similar to: Finney's fiducial confidence intervals of LD50

thanks a lot Renaud. but i was interested in Finney's fiducial confidence intervals of LD50 so to obtain comparable results with SPSS. But your reply leads me to the next question: does anybody know what is the best method (asymptotic, bootstrap etc.) for calculating confidence intervals of LD50? i could "get rid" of Finney's fiducial confidence intervals but

Hi, I have recently been attempting to find the LD50 from two predicted fits (For male and females) in a Generalised linear model which models the effect of both sex + logdose (and sex*logdose interaction) on proportion survival (formula = y ~ ldose * sex, family = "binomial", data = dat (y is the survival data)). I can obtain the LD50 for females using the dose.p() command in the MASS

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

Classification: UNCLASSIFIED Caveats: NONE A similar question has been posted in the past but never answered. My question is this: for probit analysis, how do you program a 95% confidence interval for the LD50 (or LC50, ec50, etc.), including a heterogeneity factor as written about in "Probit Analysis" by Finney(1971)? The heterogeneity factor comes into play through the chi-squared

Hi, I'm not yet familar with GLM and still learning. How can I perform a BNL (to estimate LDp values) with matrixes like this (N indicates the observed objects): data from: Kerr D, Meador J. 1996. "Modeling Dose Response Using Generalized Linear Models". Environ Toxicol Chem 15(3) conc respond n 0 0 20 0 1 19 0 1 20 0 0 20 13.5 5 23 13.5 2 20 29 9 20 29 6 20 53 12 20 53 15 20 85

Hi All! I am desperately needing some help figuring out how to calculate LD50 with a GLMM (probit link) or, more importantly, the standard error of the LD50. I conducted a cold temperature experiment and am trying to assess after how long 50% of the insects had died (I had 3 different instars (non significant fixed effect) and several different blocks (I did 4 replicates at a time)=

Quiero comparar varias dosis letales 50% (LD50) usando análisis probit. He seguido un ejemplo que viene en paquete DRC, pero no obtengo el resultado esperado. Lo que quiero es saber si las LD50s, son diferentes y si la diferencias son estadísticamente significativas. Gracias de antemano. José Arturo e-mail. jafarfan@uady.mx <grejon@uady.mx> e-mail alterno. jafarfan@gmail.com

Hello all, I have a very simple set of data and I would like to analyze them with probit analysis. The data are: X Event Trial 100 8 8 75 8 8 50 6 8 25 4 8 10 2 8 0 0 8 I want to estimate the value of X that will give a 95% hit rate (Event/Trial) and the corresponding 95% CI. Anyone can offer some help? Thanks!! -

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

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

Hi, All, I am implementing real-time Rsync on Windows 2008 system. I set up Rsync server and Rsync client on two machines. An windows service is watching all the Windows file events with FileSystemWatcher. However, the service cannot tell the exactly what happened to folders such as create, delete, or modified. So, I ignored folder event, and only catch file changing events. After I catch

At a snail's pace I keep on translating an introduction to R into italian; I have reached the section describing the glm() function, in which some example code is presented. The very last line of code, before the beginning of the section on Poisson models is: ldp <- ld50(coef(fmp)); ldl <- ld50(coef(fmp)); c(ldp, ldl) which of course gives results 43.663 and 43.663; the correct code

At a snail's pace I keep on translating an introduction to R into italian; I have reached the section describing the glm() function, in which some example code is presented. The very last line of code, before the beginning of the section on Poisson models is: ldp <- ld50(coef(fmp)); ldl <- ld50(coef(fmp)); c(ldp, ldl) which of course gives results 43.663 and 43.663; the correct code

Respected Sir/Madam, I have a question regarding calculation of LD50 (Lethal Dose) and IC50 (50% inhibitory concentration) of an antimicrobial experiment. I have used a compound isolated from a plant and observed its effect on the fungus *Fusarium oxysporum* by the food poisoning method. Solutions of the compound at concentrations of 0, 50, 100, 150, 200 and 250µg/ ml were added to

Dear all I need to obtain an estimate of the 50% lethal dose (LD50) from a logistic regression model obtained by applying the glm procedure to some binomial data. The model appears to fit the data very well. I used dope.p from MASS to try and find LD50. The following output appears: > dose.p(iso.glm.logit, cf = c(1,3), p = 1:3/4) Error in dose.p(iso.glm.logit, cf = c(1, 3), p = 1:3/4) :

The core team has decided to release R 1.5.0 on April 29th. Somewhat earlier than maybe expected, but we realized that we needed a phase-shift away from our usual cycle with main releases in June and December since it was placing feature freezes just when several members were in a creative phase due to end of teaching. The roadmap is as follows April 8 feature freeze on r-base April 15 code

I have the following problem. I have measured a dose response curve (binary response, continuous dose) on a grid of x,y positions. I would like to produce a grey-level plot that shows the LD50 at each (x,y) position. I am thinking that I have to do something like fit<-glm(resp ~ x*y + dose, family = binomial) Corrections welcome. But from here I don't know how to get LD50, and certainly

Hello, I am trying to determine LD50 and LD95 using dose.p in MASS however some of the Residual variance is larger than the degrees of freedom. Please can anyone help with any advice as to how i can correct for this? Here is the model as inputted into R y<-cbind(dead,n-dead) model<-glm(y~log(conc),binomial) summary(model) xv<-seq(min(log(conc)-1),max(log(conc)+1),0.01)