similar to: lme degrees of freedoms: SAS and R

I recently uploaded a new version of the lme contributed library to the main CRAN site. It should be available on mirrors within the next 24 hours. This version matches the beta version of NLME 3.0 for S-PLUS. It only provides methods for linear mixed-effects models in R so it is called lme rather than nlme. The other big thing that is missing in the R version is all the graphics because they

You can get around this problem by specifying the summary function used by interaction.plot : mean.rm.na <- function(x) mean(x,na.rm=T) interaction.plot( A, B, Y ) -Greg > -----Original Message----- > From: Kaiya Liu [mailto:liu.262 at osu.edu] > Sent: Friday, February 08, 2002 4:57 PM > To: Uwe Ligges > Cc: r-help at stat.math.ethz.ch > Subject: Re: [R] Problem with

Kaiya Liu wrote: > > Hi, > > I encountered a problem when I did interaction plots using the > interaction.plot() function. The labels for the trace factor appeared on > the plot when I did the plots in separate windows, but disappeared/cut off > when multiple plots were placed in one window. Can anybody help me? >From the example of ?interaction.plot: data(ToothGrowth)

I have a problem moving from multistratum aov analysis to lmer. My dataset has observations of ampl at 4 levels of gapf and 2 levels of bl on 6 subjects levels VP, with 2 replicates wg each, and is balanced. Here is the summary of this set with aov: >> summary(aov(ampl~gapf*bl+Error(VP/(bl*gapf)),hframe2)) > >Error: VP > Df Sum Sq Mean Sq F value Pr(>F) >Residuals

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

You could calculate the confidence interval of the correlation at your desired df: http://davidmlane.com/hyperstat/B8544.html The below code takes as arguments the observed correlation, N, and alpha, calculates the confidence interval and checks whether this includes 0. cor.test2=function(r,n,a=.05){ phi=function(x){ log((1+x)/(1-x))/2 } inv.phi=function(x){

Dear list, I do apologize if these are basic questions. I am fitting some GAM models using the mgcv package and following the model selection criteria proposed by Wood and Augustin (2002, Ecol. Model. 157, p. 157-177). One criterion to decide if a term should be dropped from a model is if the estimated degrees of freedom (EDF) for the term are close to their lower limit. What would be the

R2.3, WinXP Dear all, I am using the following functions: f1 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(x))/exp(log(Phi4))) f2 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(r)-log(x))/exp(log(Phi4))) subject to the residual weighting Var(e[i]) = sigma^2 * abs( E(y) )^(2*Delta) Here is my question, in steps: 1. Function f1 is separately fitted to two different datasets corresponding to

Hello, II used the logLik() function to get the log-likelihood estimate of an object. The function also prints the degrees of freedom. How can I extract the degrees of freedom and assign it to a variable. Below is the output: > logLik(fit2pl) 'log Lik.' -4842.912 (df=36) Thanks, Davood Tofighi [[alternative HTML version deleted]]

Hello, I have a little problem about degree of freedom in R. if you can help me, I will be happy. I used nlme?function to analyze my data and run the linear mixed effects model in R. I did the linear mixed effect analysis in SAS?and SPSS as well. However, R gave?the different degrees of freedom than SAS?and SPSS did. Can you help me to learn what the reason is to obtain different degrees of