Question: 30 people were randomly assigned to one of six experimental conditions. At the end of the lecture, a measure of comprehension was obtained. The data is shown here. At α = 0.05, find the test value FA × B and check the interaction of the variables using a two-way ANOVA. [ A : Presentation and B : Lecture ]
A ) 1.95, the type of lecture and the method of presentation does not affect the scores of students B ) 8.74, the type of lecture and the method of presentation affect the scores of students C ) 18.6, the type of lecture and the method of presentation affect the scores of students D ) 3.4, the type of lecture and the method of presentation does not affect the scores of students
Steps to derive
1 H0: There is no interaction effect between the type of lecture and method of presentation used on the student's score.
H1: There is an interaction effect between the type of lecture and method of presentation used on the student's score. [Hypotheses for the variables interaction.]
2a = 2, b = 3, and n = 5. The degrees of freedom(d.f.) are
Factor A: d.f.N = a - 1 = 2 - 1 = 1
Factor B: d.f.N = b - 1 = 3 - 1 = 2
Interaction (A × B): d.f.N = (a - 1)(b - 1) = (2 - 1)(3 - 1) = 2
Within(error): d.f.D = ab(n - 1) = 2·3 (5 - 1) = 24
[a, b are levels of factors A (Presentation), B (Lecture) respectively and n is the number of data values in each group.]
3 The critical value FA corresponding to α = 0.05, d.f.N = 1, and d.f.D = 24 is 4.26
4 The critical value FB corresponding to α = 0.05, d.f.N = 2, and d.f.D = 24 is 3.4
5 The critical value FA × B corresponding to α = 0.05, d.f.N = 2, and d.f.D = 24 is 3.4
6 SSA = 90.13, SSB = 808.5, SSA × B = 1719.5 and SSW = 1110.4
7 The mean squares are as follows:
MSA = SSAa-1 = 90.13
MSB = SSBb-1 = 404.4
MSA × B = SSA×B(a-1)(b-1) = 859.7
MSW = SSWab(n-1) = 46.27 [Substitute and simplify.]
8 MSA = 90.13, MSB = 404.4, MSA × B = 859.7 and MSW = 46.27
9 The F values are computed as follows:
FA = MSAMSW = 1.95
FB = MSBMSW = 8.74
FA × B = MSA×BMSW = 18.6 [Substitute and simplify.]
10 Since FA × B = 18.6 is greater than the critical value 3.4, the null hypotheses concerning the interaction effect should be rejected.
11 It can be concluded that the type of lecture and the method of presentation affect the scores of students.
