similar to: nls, the four parameter logistic equation, and prediction band

I have uploaded a datafile that contains the following two variables: time (X value) and response (Y value). This is a fairly extensive file (with > 16000 entries). I have two questions: 1. I want to use the following equation to regress Y on X: Y-hat = min + (max-min)/(1 + (X/EC50)^Hillslope). Here is my R command: nlsout <- nls(Y ~ (0 - (100-0)/(1 + (X/EC50)^hill)), start=c(EC50=125,

Greetings, We are performing a meta-analysis of mink pup survival data versus chemical concentration. We have modeled percent survival successfully using nls as shown below and the plot. What we need to do is construct a confidence interval on the concentration at which we get 50% survival (aka the EC50, although we may want other percent survivals in the future). My first question is, what seems

Hi, I've been trying to work with the gnls() function in the "nlme" package. My decision to use gnls() was so that I could fit varPower and such to some of the data. However, in working with a small dataset, I've found that the results given by gnls() don't seem to make any sense and they differ substantially from those produced by nls(). I suspect that I am just

Folks I have modified an existing function to calculate 'ec/ld/lc' 50 values and their associated Fieller's confidence limits. It is based on EC50.calc (writtien by John Bailer) - but also borrows from the dose.p (MASS) function. My goal was to make the original EC50.calc function flexible with respect to 1) probability at which to calculate the expected dose, and 2) the link

Hi all, I am trying to make a script that prints all possible models from a model I've created using nls(). It is a logisitc model which in total includes 13 variables. So its >8000 models I need to create, which I don't want to do manually. I've tried modify scripts made for linear models with no results. I've tried these scripts on a two variable model (c,a1 and a2 is what I

Dear R-Help, I have a generated file that looks like the following: ----- Begin file ----- # # Output File # float Version 2002.700000000000 int Numdays 31 int NumOFEs 1 # # Hillslope-specific variables # char HillVarNames[ 3 ] {Days In Simulation} {Hillslope: Precipitation (mm)}

Dear list, I am running some linear regressions through lapply, >lapply(c('EMAX','EC50','KOUT','GAMMA'),function(x)confint(lm(get(x)~RR0,dataset2))) I got results like [[1]] 2.5 % 97.5 % (Intercept) 0.6595789212 0.8821691261 RR0 -0.0001801771 0.0001489083 [[2]] 2.5 % 97.5 % (Intercept) -63.83694930

Dear community, I'm currently attempting to predict the occurence of an event (factor) having more than 2 levels with several continuous predictors. The model being ordinal, I was waiting the glm function to return several intercepts, which is not the case when looking to my results (I only have one intercept). I finally managed to perform an ordinal polytomous logisitc regression with the

I am trying to fit a curve to a cumulative mortality curve (logistic) where y is the cumulative proportion of mortalities, and t is the time in hours (see below). Asym. at 0 and 1 > y [1] 0.00000000 0.04853859 0.08303777 0.15201970 0.40995074 0.46444992 0.62862069 0.95885057 1.00000000 [10] 1.00000000 1.00000000 > t [1] 0 13 20 24 37 42 48 61 72 86 90 I tried to find starting values for

Hi I am having some problems running a covariate analysis with my colleage using R with the NonMem program we are using for a graduate school project. R and NonMem run fine without adding in the covariates, but the program is giving us a problem when the covariate analysis is added. We think the problem is with the R code to run the covariate data analysis. We have the control stream, R code

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

Dear all, I want to do a F test, which involves calculation of the degrees of freedom for the residuals. Now say, I have a nlme object "mod.nlme". I have two questions 1.How do I extract the degrees of freedom? 2.How is this degrees of freedom calculated in an nlme model? Thanks. Jun Shen Some sample code and data =================================================================

WinXP, Splus7 and R2.3.1. All, I have been using the SSfpl and SSlogis self-starting functions in the nlme library to fit 4PL and restricted 4PL models. I need to adapt these routines to fit the alternative model f(x) = A + (B-A)/(1 + abs(x/EC50)^C) My Question: How do I obtain good starting values for this alternative model? (The pseudo-code found on pages 517 - 520 of "Mixed

Hello all I have done some fitting of pnorm functions to dose-response data, so I could calculate EC50 values (dose where the response is 0.5). I used the nlm function for this, so I did not get any information about the confidence intervals of the fitted parameters. What would be a good way to do such a probit fit, or is there a package which I could use? Best regards Johannes Ranke

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 found that using different contrasts (e.g. contr.helmert vs. contr.treatment) will generate different fitted probabilities from multinomial logistic regression using multinom(); while the fitted probabilities from binary logistic regression seem to be the same. Why is that? and for multinomial logisitc regression, what contrast should be used? I guess it's helmert? here is an example

Dear all, As I am new to the R community - although eager to advance- I would like to pose a question to the community. I have an SPSS file which I have imported it in R (with the read.spss command) which conists of scale (continuous) variable "adiponectin" and the corresponding categorical value "death" (0=No, 1=Yes). In all there are 60 observations (among which

Shalini Raghavan 3M Pharmaceuticals Research Building 270-03-A-10, 3M Center St. Paul, MN 55144 E-mail: sraghavan at mmm.com Tel: 651-736-2575 Fax: 651-733-5096 ----- Forwarded by Shalini Raghavan/US-Corporate/3M/US on 08/16/2004 11:25 AM ----- Shalini

This data are kilojoules of energy that are consumed in starving fish over a time period (Days). The KJ reach a lower asymptote and level off and I would like to use a non-linear plot to show this leveling off. The data are noisy and the sample sizes not the largest. I have tried selfstarting weibull curves and tried the following, both end with errors. Days<-c(12, 12, 12, 12, 22, 22, 22,

Hello, I am trying to find the analytical solution to this differential equation dR/dt = k1*R (1-(R/Rmax)^k2); R(0) = Ro k1, k2, and Rmax are parameters that need to fitted, while Ro is the baseline value (which can be fitted or fixed). The response (R) increases initially at an exponential rate governed by the rate constant k1. Response has a S-shaped curve as a function of time and it