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 64 cars owned by students had an average age of 6.27 years. A sample of 40 cars owned by faculty had an average age of 7.64 years. Assume that the population standard deviation for cars owned by students is 2.64 years, while the population standard deviation for cars owned by faculty is 3.42 years. Determine the 80% confidence interval for the difference between the true mean ages for cars owned by students and faculty. Step 2 of 3 : Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

Answer 1

Margin of error = 0.81

Upper limit = -0.56

Lower limit = -2.18

Explanation:

Given Data;

Sample size Sample mean population deviation S.D

Sample 1 n₁= 64 x1-bar = 6.27 б₁ = 2.64

Sample 2; n₂ = 40 x2 -bar = 7.64 б₂ = 3.42

For 80% interval, a = 1-0.8 = 0.2

From normal standard table,

the critical value of (a/2 = 0.2/2) =0.1 is 1.28

That is, Z = 1.28

Calculating the margin of error, we have

Margin of error = Z * √(б²₁/n₁ + б²₂/n₂)

= 1.28 * √(2.64²/64 + 3.42²/40)

= 1.28 * √0.1089 + 0.29241

= 1.28 * √0.40131

=0.81

Lower and upper limit of 80% interval for the difference between the true mean ages for cars owned by student and faculty is calculated as

Lower limit = (x1 - x2) - margin of error

= (6.27-7.64) - 0.81

= -2.18

Upper limit = (x1 - x2) + margin of error

= (6.27-7.64) + 0.81

= -0.56