Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

A data set includes data from student evaluations of courses. The summary statistics are n=85​, x=3.42​, s=0.56. Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 3.50. Assume that a simple random sample has been selected. Identify the null and alternative​hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.
What are the null and alternative​ hypotheses? A.) H0​: μ=3.50 H1​: μ>3.50 B.) H0​: μ≠3.50 H1​: μ=3.50 C.) H0​: μ=3.50 H1​: μ≠3.50 D.) H0​: μ=3.50 H1​: μ<3.50
Determine the test statistic= ____ ​(Round to two decimal places as​ needed.)
Determine the​ P-value = ____ ​(Round to three decimal places as​ needed.)
State the final conclusion that addresses the original claim. ▼ Fail to reject OR Reject H0. There is ▼ sufficient OR not sufficient evidence to conclude that the mean of the population of student course evaluations is equal to 3.50 ▼ is OR not is correct.

Sagot :

- The null hypothesis Is: [tex]$H_0: \mu=4.5$[/tex]

- The alternative hypothesis Is: [tex]$H_1: \mu \neq 4.50$[/tex]

- The test statistic is: t=-0.43

- The p-value of the test is of 0.6684.

What is p-value?

The p-value Is of 0.6684>0.05, which means that we can conclude that the population of student course evaluations has a mean equal to [tex]\mathbf{4 . 5 0}$.[/tex]

We are going to test If the mean Is equals to 4.50, thus, the null hypothesis Is:

[tex]H_0: \mu=4.5[/tex]

At the alternative hypothesis, we test If the mean Is different to 4.50, that is:

[tex]H_1: \mu \neq 4.50[/tex]

Since we have the standard devlation for the sample, the t-distribution is used. The value of the test statistic Is:

[tex]t=\frac{\pi-\mu}{\frac{\sqrt{n}}{\sqrt{n}}}$[/tex]

For this problem:

[tex]$\begin{aligned}& t=\frac{4.30-1.5}{\frac{2.21}{\sqrt[3]{80}}} \\& t=-0.43\end{aligned}$[/tex]

We are testing If the mean is dlfferent from a value, thus, the p-value of test is found using a two-talled test, wlth t=-0.43 and [tex]$80-1=79 \mathrm{df}$[/tex].

Using a t-distribution calculator, the [tex]$\mathrm{p}$[/tex]-value is of [tex]$\mathbf{0 . 6 6 8 4}$[/tex].

The p-value is of 0.6684 > 0.05, which means that we can conclude that the population of student course evaluations has a mean equal to 4.50.

To learn more about hypothesis  visit:https://brainly.com/question/29519577

#SPJ4

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.