Question

in a random sample of 8 people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.3 minutes. a 98% confidence interval using the t - distribution was calculated to be (27.8,43.2). after researching commute times to work, it was found that the population standard deviation is 8.6 minutes. find the margin of error and construct a 98% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. compare the results.
the margin of error of \\( \mu \\) is
(round to two decimal places as needed.)

Explanation:

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

For a \( 98\% \) confidence interval, \( \alpha=1 - 0.98=0.02 \), so \( \alpha/2=0.01 \).
Using the standard normal distribution table or a calculator, \( z_{\alpha/2}=z_{0.01}\approx 2.33 \).

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 \( \sigma = 8.6 \), \( n = 8 \), and \( z_{\alpha/2}=2.33 \).
Substitute the values into the formula: \( E=2.33\times\frac{8.6}{\sqrt{8}} \).
First, calculate \( \sqrt{8}\approx2.828 \).
Then \( \frac{8.6}{2.828}\approx3.041 \).
Finally, \( E = 2.33\times3.041\approx7.09 \).

Answer:

\( 7.09 \)