Question

A data set includes data from student evaluations of courses.
The summary statistics are n=80​, x overbar =4.39​, s =2.29.

Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 4.50.

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.

Answer 1

Answer:

We fail to reject the null hypothesis at the significance level of 0.05.

Step-by-step explanation:

We have a larga sample size of n = 80, $\bar{x} = 4.39$ and s = 2.29. We want to test

$H_{0}: \mu = 4.50$ vs $H_{1}: \mu \neq 4.50$ (two-tailed alternative)  

Because we have a large sample, our test statistic is

$Z = \frac{\bar{X}-4.50}{s/\sqrt{n}}$ which is normal standard approximately. We have the observed value

$z_{0} = \frac{4.39-4.50}{2.29/\sqrt{80}} = -0.4296$.

The p-value is given by 2P(Z < -0.4296) = (2)(0.3337) = 0.6674 (because of the simmetry of the normal density)

With the significance level $\alpha = 0.05$, we fail to reject the null hypothesis because the p-value is greater than 0.05.

Answer 2

• The null hypothesis is: $H_0: \mu = 4.5$
• The alternative hypothesis is: $H_1: \mu \neq 4.50$
• The test statistic is: $t = -0.43$
• The p-value of the test is of $0.6684$.
• 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.

We are going to test if the mean is equals to 4.50, thus, the null hypothesis is:

$H_0: \mu = 4.5$

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

$H_1: \mu \neq 4.50$

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

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

For this problem:

$t = \frac{4.39 - 4.5}{\frac{2.29}{\sqrt{80}}}$

$t = -0.43$

We are testing if the mean is different from a value, thus, the p-value of test is found using a two-tailed test, with $t = -0.43$ and 80 - 1 = 79 df.

Using a t-distribution calculator, the p-value is of 0.6684.

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.

A similar problem is given at https://brainly.com/question/24989605