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2009 Mar 22
1
Estimating LC50 from a Weibull distribution
...reasonable
answer:
LC50 = (0.5 + 2.254)/0.135 = 20.4; i.e., higher than the highest
exposure concentration.
I'm sure I'm doing something silly--I just don't know what.
Essentially, I'm trying to do this, but with a Weibull model:
> library(MASS)
> resp <- cbind(d$ndead, nalive = d$orign - d$ndead)
> mod <- glm(resp ~ d$conc, family = binomial(link = "probit"))
> result <- dose.p(mod, p = 0.5)
> result
Dose SE
p = 0.5: 11.49053 0.1069564
Any help would be greatly appreciated.
Sincerely,
Greg.