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

Answer:

$z=\frac{7.4-7.6}{\frac{0.9}{\sqrt{130}}}=-2.534$    

$z_{\alpha}=-2.054$

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:  

$p_v =P(Z$  

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

$\bar X=7.6$ 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:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

Replacing the values we got:

$z=\frac{7.4-7.6}{\frac{0.9}{\sqrt{130}}}=-2.534$    

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:  

$p_v =P(Z$  

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.