A national organization has been working with utilities throughout the nation to find sites for large wind machines that generate electricity. Wind speeds must average more than 10miles per hour (mph) for a site to be acceptable. Recently, the organization conducted wind speed tests at a particular site under construction for a wind machine.
To determine whether the site meets the organization's requirements, consider the test, H0:u = 10 vs Ha:u >10, where u is the true mean wind speed at the site and a=.10. suppose the onserved significance level(p-value) of the tesr is calculated to be p=0.2807. imterpret this result based on the p value of 0.2807.
a. since p value greatly exceeds a=.0.10 there is strong evidence to reject the null hypothesis
b. since the p value exceeds a=.10, there is insufficient evidence to reject the null hypothesis
c. the probability of rejecting the null hypothesis is 0.2807
d. we are 71.93% confident that u=10.

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 fail reject the null hypothesis.

And the best conclusion for this case wold be:

b. since the p value exceeds a=.10, there is insufficient evidence to reject the null hypothesis

Step-by-step explanation:

Data given and notation

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

[tex]s[/tex] represent the sample standard deviation

[tex]n[/tex] sample size

[tex]\mu_o =10[/tex] represent the value that we want to test

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

t 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 the mean is higher than 10, the system of hypothesis would be:

Null hypothesis:[tex]\mu \leq 10[/tex]

Alternative hypothesis:[tex]\mu > 10[/tex]

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

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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) and we got [tex] t_{calc}[/tex]

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n-1[/tex]

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

[tex]p_v =P(t_{(n-1)}>t_{calc})=0.2807[/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 fail reject the null hypothesis.

And the best conclusion for this case wold be:

b. since the p value exceeds a=.10, there is insufficient evidence to reject the null hypothesis