similar to: Plotting confidence intervals for lme

Displaying 20 results from an estimated 3000 matches similar to: "Plotting confidence intervals for lme"

2005 Jul 13
1
Name for factor's levels with contr.sum
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
2005 Jul 27
1
Question on glm for Poisson distribution.
Good afternoon, I REALLY try to answer to my question as an autonomous student searching in the huge pile of papers on my desk and on the Internet but I can't find out the solution. Would you mind giving me some help? Please. ######################################### I'm trying to use glm with factors: > Pyr.1.glm<-glm(Pyrale~Trait,DataRav,family=poisson) If I have correctly
2005 Jun 30
2
Linear Models with mean as Intercept.
Dear advanced statisticians, *******Objectif******** I try to set up linear models with mean as intercept: Answer: y Variable: x, as factor of two modalities: x(1), x(2). I would like to have a model as: y = mean(y)+A(i)+residuals, with i in (1,2) and A(1) coefficient for x(1) and A(2) coefficient for x(2). *******Trials in R******* ## Firstly: I write in R: >Model<-lm(y~x,Data)
2006 Apr 20
3
The contrary of command %in%
Dear r-list, I've got a data base: > HData[1:10,] NumTree Site Species Date Age DBH H IdentTree 1 1 Queige Spruce 2002 184 49 33.5 Queige 1 2 2 Queige Fir 2002 NA 5 4.6 Queige 2 3 3 Queige Fir 2002 25 8 6.6 Queige 3 4 4 Queige Spruce 2002 198 47 32.5 Queige 4 5 5 Queige Fir 2002 200 59 35.3 Queige 5 6 6 Queige
2005 Jul 22
0
Significant difference of coefficients in glm with factors?
Hi and sorry to distur, ########### Setting ################# I'm trying to use glm with factors: > Pyr.1.glm<-glm(Pyrale~Trait,DataRav,family=poisson) > summary(Pyr.1.glm) Call: glm(formula = Pyrale ~ Trait, family = poisson, data = DataRav) Deviance Residuals: Min 1Q Median 3Q Max -1.7117 -0.8944 -0.6237 0.6390 1.5224 Coefficients: Estimate Std. Error z value Pr(>|z|)
2005 Dec 16
1
Vector of matrix
Dear statisticians, I would like to save results for a "for loop" in a vector previously created. My result would be of class "matrix". I tried the following script: Script: #Creation of a previous vector n.Tree<-2 VectorX<-rep(1,n.Tree) #loop for (i in 1:2) { Ti<-MatOccurTree[Tree[i],1] #number of observation for Tree i Xi<-matrix(data=1,nrow=Ti,ncol=2)
2007 Mar 09
1
Problem with ci.lmer() in package:gmodels
Dear Friends, Please note that in the following CI lower > CI higher: > require(lmer) > require(gmodels) > fm2 <- lmer(Reaction ~ Days + (1|Subject) + (0+Days|Subject), sleepstudy) > ci(fm2) Estimate CI lower CI upper Std. Error p-value (Intercept) 251.66693 266.06895 238.630280 7.056447 0 Days 10.52773 13.63372 7.389946 1.646900
2008 Mar 08
1
Installing odesolve package of R under Ubuntu (Debian) Linux
Bonjour, Je cherche à installer le package odesolve du logiciel de statistique R sous Ubuntu Linux. C'est un package qui contient des fonctions appelant du code en Fortran. A l'installation sous R via le shell, j'obtiens l'erreur suivante: Hi, I tried to install odesolve package of R under Ubuntu Linux. But I got the following error: ghislain@ghislain-laptop:~$ sudo R [sudo]
2009 Jun 25
2
Problems with subsets in NLME
I am trying to estimate models with subsets using the NLME package. However, I am getting an error in the case below (among others): > subset <- c(rep(TRUE, 107), FALSE) > fm2 <- lme(distance ~ age + Sex, data = Orthodont, random = ~ 1, subset=subset) Error in xj[i] : invalid subscript type 'closure' > fm2 <- lme(distance ~ age + Sex, data = Orthodont, random = ~ 1,
2003 Jan 30
1
as.formula(string) and augPred in lme
Using formulas constructed from strings only partially works for me in lme: library(nlme) data(Orthodont) fm2<-lme(as.formula("distance~age"),data=Orthodont,random=~1|Subject) summary(fm2) # works augPred(fm2) # fails #Error in inherits(object, "formula") : #Argument "object" is missing, with no default I assume that my use of as.formula is wrong, but
2006 Jul 24
3
standardized random effects with ranef.lme()
Using ranef() (package nlme, version 3.1-75) with an 'lme' object I can obtain random effects for intercept and slope of a certain level (say: 1) - this corresponds to (say level 1) "residuals" in MLWin. Maybe I'm mistaken here, but the results are identical. However, if I try to get the standardized random effects adding the paramter "standard=T" to the
2008 Apr 13
2
prediction intervals from a mixed-effects models?
How can I get prediction intervals from a mixed-effects model? Consider the following example: library(nlme) fm3 <- lme(distance ~ age*Sex, data = Orthodont, random = ~ 1) df3.1 <- with(Orthodont, data.frame(age=seq(5, 20, 5), Subject=rep(Subject[1], 4), Sex=rep(Sex[1], 4))) predict(fm3, df3.1, interval='prediction') # M01 M01
2006 Oct 08
2
latex and anova.lme problem
Dear R-helpers, When I try > anova(txtE2.lme, txtE2.lme1) Model df AIC BIC logLik Test L.Ratio p-value txtE2.lme 1 10 8590 8638 -4285 txtE2.lme1 2 7 8591 8624 -4288 1 vs 2 6.79 0.0789 > latex(anova(txtE2.lme, txtE2.lme1)) Error: object "n.group" not found I don't even see n.group as one of the arguments of latex() I checked to see >
2004 Dec 31
1
lme: Confusion about Variances
Dear R users! I used lme to fit a mixed model with random intercept and spatial Gaussian correlation i.e. I fitted a model of the following form: Y = X*beta + error and error = U + W(t) + Z where U is the random intercept (normally distributed), W(t) the stationary Gaussian process and Z also a normally distributed (the residual) rv. Each of these three random variables have a variance which
2012 Mar 20
1
Remove quotes from a string to use in a variable call
Hi, I have a string that I want to use in a variable call. How can I remove the quotes and/or the string properties of the string to use it in a variable call? Here's an example: library(lme) fm2 <- lme(distance ~ age, data = Orthodont, random = ~ 1) summary(fm2) I want to update the above regression to include new predictors according to what is in a string: predictors <-
2004 Aug 27
2
degrees of freedom (lme4 and nlme)
Hi, I'm having some problems regarding the packages lme4 and nlme, more specifically in the denominator degrees of freedom. I used data Orthodont for the two packages. The commands used are below. require(nlme) data(Orthodont) fm1<-lme(distance~age+ Sex, data=Orthodont,random=~1|Subject, method="REML") anova(fm1) numDF DenDF F-value p-value (Intercept) 1
1999 Jun 02
1
lme problem ?
Dear friends. I tried the session below with 10 MB in both vsize and nsize but didn't get the example work. Is this a problem in LME or in me or both or somewhere else or undefined ? R : Copyright 1999, The R Development Core Team Version 0.64.0 Patched (May 3, 1999) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type
2009 Aug 14
1
post hoc test after lme
Hi! I am quiet new with R and I have some problems to perform a posthoc test with an lme model. My model is the following: >lme1<-lme(eexp~meal+time, random=~1|id,na.action=na.omit) and then i try to get a post hoc test: >summary(glht(lme1,linfct=mcp(meal="Tukey))) but I get a warning message: Erreur dans as.vector(x, mode) : argument 'mode' incorrect Thank you for your
2006 Apr 25
1
summary.lme: argument "adjustSigma"
Dear R-list I have a question concerning the argument "adjustSigma" in the function "lme" of the package "nlme". The help page says: "the residual standard error is multiplied by sqrt(nobs/(nobs - npar)), converting it to a REML-like estimate." Having a look into the code I found: stdFixed <- sqrt(diag(as.matrix(object$varFix))) if (object$method
2004 Oct 01
4
gnls or nlme : how to obtain confidence intervals of fitted values
Hi I use gnls to fit non linear models of the form y = alpha * x**beta (alpha and beta being linear functions of a 2nd regressor z i.e. alpha=a1+a2*z and beta=b1+b2*z) with variance function varPower(fitted(.)) which sounds correct for the data set I use. My purpose is to use the fitted models for predictions with other sets of regressors x, z than those used in fitting. I therefore need to