similar to: contrasts in aov

Good morning, I used in R contr.sum for the contrast in a lme model: > options(contrasts=c("contr.sum","contr.poly")) > Septo5.lme<-lme(Septo~Variete+DateSemi,Data4.Iso,random=~1|LieuDit) > intervals(Septo5.lme)$fixed lower est. upper (Intercept) 17.0644033 23.106110 29.147816 Variete1 9.5819873 17.335324 25.088661 Variete2 -3.3794907 6.816101 17.011692 Variete3

I have some doubts about the validity of my procedure to estimeate linear contrasts ina a factorial design. For sake of semplicity, let's imagine a one way ANOVA with three levels. I am interested to test the significance of the difference between the first and third level (called here contrast C1) and between the first and the seconda level (called here contrast C2). I used the following

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

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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

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

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

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") >

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

Hi I just started using RGui.exe under widnows. I have a text file containing date arranged in columns and rows, each column has the same format, each row with different formats. 3 of the columns are something like this 1/12/2006 3:59:45 PM I need to calculate the different in seconds between 2 selected periods using their row’s index My solution: Read the file in a data frame and

Dear all, I do not seem to grasp how contrasts are set for ordered factors. Perhaps someone can elighten me? When I work with ordered factors, I would often like to be able to reduce the used polynomial to a simpler one (where possible). Thus, I would like to explicetly code the polynomial but ideally, the intial model (thus, the full polynomial) would be identical to one with an ordered factor.

Hola compañeros de la lista, qué tal. Los molesto con la siguiente duda: Tengo un experimento con dos factores A y B, cada uno de los cuales tiene los siguientes niveles (que son concentraciones de dos hormonas vegetales aplicadas a plantas): niveles del factor A: 0, 0.2, 0.5, 1 niveles del factor B: 0, 0.1, 0.2, 0.5, 1 y mi variable de respuesta es continua, todo dentro del set de datos

Dear R-users, I want to fit a linear model with Y as response variable and X a categorical variable (with 4 categories), with the aim of comparing the basal category of X (category=1) with category 4. Unfortunately, there is another categorical variable with 2 categories which interact with x and I have to include it, so my model is s "reg3: Y=x*x3". Using fit.contrast to make the

Dear all, I'm working with Cox-regression, because data could be censored. But in this particular case not. Now I have a simple example: PRO and PRE are (0,1) coded. The response is not normal distributed. We are interested in a model which could describe interaction. But my results are depending strongly in the choose of the contrast option. It is clear that there is some dependence in

In S-Plus, I can obtain polynomial contrasts for an ordered factor with contr.poly(). The function also exists in R, however is limited to factors where the levels are equally spaced. In S-Plus, one can obtain the contrasts for a set of numeric values representing unequally spaced ordered factors. Has anyone implemented this in R? I see that the S-Plus function calls another function (poly.raw())

Hello everbody, i'm currently using the anova()-test for a small data.frame of 40 rows and 2 columns. It works well, but is there any preconditions for a valid variance analysis, that i should consider? Thank you for your answer, Daniel

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

hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for "holm", "hochberg", bonferroni", but not "Tukey". Is it not possbile to get adjusted p-values after

Dear R-Community! The example "oats" in MASS (2nd edition, 10.3, p.309) is calculated for aov and lme without interaction term and the results are the same. But I have problems to reproduce the example aov with interaction in MASS (10.2, p.301) with lme. Here the script: library(MASS) library(nlme) options(contrasts = c("contr.treatment", "contr.poly")) # aov: Y ~

Hi everybody, I'm trying to construct contrasts for an ANOVA to determine if there is a significant trend in the means of my groups. In the following example, based on the type of 2x3 ANOVA I'm trying to perform, does the linear polynomial contrast generated by contr.poly allow me to test for a linear trend across groups? doi=data.frame( Group=c( rep(1, 5), rep(2, 5), rep(3, 5),