Question

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

7.6%
7.6%

14.9%
14.9%

3.3%
3.3%

0.033%

Answer 1

To calculate the margin of error for the survey results, we can use the formula for the margin of error (ME) in a proportion, which is given by:

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

Where:
- \( z \) is the z-score corresponding to the desired level of confidence (1.96 for 95% confidence),
- \( p \) is the sample proportion (the number of students who want year-round school divided by the total number of students surveyed),
- \( n \) is the sample size (the total number of students surveyed).

From the problem:
- Number of students in favor of year-round school = 19
- Total number of students surveyed = 250

Calculating \( p \):
\[
p = \frac{19}{250} = 0.076
\]

Now, calculate \( 1 - p \):
\[
1 - p = 1 - 0.076 = 0.924
\]

Next, substitute these values into the margin of error formula:

\[
ME = 1.96 \times \sqrt{\frac{0.076 \times 0.924}{250}}
\]

Calculating the fraction:
\[
\frac{0.076 \times 0.924}{250} = \frac{0.070224}{250} = 0.000280896
\]

Now calculate the square root:
\[
\sqrt{0.000280896} \approx 0.01677
\]

Finally, plug this value into the formula for the margin of error:
\[
ME = 1.96 \times 0.01677 \approx 0.0329
\]

Converting this to a percentage:
\[
0.0329 \times 100 \approx 3.29%
\]

Rounding to one decimal place, the margin of error is approximately **3.3%**.

Thus, the correct response is:

**3.3%**