In: Statistics and Probability
A data set includes data from student evaluations of courses. The summary statistics are n=85, x =3.41 , s=0.61. 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 alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses?
Solution :
This is the two tailed test .
The null and alternative hypothesis is ,
H0 : = 3.50
Ha : 3.50
= 3.41
= 3.50
s = 0.61
n = 85
Test statistic = z
= ( - ) / s / n
= (3.41 - 3.50) / 0.61 / 85
= -1.36
Test statistic = -1.36
P(z < -1.36) = 0.0869
P-value = 2 * 0.0869 = 0.1738
= 0.05
P-value >
Fail to reject the null hypothesis .
There is not enough evidence to suggest that the population of student course evaluations has a mean equal to 3.50 .