2. A tire manufacturer purchases a new machine that is designed to test the read
ID: 3333545 • Letter: 2
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2. A tire manufacturer purchases a new machine that is designed to test the readiness of tires. The machine is advertised as being able to complete one tire check in 5 minutes on average, although the amount of time it actually takes varies from tire to tire. The tire manufacturer is suspicious that the machine takes longer than advertised and is considering filing a complaint against the company from which it purchased the machine. The tire manufacturer runs a sample of 4 tires through the machine and observes the amount of time it takes to complete each test. Let represent the true amount of time it takes to complete one tire test on average and let c represent the average amount of time it takes to test the tires in the sample. (a) State the null and alternative hypotheses associated with testing the tire manufacturer's suspicion. (b) Describe, in the context of the problem, what it would mean for the tire manufacturer to make a type I error. Also describe what it would mean to make a type II error. (c) It is determined that if the advertised claim were true, (that is that it takes 5 minutes to complete a test on average) then the average amount of time it would take to test a sample of 4 tires would have the pdf and cdf shown below (units are minutes). (This is the distribution for the average amount of time it would take to test a set of 4 tires, if the advertised claim were true, not the distribution for the amount of time it would take to test an individual tire. 0.8 0.20 0.15 0.10 0.05 0.00 0.6 0.4 0.2 5 10 MeanTime 15 20 0.0 2.5 5.0 7.5 10.0 12.5 15.0 Before testing the sample of 4 tires, the company decides to file a complaint if the average tire test takes longer than 6 minutes. Approximately what is the probability of making a type I error? (d) Suppose that the advertised time is indeed wrong and that the amount of time it takes the machine to test a single tire is normally distributed with mean 7 and standard deviation 1.5. Using the criteria described in (c), what is the probability of a Type II error? (e) Suppose the company wants the probability of a type 1 error to be only 10%. Using the results from the sample of 4 tires, approximately what criteria should the manufacturer use to determine whether to reject the null hypotheses'? Using the criteria in (e) what is the probability of a type II error? Why is it larger than in (d)?Explanation / Answer
a)
H0 Null: Mean = 5
H1 Alternate: Mean > 5
Null hypothesis is mean being less than or equal to 5 minutes. Since null will not have inequality, we test it against the alternative which is mean time greater than 5 minutes.
b) Type1 error is false positive or incorrect rejection of null hypothesis. That is we prove that there is significant effect towards alternate and we reject the null, but actually the null is true.
In our case we reject that mean is equal to 5 mins and go with the alternate saying that mean is greater than 5 mins but in reality the mean is 5 minutes or less than that.
Type 2 error is false negative or failure to reject a false null hypothesis. That is we don’t find any significant effect and accept the null, but actually the null is false.
In our case we accept that mean is equal to 5 mins but in reality the mean is more than 5 minutes.
c) It is claimed the null is true, i.e. mean is 5 minutes indeed.
So when we check for the alternate hypothesis that mean is more than 5 minutes, it will turn out to be false i.e. 0 probability. However there is some chance of error to determine the alternate to be true. That is type 1 error or false positive. It depends on the significance level we assume. Most it is .05
Hence type 1 error probability is 5%
d) Claimed mean - 5
Observed mean - 6
Actual mean - 7
H0: mean = 5, H1 mean >=5
Therefore for the null hypothesis, the rejection region is on the right side with a small area whose value is 5%. Imagine a normal curve with mean = 5 and a small right side rejection region of area 5%. That value at which rejection area starts in the right hand side can be calculated as
zc = std normal (0.95) = 1.64
c = mean + zc * sig/ sqrt(n) = 5 + 1.64*1.5/2 = 6.23
the rejection region starts at value c = 6.23
Type 2 error is failure to reject null hypothesis when it is false i.e when it is less than 6.23 we fail to reject it
use this less than 6.23 area in the normal curve where mean is 7
type 2 error is the region to the left of 6.23
zc = 6.23-7/1.5/sqrt(4) = -1.0267
Prob of type 2 error is 15.23%
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