From a sample of 100 Bloemfontein car commuters, the sample mean time to commute to work daily was found to be 35.8 minutes. It is known that the population standard deviation is 20 minutes.

Find the margin of error with a 99% confidence level.

1 answer

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.