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 3.3% 14.9% 7.6% 0.033%

Answer 1

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

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

where:
- \( z \) is the z-score (1.96 for 95% confidence),
- \( p \) is the sample proportion,
- \( n \) is the sample size.

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

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

Next, we substitute \( p \) and \( n \) into the margin of error formula.

\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.076(1-0.076)}{250}}
\]

Calculating \( p(1-p) \):

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

Now, substitute this back into the formula:

\[
\text{Margin of Error} = 1.96 \times \sqrt{\frac{0.070224}{250}}
\]

Calculating the denominator:

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

Finding the square root:

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

Now calculating the margin of error:

\[
\text{Margin of Error} = 1.96 \times 0.01677 \approx 0.0329
\]

Converting to percentage:

\[
0.0329 \times 100 \approx 3.29\%
\]

Therefore, rounding to one decimal place:

**The margin of error is approximately 3.3%.**