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 6.0 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 26 engines and the mean pressure was 6.3 pounds/square inch with a standard deviation of 1.0. A level of significance of 0.025 will be used. Assume the population distribution is approximately normal. Make the decision to reject or fail to reject the null hypothesis.

Answer 1

Answer:

$t=\frac{6.3-6.0}{\frac{1}{\sqrt{26}}}=1.530$  

$df = n-1 = 26-1=25$

Since is a right-sided test the p value would be:  

$p_v =P(t_{25}>1.530)=0.0693$  

If we compare the p value and the significance level given $\alpha=0.025$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is significantly higher than 6.0

Step-by-step explanation:

Data given and notation  

$\bar X=6.3$ represent the sample mean    

$s=1.0$ represent the sample standard deviation

$n=26$ sample size  

$\mu_o =6.0$ represent the value that we want to test  

$\alpha=0.025$ 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 conduct a hypothesis in order to check if the ture mena is higher than 6.0, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 6.0$  

Alternative hypothesis:$\mu > 6.0$  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)  

t-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:  

$t=\frac{6.3-6.0}{\frac{1}{\sqrt{26}}}=1.530$  

P-value  

The degrees of freedom are given by:

$df = n-1 = 26-1=25$

Since is a right-sided test the p value would be:  

$p_v =P(t_{25}>1.530)=0.0693$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.025$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is significantly higher than 6.0

Answer 2

Answer:

The null hypothesis failed to be rejected.

There is no enough evidence to support the claim that the valve performs above the specifications.

Step-by-step explanation:

The claim is that the valve performs above the specifications. That is that the valve produce a mean pressure higher than 6.0.

Then, the null and alternative hypothesis are:

$H_0: \mu=6.0\\\\H_a:\mu>6.0$

The significance level is 0.025.

We have a sample of size n=26, in which the sample mean is 6.3 and the standard deviation is 1.0.

As the standard deviation of the population is estimated from the sample, a t-test will be performed.

The degrees of freedom are:

$df=n-1=26-1=25$

The t-statistic is:

$t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{6.3-6.0}{1.0/\sqrt{26}}=\dfrac{0.3}{0.196}=1.53$

The p-value for this right tailed test and 25 degrees of freedom is:

$P-value=P(t>1.53)=0.0693$

The P-value is greater than the significance level, so the effect is not significant. The null hypothesis failed to be rejected.

There is no enough evidence to support the claim that the valve performs above the specifications.