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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.

Respuesta :

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.

[tex]z=\dfrac{\overline{X}-\mu}{\frac{s}{\sqrt{n}}}[/tex]

where,

[tex]\overline{X}[/tex] is sample mean.

[tex]\mu[/tex] is population mean.

s is sample standard deviation.

n is the sample size.

[tex]z=\dfrac{783-750}{\frac{120}{\sqrt{73}}}[/tex]

[tex]z=2.35[/tex]

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).

[tex]P=P(z>2.35)[/tex]

[tex]P=1-P(z<2.35)[/tex]

[tex]P=1-0.9906[/tex]

[tex]P=0.0094[/tex]

Therefore the P-value is 0.0094.

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