Question

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.

Answer 1

Answer:

98%  Confidencce Interval is ( 3030.6, 7467.4 )

Step-by-step explanation:

Given that:

Sample size $n_1 =$ 71

Sample size $n_2 =$ 31

Sample mean $\overline x_1 =$ 41628

Sample mean $x_2 =$ 36,379

Population standard deviation $\sigma_1$ = 4934

Population standard deviation $\sigma_2 =$ 4180

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

Critical value at $z_{0.02/2} = 2.33$

The Margin of Error = $z \times \sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma^2_2}{n_2} }$

= $2.33 \times \sqrt{\dfrac{4934^2}{71}+\dfrac{4180^2}{31} }$

= $2.33 \times \sqrt{\dfrac{24344356}{71}+\dfrac{17472400}{31} }$

= $2.33 \times \sqrt{906504.06 }$

= 2218.40

The Lower limit = $( \overline x_1 - \overline x_2) - (Margin \ of \ error)$

= ( 41628 - 36379 ) - ( 2218.40)

= 5249 - 2218.40

= 3030.6

The upper limit = $( \overline x_1 - \overline x_2) + (Margin \ of \ error)$

= ( 41628 - 36379 ) + ( 2218.40)

= 5249 + 2218.40

= 7467.4

∴  98%  Confidencce Interval is ( 3030.6, 7467.4 )