To calculate the margin of error for a given sample size, you typically use the formula:
\[ \text{Margin of Error} = z \cdot \sqrt{\frac{p(1-p)}{n}} \]
where:
- \( z \) is the z-score corresponding to the desired confidence level,
- \( p \) is the estimated proportion of the population (if not known, you can use \( p = 0.5 \) for maximum variability),
- \( n \) is the sample size.
For a common confidence level of 95%, the z-score is approximately 1.96.
Assuming \( p = 0.5 \), the calculation for the margin of error with a sample size \( n = 120 \) would be:
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Calculate \( \sqrt{\frac{p(1-p)}{n}} \): \[ p(1-p) = 0.5(1-0.5) = 0.5 \cdot 0.5 = 0.25 \] \[ \frac{0.25}{120} = 0.00208333 \] \[ \sqrt{0.00208333} \approx 0.0455 \]
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Calculate the margin of error: \[ \text{Margin of Error} = 1.96 \cdot 0.0455 \approx 0.0893 \]
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Convert this to a percentage: \[ 0.0893 \times 100 \approx 8.93% \]
Rounding to the nearest whole number, the margin of error is approximately 9%.
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