Question

In a study, 38 random students of a sample 1 were shown positive evaluations of an instructor and 33 random students of sample 2 were shown negative evaluations of the instructor. Then all subjects were shown the same lecture video given by the instructor. Sample 1 gave a mean rating of 6.6 with a standard deviation of 0.75, while sample 2 gave a mean rating of 5.9 with a standard deviation of 1. Perform a hypothesis test at the 0.06 level of significance to determine if students of sample 1 rated the professor significantly higher than the students of sample 2.

Answer 1

To Test :-

H0 :-  

H1 :-  

Test Statistic :-


t = 3.296

Test Criteria :-
Reject null hypothesis if


DF = 58


Result :- Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 3.296 ) = 0.0008
Reject null hypothesis if P value <    level of significance
P - value = 0.0008 < 0.06 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that students of sample 1 rated the professor significantly higher than the students of sample 2 at the 0.06 level of significance.