similar to: Rsquared in summary(lm)

Dear Mac users, Hi, as you might have probably read the thread of "[R] Rsquared in summary(lm)" on May 10, it seems that Mac version of lm() seem to be working incorrectly. I enclose the script to produce the result both for lm() and manual calculation for a simple regression. Could you run the script and report with the version of R, so I don't have to go through every builds

Hi, I see a lot of folks verify the regression identity SST = SSE + SSR numerically, but I cannot seem to find a proof. I wonder if any folks on this list could guide me to a mathematical proof of this fact. Thanks. David. -- View this message in context: http://r.789695.n4.nabble.com/Regression-Identity-tp4636829.html Sent from the R help mailing list archive at Nabble.com.

Does anyone know of a literature reference, or a piece of code that can help me calculate the amount of variation explained (R2 value), in a regression constrained to have a slope of 1 and an intercept of 0? Thanks! Sebastian J. Sebastián Tello Department of Biological Sciences 285 Life Sciences Building Louisiana State University Baton Rouge, LA, 70803 (225) 578-4284 (office and lab.)

Hello, Is there any simple command to write the data.frame to the file in the format of command line input? For example, I read a table as > data <- read.table() and want to write out the 'data' object into my R file as data <- data.frame( some.column1=c( 1, 2 ), some.column2=c( 1, 2 ) row.names=c("row name1",

Hi, I am fitting a weighted least square regression and trying to compute SSE,SST and SSReg but I am not getting SST = SSReg + SSE and I dont know what I am coding wrong. Can you help please? xnam <-colnames(X) # colnames Design Matrix fmla1 <- as.formula(paste("Y ~",paste(xnam, collapse=

Hi, everybody, 3 questions about R-square: ---------(1)----------- Does R2 always increase as variables are added? ---------(2)----------- Does R2 always greater than 1? ---------(3)----------- How is R2 in summary(lm(y~x-1))$r.squared calculated? It is different from (r.square=sum((y.hat-mean (y))^2)/sum((y-mean(y))^2)) I will illustrate these problems by the following codes:

Is this how a t hypothesis test is done when I don't have the actual data, but just the summarized statistics: > #Homework 9.2.6 [1] > n<-31 > xbar<-3.10 > s_x<-1.469 > m<-57 > ybar<-2.43 > s_y<-1.35 > s_pooled<- (((n-1)*s_x^2) + ((m-1)*s_y^2)) / (n + m - 2) > s_pooled [1] 1.939521 > t_obs <- (xbar - ybar) / (s_pooled * (sqrt(1/n + 1/m)))

Hello R-help, I am using R version 2.15.1. I upgraded from R version 2.13 a few months back. Previously, I was able to plot error bars on an xy scatter plot using the errbar function: errbar(RAEthylene$TIME,RAEthylene$AVE,RAEthylene$AVE+RAEthylene$STD,RAEthylene$AVE-RAEthylene$STD,add = TRUE,lty=2,pch=17); Today, I went to update my plot. However, in R version 2.15.1 I get error code

Hello R Users, How to compute cross correlation between two time series. Data is in ASCII format. I am using R on windows. Many thanks, Regards, Yogesh [[alternative HTML version deleted]]

Hi everyone, I have a question about calculating r-squared in R. I have tried searching the archives and couldn't find what I was looking for - but apologies if there is somewhere I can find this... I carried out a droughting experiment to test plant competition under limited water. I had: - 7 different levels of watering treatment (1 -7 - from most watered to least watered/) - 15

Hi people, I want to obtain the adjusted r-square given a set of coefficients (without the intercept), and I don't know if there is a function that does it. Exist???????????????? I know that if you make a linear regression, you enter the dataset and have in "summary" the adjusted r-square. But this is calculated using the coefficients that R obtained,and I want other coefficients

Hi, when doing as.double() on an object that is already a double, the object seems to be copied internally, doubling the memory requirement. See example below. Same for as.character() etc. Is this intended? Example: % R --vanilla > x <- double(1e7) > gc() used (Mb) gc trigger (Mb) max used (Mb) Ncells 234019 6.3 467875 12.5 350000 9.4 Vcells 10103774 77.1

Hello R-users, I've been looking for a function performing one and two sample Hotelling test for testing equality of mean vectors. Has anyone implemented such a function in R? thanks a lot, Bill ============== Bill Donner Statistician

hi all a technical question for those bright statisticians. my question involves ridge regression. definition: n=sample size of a data set X is the matrix of data with , say p variables Y is the y matrix i.e the response variable Z(i,j) = ( X(i,j)- xbar(j) / [ (n-1)^0.5* std(x(j))] Y_new(i)=( Y(i)- ybar(j) ) / [ (n-1)^0.5* std(Y(i))] (note that i have scaled the Y matrix as well) k is

Hey I'm not really sure what I should put on here, but I am having trouble with my R code. I am trying to get the p-values, R^2s etc for a number of different groups of variables that are all in one dataset. This is the code: #Stand counter st<-1 #Collections stands<-numeric(67) slopes<-numeric(67) intercepts<-numeric(67) mses<-numeric(67) rsquares<-numeric(67)

How can I extract the adjusted R^2 value from an ols object (using rms package)? library(rms) x <- rnorm(10) y <- x + rnorm(10) ols1 <- ols(y ~ x) Typing "ols1" displays adjusted R^2 among other things, but how can I assign it to a variable? I tried str(ols1) but couldn't see where to go from there. Thanks, Mark Seeto

Using hccm() I got a heteroscedasticity correction factor on the diagonal of the return matrix, but I don't know how to incorporate this into my linear model: METHOD 1: > OLS1 <- lm(formula=uer92~uer+low2+mlo+spec+degree+hit) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.0623377 0.0323461 -1.927 0.057217 . uer 0.2274742 0.0758720

Dear all: I'm trying to fit the optimal Box-Cox transformation related to nls (see the code below) for the demand of money data in Green (3th Edition) but in the last step R gives the next error message. Error en `[.data.frame`(eval(object$data), , as.character(formula(object)[[2]])[2]) : undefined columns selected. ?Any idea to solve the problem? Thanks in advance,

Hello ALL! I have a problem to plot factor (lets say gender) as a line, or at least both line and point, from ols model: ols1 <- ols(Y ~ gender, data=dat, x=T, y=T) plot(ols1, gender=NA, xlab="gender", ylab="Y", ylim=c(5,30), conf.int=FALSE) If I convert gender into discrete numeric predictor, and use forceLines=TRUE, plot is not nice and true, since it shows values

Please excuse me for asking such basic questions: Here is my code > y=c(1.21,0.51,0.14,1.62,-0.8,0.72,-1.71,0.84,0.02,-0.12) > ybar=mean(y) > ll=length(y); > ybarv=rep(ybar,ll) > x=1:ll > plot(x,ybarv,pch=1) > segments(x[1],ybar,x[ll],ybar) What I get is a collection of small circles, with a segment "on top" of the circles, which is almost what I want. But I