similar to: R vs SPSS contrasts

I try to move from SPSS to R/S and am trying to reproduce the results of SPSS in R. I calculated a one-way anova with "spk" as experimental factor and erp as depended variable. The result of the Anova are the same concearning the mean square, F and p values. But I also wanted to caculate the contr.sdif(4) contrast on spk. The results are completely different now. I hope anybody can

Hello, I've fitted a parametric survival model by > survreg(Surv(Week, Cens) ~ C(Treatment, srmod.contr), > data = poll.surv.wo3) where srmod.contr is the following matrix of contrasts: prep auto poll self home [1,] 1 1 1.0000000 0.0 0 [2,] -1 0 0.0000000 0.0 0 [3,] 0 -1 0.0000000 0.0 0 [4,] 0 0 -0.3333333 1.0 0 [5,] 0 0

With the following example using contr.sum for both factors, > dd <- data.frame(a = gl(3,4), b = gl(4,1,12)) # balanced 2-way > model.matrix(~ a * b, dd, contrasts = list(a="contr.sum", b="contr.sum")) (Intercept) a1 a2 b1 b2 b3 a1:b1 a2:b1 a1:b2 a2:b2 a1:b3 a2:b3 1 1 1 0 1 0 0 1 0 0 0 0 0 2 1 1 0 0 1 0

I am not sure if this question has been asked before - but is there a procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE regression commands in SPSS? Thanks a lot! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Code: > options("contrasts") $contrasts factor ordered "contr.treatment" "contr.poly" I want to change the first entry ONLY, without retyping "contr.poly". How do I do it? I have tried various possibilities and cannot get anything to work. I found out that the response to options("contrasts") has class

This is my first post to this list so I suppose a quick intro is in order. I've been using SPLUS 2000 and R1.6.2 for just a couple of days, and love S already. I'm reading MASS and also John Fox's book - both have been very useful. My background in stat software was mainly SPSS (which I've never much liked - thanks heavens I've found S!), and Perl is my tool of choice for

I have been using generalized linear models in SPSS 18, in order to build models and to calculate the P values. When I was building models in Excel (using the intercept and Bs from SPSS), I noticed that the graphs differed from my expectations. When I ran the dataset again in R, I got totally different outcomes for both the P values as well as the Bs and the intercepts. The outcomes of R seem much

Hi all, I have been struggling with ANOVAs on R. I am new to R, so I created a simple data frame, and I do some analyses on R just to learn R and then check them on SPSS to make sure that I am doing fine. Here is the problem that I've run into: when we use the aov function, it uses SS Type I as default (on SPSS it is Type III). Then I used the Anova function under cars package using the

I've been playing around with contrasts in lm by specifying the contrasts argument. So, I want to specify a specific contrast to be tested Say: > y _ rnorm(100) > x _ cut(rnorm(100, mean=y, sd=0.25),c(-3,-1.5,0,1.5,3)) > reg _ lm(y ~ x, contrasts=list(x=c(1,0,0,-1))) > coef(reg)[2] x1 -1.814101 I was surprised to see that I get a different estimate for the

Hello R users, I am trying to run a three factor ANOVA on a data set with unequal sample sizes. I fit the data to a 'lm' object and used the Anova function from the 'car' package with the 'type=III' option to get type III sums of squares. I also set the contrast coding option to 'options(contrasts = c("contr.sum", "contr.poly"))' as

Dear all, I would like to ask one question related to statistics, for specifically on defining dummy variables. As of now, I have come across 3 different kind of dummy variables (assuming I am working with Seasonal dummy, and number of season is 4): > dummy1 <- diag(4) > for(i in 1:3) dummy1 <- rbind(dummy1, diag(4)) > dummy1 <- dummy1[,-4] > > dummy2 <- dummy1 >

dear all, i'm trying to reproduce an spss-anova in R. It is an 2x3x3 repeated measures desingn with repeated contrasts. In R i've coded a contrast matrix for all factors and made a split in the aov summary - but I can't get the repeated interaction contrasts. The output from SPSS looks like this: TaskSw * CongNow * CongBefore: SS df Mean Square F Sig. 1 vs. 2 1 vs. 2 1 vs. 2

Hi, all. I have a simple design I'm comparing to output from SPSS. the design is 1 repeated measure (session) and 1 between measure (cond). my dependent measure is rl. here is the data I'm using (in a data.frame): mig <- data.frame(subj=factor(rep(subj,3)), cond=factor(rep(cond,3)), session=factor(c(rep(1,nsubj),rep(2,nsubj),rep(3,nsubj))),

To whom it may concern I made some analysis with R using the command Anova. However, I found some problmes with the output obtained by selecting type II o type III sum of squares. Briefly, I have to do a 2x3 mixed model anova, wherein the first factor is a between factor and the second factor is a within factor. I use the command Anova in the list below, because I want to obtain also the sum

Hi, I found that using different contrasts (e.g. contr.helmert vs. contr.treatment) will generate different fitted probabilities from multinomial logistic regression using multinom(); while the fitted probabilities from binary logistic regression seem to be the same. Why is that? and for multinomial logisitc regression, what contrast should be used? I guess it's helmert? here is an example

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, I need to build a function to generate one column for each level of a factor in the model matrix created on an arbitrary formula (instead of using the available contrasts options such as contr.treatment, contr.SAS, etc). My approach to this was first to use the built-in function for contr.treatment but changing the default value of the contrasts argument to FALSE (I named this function

A common point made in discussion of contrasts, type I, II, III SS etc is that for sensible comparisons one should use contrasts that are 'orthogonal in the row-basis of the model matrix' (to quote from http://finzi.psych.upenn.edu/R/Rhelp02/archive/111550.html) Question: How would one check, in R, that this is so for a particular fitted linear model object? Steve Ellison

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Hi all, I am trying to do a ordered probit regression using polr(), replicating a result from SAS. >polr(y ~ x, dat, method='probit') suppose the model is y ~ x, where y is a factor with 3 levels and x is a factor with 5 levels, To get coefficients, SAS by default use the last level as reference, R by default use the first level (correct me if I was wrong), The result I got is a