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%
3.3%

0.033%
0.033%

14.9%
14.9%

7.6%

Answer 1

To calculate the margin of error for a proportion at a 95% confidence level, you 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 (number of students wanting year-round school divided by total number of students),
- \( n \) is the sample size.

In this case:
- Total students (n) = 250
- Students wanting year-round school = 19
- Sample proportion (p) = \( \frac{19}{250} = 0.076 \)

Now, plug these values into the formula:

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

Calculating \( 1 - p \):

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

Now we can calculate the sample variance:

\[
\frac{p(1-p)}{n} = \frac{0.076 \times 0.924}{250} = \frac{0.070224}{250} = 0.000280896
\]

Now take the square root:

\[
\sqrt{0.000280896} \approx 0.0168
\]

Now multiply by the z-score:

\[
\text{Margin of Error} = 1.96 \times 0.0168 \approx 0.032928 \text{ (or about 0.033)}
\]

To express this in percentage terms:

\[
\text{Margin of Error} \approx 0.033 \times 100 \approx 3.3\%
\]

Therefore, the margin of error is **3.3%**.