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 μ is 7.09. (round to two decimal places as needed.) a 98% confidence interval using the standard normal distribution is (□,□). (round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for confidence interval

The formula for a confidence interval when the population standard deviation \(\sigma\) is known is \(\bar{x}-E <\mu<\bar{x} + E\), where \(\bar{x}\) is the sample mean and \(E\) is the margin of error.

Step2: Identify the values of \(\bar{x}\) and \(E\)

We are given that \(\bar{x}=35.5\) (sample mean) and \(E = 7.09\) (margin of error).

Step3: Calculate the lower and upper bounds of the confidence interval

For the lower bound: \(\bar{x}-E=35.5 - 7.09=28.41\approx28.4\)
For the upper bound: \(\bar{x}+E=35.5 + 7.09=42.59\approx42.6\)

Answer:

\((28.4,42.6)\)