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?

Answer 1

To calculate the margin of error for a proportion in a survey, we can use the following formula:

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

where:
- \( z \) is the z-score corresponding to the desired level of confidence (which is 1.96 for a 95% confidence level),
- \( p \) is the proportion of the sample that has the characteristic (in this case, the proportion of students who want year-round school),
- \( n \) is the sample size.

First, we need to calculate \( p \):
- The number of students who want year-round school is 19,
- The total number of students surveyed is 250.

So, the proportion \( p \) is:

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

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

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

Next, we can substitute \( p \), \( 1 - p \), and \( n \) into the margin of error formula:

\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.076 \times 0.924}{250}}
\]

Calculating the inside of the square root:

\[
0.076 \times 0.924 = 0.070224
\]

Now, divide by \( n \):

\[
\frac{0.070224}{250} = 0.000280896
\]

Then, find the square root:

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

Finally, multiply by the z-score:

\[
\text{Margin of Error} = 1.96 \times 0.01677 \approx 0.0328
\]

Thus, the margin of error is approximately 0.0328, or 3.28%.