similar to: Anova - adjusted or sequential sums of squares?

Hi I have data from 12 subjects. The measurement is log(expression) of a particular gene and can be assumed to be normally distributed. The 12 subjects are divided into the following groups: Infected, Vaccinated, Lesions - 3 measurements Infected, Vaccintaed, No Lesions - 2 measurements Infected, Not Vaccinated, Lesions - 4 measurements Uninfected, Not Vaccinated, No Lesions - 3 measurements

Hello everybody, I have some questions on ANOVA in general and on ANOVA in R particularly. I am not Statistician, therefore I would be very appreciated if you answer it in a simple way. 1. First of all, more general question. Standard anova() function for lm() or aov() models in R implements Type I sum of squares (sequential), which is not well suited for unbalanced ANOVA. Therefore it is better

Hello Everyone, *Question: *How do you calculate the sum of squares for a two-way _unbalanced_ ANOVA? *What I have done:* I have found many useful tutorials online for running a balanced two-way ANOVA but I haven't had much luck for running a unbalanced two-way ANOVA. From what I have read, the trouble with running an unbalanced two-way ANOVA, is that things get tricky when calculating

Hi all, I've spent quite a lot of time searching through the help lists and reading about how best to run perform a 2-way ANOVA with unbalanced data. I realize this has been covered a great deal so I was trying to avoid adding yet another entry to the long list considering the use of different SS, etc. Unfortunately, I have come to the point where I feel I have to wade in and see if someone

Hi, I ran a small data set from a factorial experiment through R, Minitab and JMP... the result from R is significantly different from what Minitab or JMP give... The data set is at the following link: http://www.personal.psu.edu/nug107/Uploads/2x3_16repsANOVA.txt The first 5 columns are the factors and the next three are responses. In particular, for the response beta11MSE, two of the

I have a three-way unbalanced ANOVA that I need to calculate (fixed effects plus interactions, no random effects). But word has it that aov() is good only for balanced designs. I have seen a number of different recommendations for working with unbalanced designs, but they seem to differ widely (car, nlme, lme4, etc.). So I would like to know what is the best or most usual way to go about working

Dear R list, I have been trying to do a linear model, extracting the effect of a covariate.... and the results do not match, when I do it with other programs (e.g. minitab).... so it is obvious that I was doing something wrong. Whan I do it with minitab, I have this results: (sector is a factor and depth is the covariate): Source DF Seq SS Adj SS Adj MS F P

Hi When I use plot(hclust(dist..)...)...) etc to create a dendrogram of a hierarchial cluster analysis, I end up with a vertical tree. What do I need to do to get a horizontal tree? Also, my users are used to seeing trees who's leaves all "end" at the same place (eg. Like in minitab). Is this possible in R? Thanks Mick Michael Watson Head of Informatics Institute for Animal

I am using R to make two-way ANOVA on a number of variables using g <- aov(var ~ fact1*fact2) where var is a matrix containing the variables. However the outcome seem to be dependent on the order of fact1 and fact2 (i.e. fact2*fact1) gives a slightly (factor of 1.5) different result. Any ideas why this is? Thanks for any help Anders

Hi, I have an unbalanced dataset on which I would like to perform a one-way anova test using R (aov). According to Wannacott and Wannacott (1990) p. 333, one-way anova with unbalanced data is possible with a few modifications in the anova-calculations. The modified anova calculations should take into account different sample sizes and a modified definition of the average. I was wondering if the

Hello, This one is very perplexing. I have teacher observation data, with factors teacher ID, observer ID, component, grade and subject. When I do this, aov(data=ratings.prin.22, rating ~ obsid.f + tid.f + subject.f + grade.f + comp.f) I get this: Terms: obsid.f tid.f grade.f comp.f Residuals Sum of Squares 306.23399 221.38173 1.70000 14.52831 279.05780 Deg. of

Hello, I'm actually trying to fit a gee model with 2 nesting levels since I expect a correlation between all members of a litter at a first level and between all individuals sharing a mother at a second superior level with an exchangeable matrix. I order my dataframe by both mother and litter I try several syntaxes: id= mother*litter which give the same correlation matrix as id=

Hi, I am trying a two way anova with unequal sample sizes but results are not as expected: I take the example from Applied Linear Statistical Models (Neter et al. pp889-897, 1996) growth rate gender bone development 1.4 1 1 2.4 1 1 2.2 1 1 2.4 1 2 2.1 2 1 1.7 2 1 2.5 2 2 1.8 2 2 2 2 2 0.7 3 1 1.1 3 1 0.5 3 2 0.9 3 2 1.3 3 2 expected results are

Dear R Core Team I've a proposal to improve drop1(). The function should change the contrast from the default ("treatment") to "sum". If you fit a model with an interaction (which ist not signifikant) and you display the main effect with drop1( , scope = .~., test = "F") If you remove the interaction, then everything's okay. There is no way to fit a

Dear R-Community! For example I have a study with 4 treatment groups (10 subjects per group) and 4 visits. Additionally, the gender is taken into account. I think - and hope this is a goog idea (!) - this data can be analysed using lme as below. In a balanced design everything is fine, but in an unbalanced design there are differences depending on fitting y~visit*treat*gender or

Hi to all, So I'm also a new user. I downloaded the program last week and I think it's great. Thanks to those who have developed R. I have a special interest in GLM as a tool to describe fisheries and its variables and I'm just begging to study it. As I could understand there's two types of GLM sun of squares: in "type I" the factors are added in sequence and

Hi all I'm trying to complete a textbook example originally designed for SPSS in R, and I therefore need to find out how to compute an unbalanced ANOVA in R. I did a search on the mailinglist archives an found a post by Prof. Ripley saying one should use the lme function for (among other things) unbalanced ANOVAs, but I have not been able to use this object. My code gives me an error.. Why

Hello R-users, I am trying to obtain Type III SS for an ANOVA with subsampling. My design is slightly unbalanced with either 3 or 4 subsamples per replicate. The basic aov model would be: fit <- aov(y~x+Error(subsample)) But this gives Type I SS and not Type III. But, using the drop() option: drop1(fit, test="F") I get an error message: "Error in

I am looking for help to analyze an unbalanced AB/BA cross-over design by requesting the type III SS ! # Example 3.1 from S. Senn (1993). Cross-over Trials in Clinical Research outcome<-c(310,310,370,410,250,380,330,270,260,300,390,210,350,365,370,310,380,290,260,90,385,400,410,320,340,220) subject<-as.factor(c(1,4,6,7,10,11,14,1,4,6,7,10,11,14,2,3,5,9,12,13,2,3,5,9,12,13))

Hi all, I have been trying to calculate Type III SS in R for an unbalanced two-way anova. However, the Type III SS are lower for the first factor compared to type I but higher for the second factor (see below). I have the impression that Type III are always lower than Type I - is that right? And a clarification about how to fit Type III SS. Fitting model<-aov(y~a*b) in the base package and