Question

An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining. A sample of 65 brakes using Compound 1 yields an average brake life of 37,409 miles. A sample of 31 brakes using Compound 2 yields an average brake life of 49,036 miles. Assume the standard deviation of brake life is known to be 1341 miles for brakes made with Compound 1 and 2961 miles for brakes made with Compound 2.  Determine the 95% confidence interval for the true difference between average lifetimes for brakes using Compound 1 and brakes using Compound 2.

Step 1 of 2: Find the critical value that should be used in constructing the confidence interval.

Step 2 of 2: Construct the 95% confidence interval. Round your answers to the nearest whole number.

Answer 1

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(1798281/65 + 8767521/31)
sp = 557.2155

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      

critical value = 1.96   
                  
Margin of Error                  
ME = zc * sp                  
ME = 1.96 * 557.2155                  
ME = 1092.142                  
                  
CI = (x1bar - x2bar - tc * sp , x1bar - x2bar + tc * sp)                  
CI = (37409 - 49036 - 1.96 * 557.2155 , 37409 - 49036 - 1.96 * 557.2155                  
CI = (-12719 , -10535)