Question

Let's approach a problem from two perspectives: An engineer believes a particular component has a fatigue life of 10000. In other words, that component should be expected to fail after 10000 uses. He takes a random sample of 41 components and obtains a sample mean of 9900 with a standard deviation of 2250. Is there any evidence that he is overestimating the fatigue life of the components?

(a) What are the null and alternative hypotheses for his test? H0: Ha:

(b) What is the value of the test statistic? (3 decimal places)

(c) What is the p-value for his test? (3 decimal places) Now let's examine the problem from another perspective: A particular component has a fatigue life of 10000. An engineer took a random sample of 41 components and tested their fatigue. He obtained a sample mean of 9900 cycles with a standard deviation of 2250. Notice that the standard deviation comes from the sample, so a t distribution would be appropriate to use here.

(d) What is the center of the sampling distribution?

(e) How many standard errors from the center of the sampling distribution did the sample mean fall? Use a negative sign if it fell below the center. (3 decimal places)

(f) What was the probability of obtaining a sample mean of 9900 or less? (3 decimal places) What could we say about the answers to (c) and (f)? They are unrelated. They should be different. They should be the same.

Answer 1

a) As we are testing here whether he is overestimating the fatigue life of the components, therefore the null and the alternative hypothesis here are given as:

b) The test statistic here is computed as:

Therefore -0.285 is the required test statistic value here.

c) As this is a lower tailed test, the p-value here is computed from the t distribution tables for n - 1 = 40 degrees of freedom as:

p = P( t₄₀ < -0.285) = 0.3886

Therefore 0.389 is the required p-value here.

d) The center of the sampling distribution remains equal to the original mean which is 10,000 in this case. Therefore 10,000 is the required central point here.

e) As we still have the sample sample size, standard deviation and sample mean value, thereofore still the value of the samlpe mean is 0.285 standard errors below mean that is -0.285 is the required t score here.

f) Again we are obtaining the same probability here which is:
p = P( t₄₀ < -0.285) = 0.3886

Therefore 0.389 is the required probability here.

Answers to part c) and f) are same because we compute everything in a hypothesis test assuming the null hypothesis to be true.