Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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. â–¼ Fail to reject OR Reject H0. There is â–¼ sufficient OR not sufficient evidence to conclude that the mean of the population of student course evaluations is equal to 3.50 â–¼ 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

We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.