Question

A data set includes data from student evaluations of courses. The summary statistics are n=99​, x=3.92​, s=0.58. Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 4.00. 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

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

H0: µ = 4 versus Ha: µ ≠ 4

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 4

Xbar = 3.92

S = 0.58

n = 99

df = n – 1 = 98

α = 0.05

Critical value = - 1.9845 and 1.9845

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (3.92 – 4)/[0.58/sqrt(99)]

t = -1.3724

P-value = 0.1731

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the population of student course evaluations has a mean equal to 4.00.