confidence interval or error of x intercept of a linear
(Ted Harding) wrote:
On 24-Mar-09 03:31:32, Kevin J Emerson wrote:
...
When I have time for it (not today) I'll see if I can implement this neatly in R. It's basically a question of solving (N-2)*(1 - R(X0))/R(X0) = qf(P,1,(N-1)) for X0 (two solutions, maybe one, if any exist).
<etc.> A quick and probably not-too-dirty way is to rewrite it as a nonlinear model: x <- 1:10 Y <- 2*x - 3 + rnorm(10, sd=.1) cf <- coef(lm(Y~x)) confint(nls(Y~beta*(x-x0), start=c(beta=cf[[2]],x0=-cf[[1]]/cf[[2]])))
O__ ---- Peter Dalgaard ?ster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45) 35327907