An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 180180 engines and the mean pressure was 4.64.6 lbs/square inch. Assume the standard deviation is known to be 0.90.9. If the valve was designed to produce a mean pressure of 4.84.8 lbs/square inch, is there sufficient evidence at the 0.10.1 level that the valve performs below the specifications? State the null and alternative hypotheses for the above scenario.

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.

Step-by-step explanation:

Data given and notation

[tex]\bar X=4.6[/tex] represent the sample mean

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

[tex]n=180[/tex] sample size

[tex]\mu_o =4.8[/tex] represent the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if thetrue mean is below the specifications, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 4.8[/tex]

Alternative hypothesis:[tex]\mu < 4.8[/tex]

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

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

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98[/tex]

P-value

Since is a one sided test the p value would be:

[tex]p_v =P(Z<-2.98)=0.0014[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.