Question

What is the margin of error if the sample size is 120? Write your answer as a percentage rounded to the nearest whole number.

Answer 1

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:

1. Calculate \( \sqrt{\frac{p(1-p)}{n}} \):
\[
p(1-p) = 0.5(1-0.5) = 0.5 \cdot 0.5 = 0.25
\]
\[
\frac{0.25}{120} = 0.00208333
\]
\[
\sqrt{0.00208333} \approx 0.0455
\]

2. Calculate the margin of error:
\[
\text{Margin of Error} = 1.96 \cdot 0.0455 \approx 0.0893
\]

3. 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%**.