To calculate the margin of error for the given sample at a 99% confidence level, we can use the formula for the margin of error (ME) in the context of a population with a known standard deviation:
\[ \text{Margin of error} (ME) = z \cdot \left(\frac{\sigma}{\sqrt{n}}\right) \]
Where:
- \( z \) is the z-value corresponding to the desired confidence level.
- \( \sigma \) is the population standard deviation.
- \( n \) is the sample size.
Step 1: Identify the values
- The population standard deviation \( \sigma = 20 \) minutes.
- The sample size \( n = 100 \).
- For a 99% confidence level, we need to find the z-value. The z-value for a 99% confidence level is approximately 2.576.
Step 2: Calculate the standard error (SE) \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{100}} = \frac{20}{10} = 2 \]
Step 3: Calculate the margin of error (ME) \[ ME = z \cdot SE = 2.576 \cdot 2 = 5.152 \]
Step 4: Report the margin of error The margin of error with a 99% confidence level is approximately 5.15 minutes.