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%

3.3%
3.3%

7.6%
7.6%

14.9%
14.9%

Answer 1

To calculate the margin of error (ME) for a survey proportion, you can use the formula:

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

where:
- \( z \) is the z-score (1.96 for a 95% confidence level),
- \( p \) is the sample proportion (the number of favorable responses divided by the total number of responses),
- \( n \) is the sample size.

In this case:

1. The number of students who want year-round school is 19.
2. The total number of students surveyed is 250.

So, we first calculate the sample proportion \( p \):

\[
p = \frac{19}{250} = 0.076
\]

Next, we will use this value to find the margin of error:

\[
ME = 1.96 \times \sqrt{\frac{0.076(1 - 0.076)}{250}}
\]

Calculating the term under the square root:

\[
1 - p = 1 - 0.076 = 0.924
\]

Now, calculate \( p(1 - p) \):

\[
p(1 - p) = 0.076 \times 0.924 = 0.070224
\]

Divide by \( n \):

\[
\frac{p(1 - p)}{n} = \frac{0.070224}{250} = 0.000280896
\]

Taking the square root of that value:

\[
\sqrt{0.000280896} \approx 0.01677
\]

Now, multiply by the z-score to get the margin of error:

\[
ME = 1.96 \times 0.01677 \approx 0.0328
\]

Converting this margin of error into percentage form:

\[
ME \approx 0.0328 \times 100 \approx 3.28\%
\]

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

So, the correct answer is:

**3.3%**