Question

A data set includes data from student evaluations of courses. The summary statistics are n=95​, x=3.86​, s=0.67. 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. What are the null and alternative​ hypotheses?

Answer 1

Given:

n = 95

= 3.86

s = 0.67

Null hypothesis:

H0: The population mean,

Alternative hypothesis:

H1: The population mean,

Level of Significance:

Test-statistic:

P-value:

p-value is calculated using Excel with the function "TDIST". The inputs for the function are absolute value of t-statistic, degrees of freedom and number of tails. Absolute value of t-statistic is calculated as 2.0378, degrees of freedom = n - 1 = 95 - 1 = 94, number of tails = 2

Therefore, the final function is given as =TDIST(2.0378,94,2). p-value is calculated to be 0.0444

Conclusion:

p-value is 0.0444 which is less than the level of significance of 0.05. Therefore, the null hypothesis is rejected and conclude that the population mean of student course evaluation is not equal to 4