Question

A data set includes data from student evaluations of courses. The summary statistics are nequals=9393​ x overbarxequals=4.084.08​, sequals=0.540.54. Use a 0.010.01 significance level to test the claim that the population of student course evaluations has a mean equal to 4.254.25. 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.

b)Determine the test statistic.

c)Determine the​ P-value.

Answer 1

Given that, sample size (n) = 93, sample mean = 4.08

and sample standard deviation (s) = 0.54

a) The null and alternative hypotheses are,

H0 : μ = 4.25

Ha : μ ≠ 4.25

b) Test statistic is,

=> Test statistic = -3.036

c) Degrees of freedom = 93 - 1 = 92

Using Excel we get,

p-value = TDIST (3.036, 92, 2) = 0.0031

=> p-value = 0.0031

Note : if we used standard normal z-test, then p-value = 0.0024