similar to: Confidence intervals for estimates of all independent variables in WLS regression

I run 9 WLS regressions in R, with 7 predictors each. What I want to do now is compare: (1) The strength of predictors within each model (assuming all predictors are significant). That is, I want to say whether x1 is stronger than x2, and also say whether it is significantly stronger. I compare strength by simply comparing standardized beta weights, correct? How do I compare if one predictor is

Dear All, I've been trying to run a Weighted Least Squares (WLS) regression: Dependent variables: a 60*200 matrix (*Rit*) with 200 companies and 60 dates for each company Independent variables: a 60*4 matrix (*Ft*) with 4 factors and 60 dates for each factor Weights: a 60*200 matrix (*Wit*) with weights for 200 companies and 60 dates for each company The WLS regression I would like to run

How does one specify the weights for WLS in the systemfit command ? That is, there is a weight option in lm(), but there doesn't seem to be weight option for systemfit("WLS") Thanks!

I might be sorry for asking this question :-) I have two equations and I tried to estimate them individually with "lm" and "gls", and then in a system (using systemfit) with "OLS", "WLS" and "SUR". Quite surprisingly (for myself at least) the results are identical to the last digit. Could someone (please!) give a hint as to what am I

Hi, I'm generating the name of the variable with paste function and then using that variable name further to get the specific position value from the data.frame, here is the snippet from my code: modelResults <- extractModelParameters("C:/PilotStudy/Mplus_Input/Test", recursive=TRUE) #extractModelParameters reads all the output files from the Test folder and create the

We aim to visualize a 17*17 correlation matrix with the package *qgraph*, consisting of 16 variables. Without variable names in the input file, that works perfectly R> qgraph(data) but we'd like variable names instead of numbers for variables. In a correlation matrix, the first row and the first column usually have variable names. We've been unsuccessful so far to read such a file

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 all, I understand that gls() uses generalized least squares, but I thought that maybe optimum weights from gls might be used as weights in lm (as shown below), but apparently this is not the case. See: library(nlme) f1 <- gls(Petal.Width ~ Species / Petal.Length, data = iris, weights = varIdent(form = ~ 1 | Species)) aa <- attributes(summary(f1)$modelStruct$varStruct)$weights f2 <-

I have two regressions to perform - one with a metric DV (-3 to 3), the other with an ordered DV (0,1,2,3). Neither normal distribution not homoscedasticity is given. I have a two questions: (1) Some sources say robust regression take care of both lack of normal distribution and heteroscedasticity, while others say only of normal distribution. What is true? (2) Are there ways of using robust

I need to report exact p-values in my dissertation. Looking at my lm() results of many regressions with huge datasets I have the feeling that p-values are rounded to the smallest value of "2e-16", because this p-value is very common. Is that true or just chance? If it is true, how do I obtain the "true" unrounded p-values for these regressors? m1 <- lm(y ~ x1+x2+x3+4+x5,

R 1.6.2 for Windows, Win2k: I have fitted a weighted least squares model using the code "wls.out <- gls(y ~ x1 + x2 + x3 + x4 + x5 + x6 - 1, data = foo.frame, weights = varConstPower(form = ~ fitted(.), fixed = list(power = 0.5), const = 1))" The data has 62 rows and the response is zero when the covariates are zero. The purpose of the model was to account for the the fact that

Dear useRs, I have written a function that implements a Bayesian method to compare a patient's score on two tasks with that of a small control group, as described in Crawford, J. and Garthwaite, P. (2007). Comparison of a single case to a control or normative sample in neuropsychology: Development of a bayesian approach. Cognitive Neuropsychology, 24(4):343?372. The function (see

I have a WLS regression with 1 dependent variable and 3 independent variables. I wish to constrain the predicted values (the fitted values) so that they are greater than zero (i.e. they are positive). I do not know how to impose this constraint in R. Please respond if you have any suggestions. There are some previous postings about constraining the coefficients, but this won't accomplish

Dear list members, In sem, std.coef() will give me standardized coefficients from a sem model. But is there a trick so that path.diagram can use these coefficients rather than unstandardized ones? Thanks Steve Powell From: John Fox <jfox_at_mcmaster.ca> Date: Wed 28 Feb 2007 - 14:37:22 GMT Dear Tim, See ?standardized.coefficients (after loading the sem package). Regards, John John

Hi, I load my data set and separate it as folowing: presu <- read.table("C:/_Ricardo/Paty/qtdata_f.txt", header=TRUE, sep="\t", na.strings="NA", dec=".", strip.white=TRUE) dep<-presu[,3]; exo<-presu[,4:92]; Now, I want to use it using the wls and quantreg packages. How I change the data classes for mle and rq objects? Thanks a lot,

Hi all, I'd like to fit a multivariate regression with the variance of the error term porportional to the predictors, like the WLS in the univariate case. y_1~x_1+x_2 y_2~x_1+x_2 var(y_1)=x_1*sigma_1^2 var(y_2)=x_2*sigma_2^2 cov(y_1,y_2)=sqrt(x_1*x_2)*sigma_12^2 How can I specify this in R? Is there a corresponding function to the univariate specification lm(y~x,weights=x)??

Hi everyone, I just conducted a structural equation model for estimating a response model. This model should predict the probability that someone is responding to a direct mailing. I used the sem package for this. When I have my coefficients I want to know how well my model predicts the probability of response. How can I calculate these probabilities? I tried to use the unstandardized

Hi all, is it possible to run factor analysis in R such that the routine returns - unstandardized factor scores (according to the original scale) - rotated factor scores (these may be standardized) So far I have only found the possibility to return standardised unrotated factor scores. Thank you very much, Ursula ==================================================== NFO Infratest Ursula

Hello - Does anybody know how to get the SEM package in R to return standardized path coefficients instead of unstandardized ones? Does this involve changing the covariance matrix, or is there an argument in the SEM itself that can be changed? Thank you, Tim [[alternative HTML version deleted]]

I attempted to contact Drew Tyre, but the email I have for him is no longer in service. Hopefully someone can help. I'm using obs.error in R to model turtle occupancy in wetlands. I have 4 species and 20 possible patch and landscape variables, which I've been testing in smaller groups. > zib.out<-obs.error(y=painted,m=numvis,bp=zvars,pcovar=7) I get the following error