Question

An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 7.4 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 130 engines and the mean pressure was 7.6 pounds/square inch. Assume the standard deviation is known to be 0.9. A level of significance of 0.02 will be used. Make a decision to reject or fail to reject the null hypothesis. Make a decision.

Answer 1

`z=(7.4-7.6)/((0.9)/(√(130)))=-2.534`

`z_(\alpha)=-2.054`

If the calculated value is less than the critical value we reject the null hypothesis.

P value

The p value for this test would be:

`p_v =P(Z<-2.534)=0.0056`

Since the p value is lower than the significance level given we have enough evidence to reject the null hypothesis at the 25 of significance level given.

Explanation:

Information given

`\bar X=7.6` represent the sample mean

`\sigma=0.9` represent the population deviation

`n=130` sample size

`\mu_o =7.4` represent the value that we want to check

`\alpha=0.02` represent the significance level for the hypothesis test.

z would represent the statistic

`p_v` represent the p value for the test

System of hypothesis

We want to check if the mean pressure is less then 7.6 pounds/square inch, the system of hypothesis are:

Null hypothesis:
`\mu \geq 7.6`

Alternative hypothesis:
`\mu < 7.6`

The statistic would be:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

Replacing the values we got:

`z=(7.4-7.6)/((0.9)/(√(130)))=-2.534`

Critical value

we need to find a critical value who accumulates 0.02 of the area in the left of the normal standard distribution and we got:

`z_(\alpha)=-2.054`

If the calculated value is less than the critical value we reject the null hypothesis.

P value

The p value for this test would be:

`p_v =P(Z<-2.534)=0.0056`

Since the p value is lower than the significance level given we have enough evidence to reject the null hypothesis at the 25 of significance level given.