similar to: get response from a glm

Hi all, Can you please tell me what is the problem here. My regression eq is y = B0 + B1X1 +B2X2 +e And i am interested in coefficient B1 I am doing regression with two cases: 1) reg<-lm(y ~ X1 + X2, sam) where sam is the data 2) reg<-lm(y ~ X1 + X2, sam, na.action= na.exclude) . I have missing values in X1 but the values of coefficient is not consistent in two cases. Actually B1 in

Forgive me if this is a trivial question, but I couldn't find it an answer in former forums. I'm trying to reproduce some SAS results where they set two parameters equal. For example: y = b1X1 + b2X2 + b1X3 Notice that the variables X1 and X3 both have the same slope and the intercept has been removed. How do I get an estimate of this regression model? I know how to remove the intercept

Hi everyone, I'm trying to perform a linear regression y = b1x1 + b2x2 + b3x3 + b4x4 + b5x5 while constraining the coefficients such that -3 <= bi <= 3, and the sum of bi =1. I've searched R-help and have found solutions for constrained regression using quadratic programming (solve.QP) where the coefficients are between 0 and 1 and sum to 1, but unfortunately do not understand

Hello, I am trying to learn how to perform Multiple Regression Analysis in R. I decided to take a simple example given in this PDF: http://www.utdallas.edu/~herve/abdi-prc-pretty.pdf I created a small CSV called, students.csv that contains the following data: s1 14 4 1 s2 23 4 2 s3 30 7 2 s4 50 7 4 s5 39 10 3 s6 67 10 6 Col headers: Student id, Memory span(Y), age(X1), speech rate(X2) Now

I have a logit model with about 10 predictors and I am trying to plot the probability curve for the model. Y=1 = 1 / 1+e^-z where z=B0 + B1X1 + ... + BnXi If the model had only one predictor, I know to do something like below. mod1 = glm(factor(won) ~ as.numeric(bid), data=mydat, family=binomial(link="logit")) all.x <- expand.grid(won=unique(won), bid=unique(bid)) y.hat.new

Hi all. In a GLM in which g(mu) = b0 + b1X1 + b2X2 + b3X3 + b4X4 + b5X5 + b6X6 + b7X7, if I want to test if b1 + b5 = b2 + b6, I can use the contrast package or multicomp package. How can I do a similar test if I am fitting a GAMLSS using the gamlss package? Thank you for your help. Gustavo

Dear All, I am currently working with the coxph function within the package survival. I have the model h_ij = h_0(t) exp(b1x1 + b2x2) where the indicator variables are as follows: x1 x2 VPS 0 0 LTG 1 0 TPM 0 1 [[alternative HTML version deleted]]

From: xxgreat@hotmail.com To: r-help-bounces@r-project.org Subject: compare a vector and a row of a matrix Date: Sun, 26 Sep 2010 23:23:52 +0800 Hi Everyone: I am trying to compare a vector and rows of a matrix for example > xn <- c(1,2,4,4,5,5,5,6) >yn <- c(1,2,5,7,1,2,3,1) >mtrx <- cbind(xn, yn) when I tried, say, > c (1,4), the result was TRUE, TRUE.

Good day R-listers, My question is more a statistical question than an R related question, so please bear with me i'm currently applying the chow test of poolability in fact i'm working with panel N=17 T=5 , and my model looks like this : Yit= a0+B1X1+B2X2+B3X3+B4X4+eit My question is the following when i'm Testing for the equality of the coefficients of the unpooled data (the

Hi Dr. Snow, I am a graduate student working on analyzing data for my thesis and came across your post on an R forum: The default link function for the glm poisson family is a log link, which means that it is fitting the model: log(mu) ~ b0 + b1 * x But the data that you generate is based on a linear link. Therefore your glm analysis does not match with how the data was generated

Dear All, I have a binomial response with one continuous predictor (d) and one factor (g) (8 levels dummy-coded). glm(resp~d*g, data, family=binomial) Y=b0+b1*X1+b2*X2 ... b7*X7 how can I get the inflection point per group, e.g., P(d)=.5 I would be grateful for any help. Thanks in advance, Ren?

Dear R-Help-List, I was wondering if anyone had experience simulating case-control data in R? I've been looking through literature, and found that the main examples make heavy parametric assumptions on the distributions of the exposure (E), covariates (Z), and disease status (D). I would appreciate any guidance toward resources/examples/literature that simulate case-control data with fewer

Dear all, I would like to estimate a system of regression equations of the following form: y1 = a1 + b1 x1 + b2x2 + e1 y2 = a2 + c1 y1 + c2 x2 + c3 x3 + e2 Specifically the dependent variable in Equation 1 appears as an independent variable in Equation 2. Additionally some independent variables that appear in Equation 1 are also included in Equation 2. I assume that I cannot estimate these two

Hi, This is not an R question, but I've seen opinions given on non R topics, so I wanted to give it a try. :) How would one treat a variable that was measured once, but is known to fluctuate a lot? For example, I want to include a hormone in my regression as an explanatory variable. However, this hormone varies in its levels throughout a day. Nevertheless, its levels differ substantially

Folks I have modified an existing function to calculate 'ec/ld/lc' 50 values and their associated Fieller's confidence limits. It is based on EC50.calc (writtien by John Bailer) - but also borrows from the dose.p (MASS) function. My goal was to make the original EC50.calc function flexible with respect to 1) probability at which to calculate the expected dose, and 2) the link

Dear list, After running for a while, it crashes and gives the following error message: can anybody suggest how to deal with this? Error in if (ratio0[i] < log(runif(1))) { : missing value where TRUE/FALSE needed ################### original program ######## p2 <- function (Nsim=1000){ x<- c(0.301,0,-0.301,-0.602,-0.903,-1.208, -1.309,-1.807,-2.108,-2.71) # logdose

Hi, I don't understand why this code doesn't work (f(b2)): ///////////////// setClass("B0", representation(b0 = "numeric")) setClass("B1", representation("B0", b1 = "character")) setClass("B2", representation("B1", b2 = "logical")) f <- function(x) class(x) setMethod("f", "B0",

Dear List, I am using constrOptim to solve the following fr1 <- function(x) { b0 <- x[1] b1 <- x[2] ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 } As you can see, my objective function is ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 and I would like to solve for both b0 and b1. If I were to use optim then I would derive the gradient of the

Hi the list, I try to use callNextMethod in a setteur ([<-) but it does not work. Any raison ? Any other option ? --- 8< ------------------ ### Class B0 ### setClass("B0" , representation(b0 = "numeric")) setReplaceMethod("[","B0",function(x,i,j,value){x at b0 <- -value}) a <- new("B0") a at b0 <- 3 a a["b0"] <- 3 a

Hi, I am trying to understand the exception handler of Xen. In xen/arch/x86/x86_64/entry.S, the entry of the page fault handler is : ENTRY(page_fault) movl $TRAP_page_fault,4(%rsp) jmp handle_exception Here I did not get why to put the exception vector ($TRAP_page_fault) to stack before calling handle_exception? Thanks, -- Xinxin