Question

An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining. A sample of 125 brakes using Compound 1 yields an average brake life of 41,759 miles. A sample of 163 brakes using Compound 2 yields an average brake life of 40,361 miles. Assume that the population standard deviation for Compound 1 is 3921 miles, while the population standard deviation for Compound 2 is 3097miles. Determine the 95% confidence interval for the true difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2.

Step 3 of 3 :  

Construct the 95%95% confidence interval. Round your answers to the nearest whole number.

Answer 1

sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(15374241/125 + 9591409/163)
sp = 426.4234

Given CI level is 0.95, hence α = 1 - 0.95 = 0.05                  
α/2 = 0.05/2 = 0.025, zc = z(α/2, df) = 1.96                  
                  
Margin of Error                  
ME = zc * sp                  
ME = 1.96 * 426.4234                  
ME = 835.79                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (41759 - 40361 - 1.96 * 426.4234 , 41759 - 40361 - 1.96 * 426.4234                  
CI = (562 , 2234)