Date: Sat, 4 Apr 1998 08:42:41 -0800 (PST) From: Thomas Lumley <thomas@biostat.washington.edu> To: Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk> Subject: Re: R-beta: CI for median in funtion boxplot In-Reply-To: <x2lntlsvip.fsf@blueberry.kubism.ku.dk> On 4 Apr 1998, Peter Dalgaard BSA wrote: > Rick White <rick@stat.ubc.ca> writes: > > > > > I noticed that boxplot computes a 95% CI for the median by using > > median +/- 1.58*IQR./sqrt(n) > > > > Where does the 1.58 constant come from? > > > > Search me... However, wouldn't it be better in any case to do an exact > 95% CI based on the binomial distribution? Of course, you need at > least 6 observations to do that. > I think 1.58 is based on a Normal approximation. If the data are Normal then you can compute the asymptotic standard error of the median and use +/- 1.96 of these to get a 95% ci. 1.58*IQR/sqrt(n) is a robust estimate of 2.13sigma/sqrt(n), which is about right to be 1.96 standard errors. This will work for contaminated Normal distributions, but it won'tbe very good for genuinely long-tailed or asymmetric distributions. In any case, the distribution of the median converges to Normal rather slowly, so the CI might not be very good anyway except in large samples. The exact binomial CI would be much better. Thomas Lumley ----------------------- Biostatistics Uni of Washington Box 357232 Seattle WA 98195-7232 ------------------------ -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request@stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._