similar to: regression for several responses

Hello, I tried to install the "car" package but I can't solve the dependencies. car needs grid, lattice and dr but when I try to install grid I get this error prompt: wz3x64:/home/oggi/R/software/contrib # rpm -i R-car-1.0.R3-1.i386.rpm error: failed dependencies: R-grid is needed by R-car-1.0.R3-1 R-lattice is needed by R-car-1.0.R3-1

Hello, I have a series of questions that I hope will be simple to answer. Basically I would like a code to do the following so that I can compute the distribution free test for the slope of a postulated regression line (Theil test). As I am testing the null hypothesis that slope = 0 against the general alternative the slope does not equal 0, it should be pretty straight-forward. I have a data

hello I tried to use lars() but neither with my own data nor with the sample data it works. I get in both cases the following error prompt: > data(diabetes) > par(mfrow=c(2,2)) > attach(diabetes) > x<-lars(x,y) Error in one %*% x : requires numeric matrix/vector arguments > x<-lars(x,y, type="lasso") Error in one %*% x : requires numeric matrix/vector arguments

I've written a simple function to draw a regression line in a plot and annotate the line showing the slope with a label. It works, as I'm using it, when the horizontal variable is 'x', but gives incorrect results otherwise. What's wrong? # simple function to show the slope of a line show.beta <- function(model, x="x", x1, x2, label, col="black", ...)

Hello: In the following plot, the fitted line plots 100 percent above the points: tstDat <- data.frame(x=10^(1:3), y=10^(1:3+.1*rnorm(3))) tstFit <- lm(log(y)~log(x), tstDat) plot(y~x, tstDat, log='xy') abline(tstFit) I can get the correct line with the following: tstPredDat <- data.frame(x=10^seq(1, 3, len=2)) tstPred <- predict(tstFit, tstPredDat)

Hello R user, I have several data frames with >100 columns and I did a linear regression over time of each column df1.lm <- lapply(df1, function(x) lm(x~year)$coeff[2]) that worked fine and I get slope of each column oder time - until I divided df1 by df2 df3 <- df1/df2 > df3.lm <- lapply(df3, function(x) lm(x~year)$coeff[2]) Error in lm.fit(x, y, offset = offset, ...) :

Hello, I would like to write a loop to 1) run 100 linear regressions, and 2) compile the slopes of all regression into one vector. Sample input data are: y1<-rnorm(100, mean=0.01, sd=0.001) y2<-rnorm(100, mean=0.1, sd=0.01) x<-(c(10,400)) #I have gotten this far with the loop for (i in 1:100) { #create the linear model for each data set model1<-lm(c(y1[i],y2[i])~x)

Not really r-specific: Z = (b1 - b2) / SQRT ( SEb1^2 + SEb2^2) -------Original Message------- From: Gijsbert Stoet <stoet at volition.wustl.edu> Sent: 07/20/03 09:51 PM To: r-help at stat.math.ethz.ch Subject: [R] how to test whether two slopes are sign. different? > > Hi, suppose I do want to test whether the slopes (e.g. determined with lsfit) of two different population are

Hello, We've got a dataset with several variables, one of which we're using to split the data into 3 smaller subsets. (as the variable takes 1 of 3 possible values). There are several more variables too, many of which we're using to fit regression models using lm. So I have 3 models fitted (one for each subset of course), each having slope estimates for the predictor variables.

Hi, I'm trying to get confidence intervals to slopes from a linear model and I can't figure out how to get at them. As a cut 'n' paste example: ################# # dummy dataset - regression data for 3 treatments, each treatment with different (normal) variance x <- rep(1:10, length=30) y <- 10 - (rep(c(0.2,0.5,0.8), each=10)*x)+c(rnorm(10, sd=0.1), rnorm(10,

Hello R users, I've used the following help two compare two regression line slopes. Wanted to test if they differ significantly: Hi, I've made a research about how to compare two regression line slopes (of y versus x for 2 groups, "group" being a factor ) using R. I knew the method based on the following statement : t = (b1 - b2) / sb1,b2 where b1 and b2 are the two slope

Hi, I was hoping someone could help figure out how to write code for R to do the below. I have data that looks like below. Variables, sid and pid are strings, slope is numeric. I need R to get me the mean of slopes for all pid's nested within each sid if there are more than one pid's nested within sid. If there is only pid for a sid, like for 2.1 below, I want R to write a 0. In the

Hi All, I am confused with SMATR's test for common slope. My null hypothesis here is that all slopes are parallel (common slopes?), right? So if I get a p value < 0.05 means that we can have confidence to reject it? That slopes are different? Or the other way around? it means that we have statistical confidence that the slopes are parallel? thanks -- Eugenio Larios PhD Student University

Hello R community, I have a question regarding correlation and regression analysis. I have two variables, x and y. Both have a standard deviation of 1; thus, correlation and slope from the linear regression (which also must have an intercept of zero) are equal. I want to probe two particular questions: 1) Is the slope significantly different from zero? This should be easy with the lm

Hi, How would one go about determining if the slope terms from an analysis of covariance model are different from eachother? Based on the example from MASS: library(MASS) # parallel slope model l.para <- lm(Temp ~ Gas + Insul, data=whiteside) # multiple slope model l.mult <- lm(Temp ~ Insul/Gas -1, data=whiteside) # compare nested models: anova(l.para, l.mult) Analysis of Variance

I am interested in whether the slopes in a linear model are different from 0. I.e. I would like to obtain the slope estimates, and their standard errors, ``relative to 0'' for each group, rather than relative to some baseline. Explicitly I would like to write/represent the model as y = a_i + b_i*x + E i = 1, ..., K, where x is a continuous variate and i indexes groups (levels of a

Dear R users I use Windows XP, R2.5.1 (I have read the posting guide, I have contacted the package maintainer first, it is not homework). In a research project on renal cell carcinoma we want to compute Harrell's c index, with optimism correction, for a multivariate Cox regression and also for some univariate Cox models. For some of these univariate models I have encountered an error

Dear R users, experimenting with the lm function in R, I've encountered some behaviour I don't understand with my limited knowledge of regression. I made a data-'set' of three measurements (see syntax below). Using lm (linear model) to fit the regression-line, I expected to find an intercept of 2.0 and a slope of 0, but in fact the slope is slightly below zero. Amazed by

I have not really used lm before and I was hoping for some help on a simple problem. Here is a toy version of the problem I want to solve. y x grp -.9 1 a -.8 2 a -.7 3 a -.7 1.5 b -.5 2.5 b -.3 3.5 b -.19 2.7 c -.11 3.7 c -.41 4.7 c I want to fit a model that has one y-intercept and three slopes, one for

Hi I have (8 measures * 96 groups) = 768 datasets for which I did linear regressions using lm(). Now I want to compare the slopes for each of the 8 measures in each of the 96 groups. As I understand , I can not use > anova(lm1, ..., lm8) as the lm1 ... lm8 are based on different datasets. I also read in previous discussions in this list, that I can see if the slope +- stddev(slope)