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 78 cars owned by students had an average age of 5.04 years. A sample of 118 cars owned by faculty had an average age of 8 years. Assume that the population standard deviation for cars owned by students is 3.06 years, while the population standard deviation for cars owned by faculty is 3.24 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

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(9.3636/78 + 10.4976/118)
sp = 0.4572

Given CI level is 0.98, hence α = 1 - 0.98 = 0.02                  
α/2 = 0.02/2 = 0.01, zc = z(α/2, df) = 2.33                  
                  
Margin of Error                  
ME = zc * sp                  
ME = 2.33 * 0.4572                  
ME = 1.065                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (5.04 - 8 - 2.33 * 0.4572 , 5.04 - 8 - 2.33 * 0.4572                  
CI = (-4.03 , -1.89)