similar to: step method in glm()

Hi - How can I 'manually' reproduce the results in 'pred1' below? My attempt is pred_manual, but is not correct. Any help is much appreciated. library(splines) set.seed(12345) y <- rgamma(1000, shape =0.5) age <- rnorm(1000, 45, 10) glm1 <- glm(y ~ ns(age, 4), family=Gamma(link=log)) dd <- data.frame(age = 16:80) mm <- model.matrix( ~ ns(dd$age, 4)) pred1 <-

Hi! I would like to perform an F-Test over more than one variable within a generalized mixed model with Gamma-distribution and log-link function. For this purpose, I use the package mgcv. Similar tests may be done using the function "anova", as for example in the case of a normal distributed response. However, if I do so, the error message "error in eval(expr, envir, enclos) :

Estimados amigos, Estoy familiarizándome con los modelos lineales generalizados en R. Estoy interesado en realizar un análisis lig linear y me gustaría saber cuáles son o como extraer los valores correspondientes al chi cuadrado en el análisis para cada grupo y para las interacciones. Desde ya muchas gracias y disculpas si la pregunta es muy básica, adjunto los comandos que estoy utilizando. Si

Estimados amigos, Estoy familiarizándome con los modelos lineales generalizados en R. Estoy interesado en realizar un análisis lig linear y me gustaría saber cuáles son o como extraer los valores correspondientes al chi cuadrado en el análisis para cada grupo y para las interacciones. Desde ya muchas gracias y disculpas si la pregunta es muy básica, adjunto los comandos que estoy utilizando. Si

Estimados amigos, Estoy familiarizándome con los modelos lineales generalizados en R. Estoy interesado en realizar un análisis lig linear y me gustaría saber cuáles son o como extraer los valores correspondientes al chi cuadrado en el análisis para cada grupo y para las interacciones. Desde ya muchas gracias y disculpas si la pregunta es muy básica, adjunto los comandos que estoy utilizando. Si

Ted Harding said: > I can get the estimated RRs from > RRs <- exp(summary(GLM)$coef[,1]) > but do not see how to implement confidence intervals based > on "robust error variances" using the output in GLM. Thanks for the link to the data. Here's my best guess. If you use the following approach, with the HC0 type of robust standard errors in the

Dear all, I am running a logistic regression and this is the output: glm(formula = educationUniv ~ brncntr, family = binomial) Deviance Residuals: Min 1Q Median 3Q Max # ???? ????? ?? ???????? -0.8825 -0.7684 -0.7684 1.5044 1.6516 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.06869 0.01155 -92.487 <2e-16 *** brncntrNo

This is a question regarding the 'regTermTest' function in the 'survey' package. Imagine Z as a three level factor variable, and code ZB and ZC as the two corresponding dummy variables. X is a continuous variable. In a 'glm' of Y on Z and X, say, how do the two test specifications test.terms = c("ZB:X","ZC:X") # and test.terms = ~ ZB:X + ZC:X in

I am getting a repeated error when I try to run a logistic regression in R 2.8.1 >(glm(prop1~x1,data=glm1,family=binomial("logit"),weights=nt1)) Error in model.frame.default(formula = prop1 ~ x1, data = glm1, weights = nt1, : invalid type (list) for variable 'x1' x1 is multistate categorical (3 categories). 2 of the categories have 12 observation, one has 9. Is this what

Dear R-Users, I have a problem on extracting T-Stat and P-Value. I have written R-code below library("Matching") data("lalonde") attach(lalonde) names(lalonde) Y <- lalonde$re78 Tr <- lalonde$treat glm1 <- glm(Tr~age+educ+black+hisp+married+nodegr+re74+re75,family=binomial,data=lalonde) pscore.predicted <- predict(glm1) rr1 <-

R version 1.3.0 OS: SunOS 5.7, but I think the same problem occurs with Windows An incorrect formula seems to be used to calculate the pearson residuals for a generalized linear model with a binomial response. Here is a simple program which gives (a) the pearson residuals calculated directly, (b) the pearson residuals from glm, and (c) the deviance residuals from glm. The first and last

Hi, all I have a question on mgcv and ns. Now I want to compare the results from glm, gam and ns. Take a simple model y~x for example. glm1 = glm(y~x, data=data1) gam1 = gam(y~s(x), data=data1) ns1 = glm(y~ns(x),data=data1) In order to confirm the result from glm1 is consistent to those from gam1 and ns1, I want to define degree=1 in mgcv and ns. I am wondering if there is somebody can give me

Dear Listers, If I have several glm objects with names glm1, glm2.... and want to apply new data to these objects. Instead of typing "predict(glm1, newdata)..." 100 times, is there way I could do so in a loop? Thank you so much! wensui [[alternative HTML version deleted]]

Dear all, I am comparing logistic regression models to evaluate if one predictor explains additional variance that is not yet explained by another predictor. As far as I understand Baron and Li describe how to do this, but my question is now: how do I report this in an article? Can anyone recommend a particular article that shows a concrete example of how the results from te following simple

I am encountering an error when I attempt to reference a glm model within a function. The function uses the segmented.glm command (package = segmented). Within the segmented.glm command one specifies an object, in this case a logistic regression model, and specifies a starting threshold term (psi). I believe this is an environment problem, but I do not have a solution. Any assistance

dear list! i have run several glm analysises to estimate a mean rate of dung decay for independent trials. i would like to compare these results statistically but can't find any solution. the glm calls are: dung.glm1<-glm(STATE~DAYS, data=o_cov, family="binomial(link="logit")) dung.glm2<-glm(STATE~DAYS, data=o_cov_T12, family="binomial(link="logit")) as

Dear all, I have tried to modify the code of MASS:::stepAIC.default(), dropterm() and addterm() to use AICc instead of AIC for model selection. The code is appended below. Somehow the calculations are still not correct and I would be grateful if anyone could have a look at what might be wrong with this code... Here is a working example: ## require(nlme) model1=lme(distance ~ age + Sex, data =

Hello, I use a type III anova ("car" package) to analyse an unbalanced data design. I have two factors and I would have the effect of the interaction. I read that the result could be strongly influenced by the contrasts. I am really not an expert and I am not sure to understand indeed about what it is... Consequently, I failed to properly used the fit.contrast function (gregmisc

Hi, can anyone help with this: I am trying to analyse some data in the form of proportions with the glm function in R and S-plus. When comparing different models with an F test, I get different results from R and S-plus. Here's an example (there are two factors and an interaction in the full model "glm1<-glm(resp~time*set,family=binomial"): In R, entering

This is a question about removing the intercept in a binomial glm() model with categorical predictors. V&R (3rd Ed. Ch7) and Chambers & Hastie (1993) were very helpful but I wasn't sure I got all the answers. In a simplistic example suppose I want to explore how disability (3 levels, profound, severe, and mild) affects the dichotomized outcome. The glm1 model (see below) is