search for: budworm

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-helpers, I don't see a difference between the following two plots of effect objects, which I understand should be different. What am I missing? require(doBy) require(effects) data(budworm) m1 <- glm(ndead/20 ~ sex + log(dose), data=budworm, weight=ntotal, family=binomial) m1.eff <- all.effects(m1) plot(m1.eff, rescale.axis = FALSE, selection = 2, main = 'rescale = F') plot(m1.eff, rescale.axis = TRUE, selection = 2, main = 'rescale = T') ********************...

V&R describes binomial GLMs with mortality out of 20 budworms. Is it appropriate to use the same approach with mortality out of numbers as low as 3? I feel reticent to do so with data that is not very continuous. There are one continuous and one categorical independent variables. Would it be more appropriate to treat the response as an ordered factor wit...

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

I am not sure whether it is a design decision or just an oversight. When I ask for the standard errors of the predictions with predict(budwm.lgt,se=TRUE) where budwm.lgt is a logistic fit of the budworm data in MASS, I got Error in match.arg(type) : ARG should be one of response, terms If one is to construct a CI for the fitted binomial probability, wouldn't it be more natural to do it on the link scale before transforming to the probability scale? Of course, knowing a bit about what'...

I am trying to get a prediction interval from a glm regression. With newdat being my set of values to be fitted, and glmreg the name of my regression, I am using the following code. predict(glmreg, newdat, se.fit = TRUE, interval = "confidence", level = 0.90) The problem is that I am only getting the standard error and the fitted value, not a prediction interval. Any help would be

Dear R-Experts, I was wondering how to fit a cumulative gaussian to a set of empirical data using R. On the R website as well as in the mail archives, I found a lot of help on how to fit a normal density function to empirical data, but unfortunately no advice on how to obtain reasonable estimates of m and sd for a gaussian ogive function. Specifically, I have data from a psychometric function

...through 50% (pure guessing) at zero 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 cur...

...# 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 <- seq(0, 5, 0.1) > row.names(predict(budworm.lg, data.frame(ldose=ld, > sex=factor(rep("M", length(ld)), levels=levels(sex))), > type = "response")) [You'd want names() rather t...

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....

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

...I’ve coded the factors temperature and salinity as factors for the binomial glm, and I haven’t had any trouble fitting the model and checking assumptions. I am however having trouble with the predict.glm function. I want to create a graph using my data that is similar to the one produced by the budworm example at the bottom of the predict.glm R documentation. In my case I want temperature on the x axis, probability on the y axis, and the lines on the graph to represent the probability of the event occurring at the different salinities tested at the different temperatures. I created a smaller ve...

Hello r-help, I try to fit birds counts over years using glm. I have done (with Estate and year as factors): Model1 <- glm(Females~Estate+Year+offset = log(area)), family = quasipoisson(link = log), na.action = "na.exclude") After I have calculated the prediction using: Pred1 <- predict(Model1, type = "response", na.action = "na.exclude") My question

I am sure a common need is to plot a scatterplot with some fitted line(s) and maybe save to a file. I have this: plot.glm <- function (x, y, file = NULL, xlab = deparse(substitute(x)), ylab = deparse(substitute(y)), main = NULL) { m <- glm(y ~ x) if (!is.null(file)) pdf(file = file) plot(x, y, xlab = xlab, ylab = ylab, main = main) lines(x, y =

Dear R-users, I use glm() to do logistic regression and use stepAIC() to do stepwise model selection. The common AIC value comes out is about 100, a good fit is as low as around 70. But for some model, the AIC went to extreme values like 1000. When I check the P-values, All the independent variables (about 30 of them) included in the equation are very significant, which is impossible, because we

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!! -

Hi i would like to use some graphs or tables to explore the data and make some sensible guesses of what to expect to see in a glm model to assess if toxin concentration and sex have a relationship with the kill rate of rats. But i cant seem to work it out as i have two predictor variables~help?Thanks.:) Here's my data. >

Hello, I am trying to run a logistic regression with random effects on proportional data in glmmBUGS. I am a newcomer to this package, and wondered if anyone could help me specify the model correctly. I am trying to specify the response variable, /yseed/, as # of successes out of total observations... but I suspect that given the error below, that is not correct. Also, Newsect should be a