similar to: enhanced question / standardized coefficients

Hello everybody, a question that connect to the question of Frederik Karlsons about 'how to stand. betas' With the stand. betas i can compare the influence of the different explaning variables. What do i with the betas of factors? I can't use the solution of JohnFox, because there is no sd of an factor. How can i compare the influence of the factor with the influence of the numeric

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

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 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 R-Users, The following question is more of general nature than a merely technical one. Nevertheless I hope someone get me some answers. I have been using the mice package to perform the multiple imputations. So far, everything works fine with the standard regressions analysis. However, I am wondering, if it is theoretically correct to perform nonparametric models (GAM, spline

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

Dear R-Users, I used the box.tidwell () function of the car Package. So far everything is fine. However, if the number of dummy variables in the part not to be transformed (other.x formula) exceeds a certain level (around 70), I receive the following error message: Error in parse(file, n, text, prompt) : syntax error in What did I miss? And how can I solve this problem? I

Hi Simon, Yes, if I uses coefficients() I get the same results for lm() and lm.ridge(). So that's consistent, at least. Interestingly, the "wrong" number I get from lm.ridge()$coef agrees with the value from SPSS to 5dp, which is an interesting coincidence if these numbers have no particular external meaning in lm.ridge(). Kind regards, Nick ----- Original Message -----

Dear R-Users, I used the box.tidwell () function of the car Package. When I used the following formula: semi.sub.in.mi1.boxtidwell_h<-box.tidwell(RENT_LG ~ I(age+1)+I(age2+1)+X06A + I(X08B+1) + I(X22+1) + I(X24+1) + X31A, ~B_YEAR + C_X01 + C_X14 + C_X19 + C_X29A +C_X21 + C_X23 + D_X12 + D_X17 + D_X18 + D_X25 + D_X27 + D_X30 + D_X32 + D_X35, data = semi.sub.in.mi1) everything is

Dear R-Users, How can I put a pre-defined regression model into to an object of class lm in order to use the predict.lm function. A simplified example: I would normally run a regression analysis on a dataset, > germany<-lm(RENT~AGE1, in.mi01) > summary(germany) Call: lm(formula = RENT ~ AGE1, data = in.mi01) Residuals: Min 1Q Median 3Q Max

Hi, Here is (I hope) all the relevant output from R. > mean(s1$ZDEPRESSION, na.rm=T) [1] -1.041546e-16 > mean(s1$ZDIVERSITY_PA, na.rm=T) [1] -9.660583e-16 > mean(s1$ZMEAN_PA, na.rm=T) [1] -5.430282e-15 > lm.ridge(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)$coef ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA -0.3962254 -0.3636026

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Hello R-Users, I have the following problem with the subset function: See the following simple linear model. Here everything works fine: >germany<-lm(RENT~AGE1, in.mi01) However, if a use the same regression equation and only specify a subset, I get an error message: > berlin<-lm(RENT~AGE1, in.mi01, subset=C_X01=="Berlin") Error in lm.fit(x, y, offset

Hi John, Thanks for the comment... but that appears to mean that SPSS has a big problem. I have always been told that to include an interaction term in a regression, the only way is to do the multiplication by hand. But then it seems to be impossible to stop SPSS from re-standardizing the variable that corresponds to the interaction term. Am I missing something? Is there a way to perform the

Greetings all! Got another question for you: When doing regression, is there a way of automatically obtaining the standardized correlation coefficients? /Fredrik -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

Thanks, I was getting to try this, but got side tracked by actual work... Your analysis reproduces the SPSS unscaled estimates. It still remains to figure out how Nick got > coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)) (Intercept) ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA 0.07342198 -0.39650356

I would like to obtain Confidence Intervals for the estimates (unstandardized beta weights) of each predictor in a WLS regression: m1 = lm(x~ x1+x2+x3, weights=W, data=D) SPSS offers that output by default, and I am not able to find a way to do this in R. I read through predict.lm, but I do not find a way to get the CIs for multiple independent variables. Thank you Torvon [[alternative HTML

Hello, first of all: I'm new to R and have only used SPSS befor this (which can't do this at all...). I'm trying to output some regression results to latex. The regressions are normal OLS and I'm trying to output the results with standardized regression coefficients and t-statistics based on "Huber-White sandwich estimator for variance". The final result should be

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

Dear R-Users, Background: I have five multiple imputed datasets. For each datasets I have run a regression analysis and combined the regression coefficients according to Rubin (1987) rule. Problem: Now I want to use these combined regression coefficients on a different dataset (with the same variable names but different values) and check how good they can predict my dependent