similar to: Omnibus test for main effects in the face ofaninteraction containing the main effects.

R 2.9.1 Windows XP I am fitting a random effects ANOVA with two factors Group which has two levels and Time which has three levels: fita<-lme(Post~Time+factor(Group)+factor(Group)*Time, random=~1|SS,data=blah$alldata) I want to get the omnibus significance tests for each factor and the interaction. I believe I can get the omnibus test for the interaction by running the model:

Windows XP R2.8.1 I am running an balanced aov, in which Time has 4 levels, Group has 2 levels: fit2<-aov(Post ~ as.factor(Time) +as.factor(Group)+as.factor(Group*Time)+Error(SS), alldata) I would like to get a p value for each level of time, each level of Group and each level of Group*Time. How can I do this? I get this, which appears to be omnibus p values for the factors: Error: SS

I'm attempting to do model selection with AIC, using a glm and a lognormal distribution, but: fit1<-glm(BA~Year,data=pdat.sp1.65.04, family=gaussian(link="log")) ## gives the same result as either of the following: fit1<-glm(BA~Year,data=pdat.sp1.65.04, family=gaussian) fit1<-lm(BA~Year,data=pdat.sp1.65.04) fit1 #Coefficients: #(Intercept) Year2004 # -1.6341

I''ve noticed that the copy of Mongrel installed by the camping-omnibus gem doesn''t work with Ruby 1.8.6. Or to be more specific, cgi_multipart_eof_fix (which Mongrel is dependent upon) doesn''t work: > $ sudo gem install mongrel --source http://code.whytheluckystiff.net > > ERROR: Error installing mongrel: > cgi_multipart_eof_fix requires Ruby version

If one estimates a model using multinom, is it possible to perform the omnibus LR test ( the analogue to omnibus F in linear models ) using the output from multinom ? The residual deviance is there but I was hoping I could somehow pull out the deviance based on just using an intercept ? Sample code is below from the CAR book but I wasn't sure how to do it based on that example. Thanks

I am looking for an R package with which one can calculate an effect size for a set of contrasts given an omnibus chi-square test statistic (more than 1 degree of freedom). Is there such a package? Presumably, it would implement the procedure (or something like it) described by Roznow and Rosenthal 1996 Psychological Methods 1: 331-340. many thanks! Steven Orzack Fresh Pond Research Institute

Hello, I've been running some HLMs using the lme function quite happily; it does what I want and I'm pretty sure I understand it. The issue is that I'm currently trying to estimate a model with a 14-level "nusiance" factor as an independent variable...which makes the output quite ugly. All I'm really interested in is the question of whether these factor as a whole

Dear R-help, I'm very sorry to ask 2 questions in a week. I am using the package 'crr' and it does exactly what I need it to when I use the dataset a. However, when I use dataset b I get the following error message: Error in drop(.Call("La_dgesv", a, as.matrix(b), tol, PACKAGE = "base")) : system is computationally singular: reciprocal condition number =

Hello, I apologize for the post as I am certainly overlooking a simple solution to my difficulties with getting a legend to print on a survplot from the rms package. I am plotting the following: survplot(survest(fita), n.risk=T, conf='none', cex.n.risk=.85, dots=T, col='gray10', lty=2) survplot(survest(fit), n.risk=F, conf='none', add=T) survplot(survest(fitb), n.risk=F,

gem install camping-omnibus --source code.whytheluckystiff.net Considering the number of optional dependencies now in Camping, I''ve put together an eigengem which will load all of the recommended libraries that most devoted campers will want: ActiveRecord, Mongrel, SQLite3, RedCloth and the acts_as_versioned gem. (Those last two are used in the examples.) Stuff like mosquito and

I've been attempting to calculate the predictions from a poisson glm object, along these lines: predict(foo.glm, type = "response") and predict(foo.glm, type = "response", se.fit = TRUE) foo.glm is arrived at this way: foo.glm <- glm(Insects ~ Dad * Mum + Location, offset = log(MM), family = "poisson", data = model.df) There are two

This is very possibly not a question on R. I was under the impression that the argument that gives rise to Fisher's LSD method in ANOVA works in other situations with three-way comparisons too, given that formal logic works the same ("if the omnibus test rejects, only two of the three groups may be equal, and therefore only one hypothesis can be rejected falsely"). However, when I

I am trying to duplicate the example by Spencer Graves in the wiki, using lmer with the Nozzle data. http://wiki.r-project.org/rwiki/doku.php?id=guides:lmer-tests However the Chisq value and the fitAB values that are calculated are different compared to those in the example. I also get a warning message when I attempt the fitAB. Does anyone have any guidance as to why this might happen and how

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

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,

(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

R 2.1.1 Win 2k Would someone suggest a method (or methods) that can be used to determine ntile cutpoints of a vector, i.e. to determine values that can be used to divide a vector into thirds such as 0-33 centile, 34-66 centile, 67-100 centile. If for example I had a vector: 1,2,3,4,5,6,7,8,9 and wanted to divide the vector into thirds I would have cut-points of 3, and 6. Thanks, John John

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 want to place four plots on a page, and I would like to have all four plots share a common title. I have tried the following code, but the title is centered over the fourth graph and not centered across all four plots. Does anyone have any suggestions? R 2.1.1 windows xp oldpar<-par(mfcol =c(1,4),ask=TRUE) plot(p,varp) plot(p,SEp) plot(p,CVp) plot(p,ppval) title(paste("P and 95%CI

An apparatus exists whereby a collection of balls is displaced to the top of a stack by suction. A top level (Level 1) each ball is shifted 1 unit to the left or 1 unit to the right at random with equal probability. The ball then drops down to level 2. At Level 2, each ball is again shifted 1 unit to the left or 1 unit to the right at random. The process continues for 15 levels and the balls are