[R] Finney's fiducial confidence intervals of LD50

[Renaud Lancelot]

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 at 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

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