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 6262 feet and a standard deviation of 10.610.6 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 6868 feet and a standard deviation of 13.913.9 feet. Suppose that a sample of 7777 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 μ1μ1 be the true mean braking distance corresponding to compound 1 and μ2μ2 be the true mean braking distance corresponding to compound 2. Use the 0.050.05 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

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

STEP 1: NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

STEP 2: Test Statistics

The z-statistic is computed as follows:

z= -3.01

STEP 3: Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is zc=−1.64.

STEP 4: Since it is observed that z=−3.012<z∗=−1.64, it is then concluded that the null hypothesis is rejected.

Decision: FAIL TO REJECT NULL HYPOTHESIS H0.