similar to: Multivariable correlation

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

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 am looking for a package that will allow me to choose R (a real number) that minimizes the correlation of y and x^R, i.e. find R such that corr(y,x^R) is minimized. Any suggestions for packages I might look at would be helpful. Thanks, John John David Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology Baltimore VA

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

I have written an R function that produces multiple graphs. I use par(ask=TRUE) to allow for the inspection of each graph before the next graph is drawn. I am looking for a way to recall all graphs drawn in an R session, and a method that can be used to print all the graphs at one time. I know that I could simply print each graph after I inspect the graph, but this gets tiresome if one's

R 2.2.0 windows XP How can I perform a regression analyses using a vector of means, a variance-covariance matrix? I looked at the help screen for lm and did not see any option for using the afore mentioned structures as input to lm. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC, University of Maryland School of Medicine Claude D.

I am trying to drop rows of a dataframe based on values of the column PID, but my strategy is not working. I hope someoen can tell me what I am doing incorrectly. # Values of PID column > jdata[,"PID"] [1] 16608 16613 16355 16378 16371 16280 16211 16169 16025 11595 15883 15682 15617 15615 15212 14862 16539 [18] 12063 16755 16720 16400 16257 16209 16200 16144 11598 13594 15419 15589

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

R 2.2 on windows XP I have a dataset with multiple columns. Some of the columns represent independent variables, some represent dependent variables. I would like to run the same analyses on a fixed set of independent variables, changing only the dependent variable, e.g. y1-y2=x1+x2+x3 y3-y4=x1+x2+x3 y5-y6=x1+x2+x3, etc. I know I can write a function to perform the analyses, however in order to

Fellow Rers, Please forgive me if I have posted this to the wrong R list serve. Over the course of the years that I have used R and participated in this list server, I have noted a large number of questions and answers that direct people to specific packages. The multitude of packages is one of the great strengths of R. Unfortunately there is no (or at least I am not aware of) any single source

Please forgive a statistics question. I know that a simple bivariate linear regression, y=f(x) or in R parlance lm(y~x) can be solved using the variance-covariance matrix: beta(x)=covariance(x,y)/variance(x). I also know that a linear regression with multiple independent variables, for example y=f(x,z) can also be solved using the variance-covariance matrix, but I don't know how to do this.

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

Unbuntu 10.04 R 2.10 I am trying to install Rcmdr and receive the following messages: The downloaded packages are in ?/tmp/RtmpzhjDZG/downloaded_packages? Warning messages: 1: In install.packages("Rcmdr", dependencies = TRUE) : installation of package 'Rmpi' had non-zero exit status 2: In install.packages("Rcmdr", dependencies = TRUE) : installation of package

Fedora 10 R 2.8.1 I hope someone can tell me the meaning of error I received trying to install RODBC and how I can get around the problem. n.b. The error originally occurred then I was installing Rcmdr. I then tried to install RODBC separately and received the same error. checking sql.h usability... no checking sql.h presence... no checking for sql.h... no checking sqlext.h usability... no

Peter You question is difficult to answer without more information about the distribution of your residuals. Different residual patterns call for different transformations to stabilize the variance. One very common form of heterocedasticity is increasing variance with increasing values of an independent predictor, i.e. the variance of the residuals of y=x increase as x increases. In this case a

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),

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 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 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

Windows Vista R 2.10.1 Does the following use of groupedData and lme produce an analysis with both random intercept and slope, or only random slope? zz<-groupedData(y~time | Subject,data=data.frame(data), labels = list( x = "Time", y = "y" ), units = list( x = "(yr)", y = "(mm)") ) plot(zz)