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 alternativehypotheses, 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.

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