search for: freedoms

Hi. I got this email-address from BUGS.txt file of the samba distribution. I read it carefully. I got configured and working samba, I work with samba above one year. I still have a problem, and, according to the samba log output, I think it's probably a bug. When I try to use the shares with audit support, I have my Far disconnected and telling me that something is wrong. I have that

Hello all. I was wondering if there is any way to adjust the denominator degrees of freedom in lme(). It seems to me that there is only one method that can be used. As has been pointed out previously on the list, the denominator degrees of freedom given by lme() do not match those given by SAS Proc Mixed or HLM5. Proc Mixed, for example, offers five different options for computing the

Hello. Douglas Bates has explained in a previous posting to R why he does not output residual degrees of freedom, F values and probabilities in the mixed model (lmer) function: because the usual degrees of freedom (obs - fixed df -1) are not exact and are really only upper bounds. I am interpreting what he said but I am not a professional statistician, so I might be getting this wrong... Does

Hello, I'm running mixed models in GAMM4 with 2 (non-nested) random intercepts and I want to include a spline term for one of my exposure variables. However, when I include a spline term, I always get reported degrees of freedom of less than 1, even when I know that my spline is using more than 1 degree of freedom. For example, here is the code for my model: >

Hello, I've just carried out my first good-looking model using glmmPQL, and the output makes perfect sense in terms of how it fits with our hypothesis and the graphical representation of the data. However, please could you clarify whether my degrees of freedom are appropriate? I had 106 subjects, each of them was observed about 9 times, creating 882 data points. The subjects were in 3

It appears that the degrees of freedom reported by logLik changed between R 1.4.1 and R 1.5.1. Is this true? Detail: > I have been using the lm and logLik functions in R to develop code using > version 1.4.1. When I run it on version 1.5.1, I'm getting different > degrees of freedom with the logLik function. Version 1.5.1 seems to give > one extra degree of freedom than

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

Dear all: "predict.glm" provides an example to perform logistic regression when the response variable is a tow-columned matrix. I find some paradox about the degree of freedom . > summary(budworm.lg) Call: glm(formula = SF ~ sex * ldose, family = binomial) Deviance Residuals: Min 1Q Median 3Q Max -1.39849 -0.32094 -0.07592 0.38220 1.10375

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 friends, I have written the following lines in R console wich already exist in pdf file systemfit: data( "GrunfeldGreene" ) library( "plm" ) GGPanel <- plm.data( GrunfeldGreene, c( "firm", "year" ) ) greeneSur <- systemfit( invest ~ value + capital, method = "SUR", + data = GGPanel ) greenSur I have obtained the following incomplete

Dear R-List users, Can anyone explain exactly the difference between Weights options in lm glm and gls? I try the following codes, but the results are different. > lm1 Call: lm(formula = y ~ x) Coefficients: (Intercept) x 0.1183 7.3075 > lm2 Call: lm(formula = y ~ x, weights = W) Coefficients: (Intercept) x 0.04193 7.30660 > lm3 Call:

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 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

Dear R users, I am having problems trying to fit quasipoisson and negative binomials glm. My data set contains abundance (counts) of a species under different management regimens. First, I tried to fit a poisson glm: > summary(model.p<-glm(abund~mgmtcat,poisson)) Call: glm(formula = abund ~ mgmtcat, family = poisson) . . . (Dispersion parameter

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

Hi , Could someone pls help me about this topic, I dont know how can i extract them from my model!! Thanks, Sophie -- View this message in context: http://r.789695.n4.nabble.com/degrees-of-freedom-does-not-appear-in-the-summary-lmer-tp3741327p3741327.html Sent from the R help mailing list archive at Nabble.com.

I would like to document my findings (with a potential FIX) regarding the issue of calculation of the degrees of freedom with nlme(). The program given at the bottom of this email generates and fit 20 data sets with a mixed-effects LINEAR model, but using the function nlme() instead of lme(). In each case, the correct number of degrees of freedom for the intercept parameter is 12. However, in

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, I am new to R and my question may be trivial to you... I am doing a GLM with binomial errors to compare proportions of species in different categories of seed sizes (4 categories) between 2 sites. In the model summary the residual deviance is much higher than the degree of freedom (Residual deviance: 153.74 on 4 degrees of freedom) and even after correcting for overdispersion by

I'm running an ANOVA on some data for respiration in a forest. I am having a problem with my degrees of freedom. For one of my variables I get one fewer degrees of freedom than I should. I have 12 plots and I therefore expected 11 degrees of freedom, but instead I got 10. Any ideas? I have some code and output below: > class(Combined.Plot) [1] "character" >