search for: ldose

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 library with dose.p(mod1,c(1,2)). However I cannot find a way to determine the LD50 of males. Any help on finding this male LD50 would be appreciated....

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 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -2.9935 0.5527 -5.416 6.09e-08 *** sexM 0.1750 0.7783...

...er. I need the deviance, df, and weights for another formula, which is why I'm focused on these. My code is below. Thank you in advance for any assistance! Shelly **** # 10 record set-up group <- gl(2, 5, 10, labels=c("U","M")) dose <- rep(c(7, 8, 9, 10, 11), 2) ldose <- log10(dose) n <- c(8,8,8,8,8,8,8,8,8,8) r <- c(0,1,3,8,8,0,0,0,4,5) p <- r/n d <- data.frame(group, dose, ldose, n, r, p) SF <- cbind(success=d$r, failure=d$n - d$r) #80 record set-up dose2<-c(7,8,9,10,11) doserep<-sort(rep(dose2,8)) x<-c(doserep,dos...

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

What am I failing to understand here? The script below works fine if the dataset being used is DNase1 <- DNase[ DNase$Run == 1, ] per the example given in help(nlrob). Obviously, I am trying to understand how to use nls and nlrob to fit curves to data using R. #package=DAAG attach(codling) plot(pobs~dose) #next command returns 'step factor reduced below min factor

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

...t;Andy > >> > >> > >>>Ross Darnell > >>>-- > >>>Email: <r.darnell@uq.edu.au> > >>> > Hi Andy > > Where? > > Try predict.glm example > ## example from Venables and Ripley (2002, pp. 190-2.) > ldose <- rep(0:5, 2) > numdead <- c(1, 4, 9, 13, 18, 20, 0, 2, 6, 10, 12, 16) > sex <- factor(rep(c("M", "F"), c(6, 6))) > SF <- cbind(numdead, numalive=20-numdead) > budworm.lg <- glm(SF ~ sex*ldose, family=binomial) > ld &l...

Hello, I've got a program written in S-plus which I think is converted successfully to R with the exception of part of the opt.param function written. In S-plus it is: nlminb(start=x0, obj=negllgamma.f, scale=1, lower=c(0.01,0.0001), upper=c(10,0.9999), gamma=gamma, maxlik=maxlik, y=ldose, s=lse, max.iter = 1000, max.fcal = 1000)$par and so far with R I've got to: optim(par=x0, fn=negllgamma.f, method="L-BFGS-B", lower=c(0.01,0.0001), upper=c(10,0.9999), gamma=gamma, maxlik=maxlik, y=ldose, s=lse, control=list(maxit = 1000))$par however I've failed t...

...and normally approaches 100% at large difference. To compare two psychometric curves, the conventional way is to fit two logistic curves and compare the 75% correct "threshold" values (whatever threshold means). I want to handle the case similar to the budworm example in MASS (glm(SF~sex*ldose, family=binomial)). My basic idea is that the 2AFC forced choice psychometric curve, normally only defined for positive stimuls differences, could conceptually be continued to negative values by mirroring the values at (0,0.5) to get the whole binomial/logistic curve. As far as I can see, the resul...

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

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

Hello running R 1.7.1 on Windows 2000 I have a model notmar1 <- glm(yprisx~age+harddrug+sex, subset = marcom == 0, family = quasipoisson) and summary(notmar1) gives (as it should) 433 df for the null model but when I run predict(notmar1 <- glm(yprisx~age+harddrug+sex, subset = marcom == 0, family = quasipoisson)) I get preditions for 528 people (the full data set, not the subset)