To calculate the margin of error for a poll at a given confidence level, we can use the formula:
\[ ME = z \times \sqrt{\frac{p(1-p)}{n}} \]
where:
- \(ME\) is the margin of error,
- \(z\) is the z-score corresponding to the confidence level,
- \(p\) is the sample proportion (as a decimal), and
- \(n\) is the sample size.
Step 1: Find the values
- The sample size \(n\) is 500.
- The proportion \(p = 56%\) means \(p = 0.56\).
- The complement of the proportion \(1 - p = 1 - 0.56 = 0.44\).
Step 2: Find the z-score for 99% confidence level
For a 99% confidence interval, the z-score (z) is approximately 2.576 (you can find this value in a z-table or from standard normal distribution values for a two-tailed test).
Step 3: Plug the values into the formula
Now we can substitute our values into the margin of error formula:
\[ ME = 2.576 \times \sqrt{\frac{0.56 \times 0.44}{500}} \]
Calculating the inside of the square root first:
\[ 0.56 \times 0.44 = 0.2464 \]
Now, divide by the sample size \(n\):
\[ \frac{0.2464}{500} = 0.0004928 \]
Now, take the square root:
\[ \sqrt{0.0004928} \approx 0.022187 \]
Finally, multiply by the z-score to find the margin of error:
\[ ME = 2.576 \times 0.022187 \approx 0.057228 \]
Step 4: Rounding
Rounding to three decimal places gives us a margin of error of:
\[ \text{Margin of Error} \approx 0.057 \]
Thus, the final answer for the margin of error at the 99% confidence level is:
\[ \boxed{0.057} \]