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 76 feet and a standard deviation of 8.9 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 80 feet and a standard deviation of 14.6 feet. Suppose that a sample of 45 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 μ2 be the true mean braking distance corresponding to compound 2. Use the 0.1 level of significance.

Step 1 of 4: State the null and alternative hypotheses for the test.

Step 2 of 4: Compute the value of the test statistic. Round your answer to two decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places.

Step 4 of 4: Make the decision for the hypothesis test. Fail or reject to fail.

Answer 1

μ1 be the true mean braking distance corresponding to compound 1 and μ2 be the true mean braking distance corresponding to compound 2.  

Claim : braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used

1) H₀ : µ₁ = µ₂    vs H_a : µ₁ < µ₂   

2) = 76 , S₁ = 8.9 , n₁ = 45 and   = 80,S₂= 14.6, n₂ = 45

Pooled estimate of standard deviaion :

S =

S =

S = 12.0907

Test statistic:

t =

t =

t = -1.57

We are given α =0.1

Test statistic follows t distribution with degrees of freedom (df) = n₁+ n₂ - 2 = 45+45-2 = 88

We can find critical value using excel function =TINV( α , d.f )

=TINV( 0.1 , 88 ) = 1.29

Since this is left tail test , critical value would be negative , -1.29

3) Critical region: Reject H₀, if t ≤ -1.29 Or fail to reject H₀ , if t > -1.29

4) Decision : We have t = -1.57,

As t < -1.29 we reject H₀

Conclusion : We have significant evidence that mean braking distance for SUVs equipped with tires using compound 1 is shorter than the mean braking distance when compound 2 is used