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)
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Determine the z-score: For a 95% confidence level, the z-score is approximately \(1.96\).
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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:
\[ \sqrt{0.000024} \approx 0.004899 \]
- 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.