Answer:
a. P value = 0.0094, b. 0.94% of the distribution
Step-by-step explanation:
From the given information, let X be the lifetimes of parts manufactured from a certain aluminum alloy
Total No. of parts n = 73, Mean number of kilocycles to failures (x’) = 783
Standard deviation s = 120
Let μ be the mean number of kilocycles to failure for parts
a. Null hypothesis, The mean number of kilocycles to failure is less than or equal to 750, H₀ : μ ≥ 750
Alternative hypothesis, The mean number of kilocycles to failure is greater than 750, H1 : μ > 750
For large samples, the sample standard deviation (s) is approximate to population standard deviation (σ).
Use the following formula to compute the test statistic
z = (X’ – μ₀)/(s/√n)
z= (783 - 750)/(120/sq73) = 2.35
Th test is one- tailed, therefore the p value is probability of observing a sample mean greater than to 783
P = P(X’>783)
= P(z > 2.35), (since, z = (X’ – μ₀)/(s/√n))
= 1 – 0.9906 = 0.0094
b. From the above output, it is observed that 0.94% of samples are greater than 750. Therefore, either the mean number of kilocycles to failure is greater than 750, or the sample is in the most extreme 0.94% of its distribution
