similar to: Prediction with multiple zeros in the dependent variable

Daniel, When Group is entered as a factor, and the factor has two levels, the ANOVA table gives a p value for each level of the factor. What I am looking for is the omnibus p value for the factor, i.e. the test that the factor (with all its levels) improves the prediction of the outcome. You are correct that normally one could rely on the fact that the model

windows Vista R 2.10.1 Is it possible to get p values from gee? Summary(geemodel) does not appear to produce p values.: > fit4<- gee(y~time, id=Subject, data=data.frame(data)) Beginning Cgee S-function, @(#) geeformula.q 4.13 98/01/27 running glm to get initial regression estimate (Intercept) time 1.1215614 0.8504413 > summary(fit4) GEE: GENERALIZED LINEAR MODELS FOR

R 2.8.1 Windows XP I am trying to plot the results of a coxph using plot(survfit()). The plot should, I believe, show two lines one for survival in each of two treatment (Drug) groups, however my plot shows only one line. What am I doing wrong? My code is reproduced below, my figure is attached to this EMail message. John > #Create simple survival object >

I have written a function that contains runs lm() vif() and glm() When the glm() blows up with an error message, I don't get the output from either the lm() or vf() even thought neither lm() nor vif() have any problems . How can I force the function to print sequential results rather than wait for the entire function to complete before listing the functhion's output? Thanks, John

R 2.8.1 windows XP I am getting an error message that I don't understand when I try to run GLM. The error only occurs when I have all independent variables in the model. When I drop one independent variable, the model runs fine. Can anyone help me understand what the error means and how I can correct it? Thank you, John > fit11<-glm(AAMTCARE~BMI+BMIsq+SEX+jPHI+jMEDICAID+factor(AgeCat)+

I'm looking for an icon to represent an R package. Perhaps something like http://cdn2.iconfinder.com/data/icons/DarkGlass_Reworked/128x128/apps/package.png but with the R logo rather than KDE. -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Street Web:

How do you get the design matrix R used when calculating ANOVA? -- View this message in context: http://www.nabble.com/how-to-get-design-matrix--tp23464638p23464638.html Sent from the R help mailing list archive at Nabble.com.

I would like to plot 3-dimensional data on a two-dimensional scatter-plot. Is there a way I can automatically modify the plot symbol (e.g. changing size or color) to indicate the value of a third variable? E.g. How can I plot weight vs. age and indicate the value of muscle mass for each value weight-age pair by making the plot point proportional to the subject's muscle mass? Thanks, John

I am having trouble selecting rows of a dataframe that will be included in a regression. I am trying to select those rows for which the variable Meno equals PRE. I have used the code below: difffitPre<-lm(data[,"diff"]~data[,"Age"]+data[,"Race"],data=data[data[,"Meno"]=="PRE",]) summary(difffitPre) The output from the summary indicates that

Forgive my resending this post. To data I have received only one response (thank you Bert Gunter), and I still do not have an answer to my question. Respectfully, John Windows XP R 2.12.1 contrast package. I am trying to understand how to create contrasts for a model that contatains an interaction. I can get contrasts to work for a model without interaction, but not after adding the

HI, I am using Windows XP and R version 2.9.2. I have a data frame written by R similar to the following: Lab_ID Analysis_Soil Results -4MAD -2.5MAD +2.5MAD +4MAD 55003 Calcium-2008-116 900 961 1121.5 1656.5 1817 55003 Calcium-2008-117 3321 2175 2380.5 3065.5 3271 55003

I hope one and all will allow a stats question: When running a cox proportional hazards model ,there are two ways to deal with age, including age as a covariate, or to include age as part of the follow-up time, viz, Age as a covariate: tetest1 <- list(time= c(4, 3,1,1,2,2,3), status=c(1,NA,1,0,1,1,0), age= c(0, 2,1,1,1,0,0),

R2.1.1 Win 2k I have a function, B, within a function, A. I would like to have B print the name of the argument passed to it (not the value of the arguments). i.e., A<-function() { B<-function(x,y) { fit1<-lm(y~x,data=jo) print(summary(fit1) I want B to print the string "age" and the string "height". } B(age,height) }

Windows XP R 2.8.1 Colleagues, I am trying to run a function testone() and if the function completes without error do one set of instructions, and if the function generates either a warning or an error run another set of instructions. I have read try, and tryCatch help screens at least 20 times, have tried to experiment with code, but I can't understand how to accomplish my desired task.

Hello, I have got a linear model that looks like this: lm(criterion ~ variable.A*variable.a + variable.B*variable.b + variable.C *variable.c) The output computed with stdCoeff() seems to be all right, but it does not show the coefficients of the interaction of the first pair of variables. Instead, it shows "NA": (Intercept) NA

Widows XP R 2.3.1 I have been trying to make a data structure that will contain both the coefficients from a linear regression along with column and row titles AND the call, i.e. myreg<-lm(y~x+y+z) whatIwant<-cbind(c(summary(myreg)$call,"",""),summary(myreg)$coefficients) Neither the statement above, nor any one of twenty variations I have tried work. I would appreciate

(1) I know that \n when used in cat, e.g. cat("\n") produces a line feed (i.e. skips to the next line). Is there any escape sequence that will go to the top of the next page? (2) I know that control L will clear the console. Is there an equivalent function or other means that can be used in R code to clear the console? Thanks, John John David Sorkin M.D., Ph.D. Chief, Biostatistics

I am trying to model data in which subjects are followed through time to determine if they fall, or do not fall. Some of the subjects fall once, some fall several times. Follow-up time varies from subject to subject. I know how to model time to the first fall (e.g. Cox Proportional Hazards, Kaplan-Meir analyses, etc.) but I am not sure how I can model the data if I include the data for those

I am using read.csv to read a CSV file (produced by saving an Excel file as a CSV file). The columns containing dates are being read as factors. Because of this, I can not compute follow-up time, i.e. Followup<-postDate-preDate. I would appreciate any suggestion that would help me read the dates as dates and thus allow me to calculate follow-up time. Thanks John John Sorkin M.D., Ph.D. Chief,

??I would like to apply a function, fract, to the columns of a dataframe. I tried the following apply(data5NonEventEpochs,2,fract) but, no surprise it did not work as apply works on matrices not data frames. How can I apply a fuction to the columns of a data frame? (I can't covert data5NonEventsEpochs to a matrix as it contains character data). Thank you, John John David Sorkin M.D., Ph.D.