To calculate the margin of error for the survey results, we can use the formula for the margin of error (ME) in a proportion, which is given by:
\[ ME = z \times \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \( z \) is the z-score corresponding to the desired level of confidence (1.96 for 95% confidence),
- \( p \) is the sample proportion (the number of students who want year-round school divided by the total number of students surveyed),
- \( n \) is the sample size (the total number of students surveyed).
From the problem:
- Number of students in favor of year-round school = 19
- Total number of students surveyed = 250
Calculating \( p \): \[ p = \frac{19}{250} = 0.076 \]
Now, calculate \( 1 - p \): \[ 1 - p = 1 - 0.076 = 0.924 \]
Next, substitute these values into the margin of error formula:
\[ ME = 1.96 \times \sqrt{\frac{0.076 \times 0.924}{250}} \]
Calculating the fraction: \[ \frac{0.076 \times 0.924}{250} = \frac{0.070224}{250} = 0.000280896 \]
Now calculate the square root: \[ \sqrt{0.000280896} \approx 0.01677 \]
Finally, plug this value into the formula for the margin of error: \[ ME = 1.96 \times 0.01677 \approx 0.0329 \]
Converting this to a percentage: \[ 0.0329 \times 100 \approx 3.29% \]
Rounding to one decimal place, the margin of error is approximately 3.3%.
Thus, the correct response is:
3.3%