similar to: Finding LD50 from an interaction Generalised Linear model

Dear all: "predict.glm" provides an example to perform logistic regression when the response variable is a tow-columned matrix. I find some paradox about the degree of freedom . > summary(budworm.lg) Call: glm(formula = SF ~ sex * ldose, family = binomial) Deviance Residuals: Min 1Q Median 3Q Max -1.39849 -0.32094 -0.07592 0.38220 1.10375

I am working with Probit regression (I cannot switch to logit) can anybody help me in finding out how to obtain with R Finney's fiducial confidence intervals for the levels of the predictor (Dose) needed to produce a proportion of 50% of responses(LD50, ED50 etc.)? If the Pearson chi-square goodness-of-fit test is significant (by default), a heterogeneity factor should be used to calculate

> From: Ross Darnell > > Liaw, Andy wrote: > >>From: Liaw, Andy > >> > >> > >>>From: Ross Darnell > >>> > >>>A good point but what is the value of storing a large set of > >>>predicted > >>>values when the values of the explanatory variables are lost > >>>(predicted >

Dear R-help list members, I am currently trying to fit a generalized linear model using a binomial with the canonical link. The usual solution is to use the R function glm() in the package "stats". However, I run into problem when I want to fit a glm without an intercept. It is indicated that the solution is in changing the function glm.fit (also in "stats"), by specifying

Hi. I'm trying to plot the ratio of used versus unused bird houses (coded 1 or 0) versus a continuous environmental gradient (proportion of urban cover [purban2]) that I would like to convert into bins (0 - 0.25, 0.26 - 0.5, 0.51 - 0.75, 0.76 - 1.0) and I'm not having much luck figuring this out. I ran a logistic regression and purban2 ends up driving the probability of a box being

Hi - I am fitting a probit model using glm(), and the deviance and residual degrees of freedom are different depending on whether I use a binary response vector of length 80 or a two-column matrix response (10 rows) with the number of success and failures in each column. I would think that these would be just two different ways of specifying the same model, but this does not appear to be the case.

Hello All,  I have used logistic regression glm in R and I am evaluating two models both learned with glm but with different predictors. model1 <- glm (Y ~ x4+ x5+ x6+ x7, data = dat, family = binomial(link=logit))model2 <- glm (Y~ x1 + x2 +x3 , data = dat, family = binomial(link=logit))  and I would like to compare these two models based on the prediction that I get from each model: pred1 =

Greets, I'm using a data frame that looks like: > head(pr2) X1 X2 X3 X4 Y fit res 1 44 33.2 5 30 41.2 39.22201 1.977991 2 43 33.8 4 41 31.7 38.48476 -6.784761 3 48 40.6 3 38 39.4 44.78278 -5.382783 4 52 39.2 7 48 57.5 51.48134 6.018656 5 71 45.5 11 53 74.8 68.25585 6.544153 6 44 37.5 9 65 59.8 53.27743 6.522569 Along with the command: >

Hello, I would like to know if there is a clever way to avoid the problem illustrated below within the xyplot function. x <- seq(1:10) y <- seq(1:10) pr1 <- xyplot(x ~ y) u <- seq(1:12) v <- seq(1:12) pr2 <- xyplot(u ~ v, col = "red", more = FALSE) prts <- list(pr1, pr2) for(i in prts) print(i, more = TRUE) I realize that one possibility is to

Dear R users, I'm having a hard time with some very simple things. I have a time series where the dates are in the format 7-Oct-04. I imported the file with read.csv so the date column is a factor. The series is rather long and I want to plot it piece by piece. The function below works fine, except that the labels for date are meaningless (ie 9.47e+08 or 1098000000 - apparently the number of

Hi, When attempting to restore to LINUX RHEL5 - OCFS2 filesystem received the following error during RMAN restore for nearly all of the datafiles attempted to restore with exception of a couple of smaller datafiles which were smaller < 2GB. ORA-19870: error reading backup piece /db/dumps/TR1_1/rmanbackup/TR1_88_1 ORA-19502: write error on file "/db/devices/db1/PR2/pr2_1/pr2.data1",

Dear r helpers I run the following code for nested logit and got a message that Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite What does this mean? and how can I correct it? Thank you June > yogurt = read.table("yogurtnp.csv", header=F,sep=",")> attach(yogurt)>

Dear Rhelpers I have the problem with initial values, could you please tell me how to solve it? Thank you June > p = summary(maxLik(fr,start=c(0,0,0,1,0,-25,-0.2))) Error in maxRoutine(fn = logLik, grad = grad, hess = hess, start = start, : NA in the initial gradient > p = summary(maxLik(fr,start=c(0,0,0,1,0,-25,-0.2),method="BFGS")) Error in optim(start, func, gr =

Dear R experts Sorry for this question M1 <- 1:10 lcd1 <- c(11, 22, 33, 44, 11, 22, 33, 33, 22, 11) lcd2 <- c(22, 11, 44, 11, 33, 11, 22, 22, 11, 22) lcd3 <- c(12, 12, 34, 14, 13, 12, 23, 23, 12, 12) #generating variables through sampling pvec <- c("PR1", "PR2", "PR3", "PR4", "PR5", "PR6", "PR7",

Help needed with lm function: Dear R's, Could anyone tell me how to replace the link function (probit logit, loglog etc.) in lm with an abitrary user-defined function? The task is to perform ML Estimation of betas for a dichotome target variable. Maybe there is already a package for this (I did not find one). Any hints or a code excerpt would be welcome! Thank you -Jeff jeff.pr2 (at)

Dear R Users on a self-written function for calculating maximum likelihood probability (plz check function code at the bottom of this message), one value, wden, suddenly jump to zero. detail info as following: w[11]=2.14 lnw =2.37 2.90 3.76 ... regw =1.96 1.77 1.82 .... wden=0.182 0.178 0.179... w[11]=2.14 lnw=2.37 2.90 3.76 ... regw =1.96 1.77 1.82 .... wden=0.182

Dear R-Listers, to measure the psychometric curve of pitch discrimination, one sequentially presents two tones of slightly different pitch to an observer (animal will do), and asks "which is higher". The pschometric curve is the fraction of correct responses plotted against the pitch difference. It passes through 50% (pure guessing) at zero and normally approaches 100% at large

Dear all, How to obtain the odds ratio (OR) and 95% confidence interval (CI) with 1 standard deviation (SD) change of a continuous variable in logistic regression? for example, to investigate the risk of obesity for stroke. I choose the happening of stroke (positive) as the dependent variable, and waist circumference as an independent variable. Then I wanna to obtain the OR and 95% CI with

Dear R-list, I have a data set from 20 pigs, each of which is tested at crossed 9 doses (logdose -4:4) and 3 skin treatment substances when exposed to a standard polluted environment. So there are 27 patches on each pig. The response is irritation=yes/no. I want to determine "equally effective 50% doses" (similar to old LD50), and to test the treatments against each other. I am looking

Hey, R users I do not know how to describe my question. I am a new user for R and write the following?code for a dynamic labor economics?model and use OPTIM to get optimizations and parameter values. the following code does not work due to the?equation: ?? wden[,i]<-dnorm((1-regw[,i])/w[5])/w[5] where w[5]?is one of the parameters (together with vector a, b and other elements in vector