In: Advanced Math
For the following, A = attractiveness and B=time; same levels. Conduct a two-factor fixed effects ANOVA (alpha=.05).
A₁B₁: 10, 8, 7, 3
A₁ B₂: 15, 12, 21, 13
A₂ B₁: 13, 9, 18, 12
A₂B₂: 20, 22, 24, 25
A₃ B₁: 24, 29, 27, 25
A₃ B₂: 10, 12, 21, 14
A₄ B₁: 30, 26, 29, 28
A₄ B₂: 22, 20, 25, 15
Answer For the Fixed Effects Model of the above problem
Factors are
1 . Attracttiveness A: 4 levels(a) - A1, A2, A3 and A4
2. Time B 2 levels (b) B1 and B2
There are 4 replications(n) .
Hence for Anova Effect and no of degrees of freed om will be
Effect Degrees of Freedom
A a-1= 4-1 = 3
B b-1= 2-1 =1
AB interaction (a-1)*(b-1)= 3*1 = 3
Error a*b*(n-1)= 4*2*(4-1)=24
Total 4*2*4-1 =31
ANOVA table can be derived using Minitab based on the data /or manually calculated (lengthy)
in mintab use following steps to get the ANOVA results
1. put data as per following format in worksheet.
2.choose Stat > ANOVA > Two-Way.
3. Select Response as Response ,Row Factor as "Attractiveness and Column Factor as "Time"
Check confidence level at 95%, Click OK . you will get ANOVA on session window output. The same is given below
Two-way ANOVA: Response versus Attractiveness, Time
Source DF SS MS F P
Attractiveness 3 738.59 246.198 21.35 0.000
Time 1 1.53 1.531 0.13 0.719
Interaction 3 732.84 244.281 21.18 0.000
Error 24 276.75 11.531
Total 31 1749.72
S = 3.396 R-Sq = 84.18% R-Sq(adj) = 79.57%
The above Table is the ANOVA table ,
As p value is lower for Attractiveness and Interaction it can be said that there is significant effect for attractiveness and intercation of attractivess and time interaction. and effect of time is not significant.
Datasheet ----------------------
Attractveness Time Response
A1 B1 10
A1 B1 8
A1 B1 7
A1 B1 3
A1 B2 15
A1 B2 12
A1 B2 21
A1 B2 13
A2 B1 13
A2 B1 9
A2 B1 18
A2 B1 12
A2 B2 20
A2 B2 22
A2 B2 24
A2 B2 25
A3 B1 24
A3 B1 29
A3 B1 27
A3 B1 25
A3 B2 10
A3 B2 12
A3 B2 21
A3 B2 14
A4 B1 30
A4 B1 26
A4 B1 29
A4 B1 28
A4 B2 22
A4 B2 20
A4 B2 25
A4 B2 15