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?
Responses

0.033%
0.033%

3.3%
3.3%

14.9%
14.9%

7.6%

Answer 1

To calculate the margin of error for the survey results, you 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 confidence level (1.96 for 95% confidence),
- \( p \) is the sample proportion (the number of respondents who want year-round school divided by the total number of respondents),
- \( n \) is the total number of respondents.

1. Calculate the sample proportion \( p \):
\[
p = \frac{19}{250} = 0.076
\]

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

3. Now substitute \( p \), \( 1 - p \), and \( n \) into the margin of error formula:
\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.076 \times 0.924}{250}}
\]

4. Calculate the expression inside the square root:
\[
0.076 \times 0.924 = 0.070224
\]
\[
\frac{0.070224}{250} = 0.000280896
\]
\[
\sqrt{0.000280896} \approx 0.01677
\]

5. Now calculate the margin of error:
\[
\text{Margin of Error} = 1.96 \times 0.01677 \approx 0.0328
\]

6. Convert the margin of error into a percentage:
\[
0.0328 \times 100 \approx 3.28\%
\]

Rounding this gives approximately **3.3%**.

So the correct response for the margin of error is:
**3.3%**