Question

1. In an experiment to measure the lifetimes of parts manufactured from a certain aluminum alloy, 73 parts were loaded cyclically until failure. The mean number of kilocycles to failure was 783, and the standard deviation was 120. Let μ represent the mean number of kilocycles to failure for parts of this type. A test is made of H0 :μ ≤ 750 versus H1 :μ > 750. a. Find the P-value. b. Either the mean number of kilocycles to failure is greater than 750, or the sample is in the

Answer 1

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