similar to: a kinder view of Type III SS

I am trying to reproduce an analysis done in SAS from which Type III Sums of squares are used. How can you get Type III sums of squares in an aov similar to what SAS provides? With best wishes and kind regards I am Sincerely, Corey A. Moffet Instructor Department of Range, Wildlife, and Fisheries Management Mail Stop 2125 Texas Tech University Lubbock, Texas 79409-2125 (806) 742-2843 Voice

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

Hello, I have come across a result that I cannot explain, and am hoping that someone else can provide an answer. A student fitted a mixed model using the lme function: out<- lme(fixed=Y~A+B+A:B, random=~1|Site). Y is a continuous variable while A and B are factors. The data set is balanced with the same number of observations in each combination of A and B. There are two hierarchical

Hi, When doing a two-ways anova in R and comparing my same results with an SPSS output, I noticed that R calculated type I Sum of Squares. Is it possible to use Type III Sum of Squares? Thanks, S?bastien Plante Institut des Sciences de la Mer de Rimouski (ISMER) -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Hello, as far as I see, R reports type I sums of squares. I'd like to get R to print out type III sums of squares. e.g. I have the following model: vardep~factor1*factor2 to get the type III sum of squares for factor1 I've tried anova(lm(vardep~factor2+factor1:factor2),lm(vardep~factor1*factor2)) but that didn't yield the desired result. Could anyone give me a hint how to proceed?

Hi, For the package "pls", I need to understand the algorithm for "simpls.fit" for Partial Least Squares. I'm not sure if simpls.fit tries to find the weight vectors (loadings) to maximize which of the two: Cov(Xw, y) or maximize Cov^2(Xw,y)? Are these objective functions equivalent? (in some texts, they use the first and in other texts, they use the second obj.

I should hope that there is trouble, since "type III" is an undefined concept for a Cox model. Since SAS Inc fostered the cult of type III they have recently added it as an option for phreg, but I am not able to find any hints in the phreg documentation of what exactly they are doing when you invoke it. If you can unearth this information, then I will be happy to tell you whether

To avoid angry replies, let me first say that I know that the use of Type III sums of squares is controversial, and that some statisticians recommend instead that significance be judged using the non-marginal terms in the ANOVA. However, given that type III SS is also demanded by some… is there a function (equivalent to drop1 for lm) to obtain type III sums of squares for mixed models using the

Hi everyone, Can someone explain me how to calculate SAS type III sums of squares in R? Not that I would like to use them, I know they are problematic. I would like to know how to calculate them in order to demonstrate that strange things happen when you use them (for a course for example). I know you can use drop1(lm(), test="F") but for an lm(y~A+B+A:B), type III SSQs are only

Peter Dalgaard writes (in response to a question about 2-way ANOVA with imbalance): > ... There are various > boneheaded ways in which people try to use to assign some kind of > SumSq to main effects in the presence of interaction, and they are all > wrong - although maybe not very wrong if the unbalance is slight. People keep saying this

Hello, I believe the aov() function in R uses a "Type-I sum-of-squares" by default as against "Type-III". This is relevant for me because I am trying to understand ANOVA in R using my knowledge of ANOVA in SPSS. I can only reproduce the results of an ANOVA done using R through SPSS if I specify that SPSS uses a Type-I sum-of-squares. (And yes, I know that when the sample

Hi Duncan I tried Ecdf and/or wtd.quantile from Hmisc and it is working (probably). Ecdf(x, q=.5) Ecdf(x, weights=xw,col=2, add=T, q=.5) wtd.quantile(x) 0% 25% 50% 75% 100% 10 10 10 100 300 wtd.quantile(x, weights=xw, type="i/n") 0% 25% 50% 75% 100% 10.0000 138.8667 192.5778 246.2889 300.0000 But could you please be more specific in this? >

Folks, Normally, in a data frame, one observation counts as one observation of the distribution. Thus one can easily produce a CDF and (in Splus atleast) use cdf.compare to compare the CDF (BTW: what is the R equivalent of the SPlus cdf.compare() function, if any?) However, if each point should not count equally, how can I weight the points before comparing the distributions? I was thinking of

I am running a two-way anova with Type III sums of squares and would like to be able to understand what the different SS mean when I use different contrasts, e.g. treatment contrasts vs helmert contrasts. I have read John Fox's "An R and S-Plus Companion to Applied Regression" approach -p. 140- suggesting that treatment contrasts do not usually result in meaningful results with Type

Dear R People: I'm having trouble with something that should be very simple. I'm setting up a matrix outside of a loop and writing items into it during the loop. Here is the output: > glob3b("sites.info") dim 27 3 [1] "/raid1/osg-app" Error in xy[i, ] : incorrect number of dimensions Here is the function: > glob3b function(xx) { x.df <-

Hi Petr, I think that Duncan suggests something like this: x<- c(rep(10,20), rep(300,5), rep(100, 10)) tx <- table(x) prop.x <- tx / sum(tx) vx <- as.integer(names(tx)) prop.wx <- tx * vx / sum(tx * vx) plot(ecdf(x)) plot(vx, cumsum(prop.x), ylim = 0:1) plot(vx, cumsum(prop.wx), ylim = 0:1) Best regards, Thierry ir. Thierry Onkelinx Statisticus / Statistician Vlaamse

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

HI All: Can anyone give me the formulae/steps for calculating the type III sum of squares for an unbalanced 2-way ANOVA design? Eg. we are looking at 8 treatments x 4 doses, with unequal numbers of replications within the groups. I really need the stepwise calculation, as I would try to put it in my own code (possibly in Visual Basic) to automate the task. Thanks very much. Karth.

I've recently run into the problem of using aov with nested factors, and wanting to get the type II and III sums of squares. Normally Anova from the car package would do fine, but it doesn't like having an Error included, so my.aov <-aov(Response ~ Treatment + Error(Treatment:Replicate)) Anova(my.aov, type="II") yields Error in Anova(nested.anova) : no applicable method

Hello, is it possible to obtain type III sums of squares for a nested model as in the following: lmod <- lm(resp ~ A * B + (C %in% A), mydata)) I have tried library(car) Anova(lmod, type="III") but this gives me an error (and I also understand from the documentation of Anova as well as from a previous request (http://finzi.psych.upenn.edu/R/Rhelp02a/archive/64477.html) that it is