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.

Hence the confidence interval (2.2, 3.52).

Step-by-step explanation:

Hence,

The point estimate =

= 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 = ( - tc x sp , + 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)

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

subtract 112.49 from both sides we get;

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt

yes because they have the same amount of green and blur marbles.

step-by-step explanation: