search for: ed50

I am using semiparametric Model  library(mgcv) sm1=gam(y~x1+s(x2),family=binomial, f) How should I  find out standard error for ed50 for the above model ED50 =( -sm1$coef[1]-f(x2)) / sm1$coef [2]   f(x2) is estimated value for non parametric term.   Thanks [[alternative HTML version deleted]]

Hi, I am trying to use drc package to calculate IC50 value. The ED50 calculated in some models (LL4 for example) as a response half-way between the upper and lower limit, which is the definition of the relative IC50 value. Does that mean the ED50 in drc package is IC50? How the ED function in drc package distinguish to estimate ED or IC values? Thanks a lot [[al...

Hi R-Users, I have plotted a region whose polygon coordinates are given in shp format ED50 UTM (zone=30) ) using "readShapePoly" in library(maptools). Now I need to plot a set of points in that region (my.dataframe, with X and Y geographic coordinates), which have been read using GPS in Longitud-Latitud form (using WGS84 system), so I first need to convert these Longitud-Lati...

...Treatment B (B-/B+). A single muscle was used for each experiment - a full dose response curve and one treatment from the matrix A*B (A-/B-, A+/B-, A-/B+ and A+,B+). There are 8 replicates for each combination of treatments We fit a dose response curve to each experiment with parameters upper, ed50 and slope; we expect treatment A to change upper and ed50. We want to know if treatment B blocks the effect of treatment A and if so to what degree. This is similar to the Ludbrook example in Venables and Ripley, however they only had one treatment and I have two. my approach The dataframe is...

Dear R Community, I am using the package DRC ( to fit a boltzman model to my data. I can fit the model and extract the lower limit, upper limit, and ED50 (aka V50), but I cannot figure out how to get the slope of the curve at ED50. Is there a simple way to do this? I've searched the mailing list and looked through the package documentation, but could not find anything. I am new to r, and especially the DRC package, so any help would be apprec...

...curve represents an independent experiment). The data are stored as follows: expt treatA treatB dose force I use a groupedData object mydata=groupedData(force ~ dose | expt) I used an nlme obect to model the data as follows (pseudocode): myfit <- nlme(force ~ ssThreeParLogistic(dose, upper, ed50,slope), fixed=list(ed50~factor(treatmentA)*factor(treatmentC))) The ThreeParLogistic is a properly debugged and fully functional selfstarting object that I wrote- no problem here. I also included terms for the other terms; upper and slope, but my main focus is on the ed50 so that's all I&...

Hi! I want to use the DRC package in order to calculate the IC50 value of an enzyme inhibition assay. The problem is that the estimated ED50, is always out of the fitted curve. In the example below, I had a ED50 value of 2.2896, But when I predict the response level for this concentration I get a value of 45.71 instead of the expected value of 50. This is my data: #Dose unit is concentration (mM); response unit is % of inhibition...

Hi, I was wondering if someone could please help me. I am doing a logistic regression to compare size at maturity between 3 seasons. My model is: fit <- glm(Mature ~ Season * Size - 1, family = binomial, data=dat) where Mature is a binary response, 0 for immature, 1 for mature. There are 3 Seasons. The Season * Size interaction is significant. I would like to compare the size at 50%

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

...the following: > FourP A 'drc' model. Call: drm(formula = Response ~ Expected, data = SC, fct = LL.4()) Coefficients: b:(Intercept) c:(Intercept) d:(Intercept) e:(Intercept) 1.336 6.236 85.521 59.598 > summary(FourP) Model fitted: Log-logistic (ED50 as parameter) (4 parms) Parameter estimates: Estimate Std. Error t-value p-value b:(Intercept) 1.33596 0.15861 8.42309 0.0011 c:(Intercept) 6.23557 3.18629 1.95700 0.1220 d:(Intercept) 85.52140 2.15565 39.67313 0.0000 e:(Intercept) 59.59835 5.18781 11.48815 0.00...

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 the limits. Response<-c(0,7,26,27,0,5,13,29,0,4,11,25) Tot<-rep(30.5,12) Dose<-rep(c(10,40,160,640),3) probit<-glm(formula = Response/Tot~...

Quiero comparar varias dosis letales 50% (LD50) usando análisis probit. He seguido un ejemplo que viene en paquete DRC, pero no obtengo el resultado esperado. Lo que quiero es saber si las LD50s, son diferentes y si la diferencias son estadísticamente significativas. Gracias de antemano. José Arturo e-mail. jafarfan@uady.mx <grejon@uady.mx> e-mail alterno. jafarfan@gmail.com

...6, 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:(Intercept) 1.0955e+01 2.2546e-02 4.8591e+02 4.364e-35 top:(Intercept) 1.0502e+02 9.0935e-02 1.1549e+03 4....

...gt; 2006/8/21, carlos riveira : > > 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 the limits. > > > > Response<-c(0,7,26,27,0,5,13,29,0,4,11,25) > > Tot<-rep(30.5,12) > > Dose<-rep(c(10,40,160,640),3) &gt...

I have the following problem. I have measured a dose response curve (binary response, continuous dose) on a grid of x,y positions. I would like to produce a grey-level plot that shows the LD50 at each (x,y) position. I am thinking that I have to do something like fit<-glm(resp ~ x*y + dose, family = binomial) Corrections welcome. But from here I don't know how to get LD50, and certainly

...vels 1-4. Subject will be fitted as a random variable. I have started with the 3 parameter Emax model using nls, with no other covariates and no random term. The syntax of this I know to be correct. When I progress to nlme, firstly fitting subject as random, I get nonsensical answers for Emax and ED50. Even if I was getting reasonable answers at this point, I am not sure how I specify period, a co-factor in the model? Hoping someone can help. Many thanks in advance. B Surujbally LEGAL NOTICE Unless expressly stated otherwise, this message is confidential and may...

...38.88889 26 0.00 MC 100.00000 27 1.23 MC 81.81818 28 2.46 MC 54.54545 29 3.69 MC 45.45455 30 4.92 MC 31.81818 > mydata1 <- multdrc (Survival ~ Dose, Herbicide, fct = l4(names = c("Slope", "Upper Limit", "Lower Limit", "ED50")), data=mydata) There were 36 warnings (use warnings() to see them) > warnings() Warning messages: 1: NaNs produced in: log(x) 2: NaNs produced in: log(x) 3: NaNs produced in: log(x) 4: NaNs produced in: log(x) 5: NaNs produced in: log(x) 6: NaNs produced in: log(x) 7: NaNs produced...

...(Weiwei Shi) > 43. Re: to draw a smooth arc (Paulo Barata) > 44. Re: to draw a smooth arc (Paul Murrell) > 45. Percentage area of a distribution (Alan Gibson) > 46. Re: Percentage area of a distribution (Alan Gibson) > 47. optimising fitted distributions (Floris Van Ogtrop) > 48. ED50 from logistic model with interactions (Kate Stark) > 49. Re: Concepts question: environment, frame, search path > (graham wideman) > 50. the Surv function (Jennifer Dillon) > 51. Re: the Surv function (Christos Hatzis) > 52. ? R 2.5.0 alpha bug (Inman, Brant A. M.D.) > 53....

...g Subject: Re: [R] Probit Analysis: Confidence Interval for the LD50 using Fieller's and Heterogeneity (UNCLASSIFIED) Message-ID: <AANLkTinRodwVPCrXm8SnfAfcn3QQ6-qJg4GnaNZMG0B3 at mail.gmail.com> Content-Type: text/plain Hi: The MASS package has a function dose.p() to produce a CI for ED50, ED90 or EDp in general (0 < p < 100). It takes a model object (presumably from a suitable logistic regression) as input. You could always take the code already available and adapt it to your situation or you could investigate one or more of the packages devoted to dose-response models. A u...