Question

Suppose you work at a large tire distribution center. The tires’ average tread life has been 50,000 miles with a standard deviation of 5,000 miles. At the end of the year, the company can reevaluate their supply contract. There are four supply options for the next contract: the current supplier or one of three competitors. The current supplier produces tires with an average tread life of 50,000 miles with a standard deviation of 5,000 miles. Competitor A claims to produce tires with an average tread life of 52,000 miles with a population standard deviation of 8,000 miles. Competitor B claims to produce tires with an average tread life of 50,000 miles with a population standard deviation of 3,000 miles. Competitor C claims to produce tires with an average tread life of 60,000 miles with a population standard deviation of 12,000 miles. Question 1:Based on Competitor A's claim, what is the z-score associated with the mean of the sample of their tires (54,028.400 miles)? Question 2:Competitor A's claim is less than the sample mean. Thus, testing their claim requires us to calculate P(Z>z). Assuming the competitor's claim is the true mean, what is the probability that we would observe a sample mean greater than the 54,028.400 miles (the mean of our sample)? Question 3:Based on Competitor B's claim, what is the z-score associated with the mean of the sample of their tires (50,407.686 miles)? Question 4:Competitor B's claim is less than the sample mean. Thus, testing their claim requires us to calculate P(Z>z). Assuming the competitor's claim is the true mean, what is the probability that we would observe a sample mean greater than the 50,407.686 miles (the mean of our sample)? Question 5:Based on Competitor C's claim, what is the z-score associated with the mean of the sample of their tires (63,478.371 miles)? Question 6:Competitor C's claim is less than the sample mean. Thus, testing their claim requires us to calculate P(Z>z). Assuming the competitor's claim is the true mean, what is the probability that we would observe a sample mean greater than the 63,478.371 miles (the mean of our sample)?

Answer 1

1. z =

z = (54028.400 - 52000) / 8000 = 0.2535

2. P(Z>z) = P(Z>0.2535) = 1 - P(Z<0.2535) = 1 - 0.0987 = 0.9013

3. z = = (50407.686 - 50000) / 3000 = 0.1359

4. P(Z>z) = P(Z>0.1359) = 1 - P(Z<0.1359) = 1 - 0.0557 = 0.9443

5. z =

=> z = (63478.371 - 60000) / 12000 = 0.2898

6. P(Z>z) = P(Z>0.2898) = 1 - P(Z<0.2898) = 1 - 0.1141 = 0.8859