To calculate the margin of error (ME) for a population mean at a 99% confidence level, you can use the following formula:
\[ ME = Z \times \frac{\sigma}{\sqrt{n}} \]
Where:
- \( Z \) is the critical value for the desired confidence level
- \( \sigma \) is the standard deviation of the population
- \( n \) is the sample size
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Determine the critical value \( Z \) for a 99% confidence level. From the z-table, the critical value is approximately \( Z = 2.576 \).
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Plug in the values:
- \( \sigma = 56 \)
- \( n = 120 \)
Now, calculating the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{56}{\sqrt{120}} \approx \frac{56}{10.9545} \approx 5.11 \]
- Calculate the margin of error:
\[ ME = Z \times SE = 2.576 \times 5.11 \approx 13.18 \]
Thus, rounding to the nearest hundredth, the margin of error is:
\[ \text{Margin of error} \approx 13.18 \]
So the final answer is:
\[ \text{The margin of error is } 13.18 \]