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 78 feet and a standard deviation of 14.9 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 86 feet and a standard deviation of 5.9 feet. Suppose that a sample of 7979 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 ?1 be the true mean braking distance corresponding to compound 1 and ?2 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.

Answer 1

Step 1 of 4:
H0: Null Hypothesis: 1 2

HA: Alternative Hypothesis: 1 < 2

Step 2 of 4:

Test statistic is:
Z = (1 - 2)/SE

= (78 - 86)/1.8030 = - 4.4370

Step 3 of 4:

= 0.10

One Tail - Left Side test

From Table, critical value of Z = - 1.28

Decision Rule for rejecting the null hypothesis H0:

Reject H0 if

Z < - 1.28.

Step 4 of 4:
Since the calculated value of Z = - 4.4370 is less than critical value of Z = - 1.28, Reject H0.

Conclusion:
The data support the claim that the braking distance of SUVs equipped with tires using compound 1 is shorter than the braking distancxe when compound 2 is used.