Re: [R] Finney's fiducial confidence intervals of LD50

From: Renaud Lancelot <renaud.lancelot_at_gmail.com>
Date: Tue 22 Aug 2006 - 00:02:26 EST

I don't know what Finney's fiducial confidence interval is but if your problem is to handle the output of dose.p (from MASS), you can do as follows:

> library(MASS)
> 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)
> fm <- glm(Response/Tot ~ log10(Dose), family = quasibinomial(link = probit))
> logD50 <- dose.p(fm, cf = 1:2, p = 0.5)
> D50 <- 10^c(logD50 + c(1, -1.96, 1.96) * attr(logD50, "SE"))
> names(D50) <- c("D50", "lower", "upper")
> D50

     D50 lower upper
164.9506 103.3171 191.9777

Best,

Renaud

2006/8/21, carlos riveira <carlos.riveira@yahoo.com>:
> 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~ log10(Dose), family=quasibinomial
> (link=probit))
> D50<- round(10^(dose.p(probit,cf=1:2,p=0.5)))
>
> #This is what SPSS calculates. I would like to reproduce these results with R:
> #SPSS RESULTS:
> #PRNT50= 140,83525
> #CI = 98,37857;205,34483
> #Regr.coeff= 1,91676 (S.E.=0,16765)
> #Intercept=-4,11856 (S.E.=0,36355)
> Thanks a lot for your help.
>
> Carlos
>
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-- 
Renaud LANCELOT
Département Elevage et Médecine Vétérinaire (EMVT) du CIRAD
Directeur adjoint chargé des affaires scientifiques

CIRAD, Animal Production and Veterinary Medicine Department
Deputy director for scientific affairs

Campus international de Baillarguet
TA 30 / B (Bât. B, Bur. 214)
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Received on Tue Aug 22 01:29:47 2006

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