search for: ec50

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 function. I would appreciate comments about the validity of...

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 76.54...

...1=GUT, 2=CENTRAL ;DV =PK ORDERED BY ID, TIME $DATA ../projectRSD1.csv IGNORE=C $SUBROUTINE ADVAN2 TRANS2 INFN=../MItoolsRunlogNM6.for $ABB COMRES=8 $PK FLA2=FLAG TVKA=THETA(1) TVCL=THETA(2) TVV=THETA(3) CL=TVCL*EXP(ETA(1)) V=TVV*EXP(ETA(2)) KA=TVKA*EXP(ETA(3)) S2=V $ERROR EMAX=THETA(4)*EXP(ETA(4)) EC50=THETA(5)*EXP(ETA(5)) N = THETA(6)*EXP(ETA(6)) EFF= (EMAX*F**N)/(EC50**N+F**N) + ERR(2) CONC= F + F*ERR(1) Y = CONC*(2-FLAG) + EFF*(FLAG-1) IPRED=Y " LAST " COM(1)=G(1,1) " COM(2)=G(2,1) " COM(3)=G(3,1) " COM(4)=G(4,1) " COM(5)=G(5,1) &...

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 -.-.-.-.-.-....

....924, 16.832, 6.832, 11.118, 1.319, 5.495, -3.352, 102.464, 83.114, 50.631, 22.792, 18.348, 19.066, 27.794, 14.682, 11.992, 12.868)) m<- drm(response ~ (log10(dose*1e6)), data = d, fct = l4(fixed = c(NA,NA,NA,NA), names = c("hs", "bottom", "top", "ec50")), logDose = 10, control = drmc(useD = T)) summary(m) results in: Model fitted: Log-logistic (ED50 as parameter) (4 parms) Parameter estimates: Estimate Std. Error t-value p-value hs:(Intercept) -9.8065e-01 2.5821e-03 -3.7979e+02 2.248e-33 bottom:(Intercep...

...d.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 ================================================================= mod.nlme<-nlme(RESP~E0+(Emax-E0)*CP**gamma/(EC50**gamma+CP**gamma),data=Data, fixed=E0+Emax+gamma+EC50~1, random=list(pdDiag(EC50+E0+gamma~1)), groups=~ID, start=list(fixed=c(E0=1,Emax=100,gamma=1,b=50)) ) The "Data" object structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,...

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, hill=-1)) However, I get the following error message: Error in numericDeriv(form[[3L]], names(ind), env) : Missing value or an infinity produced when evaluating the model Could someone...

...loaded 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, > hill=-1)) > > However, I get the following error message: > > Error in numericDeriv(form[[3L]], names(ind), env) : > Missing value or an infinit...

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]]

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

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 test for homogeneity and is equal to h=chi^2/(k-2), where k is the number of doses and k-2 are the degrees of freedom. I have...

...reetings, 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 like the best method for determining such a confidence interval? Is the model we have chosen being estimated in the best way? We could use the number of survival and the sample size and use the binomial g...

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 Effects Models in S and Splus" is not applicable to my problem because it deals with f(x)=A+(B-A)/(1+exp(EC50-x)/C)) which is not the same function.)...

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

[This email is either empty or too large to be displayed at this time]

...e.ds.army.mil> Content-Type: text/plain; charset="us-ascii" 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 test for homogeneity and is equal to h=chi^2/(k-2), where k is the number of doses and k-2 are the degrees of freedom. I have...