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 5.1 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 210 engines and the mean pressure was 5.2 pounds/square inch. Assume the standard deviation is known to be 0.6. A level of significance of 0.1 will be used. Determine the decision rule.

Answer 1

Answer:

$z=\frac{5.2-5.1}{\frac{0.6}{\sqrt{210}}}=2.415$  

The critical region would be $(1.28; \infty)$  

$p_v =P(Z>2.415)=0.00786$  

If we compare the p value and a significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 5.1 pounds/square inch at 10% of significance.  

Step-by-step explanation:

Data given and notation  

$\bar X=5.2$ represent the average score for the sample  

$\sigma=0.6$ represent the population standard deviation  

$n=210$ sample size  

$\mu_o =5.1$ represent the value that we want to test  

$\alpha=0.1$ represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to apply a one upper tailed test.  

What are H0 and Ha for this study?  

Null hypothesis: $\mu \leq 5.1$  

Alternative hypothesis :$\mu > 5.1$  

Compute the test statistic  

The statistic for this case is given by:  

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

$z=\frac{5.2-5.1}{\frac{0.6}{\sqrt{210}}}=2.415$  

Critical value

On this case since we have a right tailed test we need to look into the normal standard distribution and find a value that accumulates 0.1 of the area on the right of the distribution or 0.9 of the area on the left and for this case we see that $z_{critical}=1.28$

The critical region would be $(1.28; \infty)$  

Give the appropriate conclusion for the test  

Since is a one side upper tailed test the p value would be:  

$p_v =P(Z>2.415)=0.00786$  

Conclusion  

If we compare the p value and a significance level given $\alpha=0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 5.1 pounds/square inch at 10% of significance.