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)

Answer 1

To calculate the margin of error for a proportion, you can use the following formula:

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

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 find the sample proportion \( p \):

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

Next, we calculate \( 1 - p \):

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

Now we can plug these values into the margin of error formula:

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

Calculating the fraction inside the square root:

\[
\frac{0.076 \cdot 0.924}{250} = \frac{0.070224}{250} = 0.000280896
\]

Now take the square root:

\[
\sqrt{0.000280896} \approx 0.01675
\]

Now, multiply by the z-score:

\[
\text{ME} \approx 1.96 \cdot 0.01675 \approx 0.0328
\]

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