similar to: Accounting for within family correlation in genetic analysis

R 2.5 windows XP I am getting an error from glm() that I don't understand. Any help or suggestions would be appreciated. N.B. 1<=AAMTCAREJ<=327900 > summary(data$AAMTCAREJ) Min. 1st Qu. Median Mean 3rd Qu. Max. 1.0 404.3 1430.0 6567.0 5457.0 327900.0 > fitglm<-glm(AAMTCAREJ~sexcat+H_AGE+SmokeCat+InsuranceCat+MedicadeCat+ +

Hello all: A question from a new user. I have data on a geo-referenced curvilinear grid. This is a grid with 75 rows and 51 columns, is not aligned north-south, and the rows and columns are not straight. (And the coordinates are in meters.) I want to make image plots of this data, but where the grid is deformed according to the correct locations of the grid points, instead of coming out

Dear useRs, I'm trying to simply fill in the area under a curve using RGL. Here' the set up: x <- c(0.75,75.75,150.75,225.75,300.75,375.75,450.75,525.75,600.75,675.75, 0.5,50.5,100.5,150.5,200.5,250.5,300.5,350.5,400.5,450.5, 0.25,25.25,50.25,75.25,100.25,125.25,150.25,175.25,200.25,225.25) y <- c(0.05,4.91,9.78,14.64,19.51,24.38,29.24,34.11,38.97,43.84,

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),

Can anyone suggest a good non-parametric test, and an R implementation of the test, that allows for two or more factors? Wilcoxon signed rank allows for only one. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC and University of Maryland School of Medicine Claude Pepper OAIC University of Maryland School of Medicine Division of

R 2.7.2 Windows XP I am using apply to compute a series of Student's t-test from two matrices, sample1 and sample2. boo<-apply(sample1,1,t.test,sample2) I want to pick of the p-values from the tests, but can't seem to get it to work. I have tried several methods to get the values including: boo<-apply(sample1,1,t.test$t.test,sample2) boo<-apply(sample1,1,t.test,sample2)$t.test

Is there any way to load a package using R code rather than the Load Package command in the Package menu? I would like to load the boot package. R.2.1.0 Patched Win 2k. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC and University of Maryland School of Medicine Claude Pepper OAIC University of Maryland School of Medicine Division of

I am having trouble with apply. Could someone suggest changes to the code below that will eliminate the for loops? R 2.1.1 patched Windows 2k Thanks, John function () { NaCon<-array(dim=c(10,10)) for (i in 1:10) {NaCon[i,]<-rnorm(10,10.77,0.02)} MeanNaCon<-vector(mode="numeric",length=10) for (j in 1:10) {MeanNaCon[j]<-mean(NaCon[j,1:10])} print(MeanNaCon)

I am trying to fit lme using R 2.1.1 under Windows 2k. I am getting the following message noted below. Any suggestions that would help me correct my error would be greatly appreciated. Thanks, John > fit1lme<-lme(Velocity~time,data=gate) Error in getGroups.data.frame(dataMix, groups) : Invalid formula for groups John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics

R 2.1.1 on Windows 2K I hope one and all will allow both an R question and a general stats question: (1) Is there any function that will perform non-parametric tests such as a sign test or a signed rank test? I know I could program both but I would prefer not re-inventing the wheel. (2) When performing a sign test, the usual practice is to drop zero values. Is there any theoretical reason

R 2.3.1 Windows XP Question: How can I obtain the values of BIAS and STD. ERROR from a bootstrap. Background: I am running a bootstrap: result2<-boot(1:400,regSEvssample,R=5000) and obtain the following results: Bootstrap Statistics : original bias std. error t1* 1.876602 -0.0001368616 0.1630380 I would like to get the values of ORIGINAL and BIAS. I can get the value of

Given a square matrix of variables X, is there any way to compute a multivariable correlation among all the variables in the array? It is possible to calculate the correlation of all pairs of variables in the array, but I want to know the correlation of all the variables taken together, i.e. for the matrix X=x1 x2 x3 x4 x5 x6 x7 x8 x9 I don't want the pair-wise correlations

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) }

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

Is there an R function that can be used to calculate a sample size for a linear regression: i.e. Given a linear regression: y=f(x,z) i.e. fit1<-lm(y~x+z) is there a function which can be passed coef(z), SE(z), and other parameters to determine a sample size based on z, SEz and other information from the regression? Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics

I think I am doing something wrong when I try to bootstrap R square obtained from lm. My code is included below. No matter how many times I run the simulation, I always get exactly the same result, the bias and std.error are always zero. I would think that these values should be non-zero. I would appreciate any suggestions as to what I am doing wrong, or perhaps what I fail to understand. R 2.1.0

Can someone tell me why apply (and apply) are faster in performing repeated operations than a for (or do) loop? I am looking for a technical explanation. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC and University of Maryland School of Medicine Claude Pepper OAIC University of Maryland School of Medicine Division of Gerontology

Timothy, I believe you are mistaken. Fisher's exact test give the correct answer even in the face of small expected values for the cell counts. Pearson's Chi-square approximates Fisher's exact test and can give the wrong answer when expected cell counts are low. Chi-square was developed because it is computationally "simple". Fisher's exact test, particularly with tables

At the suggestion of many people, I have installed emacs on my linux (Fedora 8.0) computer with the intention of using emacs as window interface to R (2.6.0). I have gone though the emacs tutorial and don't see any information about how I should use emacs to run R. Can anyone suggest a document that I might read? In the past I have used R on a Windows XP system and used the built-in windowing

I am trying to install R.2.1.1 under Linux FC4 using RPM. During the installation I get the following message: The following package could not be found on your system. Installation cannot continue until it is installed. The package in question of libblas.so.3. Does anyone know where I can get the package? Can I use RPM to intall the package? Thanks, John John Sorkin M.D., Ph.D. Chief,