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. The mean braking distance for SUVs equipped with tires made with compound 1 is 56 feet, with a population standard deviation of 13.0. The mean braking distance for SUVs equipped with tires made with compound 2 is 61 feet, with a population standard deviation of 5.3. Suppose that a sample of 50 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 5: State the null and alternative hypotheses for the test. Step 2 of 5: Compute the value of the test statistic. Round your answer to two decimal places. Step 3 of 5: Find the p-value associated with the test statistic. Round your answer to four decimal places. Step 4 of 5: Make the decision for the hypothesis test. Step 5 of 5: State the conclusion of the hypothesis test.
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. The mean braking distance for SUVs equipped with tires made with compound 1 is 56 feet, with a population standard deviation of 13.0. The mean braking distance for SUVs equipped with tires made with compound 2 is 61 feet, with a population standard deviation of 5.3. Suppose that a sample of 50 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.
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