Date: Mon, 6 Apr 1998 10:21:15 +0200 Message-Id: <199804060821.KAA00495@sophie.ethz.ch> From: Martin Maechler <maechler@stat.math.ethz.ch> To: p.dalgaard@biostat.ku.dk Subject: Re: R-beta: CI for median in funtion boxplot >>>>> "PD" == Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk> writes: PD> 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? >> PD> Search me... However, wouldn't it be better in any case to do an PD> exact 95% CI based on the binomial distribution? Of course, you PD> need at least 6 observations to do that. No, please not yet another definition of the boxplot! People looking at boxplots should be able to rely on their knowledge of what a boxplot is. I don't know the exact history; in any case, John Tukey devised the boxplot, including the notches, and ``1.58 is THE number''. A very accessible reference on how 1.58 was construed is Section 3.12, p.79--81 of @Book{VelPH81, author = {Paul F. Velleman and David C. Hoaglin}, title = {Applications, Basics, and Computing of Exploratory Data Analysis}, publisher = {Duxbury Press, Boston, Massachusetts}, year = 1981 } Here a ``compact'' summary (if you really want to know ...) Comparing two normal populations, there are two extreme cases: In the first one, the variances are about equal, in the other, one variance is much higher than the other. The corresponding z-Tests are abs(mean(x1) - mean(x2)) - 1.96 sqrt(2) sigma_xbar and abs(mean(x1) - mean(x2)) - 1.96 sigma_xbar (the big one). Where the first corresponds to a CI of mean(x) +/- 1.96 sqrt(2) / 2 sigma_xbar = = mean(x) +/- 1.39 sigma_xbar the second one must have mean(x1) +/- 1.96 sigma_xbar(x1) and the same for x2. An omnibus compromise factor is (1.39 + 1.96) / 2 ~= 1.7 [``exact'' would be qnorm(.975)*(1 + sqrt(2)/2)/2 = 1.672934]. Now, we also have sigma = 1.349 * IQR, [[exact: 2*qnorm(3/4) * IQR ]] and var(median) = pi/2 * var(arith.mean) The three things put together: "notch length" = (IQR/1.349) * sqrt(pi/2) * (1.7 / sqrt(n)) = = 1.58 * IQR / sqrt(n), i.e. 1.58 = sqrt(pi/2)*1.7/1.349 (= 1.579417) Instead, the ``exact'' value for 1.58 would be 1/(2*qnorm(.75))* sqrt(pi/2) * (qnorm(.975)*(1 + sqrt(2)/2)/2) = 1.554295 --- So, 1.58 ``should be'' 1.554 instead, but of course, the big deal is the compromise of the two extreme situations, anyway. Rounding up leads to the slightly increased factor which may be somewhat more realistic for long-tailed nonnormal situations. ---------- PS: Should the above go into the online documentation? Martin Maechler <maechler@stat.math.ethz.ch> <>< Seminar fuer Statistik, ETH-Zentrum SOL G1; Sonneggstr.33 ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND phone: x-41-1-632-3408 fax: ...-1086 http://www.stat.math.ethz.ch/~maechler/ -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- 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 _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._