An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining. A sample of 71 brakes using Compound 1 yields an average brake life of 41,628 miles. A sample of 31 brakes using Compound 2 yields an average brake life of 36,379 miles. Assume the standard deviation of brake life is known to be 4934 miles for brakes made with Compound 1 and 4180 miles for brakes made with Compound 2. Determine the 98% 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.

98%  Confidencce Interval is ( 3030.6, 7467.4 )

Step-by-step explanation:

Given that:

Sample size 71

Sample size 31

Sample mean 41628

Sample mean 36,379

Population standard deviation = 4934

Population standard deviation 4180

At 98% confidence interval level, the level of significcance = 1 - 0.98 = 0.02

Critical value at

The Margin of Error =

=

=

=

= 2218.40

The Lower limit =

= ( 41628 - 36379 ) - ( 2218.40)

= 5249 - 2218.40

= 3030.6

The upper limit =

= ( 41628 - 36379 ) + ( 2218.40)

= 5249 + 2218.40

= 7467.4

∴  98%  Confidencce Interval is ( 3030.6, 7467.4 )

answer: 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

step-by-step explanation: