Question

in a random sample of six people, the mean driving distance to work was 23.9 miles and the standard deviation was 7.2 miles. assuming the population is normally distributed and using the t - distribution, a 99% confidence interval for the population mean μ is (12.0, 35.8) (and the margin of error is 11.9). through research, it has been found that the population standard deviation of driving distances to work is 6.3. using the standard normal distribution with the appropriate calculations for a standard deviation that is known, find the margin of error and construct a 99% confidence interval for the population mean μ. interpret and compare the results

identify the margin of error.
miles
(round to one decimal place as needed.)

Explanation:

Step1: Find the critical value \( z_{\alpha/2} \)

For a 99% confidence interval, \( \alpha=1 - 0.99=0.01 \), so \( \alpha/2=0.005 \).
From the standard normal distribution table, \( z_{0.005} = 2.576 \).

Step2: Calculate the margin of error \( E \)

The formula for the margin of error when the population standard deviation \( \sigma \) is known is \( E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \).
Given \( \sigma = 6.3 \), \( n = 6 \), and \( z_{\alpha/2}=2.576 \).
Substitute the values into the formula: \( E=2.576\times\frac{6.3}{\sqrt{6}} \).
First, calculate \( \sqrt{6}\approx2.45 \), then \( \frac{6.3}{2.45}\approx2.571 \).
Then \( E = 2.576\times2.571\approx6.6 \).

Answer:

\( 6.6 \)