An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining. A sample of 263 brakes using Compound 1 yields an average brake life of 41,270 miles. A sample of 291 brakes using Compound 2 yields an average brake life of 40,579 miles. Assume that the population standard deviation for Compound 1 is 4540 miles, while the population standard deviation for Compound 2 is 2990 miles. Determine the 90% 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 90% confidence interval. Round your answers to the nearest whole number.
Compound 1:
n=263, mean=41270, standard deviation=4540, z value (for 90%CI)=1.645
C.I=meanz*s/sqrt(n)
=412701.645*4540/sqrt(263)
CI=41270460.0517
C.I=(41730.0517,40809.9483)
Compound 2:
n=291, mean=40579, standard deviation=2990, z value (for 90%CI)=1.645
C.I=meanz*s/sqrt(n)
=405791.645*2990/sqrt(291)
=40579288.3308
C.I=(40867.3308,40290.6692)
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