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 53 feet and a standard deviation of 8.2 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 59 feet and a standard deviation of 12.1 feet. Suppose that a sample of 47 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 0.01 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.

Answer 1

1)We have to test

2)The level of significance is . The sample sizes are . The degrees of freedom is

Here, the sample means are . The sample standard deviations are

The test statistic is

3) Here the Decision rule is reject null hypothesis if  . (Critical value method).

4) So we reject the null hypothesis. The decision is we are 99% confident that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used.