In: Statistics and Probability
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 74 feet, with a population standard deviation of 13.4. The mean braking distance for SUVs equipped with tires made with compound 2 is 77 feet, with a population standard deviation of 14.3. Suppose that a sample of 41 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.
Solution:
Given:
Compound 1:
n1 = 41
Compound 2:
n2 = 41
Claim: the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used.
Step 1) State H0 and H1:
Vs
Left tailed test.
Step 2) Test statistic:
Step 3) z critical value:
Level of significance = 0.05
Look in z table for Area = 0.0500 or its closest area and find z value
Area 0.0500 is in between 0.0495 and 0.0505 and both the area are at same distance from 0.0500
Thus we look for both area and find both z values
Thus Area 0.0495 corresponds to -1.65 and 0.0505 corresponds to -1.64
Thus average of both z values is : ( -1.64+ - 1.65) / 2 = -1.645
Thus z critical value = -1.645
Step 4) Decision Rule:
Reject null hypothesis ,if z test statistic value < z
critical value = -1.645, otherwise we fail to reject H0.
Since z test statistic value = > z critical value = -1.645, we fail to reject H0.
Step 5) Conclusion:
At 0.05 level of significance, we do not have 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.