Question

A data set includes data from student evaluations of courses. The summary statistics are n=97​, x=4.18​, s=0.54. Use a 0.01 significance level to test the claim that the population of student course evaluations has a mean equal to 4.25. 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.

also, P-value

Answer 1

Explanation:

The null and alternative hypotheses are:

Null hypothesis: The population mean of student course evaluations is equal to 4.25.

Alternative hypothesis: The population mean of student course evaluations is not equal to 4.25.

The significance level is 0.01, so the test will be a two-tailed test.

The test statistic is calculated as:

t = (x - μ) / (s / √n) = (4.18 - 4.25) / (0.54 / √97) = -2.63

The degrees of freedom for the t-distribution is n-1 = 96. Using a t-table or a calculator, the P-value for a two-tailed test with 96 degrees of freedom and a t-value of -2.63 is approximately 0.010.

Since the P-value (0.010) is less than the significance level (0.01), we reject the null hypothesis. We have sufficient evidence to conclude that the population mean of student course evaluations is not equal to 4.25.

In other words, the data provides strong evidence to suggest that the average student course evaluation rating is significantly lower than 4.25.