similar to: degrees of freedom in lme

I want to compute confidence intervals for the random effect estimates for each subject. From checking on postings, this is what I cobbled together using Orthodont data.frame as an example. There was some discussion of how to properly access lmer slots and bVar, but I'm not sure I understood. Is the approach shown below correct? Rick B. # Orthodont is from nlme (can't have both nlme and

I can ask my question using and example from Chapter 1 of Pinheiro & Bates. > # 1.4 An Analysis of Covariance Model > > OrthoFem <- Orthodont[ Orthodont$Sex == "Female", ] > fm1OrthF <- + lme( distance ~ age, data = OrthoFem, random = ~ 1 | Subject ) > summary( fm1OrthF ) Linear mixed-effects model fit by REML Data: OrthoFem AIC BIC

Hello, I am running this script from Pinheiro & Bates book in R Version 2.1.1 (WinXP). But, I can't plot Figure 2.3. What's wrong? TIA. Rod. --------------------------------------------------------- >library(nlme) > names( Orthodont ) [1] "distance" "age" "Subject" "Sex" > levels( Orthodont$Sex ) [1] "Male"

Hi all, To follow up on an older thread, it was suggested that the following would produce confidence intervals for the estimated BLUPs from a linear mixed effect model: OrthoFem<-Orthodont[Orthodont$Sex=="Female",] fm1OrthF. <- lmer(distance~age+(age|Subject), data=OrthoFem) fm1.s <- coef(fm1OrthF.)$Subject fm1.s.var <- fm1OrthF. at bVar$Subject fm1.s0.s <-

Dear r-helpers, After successfully running require(nlme) vfr.lmL <- lmList( estimate ~ (slant + respType + visField + hand)^2 | subject, vfr ) pairs(vfr.lmL, id = 0.01, adj = -0.5) # Pinheiro & Bates (p. 141) produces the following error: Error in sprintf(gettext(fmt, domain = domain), ...) : object "form" not found Any guesses as to what I may have done wrong?

Hello, I am looking for a way to obtain standard errors for emprirical Bayes estimates of a model fitted with lmer (like the ones plotted on page 14 of the document available at http://www.eric.ed.gov/ERICDocs/data/ericdocs2/content_storage_01/0000000b/80/2b/b3/94.pdf). Harold Doran mentioned (http://tolstoy.newcastle.edu.au/~rking/R/help/05/08/10638.html) that the posterior modes' variances

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

Dear R users, I want to explain binomial data by a serie of fixed effects. My problem is that my binomial data are spatially correlated. Naively, I thought I could found something similar to gls to analyze such data. After some reading, I decided that lmer is probably to tool I need. The model I want to fit would look like lmer ( cbind(n.success,n.failure) ~ (x1 + x2 + ... + xn)^2 ,

Dear all, I am experiencing problems with glmmmPQL. I am trying to analyze binomial data with some spatial autocorrelation. Here is my code and some of the outputs > colnames(d.glmm) [1] "BV" "Longitude" "Latitude" "nb_pc_02" "nb_expr_02" [6] "pc_02" "nb_pc_07" "nb_expr_07"

Dear list members, I have 982 quotations of a given stock index and I want to run a Ljung-Box test on these data to test for autocorrelation. Later on I will estimate 8 coefficients. I do not know how many degrees of freedom should I assume in the formula for Ljung-Box test. Could anyone tell me please? Below the formula: Box.test(x, lag = ????, type = c("Ljung-Box"), fitdf = 0)

Hello everyone, I am trying to regress applicants' performance in an assessment center (AC) on their gender (individual level) and the size of the AC (group level) with a multi-level model: model.0 <- lme(performance ~ ACsize + gender, random = ~1 | ACNumber, method = "ML", control = list(opt = "optim")) I have 1047 applicants in 118 ACs: >

Hello, I want to compute a correlation test but I do not want to use the degrees of freedom that are calculated by default but I want to set a particular number of degrees of freedom. I looked in the manual, different other functions but I did not found how to do it Thanks in advance for your answers Yours Florence Dufour PhD Student AZTI Tecnalia - Spain

Dear all, How can I extract the total and residual d.f. from a gnls object? I have tried str(summary(gnls.model)) and str(gnls.model) as well as gnls(), but couldn?t find the entry in the resulting lists. Many thanks! Best wishes Christoph -- Dr. rer.nat. Christoph Scherber University of Goettingen DNPW, Agroecology Waldweg 26 D-37073 Goettingen Germany phone +49 (0)551 39 8807 fax +49

Hi there, some statistical programs (e.g. SPSS) calculate a correction of the degrees of freedom in a repeated measure analysis of variance (see Greenhouse-Geisser (1958) or Huynh-Feld (1976)) by a factor epsilon. This factor is used to correct the deg. of freedom to get a corrected f-test. Is this also possible with R? Thanks, Sven P.S.: I read in the lm help page: singular.ok logical,

Hi all, Could someone please help me to calculate the P-value by using Chi-square test with a certain number of degrees of freedom? I have a data set to be calculated here: observed: 224, 64, 6 expected: 222.9, 66.2, 4.9 degrees of freedom: 1 I have been reading the documentations for three days, and can't find the answers. Please help.Thanks in advance. Regards, Frank

Hi, I have trouble to figure out how the df is derived in LME. Here is my model, lme(y~x+log(den)+sex+dep,data=lwd,random= list(group=~x)) Number of total samples (N) is 3237 number of groups (J) is 26 number of level-1 variables (Q1) is 3, i.e., x, log(den) and sex number of level-2 variables (Q2) is 1, i.e., dep x and den are continuous variable sex is associated with individual samples

I'm having hard time understanding the computation of degrees of freedom when runing nlme () on the following model: > formula(my data.gd) dLt ~ Lt | ID TasavB<- function(Lt, Linf, K) (K*(Linf-Lt)) my model.nlme <- nlme (dLt ~ TasavB(Lt, Linf, K), data = my data.gd, fixed = list(Linf ~ 1, K ~ 1), start = list(fixed = c(70, 0.4)), na.action= na.include,

Hello all, I'm trying to do a nested linear model with a dataset that incorporates an observation for each of several classes within each of several plots. I have 219 plots, and 17 classes within each plot. data.frame has columns "plot","class","age","dep.var" With lm(dep.var~class*age), The summary(lm) function returns t-test and F-test values

After an RSiteSeach("Box.test") I found some discussion regarding the degrees of freedom in the computation of the Ljung-Box test using Box.test(), but did not find any posting about the proper degrees of freedom. Box.test() uses "lag=number" as the degrees of freedom. However, I believe the correct degrees of freedom should be "number-p-q" where p and q are

I wonder whether any of you know of an efficient way to calculate the approximate degrees of freedom of a gamm() fit. Calculating the smoother/projection matrix S: y -> \hat y and then its trace by sum(eigen(S))$values is what I've been doing so far- but I was hoping there might be a more efficient way than doing the spectral decomposition of an NxN-matrix. The degrees of freedom