similar to: lm function

Dear all R users, Can anyone tell me how I can get estimate of SE of coefficients from, lm() function? I tried following : x = 1:10 lm(x[-1]~x[-10]-1)$coefficients Here I got the est. of coefficient, however I also want to get some "automated" way to get estimate of SE. Regards, [[alternative HTML version deleted]]

Full_Name: Christian Lajaunie Version: 2.5.1 OS: Fedora fc6 Submission from: (NULL) (193.251.63.39) confint() does not use the appropriate variance term when the design matrix contains a zero column (which of course should not happen). Example: A 10x2 matrix with trivial column 1: > junk <- data.frame(x=rep(0,10), u=factor(sample(c("Y", "N"), 10, replace=T))) The

Hello, I'd like to ask for the exact definition of the t-value, which R uses in its summaries of a linear model for judging the importance of an independent variable in explaining the dependent variable. I searched the documentation, some groups, and the web for quite a long time, but the best I could come up with is the following from www.answers.com/topic/value which reads: Measure of

Dear R helpers I have following data stocks <- c("ABC", "DEF", "GHI", "JKL") prices_df <- data.frame(ABC = c(17,24,15,22,16,22,17,22,15,19),                                          DEF = c(22,28,20,20,28,26,29,18,24,21),                                           GHI = c(32,27,32,36,37,37,34,23,25,32),                                          

Hi all: I have a question about lm on t-test. data(sleep) I wanna perform t-test to test the difference between the 2 groups: I can use: t.test(extra~group) The t.test result shows that:t = -1.8608; mean1=0.75,mean2=2.33 But I still wanna use: summary(lm(extra~group)) Intercept=0.75,which is mean1,just the same as t.test. group2=1.58 means the difference of the 2 groups,so

I tried multiple imputation with aregImpute() and fit.mult.impute() in Hmisc 3.8-3 (June 2010) and R-2.12.1. The warning message below suggests that summary(f) of fit.mult.impute() would only use the last imputed data set. Thus, the whole imputation process is ignored. "Not using a Design fitting function; summary(fit) will use standard errors, t, P from last imputation only. Use

Hi, I would like to extract the coefficients of a logistic regression (estimates and standard error as well) in lrm as in glm with summary(fit.glm)$coef Thanks David

Hello, I am work with a linear regression model: y=ax+b with the function of lm. y= observed migration distance of butterflies x= predicted migration distance of butterflies Usually the result will show if the linear term a is significantly different from zero based on the p-value. Now I would like to test if the linear term is significantly different from one. (because I want to know

Hi, I am trying to replicate Lambert (1992)'s simulation with zero-inflated Poisson models. The citation is here: @article{lambert1992zero, Author = {Lambert, D.}, Journal = {Technometrics}, Pages = {1--14}, Publisher = {JSTOR}, Title = {Zero-inflated {P}oisson regression, with an application to defects in manufacturing}, Year = {1992}} Specifically I am trying to recreate Table 2. But my

Hi R users I used R to get the results of a linear regression reg<-lm(y~x) here are the results: # Call: # lm(formula = donnees$txi7098 ~ donnees$txs7098) # # Residuals: # Min 1Q Median 3Q Max #

Hello Is there any function to calculate White's standard errors in R in an NLS regression. The sandwich and car package do it but they need an lm object to calculate the error's. Does anyone have idea how to do it for an NLS object ? Regards The woods are lovely, dark and deep But I have promises to keep And miles before I go to sleep And miles before I go to sleep ----- [[alternative

Dear All,   is there a function in R that would help me convert a covariance matrix built based on arithmetic returns to a covariance matrix from log-returns?   As an example of the means and covariance from arithmetic:   mu <-c(0.094,0.006,1.337,1.046,0.263) sigma

Hello, I am modeling some survival data wih cph (Design). I have modeled a predictor which showed non linear effect with restricted cubic splines. I would like to retrieve the se(coef) for other, linear, predictors. This is just to make nice LateX tables automatically. I have the coefficients with coef(). How do I do that? Thanks, David Biau. [[alternative HTML version deleted]]

Hi, I am using R interactively for estimation and even data manipulation. I want to use lag() function within lm() function in a way looks like: mymodel<- lm(y~x+I(lag(y,-1)-1, data=anydata) What I get is always perfect fit; (that is, coefficient=1) which is not true. If the model looks like mymodel<- lm(y~z+x+I(lag(x,-1)-1, data=anydata) summary returns only one coefficient for x and

Hi All, I have tried every fix on my try or tryCatch that I have found on the internet, but so far have not been able to get my R code to continue with the "for loop" after the lmer model results in an error. Here is two attemps of my code, the input is a 3D array file, but really any function would do.... metatrialstry<-function(mydata){ a<-matrix(data=NA, nrow=dim(mydata)[3],

A copy of this question can be found on Cross Validated: https://stats.stackexchange.com/questions/645362 I am estimating a system of seemingly unrelated regressions (SUR) in R. Each of the equations has one unique regressor and one common regressor. I am using `gmm::sysGmm` and am experimenting with different weighting matrices. I get the same results (point estimates, standard errors and

Dear R-Helpers, I want to compare the results of outputs from glmmPQL and lmer analyses. I could do this if I could extract the coefficients and standard errors from the summaries of the lmer models. This is easy to do for the glmmPQL summaries, using > glmm.fit <- try(glmmPQL(score ~ x*type, random = ~ 1 | subject, data = df, family = binomial), TRUE) > summary(glmmPQL.fit)$tTable

Hi, I am writing a basic function to extract the z scores for some linear regression coefficients: zscore<-function( y, x) { lm<-lm( y ~ x ) z <- coef(lm)/sqrt(diag(vcov(lm))) return(z) } I would like to pass a dataframe to the function as a argument so the function call changes from zscore(df$y1,df$x1) to zscore(y1,x1,data=df) but I am not sure how to reference the data

Hi, I have 20*60 data matrix (with some NAs) and I wish to perfom a Pearson correlation coefficient matrix as well as simple linear regression equation and coefficient of determination (R2) for every possible combination. Any tip/idea/library/script how do to so. Thanks, As hz -- View this message in context:

Hi, I'm trying to calculate 95% confidence interval of C statistic of logistic regression model using rcorr.cens in rms package. I wrote a brief function for this purpose as the followings; CstatisticCI <- function(x) # x is object of rcorr.cens. { se <- x["S.D."]/sqrt(x["n"]) Low95 <- x["C Index"] - 1.96*se Upper95 <- x["C