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)

Question

Responses

3.3%
3.3%

7.6%
7.6%

14.9%
14.9%

0.033%

Answer 1

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

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

where:

• \( z \) is the z-score corresponding to the desired confidence level,
• \( p \) is the sample proportion,
• \( n \) is the sample size.

In this case:

• The sample size \( n = 250 \).
• The number of students who want year-round school \( x = 19 \).
• Therefore, the sample proportion \( p = \frac{x}{n} = \frac{19}{250} = 0.076 \).

Now we can calculate the margin of error:

1. Calculate \( 1 - p \): \[ 1 - p = 1 - 0.076 = 0.924 \]

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

3. Take the square root: \[ \sqrt{0.000280896} \approx 0.01677 \]

4. Calculate the margin of error: \[ \text{Margin of Error} = 1.96 \times 0.01677 \approx 0.0328 \]

To express this as a percentage: \[ 0.0328 \times 100 \approx 3.28% \]

So, rounding to one decimal place, the margin of error is approximately 3.3%.

Therefore, the correct answer is: 3.3%.