similar to: Different results in glm() probit model using vector vs. two-column matrix response

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: "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

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

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

> 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 >

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,

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

Dear R helpers, I have been using the loadings function from the multiv library and I get the typical output (see below). When I try to export these results to a file using a write.table() I get the following error message "Error in as.data.frame.default(x[[i]], optional = TRUE) : can't coerce loadings into a data.frame" Any idea why write.table is doing that and any

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)

Hi all, I'm not familiar with R programming and I'm trying to reproduce a result from a paper. Basically, I have a dataset which I would like to model in terms of successive increments, i.e. (y denote empirical values of y) y_1 = y1, y_2 = y1 + delta1, y_3 = y1 + delta1 + delta2. ... y_m = y1 + sum_2^m delta j where delta_j donote successive increments in the y-values, i.e. delta

Hi there, I've sent this e-mail to the list twice but didn't get it back from the list. Have it reach list members? cheers, Ronaldo ---------- Forwarded message ---------- From: Ronaldo Prati <rcprati at gmail.com> Date: 14/12/2006 11:59 Subject: Model formula question To: r-help at stat.math.ethz.ch Hi all, I'm not familiar with R programming and I'm trying to

Hi, I have two groups of data of different size: group A: x1, x2, ...., x_n; group B: y1, y2, ...., y_m; (m is not equal to n) The two groups are independent but observations within each group are not independent, i.e., x1, x2, ..., x_n are not independent; but x's are independent from y's I wonder if randomization test is still applicable to this case. Does R have any function

We are starting the process for purchasing a server with Windows Server 2008 OS. We are mainly interested in storage for multiple users, but would like to have the capability to run R simulations on the server. Are there any issues we should be aware of for installing R on the server or with having multiple users run R simultaneously? Thanks much! <http://www.uams.edu/biostat>

Good morning gentlemen! How using a weighted model in nls2? Values with the nls are logical since values with nls2 are not. I believe that this discrepancy is due to I did not include the weights argument in nls2. Here's an example: MOISTURE <- c(28.41640, 28.47340, 29.05821, 28.52201, 30.92055, 31.07901, 31.35840, 31.69617, 32.07168, 31.87296, 31.35525, 32.66118, 33.23385,