similar to: 2^k designs "anova"

Is names(model1$coef) what you're looking for? -----Original Message----- From: Kenneth Cabrera [mailto:krcabrer at epm.net.co] Sent: 29 November 2002 10:36 Cc: R-help at stat.math.ethz.ch Subject: [R] Obtaining the variable names of a glm object Hi, R users! Suppose I make a model like this:

Dear R users: I want to have in the same column both symbols, line and points, one for a data, and the other for the model. How can I do that? Or is there other better form to show both data and model in the same graphic? I need to make the difference on time of the same subject (repeated measurements on time) Problem example: ###########################################################

Hi R users: I will like to know if somebody works on insurance statistics (actuarial problems) and had use TRICAST, and can tell me if with all the R tools it can be build a solution like TRICAST or similar. In a word: Do you think that R has all the statistical tools (I mean modeling tools) to make a job similar to TRICAST? Does TRICAST has modeling tools that are not implemented on R yet?

Hello All: I am trying to replicate the results of an example found in Alan Agresti's "Categorical Data Analysis" on pages 267-269. The example is one of a 2 x 2 cross-classification table of ordinal counts: job satisfaction and income. I am able to get Agresti's results for the independence model (G^2 = 12.03 with df = 9) assuming as he does that the data is nominal, but

Hi R users: I will like to have a legend with math symbols and also with the value of a variable. But I cannot obtain both at the same time (symbol + value of a variable): Here is a reproducible example: m1<-5 m2<-12 plot(1:5,1:5,type="n") legend("topleft",legend=c(paste(expression(mu),"=",m1),expression(paste(mu,"=",m2))),lty=1:2) Thank you for

Hi R users: I will like to have a legend with math symbols and also with the value of a variable. But I cannot obtain both at the same time (symbol + value of a variable): Here is a reproducible example: m1<-5 m2<-12 plot(1:5,1:5,type="n") legend("topleft",legend=c(paste(expression(mu),"=",m1),expression(paste(mu,"=",m2))),lty=1:2) Thank you for

Hi R users: I got an error that "l1" is not found. This is a sample code: f1<-function(i,l1){ print(l1[[i]]) l1[[i+1]]<<-i return(i) } z1<-list(x=100,NULL,a=c(1,1)) lapply(1:3,f1,l1=z1) My final goal is to update the value of each list component to use it in the following step. I see that print function founds l1[[1]], but the assign instruction not. Thank you

Hi R users: I got this code to generate a graphic for each member of a lists. list1<-list(A=data.frame(x=c(1,2),y=c(5,6)),B=data.frame(x=c(8,9),y=c(12,6))) names1<-names(list1) sapply(1:length(list1),function(i) with(list1[[i]],plot(x,y,type="l",main=paste("Graphic of",names1[i])))) Is there a more elegant solution for not to use two separate lists? I would like to

Hallo all Suppose I have a summary amount of various colours used in different areas colour plastic paint black 15.5 173.8 brown 6.0 523.2 green 2.5 6.4 red 77.1 237.4 yellow 144.6 77.3 It seems that there are some preferable colours in these two areas (yellow in plastic, brown an red in paint). The problem seems to me similar to chi square test but it is intended for testing counts or

Hi, I ran a small data set from a factorial experiment through R, Minitab and JMP... the result from R is significantly different from what Minitab or JMP give... The data set is at the following link: http://www.personal.psu.edu/nug107/Uploads/2x3_16repsANOVA.txt The first 5 columns are the factors and the next three are responses. In particular, for the response beta11MSE, two of the

Hi R users: I want to know if any one of you had the problem with Tinn-R, when you call the library svIDE on the new R 2.5.0, (because in the old R 2.4.1 works with out any problem). I got this message: > library(svIDE) Loading required package: tcltk Loading Tcl/Tk interface ... done Warning messages: 1: '\A' is an unrecognized escape in a character string 2: no se reconoce el valor

Hello everybody: Does any one has a function to build Chernoff faces? Thank you very much Kenneth Roy Cabrera Torres krcabrer at epm.net.co -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not

This can be accomplished with the front-end "Emacs" / "ESS". http://cran.r-project.org/doc/FAQ/R-FAQ.html#R%20and%20Emacs -----Original Message----- From: Kenneth Cabrera [mailto:krcabrer at perseus.unalmed.edu.co] Sent: 26 July 2002 15:30 To: r-help at stat.math.ethz.ch Subject: [R] Frame o windows result. Helllo Dear R-user! I want to know how can I open and other text

I have a probably basic problem concerning the R language. I will write some small R programs but don't really know how to start. Do I have to create my files in some editor like emacs or can I write my code in the R prompt? How to compile the code? I haven't found any good manual parts about this subject. regards A.P

Hi, I'm using: platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 1 minor 5.1 year 2002 month 06 day 17 language R and I would like to apply: > update.packages() trying URL

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

I want to specify a two-factor model in lme, which should be easy? Here's what I have: factor 1 - treatment FIXED (two levels) factor 2 - genotype RANDOM (160 genotypes in total) I need a model that tells me whether the treatment, genotype and interaction terms are significant. I have been reading 'Mixed effects models in S' but in all examples the random factor is not in the main

Hi everybody, I'm trying to construct contrasts for an ANOVA to determine if there is a significant trend in the means of my groups. In the following example, based on the type of 2x3 ANOVA I'm trying to perform, does the linear polynomial contrast generated by contr.poly allow me to test for a linear trend across groups? doi=data.frame( Group=c( rep(1, 5), rep(2, 5), rep(3, 5),

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

Hi, I'm working on fitting a glm model to my data using Gamma error structure and reciprocal link. I've been using dropterm (MASS) in the model simplification process, but the F values from analysis of deviance tables reported by dropterm and anova functions are different - sometimes significantly so. However, the reported residual deviances, degrees of freedom, etc. are not different.