similar to: Doing partial-f test for stepwise regression

Dear all: I have a regression model that has collinearity problems (between three regressor variables). I need a F-test that will allow me to compare between full (with all variables) and partial models (minus 1=< variables). The general F-test formula I'm using is: F = {[SS(full model) - SS(reduced model)] / (#variables taken out)} / MSS(full model) Unfortunately, the ANOVA table

Dear R users, I am using lme and nlme to account for spatially correlated errors as random effects. My basic question is about being able to correct F, p, R2 and parameters of models that do not take into account the nature of such errors using gls, glm or nlm and replace them for new F, p, R2 and parameters using lme and nlme as random effects. I am studying distribution patterns of 50 tree

I would like to try a likelihood ratio test in place of waldtest. Ideally I'd like to provide two glm models, the second a submodel of the first, in the style of lrt (http://www.pik-potsdam.de/~hrust/tools/farismahelp/lrt.html). [lrt takes farimsa objects] Does anyone know of such a likelihood ratio test? Chris Elsaesser, PhD Principal Scientist, Machine Learning SPADAC Inc. 7921

Hi everyone, I have recently experienced a strange behavior (strange from my knowledge) in rails. In my controllers ''new'' action, I am creating a few instance variables in the following manner : @controllerModel = ControllerModel.new @model1 = Model1.all @model2 = Model2.all in my ''new'' view, I am using the @controllerModel to create the form for new and I

Dear R users, i'm using a custom function to fit ancova models to a dataset. The data are divided into 12 groups, with one dependent variable and one covariate. When plotting the data, i'd like to add separate regression lines for each group (so, 12 lines, each with their respective individual slopes). My 'model1' uses the group*covariate interaction term, and so the coefficients

R gurus, I'm working on data analysis for a small project. My response variable is total vines per tree (median = 0, mean = 1.65, min = 0, max = 24). My predictors are two categorical variables (four sites and four species) and one continuous (tree diameter at breast height (DBH)). The main question I'm attempting to answer is whether or not the species identity of a tree has

Hello, I am trying to use the multi_search method, but I keep getting type error on nil objects, I send it [Model1,Model2] and it seems as though the class names keep getting clobbered and turn to nil, somewhere along the multi_index area. I tried to trace what was going on, but I got nothing, also, this only happens when there are actually hits(thank god, most of the time). Perhaps some insight?

Hello, I have a problem concerning logistic regressions. When I add a quadratic term to my linear model, I cannot draw the line through my scatterplot anymore, which is no problem without the quadratic term. In this example my binary response variable is "incidence", the explanatory variable is "sun": > model0<-glm(incidence~1,binomial) >

This has got to be incredibly simple but I nevertheless can't figure it out as I am apparently brain dead. I just want to convert the elements of a character vector to variable names, so as to then assign formulas to them, e.g: z = c("model1","model2"); I want to assign formulas, such as lm(y~x[,1]) and lm(y~x[,2]), to the variables "model1" and

Hello R-Help, ----------------------------------------------------------------------------------------------------------------------------------------- Issues (there are 2): 1) Possible bug when using lmtest::encomptest() with a linear model created using nlme::lmList() 2) Possible modification to lmtest::encomptest() to fix confusing fail when models provided are, in fact, nested. I have

Dear r-helpers, Spencer Graves and Manual Morales proposed the following methods to simulate p-values in lme4: ************preliminary************ require(lme4) require(MASS) summary(glm(y ~ lbase*trt + lage + V4, family = poisson, data = epil), cor = FALSE) epil2 <- epil[epil$period == 1, ] epil2["period"] <- rep(0, 59); epil2["y"] <- epil2["base"]

Full_Name: Menelaos Stavrinides Version: 1.3. 1 OS: Windows 98 Submission from: (NULL) (193.129.76.90) When model simplification is used in glm (binomial errors) and anova is used two compare two competitive models one can use either an "F" or a "Chi" test. R always performs an F test (Although when test="Chi" the test is labeled as Chi, there isn't any

I''d like to create scaffolds for each of the tables in my database, but would like to access them below the path: localhost:3000/admin For example localhost:3000/admin/table1 localhost:3000/admin/table2 Where table1 and table2 would be the controller names respectively, calling their own list.rhtml, show.rhtml, edit.rhtml, etc.. Any recommendations on how/where to

Hi, I know that is not possible make a stepwise procedure using REML in R, I can use ML for this. For nested design it may be very dangerous due the difference in variance structure, mainly in a splitplot design. ML make significative variables that REML dont make. I read an article that is made a stepwise procedure using GENSTAT. from article: "Terms were dropped from a model in a

Is there any way to put an argument into an object name. For example, say I have 5 objects, model1, model2, model3, model4 and model5. I would like to make a vector of the r.squares from each model by code such as this: rsq <- summary(model1)$r.squared for(i in 2:5){ rsq <- c(rsq, summary(model%i%)$r.squared) } So I assign the first value to rsq then cycle through models 2 through

Hi, I'm trying to compare models, one of which has all parameters fixed using offsets. The log-likelihoods seem reasonble in all cases except the model in which there are no free parameters (model3 in the toy example below). Any help would be appreciated. Cheers, Jarrod x<-rnorm(100) y<-rnorm(100, 1+x) model1<-lm(y~x) logLik(model1) sum(dnorm(y, predict(model1),

I am using anova.lm to compare 3 linear models. Model 1 has 1 variable, model 2 has 2 variables and model 3 has 3 variables. All models are fitted to the same data set. anova.lm(model1,model2) gives me: Res.Df RSS Df Sum of Sq F Pr(>F) 1 135 245.38 2 134 184.36 1 61.022 44.354 6.467e-10 *** anova.lm(model1,model2,model3) gives

Hello, I want to understand how to tell if a model is significant. For example I have vectX1 and vectY1. I seek first what model is best suited for my vectors and then I want to know if my result is significant. I'am doing like this: model1 <- lm(vectY1 ~ vectX1, data= d), model2 <- nls(vectY1 ~ a*(1-exp(-vectX1/b)) + c, data= d, start = list(a=1, b=3, c=0)) aic1 <- AIC(model1)

Hello, I have a simple theorical question about regresion... Let's suppose I have this: Model 1: Y = B0 + B1*X1 + B2*X2 + B3*X3 and Model 2: Y = B0 + B2*X2 + B3*X3 I.E. Model1 = lm(Y~X1+X2+X3) Model2 = lm(Y~X2+X3) The Ajusted R-Square for Model1 is 0.9 and the Ajusted R-Square for Model2 is 0.99, among many other significant improvements. And I want to do the anova test to choose the best

Dear community, I've check it while working, but just to reassure myself. Let's say we have 2 models: model1 <- lm(vdep ~ log(v1) + v2 + v3 + I(v4^2) , data = mydata) model2 <- lm(vdep ~ log(v1) + v2 + v3 + v4^2, data = mydata) So in model1 you really square v4; and in model2, v4*^2 *doesn't do anything, does it? Model2 could be rewritten: model2b <- lm(vdep ~