similar to: Fieller's Conf Limits and EC50's

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 everybody, In Splus, the following works fine. foo1 <- function(x, const) x + const foo2 <- function(x, const) x * const const1 <- 1 const2 <- 2 foo <- substitute(function(x) { a <- foo1(x, const1) b <- foo2(a, const2) b }, list(const1=const1, const2=const2)) foo(1:3) In R, the code doesn't work, as foo is of mode "call", and

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

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 =================================================================

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

Hello I am fitting a Cox PH model using the function coxph(). Does anyone know how to obtain the estimate of the covariance matrix under the null hypothesis. The function coxph.detail() does not seem to be useful for this purpose. Thanks, KD. [[alternative HTML version deleted]]

The error msg is telling you that R cannot evaluate the loss function, so you should not expect answers. You might try examining the data -- Are there NA or Inf entries? Or prepare a dataframe with just X and Y, sort by X and graph. Then check the nls computations by sampling, say, every 100 X's to give you a dataset with about 160 observations. If that doesn't work, it is at least

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,

Hello, I have this data: Time AMP 0 0.2000000 10 0.1958350 20 0.2914560 40 0.6763628 60 0.8494534 90 0.9874526 120 1.0477692 where AMP is the concentration of this metabolite with time. If you plot the data, you can see that it could be fitted using a logistic regression. For this purpose, I used this code: AMP.nls <- nls(AMP~SSlogis(Time,Asym, xmid, scal), data

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

I have a question regarding significance testing for the difference in the ratio of means. The data consists of a control and a test group, each with and without treatment. I am interested in testing if the treatment has a significantly different effect (say, in terms of fold-activation) on the test group compared to the control. The form of the data with arbitrary n and not assuming equal

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

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

Take a look at this message thread. I think it would be interesting/nice if our constants and enums also returned a nice name when inspected. Thoughts? Roy -------- Original Message -------- Subject: Re: [Swig-user] wrapping enums for python Date: Wed, 13 Sep 2006 17:14:50 +0200 From: Nitro <nitro@dr-code.org> To: K M <intra611@gmail.com>, swig-user@lists.sourceforge.net

I've been working on some basic library call optimizations, the SimplifyLibCalls pass (lib/Transforms/IPO/SimplifyLibCalls.cpp). Tonight I conjured up a list of the potential libcall simplifications that could be done. There's a lot of them. I could use some help if anyone wants to pitch in. The individual optimizations are self-contained and fairly straight forward to write. They range

Dear R-help subscribers Would you tell me how to apply SSfpl with binary data as below? Unfortunately, there is not the EXAMPLE in help(SSfpl) for binary data but for quantitative data(Chick). V1: dose V2: log-transformed dose V3: response (rate) V1 V2 V3 1 0.775 -0.2548922 0.1666667 2 5.000 1.6094379 0.8148148 3 10.000 2.3025851 0.5000000 4 20.000 2.9957323

Dear R-experts, I have 28 data points that I would like to fit with a non linear broken-stick -- with three fitted parameters. When I view trace -- and use the final values as lines on the graph of data -- it looks pretty good. Q1. Why am I getting singular convergeance? Q2. Can you suggest another approach that may prove more satisfying? I have read previous examples on nls and this sort of

Hello all, This is something that I am sure has a really suave solution in R, but I can't quite figure out the best (or even a basic) way to do it. I have a simple linear regression that is fit with lm for which I would like to estimate the x intercept with some measure of error around it (confidence interval). In biology, there is the concept of a developmental zero - a temperature under