To calculate the margin of error (ME) for a proportion at a specified confidence level, you can use the formula:
\[ ME = z \times \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \( z \) is the z-score corresponding to the desired confidence level.
- \( p \) is the sample proportion (the number of students in favor divided by the total number of students surveyed).
- \( n \) is the sample size.
In this case:
- The total number of students surveyed (\( n \)) = 250
- The number of students wanting year-round school = 19
- Thus, the sample proportion (\( p \)) = \( \frac{19}{250} = 0.076 \)
Now, let's calculate the margin of error using the given z-score of 1.96.
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Calculate \( p(1-p) \): \[ p(1-p) = 0.076(1 - 0.076) = 0.076 \times 0.924 = 0.070224 \]
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Now calculate the standard error (SE): \[ SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.070224}{250}} = \sqrt{0.000280896} \approx 0.01677 \]
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Now calculate the margin of error (ME): \[ ME = z \times SE = 1.96 \times 0.01677 \approx 0.0328 \]
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To express this as a percentage, we multiply by 100: \[ ME \approx 0.0328 \times 100 \approx 3.28% \]
So, the margin of error, rounded to one decimal point, is approximately 3.3%.