Question

A student researcher compares the ages of cars owned by students and cars owned by faculty at a local state college. A sample of 187 cars owned by students had an average age of 7.9 years. A sample of 221 cars owned by faculty had an average age of 5.04 years. Assume that the population standard deviation for cars owned by students is 3.07 years, while the population standard deviation for cars owned by faculty is 2.53 years. Determine the 98% confidence interval for the difference between the true mean ages for cars owned by students and faculty. Step 3 of 3 : Construct the 98% confidence interval. Round your answers to two decimal places.

Answer 1

Hence the confidence interval (2.2, 3.52).

Explanation:

Hence,

The point estimate =
`\bar x_(1) - \bar x_(2)`

= 7.9 - 5.04

= 2.86

Given CI level is 0.98, hence α = 1 - 0.98 = 0.02

α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.326

Margin of Error

ME = tc x sp

ME = 2.326 \ 0.2817

ME = 0.6552

CI = (
`\bar x_(1) - \bar x_(2)` - tc x sp ,
`\bar x_(1) - \bar x_(2)` + tc x sp)

CI = (7.9 - 5.04 - 2.326 x 0.2817 , 7.9 - 5.04 - 2.326 x 0.2817

CI = (2.2 , 3.52)