Question

In a survey of 250 students, 19 said they want year-round school. Given a 95% level of confidence with a z-score of 1.96, what is the margin of error?(1 point)
Responses

0.033%
0.033%

14.9%
14.9%

3.3%
3.3%

7.6%

Answer 1

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

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

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

1. **Calculate the sample proportion \( p \)**:
- \( p = \frac{x}{n} = \frac{19}{250} = 0.076 \)

2. **Calculate \( 1 - p \)**:
- \( 1 - p = 1 - 0.076 = 0.924 \)

3. **Calculate the sample size \( n \)**:
- \( n = 250 \)

4. **Calculate the margin of error**:
\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.076 \times 0.924}{250}}
\]

5. **Calculating the inside of the square root**:
- \( 0.076 \times 0.924 = 0.070224 \)
- \( \frac{0.070224}{250} = 0.000280896 \)

6. **Taking the square root**:
- \( \sqrt{0.000280896} \approx 0.0168 \)

7. **Final calculation**:
\[
\text{Margin of Error} = 1.96 \times 0.0168 \approx 0.0330
\]

8. **Convert the margin of error into a percentage**:
- \( 0.0330 \times 100 = 3.3\% \)

Therefore, the margin of error is **3.3%**.

The correct response is:

**3.3%**