similar to: reference on contr.helmert and typo on its help page.

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

Hi, perhaps this is a stupid question, but i need some help about Helmert contrasts in the Cox model. I have a survival data frame with an unordered factor `group' with levels 0 ... 5. Calculating the Cox model with Helmert contrasts, i expected that the first coefficient would be the same as if i had used treatment contrasts, but this is not true. I this a error in reasoning, or is it

On 6/23/05, RenE J.V. Bertin <rjvbertin at gmail.com> wrote: > Hello, > > I was just having a look at the aov function source code, and see that when the model used does not have an Error term, Helmert contrasts are imposed: > > if (is.null(indError)) { > ... > } > else { > opcons <- options("contrasts") >

Is there any logical reason why glm prints out the labels of factor levels after variable names when baseline contrasts (contr.treatment) are used but the codes for the levels when mean contrasts (contr.sum) are used? Jim -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",

Hi! I would like to suggest to make it possible, in one way or another, to get meaningful contrast names when using contr.sum(). Currently, when using contr.treatment(), one gets factor levels as contrast names; but when using contr.sum(), contrasts are merely numbered, which is not practical and can lead to mistakes (see code at the end of this message). This issue was discussed quickly in 2005

I'm running a series of simple bivariate linear regressions on grouped data. I want to test the slopes to see if they are parallel. I normally use analysis of covariance to do so, looking at interaction between the covariate and the factor to make this determination. VR3 pp.149 - 154 has a very nice example of an ANOCOVA, ending with a discussion of this very operation. My question has

I am using RW 1.2.3. on an IBM PC 300GL. Using the data bp.dat which accompanies Helen Brown and Robin Prescott 1999 Applied Mixed Models in Medicine. Statistics in Practice. John Wiley & Sons, Inc., New York, NY, USA which is also found at www.med.ed.ac.uk/phs/mixed. The data file was opened and initialized with > dat <- read.table("bp.dat") >

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

Hi Folks, I've encountered something I hadn't been consciously aware of previously, and I'm wondering what the explanation might be. In (on another list) using R to demonstrate the difference between different contrasts in 'lm' I set up an example where Y is sampled from three different normal distributions according to the levels ("A","B","C")

Hi all, I have dataset with 2 independent variable, one (x1) is continuous, the other (x2) is a categorical variable with 2 levels. The dependent variable (y) is continuous. When I run linear regression y~x1*x2, I found that the p value for the continuous independent variable x1 changes when different contrasts was used (helmert vs. treatment), while the p values for the categorical x2 and

The following is a data frame > "jjd" <- structure(list(Observations = c(6.8, 6.6, 5.3, 6.1, 7.5, 7.4, 7.2, 6.5, 7.8, 9.1, 8.8, 9.1), LevelA = structure(c(1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3), .Label = c("A1", "A2", "A3"), class = "factor"), LevelB = structure(c(1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2), .Label =

Hi all, I'm working with a dataset from 10 treatments, each treatment with 30 subjects, each subject measured 5 times. The plot of the dataset suggests that a 3-parameter logistic could be a reasonable function to describe the data. When I try to fit the model using gnls I got the message 'Step halving factor reduced below minimum in NLS step'. I´m using as the initial values of the

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

I was surprised by this (in R 2.0.1): > a <- ordered(-1:1) > a [1] -1 0 1 Levels: -1 < 0 < 1 > model.matrix(~ a) (Intercept) a.L a.Q 1 1 -7.071068e-01 0.4082483 2 1 -9.073800e-17 -0.8164966 3 1 7.071068e-01 0.4082483 attr(,"assign") [1] 0 1 1 attr(,"contrasts") attr(,"contrasts")$a [1]

I am new to statistics, R, and this list, so apologies in advance for the errors etiquette I am certain to make (in spite of reading the posting guide, help on various commands, etc.). ?Any help is greatly appreciated. Here is my data: fatigue = c(3,2,2,3,2,3,4,3,2,4,5,3,3,2,4,5,4,5,5,6,4,6,9,8,4,3,5,5,6,6,6,7,9,10,12,9) n <- 3 train <- gl(3, 4*n, labels=c("6wks",

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

Hi, I have 6 career types, represented as a factor in R, coded from 1 to 6. I need to use the effect coding (also known as deviation coding) which is normally done by contr.sum, e.g. contrasts(career) <- contr.sum(6) However, this results in the 6th category being the reference, that is being coded as -1: $contrasts [,1] [,2] [,3] [,4] [,5] 1 1 0 0 0 0 2 0 1 0

Dear R-users: I can't run the following code from "Mixed Effects Models in S and S-plus". library( nlme ) options( width = 65, digits = 5 ) options( contrasts = c(unordered = "contr.helmert", ordered = "contr.poly") ) # Chapter 5 Extending the Basic Linear Mixed-Effects Models # 5.1 General Formulation of the Extended Model data( Orthodont ) vf1Fixed

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

Dear R-devel list members, I'd like to suggest a more flexible procedure for generating contrast names. I apologise for a relatively long message -- I want my proposal to be clear. I've never liked the current approach. For example, the names generated by contr.treatment paste factor to level names with no separation between the two; contr.sum simply numbers contrasts (I recall an