Question

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

95% confidence; the sample size is 10,000, of which 40% are successes

A) 0.0096
B) 0.0126
C) 0.0072
D) 0.0110

Answer 1

To calculate the margin of error (E) for estimating a population proportion at a 95% confidence level, we can use the formula for the margin of error related to proportions:

\[
E = z \cdot \sqrt{\frac{p(1 - p)}{n}}
\]

where:
- \(z\) is the z-score corresponding to the desired confidence level,
- \(p\) is the sample proportion,
- \(n\) is the sample size.

Given:
- Confidence level = 95%
- Sample size \(n = 10000\)
- Sample proportion \(p = 0.40\) (since 40% are successes)

1. **Determine the z-score**: For a 95% confidence level, the z-score is approximately \(1.96\).

2. **Plug the values into the formula**:

\[
E = 1.96 \cdot \sqrt{\frac{0.40(1 - 0.40)}{10000}}
\]

Calculating the components inside the square root:

\[
1 - p = 1 - 0.40 = 0.60
\]

\[
p(1 - p) = 0.40 \cdot 0.60 = 0.24
\]

\[
\frac{p(1 - p)}{n} = \frac{0.24}{10000} = 0.000024
\]

Now taking the square root:

\[
\sqrt{0.000024} \approx 0.004899
\]

3. **Calculate the margin of error**:

\[
E = 1.96 \cdot 0.004899 \approx 0.0096
\]

Finally, rounding to four decimal places:

\[
E \approx 0.0096
\]

Thus, the margin of error is **0.0096**.

So the correct answer is **A) 0.0096**.