In: Statistics and Probability
The following partial ANOVA table was based on an experiment performed as a two-factor design with 2 levels of factor A, 3 levels of factor B, and 5 observations on each treatment. It was determined that treatment groups were independently selected from a normal distribution with the same variance for each treatment.
Source of variation | df | Sum of Squares | Mean Square | F |
Factor A | 27 | |||
Factor B | 2 | 32 | ||
Interaction | ||||
Error | 108 | |||
Total | 29 | 227 |
a) Fill in the missing entries in the ANOVA table.
b) Test the hypothesis of no interaction between the factors at a 5% level of significance. Your answer should show assumptions, hypotheses, and conclusions.
c) Indicate whether testing for the main effects is appropriate, and if so show, the results of the tests; if not, indicate why testing for the main effects is not justified. Your answer should show assumptions, hypotheses, and conclusions.
The complete ANOVA Table is Give below
Source of variation | df | Sum of Squares | Mean Square | F | P.Value |
Factor A | 1 | 27 | 27 | 6.00 | 0.022 |
Factor B | 2 | 32 | 16 | 3.56 | 0.044 |
Interaction | 2 | 60 | 30 | 6.67 | 0.005 |
Error | 24 | 108 | 4.5 | ||
Total | 29 | 227 |
DF for Factor A =( Levels of A - 1) = (2-1) = 1
DF for Interaction =( DF for Factor A * DF for Factor B) = (2*1) = 2
DF for Error = (DF for Total - (DF for Factor A + DF for Factor B+DF for Interaction))
b) We will set up the null hypothesis that
There is no interaction between the factors.
There is interaction between the factors.
Since calculated P . Value for interaction = 0.005 which is less then 0.05 therefore we accept our null hypothesis and concludes that there is interaction between the factors.
c) We will set up the null hypothesis that
There is no effect of factors.
There is significant effect between the factors.
Since calculated P . Value for factor A = 0.022 which is less then 0.05 therefore we accept our null hypothesis and concludes that there is significant effect between the factors.
Also we have calculated P . Value for factor A = 0.0444 which is less then 0.05 therefore we accept our null hypothesis and concludes that there is significant effect between the factors.