To calculate the margin of error for the population mean at a 99% confidence level, you can use the formula:
\[ E = z \times \left(\frac{\sigma}{\sqrt{n}}\right) \]
Where:
- \(E\) is the margin of error.
- \(z\) is the critical value for the confidence level.
- \(\sigma\) is the standard deviation.
- \(n\) is the sample size.
First, let's find the critical value \(z\) for a 99% confidence level. For a 99% confidence level, the critical z-value is approximately 2.576.
Now plug in the values into the formula:
- \(\sigma = 56\)
- \(n = 120\)
-
Calculate the standard error (\(SE\)): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{56}{\sqrt{120}} \approx \frac{56}{10.954} \approx 5.11 \]
-
Now calculate the margin of error \(E\): \[ E = z \times SE = 2.576 \times 5.11 \approx 13.18 \]
Rounding to the nearest hundredth, the margin of error is:
\[ \text{Margin of Error} \approx 13.18 \]
Thus, the margin of error for the population mean is $13.18.