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 56 feet and a standard deviation of 6.4 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 58 feet and a standard deviation of 6.6 feet. Suppose that a sample of 80 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 0.05 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 H0H0. Round the numerical portion of your answer to three decimal places.

Step 4 of 4:

Make the decision for the hypothesis test.

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.

Since , the population standard deviations are not equal and unknown.

We replace the and by their corresponding sample estimates.

Therefore , use the Z-test for two means.

Step 1 or 4 : The null and alternative hypothesis is ,

Hypothesis : VS

Step 2 of 4 : The value of the test statistic is ,

Step 3 or 4 :

Decision rule : If , then reject the null hypothesis , accept otherwise

Step 4 of 4 :

Decision : Here ,

Therefore , reject the null hypothesis.

Conclusion : Hence , there is sufficient evidence to support 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.