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% 7.6% 7.6% 3.3%

Answer 1

To calculate the margin of error (ME) for the proportion of students who want year-round school, 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
- \( n \) is the sample size

First, we need to calculate the sample proportion \( p \):

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

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

\[
p(1-p) = 0.076 \times (1 - 0.076) = 0.076 \times 0.924 = 0.0702
\]

Next, we can substitute into the margin of error formula:

\[
ME = 1.96 \times \sqrt{\frac{0.0702}{250}}
\]

Calculating the standard error:

\[
\sqrt{\frac{0.0702}{250}} \approx \sqrt{0.0002808} \approx 0.01674
\]

Now, substituting back into the margin of error equation:

\[
ME = 1.96 \times 0.01674 \approx 0.0328
\]

Finally, expressing the margin of error as a percentage:

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

The closest match from the options you've provided is **3.3%**.