Question

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 784, 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. Find P-Value.

Answer 1

Answer:

0.0094.

Step-by-step explanation:

For the given experiment, let X be the lifetimes of parts manufactured from a certain aluminum alloy.

sample size =73

Sample mean = 784

Sample standard deviation = 120

Let µ represent the mean number of kilocycles to failure for parts of this type.

Null hypothesis, H0 : µ ≤ 750

Alternative hypothesis, H1 : µ > 750.

$z=\dfrac{\overline{X}-\mu}{\frac{s}{\sqrt{n}}}$

where,

$\overline{X}$ is sample mean.

$\mu$ is population mean.

s is sample standard deviation.

n is the sample size.

$z=\dfrac{783-750}{\frac{120}{\sqrt{73}}}$

$z=2.35$

It is a right tailed test because alternative hypothesis is µ > 750. So, P-value is the probability of observing a sample mean greater that 783 or the probability of P(z>2.35).

$P=P(z>2.35)$

$P=1-P(z$

$P=1-0.9906$

$P=0.0094$

Therefore the P-value is 0.0094.