To calculate the margin of error for a poll, we can use the formula for the margin of error (ME) in proportion estimates, which is:
\[ ME = Z \times \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \(p\) is the proportion (in decimal form) of respondents preferring the computer program (39% or 0.39).
- \(n\) is the sample size (350).
- \(Z\) is the Z-score corresponding to the desired confidence level (for a 95% confidence level, \(Z \approx 1.96\)).
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Calculate \(p(1-p)\): \[ p(1-p) = 0.39 \times (1 - 0.39) = 0.39 \times 0.61 = 0.2379 \]
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Calculate \(\frac{p(1-p)}{n}\): \[ \frac{p(1-p)}{n} = \frac{0.2379}{350} \approx 0.000678 \]
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Calculate the square root: \[ \sqrt{\frac{p(1-p)}{n}} \approx \sqrt{0.000678} \approx 0.0260 \]
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Calculate the Margin of Error (ME): \[ ME = 1.96 \times 0.0260 \approx 0.0510 \text{ (or 5.10% when converted to percentage)} \]
So, rounding up, the margin of error is approximately 5.1%.
From your given options (39%, 5.3%, 18.7%, 61%), the closest option to our calculated margin of error of 5.1% is 5.3%.