similar to: How to get the P-values from GLM results?

Hello, R-i-zens. I'm working on an data set with a factorial ANOVA that has a significant interaction. I'm interested in seeing whether the simple effects are different from 0, and I'm pondering how to do this. So, I have my.anova<-lm(response ~ trtA*trtB) The output for which gives me a number of coefficients and whether they are different from 0. However, I want the

Dear R People: I have an AOV model that I get confidence intervals from via > confint(chick1.aov1) 2.5 % 97.5 % trtA 1.472085 1.607915 trtB 1.512085 1.647915 trtC 1.328751 1.464582 > I am using R2HTML to produce HTML output. However, the HTML code itself just has rounded values, i.e., 1.5 and 1.6. Has anyone run across this, please? Any suggestions would be much appreciated.

Dear list, I asked some questions about multilevelanalysis a couple of months ago. In the meantime I did some reading about the subject. Now I'd like to check, if I understood it all correctly. If you think my questions are not appropriate for this list, please tell me so and i will immediatly stop asking them. I have a dataset with one predicted variable (y), two explanatory variables

Hello, I am trying to do a logistic regression and have one predictor variable (x) that is ratio and two predictor variables (y and z) that are categorical. These have three levels each which I have called "High", "Medium" and "Low". My question: do I need to use a numerical coding scheme for the categorical variables as required by some statistical software

Hi, why don't you try try ks.test(VeriSeti1, VeriSeti2)$p.value All the best Jenny >How can i print only the P-Value of the kolmogorov smirnov test? > > >> ks.test(VeriSeti1, VeriSeti2) > > Two-sample Kolmogorov-Smirnov test > >data: VeriSeti1 and VeriSeti2 >D = 0.5, p-value = 0.4413 >alternative hypothesis: two-sided > > >This expression

Dear all, I have a question concerning the p-value. When running coxph I get a p-value = 0. :confused: Can this be true? Why aren?t there decimal points? Is there a way to find out the exact p-value? Here is the output: ---------------------------------------------------------------------------------------------------- Call: coxph(formula = Surv(start, stop, status) ~ Albumin_gproL, data = dial,

Dear all, For an introductory course on glm?s I would like to create an example to show the difference between glm and transformation of the response. For this, I tried to create a dataset where the variance increases with the mean (as is the case in many ecological datasets): poissondata=data.frame( response=rpois(40,1:40), explanatory=1:40) attach(poissondata) However, I have run into

Dear all, I coded a function called u.glm u.glm <- function (x,ahi,age,bmiz,gender) { library(nortest) lil.rslt <- lillie.test(x) if (lil.rslt$p.value >0.05) { cat("Logtrans=0, lillie=",lil.rslt$p.value,"\n") xmodel<-glm(x~ahi+age+bmiz+as.factor(gender)) summary(xmodel) confint(xmodel) } else { cat("Logtrans=1,

Dear R-users, I would like to know if there is any way to constraints optimized parameters using the function lm, glm or others that are written in the form: Lm( formula, data ...) As I understand, formula are of the type y ~ X1 +X2+ ... Xi (where Y, X1, X2 ..Xi are vectors). In my case I would like the estimates of this linear combination computed with lm to be positives. I haven't found

Hello, I am wondering if there is an easy way to combine loess() with glm() to produce a locally fitted generalised regression. I have a data set of about 5,000 observations and 5 explanatory variables, with a binary outcome. One of the explanatory variables (lets call it X) is much more predictive than the others. A single glm() regression over the entire data set produces rather poor results,

Quick question about the usage of glht. I'm working with a data set from an experiment where the response is bounded at 0 whose variance increases with the mean, and is continuous. A Gamma error distribution with a log link seemed like the logical choice, and so I've modeled it as such. However, when I use glht to look for differences between groups, I get significant

Hi, I am trying to use glm to fit my data, wondering if there is a easy way to fit a glm without typing all the explanatory variable names. For example, if I have 100 explanatory variables x1, x2, ..., x100 and response variable is y, I don't want to do something like glm1 <- glm(y ~ x1 + x2 + ... + x100, family = gaussian, data = dataA) since it would be a lot of typing. Many thanks,

Dear all, I am running a panel regression with time and location fixed effects: ### reg1 <- lm(lny ~ factor(id) + factor(year) + x1+ I(x1)^2 + x2+ I(x2)^2 , data=mydata, na.action="na.omit") ### My goal is to use the estimation for prediction. However, I have 8,500 IDs, which is resulting in very slow computation. Ideally, I would like to do the following: ### reg2 <-

Dear R Gurus, I have a problem related to plot. For example, I have two variables, pre and post. pre <- c(1,2,3,4,5) post <- c(2,5,7,2,3) How can I plot a line graph similar to this one? http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=1847566&rendertype=figure&id=F1 Would you please provide me a keyword so that I can search. I don't know what is the name of this

Dear R-list. I'm doing af logistic analyses using gml. The model explaines variations in Adverse events infections (0 og 1) using age as explanatory variable. model2d<-glm(formula=AEorSAEInfecBac~Age,family=binomial("logit"),data=emrisk) I want to get predictions with 95% confidence limits for age 30 and age 60. I've been reading the "google" and "search

Hi everybody, somehow i dont get the shapiro wilk test for normality. i just can?t find what the H0 is . i tried : shapiro.test(rnorm(5000)) Shapiro-Wilk normality test data: rnorm(5000) W = 0.9997, p-value = 0.6205 If normality is the H0, the test says it?s probably not normal, doesn ?t it ? 5000 is the biggest n allowed by the test... are there any other test ? ( i know qqnorm

Hi, I have some data on the effect of cycle shape (categorical) and frequency (continuous) on the efficiency of muscle contraction. My minimum adequate model is: m15<-glm(efficiency~cycle.shape*freq, family=quasipoisson) However, I wish to know where significant differences lie between specific combinations of treatments. I guess I want an equivalent of a post hoc test following an

LS, How large a dataset can glm fit with a binomial link function? I have a set of about 100.000 observations and about 8000 explanatory variables (a factor with 8000 levels). Is there a way to find out how large datasets R can handle in general? Thanks in advance, geelman

Hi all R users I did a logistic regression with my binary variable Y (0/1) and 2 explanatory variables. Now I try to draw my nomogram with predictive value. I visited the help of R but I have problem to understand well the example. When I use glm fonction, I have a problem, thus I use lrm. My code is: modele<-lrm(Y~L+P,data=donnee) fun<- function(x) plogis(x-modele$coef[1]+modele$coef[2])

Hello all, I've been having trouble with assessing a breakpoint in a logistic GLM with two explanatory variables. For this analysis I've been using the 'segmented' package version 0.2-9.1. But I keep getting an error and I don't see where I would be going awry. The situation is the following: Two explanatory variables: bedekking - a variable with possible values between 0 and