Hi I am performing an analysis of variance with two factors, each with two levels. I have differing numbers of observations in each of the four combinations, but all four combinations *are* present (2 of the factor combinations have 3 observations, 1 has 4 and 1 has 5)
I have used both anova(aov(...)) and anova(lm(...)) in R and it gave the same result - as expected. I then plugged this into minitab, performed what minitab called a General Linear Model (I have to use this in minitab as I have an unbalanced data set) and got a different result. After a little mining this is because minitab, by default, uses the type III adjusted SS. Sure enough, if I changed minitab to use the type I sequential SS, I get exactly the same results as aov() and lm() in R. So which should I use? Type I adjusted SS or Type III sequential SS? Minitab help tells me that I would "usually" want to use type III adjusted SS, as type I sequential "sums of squares can differ when your design is unbalanced" - which mine is. The R functions I am using are clearly using the type I sequential SS. Any help would be very much appreciated! Thanks Mick ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
