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.

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.

Step-by-step explanation:

Information given

[tex]\bar X=7.6[/tex] represent the sample mean

[tex]\sigma=0.9[/tex] represent the population deviation

[tex]n=130[/tex] sample size

[tex]\mu_o =7.4[/tex] represent the value that we want to check

[tex]\alpha=0.02[/tex] represent the significance level for the hypothesis test.

z would represent the statistic

[tex]p_v[/tex] 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:[tex]\mu \geq 7.6[/tex]

Alternative hypothesis:[tex]\mu < 7.6[/tex]

The statistic would be:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

Replacing the values we got:

[tex]z=\frac{7.4-7.6}{\frac{0.9}{\sqrt{130}}}=-2.534[/tex]

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:

[tex]z_{\alpha}=-2.054[/tex]

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:

[tex]p_v =P(Z<-2.534)=0.0056[/tex]

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.