similar to: LD50 contrasts with lmer/lme4

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 I would like to produce interaction boxplots and this seems to work: par(mfrow=c(2,2)) A=sample(rnorm(50,50,10)) B=sample(rnorm(50,100,10)) Test=merge(A,B,by=0)#by=0 where 0 is the row.names TreatA=(gl(2,50,100,labels=c("High","Low"))) TreatB=rep(gl(2,25,50,labels=c("High","Low")),2) Newdata=data.frame(TreatA,TreatB,Test)

Hi all, Here is my problem: I performed bioassays using a unique insecticide on 9 different genotypes and got their mortality depending on the dose of insecticide used. Now, I want to see wether some genotypes are different or not in their responses to insecticide. My problem is that I have up to four replicates for some genotypes, but only one for other... Due to this unbalanced design, I

In an intervention study with subjects randomly allocated to two treatments (treat A and B) and three time points (time) plus an additional baseline measurement (dv_base), I've set up the following model to test for differences in temporal courses of treatments for the outcome (dv), thereby allowing for individual intercepts and slopes: lmer(dv ~ dv_base+treat*time+(1+time|subject)) Fixed

termplot() does not carry subsetting over to carriers that are in the environment but not in the data frame. This generates a "subscript out of bounds" error. > data(ToothGrowth) > logdose <- log(ToothGrowth$dose) > tooth.lm <- lm(len ~ logdose, data=ToothGrowth) > termplot(tooth.lm) ## Works fine > toothVC2.lm <- lm(len ~ poly(dose,2),

Gday, This is a repost since I only had one direct reply and I remain mystified- This may be stupidity on my part but it may not be so simple. In brief, my problem is I'm not sure how to extract parameter values/effect sizes from a nonlinear regression model with a significant interaction term. My data sets are dose response curves (force and dose) for muscle that also have two

Hello, We use drc to fit dose-response curves, recently we discovered that there are quite different standard error values returned for the same dataset depending on the drc-version / R-version that was used (not clear which factor is important) On R 2.9.0 using drc_1.6-3 we get an IC50 of 1.27447 and a standard error on the IC50 of 0.43540 Whereas on R 2.7.0 using drc_1.4-2 the IC50 is

Hi, I am newbie of R. I a currently using multdrc object to generate fitting curve and IC50. My 384 well format raw data contains multi dose response curves. My script goes through set of data then produce curve and ic50. Here is my sudo code: For (plateid in platelist) { Input data (plateid) as matrix Curve fitting model4logistic <- multdrc(rdata ~ ld, logDose=10) }

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

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

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 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)=

G'day, My problem is I'm not sure how to extract effect sizes from a nonlinear regression model with a significant interaction term. My data sets are multiple measurements of force response to an agonist with two superimposed treatments each having two levels. This is very similar to the Ludbrook example in Venables and Ripley. The experiment is that a muscle is exposed to an agonist

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

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

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) :