similar to: Degree of freedom in the linear mixed effect model using lme function in R

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

Is there code available to decompose interactions involving at least one nominal factor with more than 2 levels as described, e.g., by Tukey or by Mandel (1971, Technometrics, 13: 1-18)? Tukey's model: E(y[i,j]) = mu0 + a[i] + b[j] + c*a[i]*b[j], estimating a, b, and c so sum(a) = sum(b)= 0. Mandel essentially describes a singular value decomposition of the interaction. Thanks,

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 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

Hi, I'm trying to analyse some data I have imported into R from a .csv file but when I carry out the aov command the results show only one degree of freedom when there should be 14. Does anyone know why? I'd really appreciate some help, the data is pasted below. /The imported table looks ike this this:/ Order Transect Sample Abundance 1 Coleoptera 1 1 13 2

Hi, I have a problem with the df. I read in a big csv file. Tabelle <- read.csv("C:\\Users\\Public\\Documents\\Bachelorarbeit\\eingabe8_durchnummeriert.csv" , header = T , sep=";") then I try this: > ygamma <- glm(Tabelle$sb_ek_ber ~1+ Tabelle$FAHRL_C + Tabelle$NUTZKREIS + Tabelle$schw_drittel_c   , family = Gamma) >  anova(ygamma, test="Chisq")

Hello, I have a bigger problem in calculating the Maximum Likelihood Estimator regarding the degree of freedom of a multivariate skew-t copula. First of all I would like to describe what this is all about, so that you can understand my problem: I have 2 time series with more than 3000 entries each. I would like to calculate a multivariate skew-t Copula that fits this time series. Notice:

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