similar to: A Contrast Question

Hello, I can`t figure out how can increase the velocity of the fitting data by nls. I have a long data .csv I want to read evry time the first colunm to the other colunm and analisy with thata tools setwd("C:/dati") a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F) for (i in 1:dim(a[[2]]]) { #preparazione dati da analizzare

I have a model (run with glm) that has a factor, type. Type has two levels, "general" and "regional". I am trying to get estimates (and SEs) for the model with type="general" and type ="regional" using fit.contrast but I can't get the syntax of the coefficients to use in fit.contrast correct. I hope someone can show me how to use fit.contrast, or some

> On Oct 22, 2017, at 6:04 AM, Sorkin, John <jsorkin at som.umaryland.edu> wrote: > > I have a model (run with glm) that has a factor, type. Type has two levels, "general" and "regional". I am trying to get estimates (and SEs) for the model with type="general" and type ="regional" using fit.contrast ?fit.contrast No documentation for

> On Oct 22, 2017, at 3:56 PM, Sorkin, John <jsorkin at som.umaryland.edu> wrote: > > David, > Thank you for responding to my post. > > Please consider the following output (typeregional is a factor having two levels, "regional" vs. "general"): > Call: > glm(formula = events ~ type, family = poisson(link = log), data = data, > offset =

David, Thank you for responding to my post. Please consider the following output (typeregional is a factor having two levels, "regional" vs. "general"): Call: glm(formula = events ~ type, family = poisson(link = log), data = data, offset = log(SS)) Deviance Residuals: Min 1Q Median 3Q Max -43.606 -17.295 -4.651 4.204 38.421 Coefficients:

> On Oct 22, 2017, at 5:01 PM, Sorkin, John <jsorkin at som.umaryland.edu> wrote: > > David, > Again you have my thanks!. > You are correct. What I want is not technically a contrast. What I want is the estimate for "regional" and its SE. There needs to be a reference value for the contrast. Contrasts are differences. I gave you the choice of two references

David, Again you have my thanks!. You are correct. What I want is not technically a contrast. What I want is the estimate for "regional" and its SE. I don't mind if I get these on the log scale; I can get the anti-log. Can you suggest how I can get the point estimate and its SE for "regional"? The predict function will give the point estimate, but not (to my knowledge)

David, predict.glm and se.fit were exactly what I was looking for. Many thanks! John John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax)

Hi all, This is probably going to come off as unnecessary (and show my ignorance) but I am trying to understand the parameter estimates I am getting from R when doing an ANCOVA. Basically, I am accustomed to the estimate for the categorical variable being equivalent to the respective cell means minus the grand mean. I know is the case in JMP - all other estimates from these data match the

Hello, R experts, If I understand Ted's anwser correctly, then I can not contrast the mean yields between sections 1-8 and 9-11 under "Trt" but I can contrast mean yields for sections 1-3 and 6-11 because there exists significant interaction between two factors (Trt:section4, Trt:section5). Could I use the commands below to test the difference between sections 1-3 and 6-11 ?

#I am trying to understand how R fits models for contrasts in a #simple one-way anova. This is an example, I am not stupid enough to want #to simultaneously apply all of these contrasts to real data. With a few #exceptions, the tests that I would compute by hand (or by other software) #will give the same t or F statistics. It is the contrast estimates that R produces #that I can't seem to

Hi Folks, I'm comparing some output from R with output from SPSS. The coefficients of the independent variables (which are all factors, each at 2 levels) are identical. However, R's Intercept (using default contr.treatment) differs from SPSS's 'constant'. It seems that the contrasts were set in SPSS using /CONTRAST (varname)=Simple(1) I can get R's Intercept to match

I've been playing around with contrasts in lm by specifying the contrasts argument. So, I want to specify a specific contrast to be tested Say: > y _ rnorm(100) > x _ cut(rnorm(100, mean=y, sd=0.25),c(-3,-1.5,0,1.5,3)) > reg _ lm(y ~ x, contrasts=list(x=c(1,0,0,-1))) > coef(reg)[2] x1 -1.814101 I was surprised to see that I get a different estimate for the

Hi Folks, Given a series of n independent Bernoulli trials with outcomes Yi (i=1...n) and Prob[Yi = 1] = Pi, I want P = Prob[sum(Yi) = r] (r = 0,1,...,n) I can certainly find a way to do it: Let p be the vector c(P1,P2,...,Pn). The cases r=0 and r=n are trivial (and also are exceptions for the following routine). For a given value of r in (1:(n-1)), library(combinat) Set <- (1:n)

Hello, from Verzani, simpleR (pdf), p. 80, I created the following function to test the coefficient of lm() against an arbitrary value. coeff.test <- function(lm.result, var, coeffname, value) { # null hypothesis: coeff = value # alternative hypothesis: coeff != value es <- resid(lm.result) coeff <- (coefficients(lm.result))[[coeffname]] # degrees of freedom = length(var) -

Greetings! I suspect that there is an error in the code for the function ci.pd() in the Epi package. This function is for computing confidence intervals for a difference of proportions between two independent groups of 0/1 responses, and implements the Newcombe ("Nc") method and the Agrasti-Caffo "AC" method. I think there is an error in the computation for the Newcombe

Hello, I use a type III anova ("car" package) to analyse an unbalanced data design. I have two factors and I would have the effect of the interaction. I read that the result could be strongly influenced by the contrasts. I am really not an expert and I am not sure to understand indeed about what it is... Consequently, I failed to properly used the fit.contrast function (gregmisc

Hello. I am getting results from sem that are not correct (that's assuming that the results from my AMOS 4.0 software are correct). sem does not vary some of the parameters substantially from their starting values, and the final estimates of those parameters as well as the model chisquare value are incorrect. I've attached some code that replicates the problem. The parameters in

Hi Folks, I'm using the 'norm' package (based on Shafer's NORM) on some data. In outline, (X,Y) are bivariate normal, var(X)=0.29, var(Y)=24.4, cov(X,Y)=-0.277, there are some 900 cases, and some 170 values of Y have been set "missing" (NA). The puzzle is that, repeating the multiple imputation starting from the same random seed, I get different answers from the repeats

Dear all, I am trying to calculate the fitted values using a ridge model (lm.ridge(), MASS library). Since the predict() does not work for lm.ridge object, I want to get the fitted_value from the coefficients information. The following are the codes I use: fit = lm.ridge(myY~myX,lambda=lamb,scales=F,coef=T) coeff = fit$coef However, it seems that "coeff" (or "fit$coef") is