times \sqrt{\frac{0.24 \times 0.76}{75}}
\]

3. Calculate \( 0.24 \times 0.76 \):

\[
0.24 \times 0.76 = 0.1824
\]

4. Then calculate \( \frac{0.1824}{75} \):

\[
\frac{0.1824}{75} = 0.002432
\]

5. Now calculate the square root:

\[
\sqrt{0.002432} \approx 0.04932
\]

6. Finally, calculate the margin of error:

\[
\text{ME} = 1.96 \times 0.04932 \approx 0.0967
\]

Thus, the margin of error (rounded) is approximately \( 0.097 \) or \( 9.7\% \).

Given your options: **10%, 47%, 24%, 0.096%**, the closest choice is **10%**.