Question

A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. SUVs equipped with tires using compound 1 have a mean braking distance of 62 feet and a standard deviation of 10.6 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 6868 feet and a standard deviation of 13.9 feet. Suppose that a sample of 77 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1 be the true mean braking distance corresponding to compound 1 and μ2μ2 be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance.

Step 1 of 4 :  

State the null and alternative hypotheses for the test.

STEP 2:Compute the value of the test statistic. Round your answer to two decimal places.

STEP 3:Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to two decimal places. (Reject H0 if)

STEP 4:Make the decision for the hypothesis test. (Reject or Fail to reject)

Answer 1

step 1:

null hypothesis:              Ho:μ1 >= μ2

Alternate hypothesis: Ha:μ1 < μ2

step 2:

		compound 1	compound 2
sample mean x =		62.00	68.00
std deviation s=		10.60	13.90
sample size n=		77	77
std error σx1-x2=√(σ21/n1+σ22/n2)    =			1.992
test stat z =(x1-x2-Δo)/σx1-x2     =			-3.01

STEP 3:

Decision rule : reject Ho if test statistic z<-1.64

step 4:

reject Ho