similar to: estimating degrees of freedom iof student t

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

Sorry, by mistake I sent this to Professor Bates instead of r-help. Han -------- Original Message -------- Subject: Re: [R] degrees of freedom for t-tests in lme Date: Thu, 25 Apr 2002 09:16:16 -0700 From: Han-Lin Lai <Han-Lin.Lai at noaa.gov> To: Douglas Bates <bates at stat.wisc.edu> References: <3CC6E87F.5400277D at noaa.gov> <6rg01lottu.fsf at franz.stat.wisc.edu>

Hello R-users: Could you please tell me how can I extract the "df (degree of freedom)" and "estfun (estimating functions)" for the following lmer (or lmer2) model? wtd.mixed<-lmer(ddimer~race+steroid+psi+sofa+apache + (1|subject), method="ML", data=final, cluster="id", weights=w) I tried the following codes: - for the degree of freedom (erorr

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

Good day all! I administer two Samba servers (RHEL 4.5) which, up to recently, had been working well. Our security officials changed the LAN Manager group policy for the new Win7 systems from 'Send NTLMv2 response only; Refuse LM' to 'Send NTLMv2 response only; Refuse LM & NTLM'. We were running samba 3.0.33. I have upgraded to 3.6.8-44. I have tried a variety of

Hello, I am struggling to understand how denominator degrees of freedom and subsequent significance testing based upon them works in nlme models. I have a data set of 736 measurements (weight), taken within 3 different age groups, on 497 individuals who fall into two morphological catagories (horn types). My model is: Y ~ weight + horn type / age group, random=~1|individual I am modeling

Dear all Somebody may have asked this before but I could not find any answers in the web so let me ask a question on lme. When I have a fixed factor of, say, three levels (A, B, C), in which each level has different size (i.e. no. of observations; e.g. A>B>C). When I run an lme model, I get the same degree of freedom for all the contrast t-tests (e.g. AvsB or BvsC). I have tried this to

Hello all. It seems that the pf() function when used with noncentral parameters can behave badly at times. I've included some examples below, but what is happening is that with some combinations of df and ncp parameters, regardless of how large the quantile gets, the same probability value is returned. Upon first glance noncentral values greater than 200 may seem large, but they are in

Gabrielle, The way I see it, its pretty much the same thing... Conversion to LLVM-IR of a custom bytecode is the same as conversion to LLVM-IR of a custom language. The syntax of the 'custom language' just happens to be binary bytecode. On Sun, Apr 5, 2009 at 1:15 PM, Gabriele Farina <gabriele at sephiroth.it>wrote: > Hi, > > Isn't it intended to explain how to build

Well, you are right :) In fact I've started porting the VM in the spare time and it is working fine. I'm still having some issues to understand the garbage collector, but I'll delve more into it as soon as the other features will be complete Gabriele Il giorno 07/apr/09, alle ore 08:22, someguy ha scritto: > Gabrielle, > > The way I see it, its pretty much the same

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

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

Dear R-ers, I am currently running some Wilcoxon tests in R-64. How do I find the degrees of freedom in the output I am receiving? > wilcox.test(good$TRUE, good$x4a, paired=FALSE) Wilcoxon rank sum test with continuity correction data: good$TRUE and good$x4a W = 2455, p-value < 2.2e-16 alternative hypothesis: true location shift is not equal to 0 Thank you, Stephen.

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)

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

I have a question about the computation of the degrees of freedom in a linear model: x <- runif(20); y <- runif(20) f <- lm(y ~ x) logLik(f) 'log Lik.' -1.968056 (df=3) The 3 is coming from f$rank + 1. Shouldn't it be f$rank? This affects AIC(f). Thanks Frank ----- Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context:

Dear R users, we proudly announce the latest release of our R package plsdof: Degrees of Freedom and Statistical Inference for Partial Least Squares. Features include: * Degrees of Freedom estimates for Partial Least Squares (PLS) Regression * Model selection for PLS based on various information criteria and on cross-validation * approximate confidence intervals and significance tests for PLS

Dear R users, we proudly announce the latest release of our R package plsdof: Degrees of Freedom and Statistical Inference for Partial Least Squares. Features include: * Degrees of Freedom estimates for Partial Least Squares (PLS) Regression * Model selection for PLS based on various information criteria and on cross-validation * approximate confidence intervals and significance tests for PLS

Greetings, I am performing a simple Pearson's correlation test. Length of both vectors is 40, therefore the resulting df is 38. Nevertheless, a colleague is asking me for the "effective degrees of freedom". As far as I understand, those degrees of freedom have to be estimated for more complex regressions, but I was not able to find detailed information about it. Does any one of