Author: Joseph Ciarrochi
Date: 02:41:22 01/17/06
Go up one level in this thread
Another issue is that players may vary in how variable their play is, and more variable players would have larger standard errors for a given n size I used a stat package to generate the following statistical output. player 1 and 2 have the same mean level of points and the same N, but player 1 tends to win and lose alot, wherease player 2 tends to draw alot. AS you can see, standard error is bigger for player 1. This sort of analysis assumes that the variable "mean" is contionous, though really we have three categories(win, draw, loss). Still, with large samples, it is probably ok. I guess I could just use the formula for confidence intervals around means. The formula you suggest is useful in that it does give you the confidence interval for win verus everything else, which is an important statistic PLAYER Mean N Std. Deviation Std. Error of Mean 1.00 .5000 300 .48385 .02794 2.00 .5000 300 .40893 .02361 Total .5000 600 .44759 .01827 On January 17, 2006 at 05:23:52, Joseph Ciarrochi wrote: >yes, I can see that would be a useful of looking at it. and certainly allows us >to use binomial distribution. > >But...would it allow you to distinguish between one player who had fifty draws >and fifty losses versus another player who has had 10 draw and 90 losses. They >both have 0 wins?? but player 1 has 25 points, and player 2 has only 5 points. > >best >Joseph > > > >On January 17, 2006 at 05:18:18, Renze Steenhuisen wrote: > >>It is my opinion that you have winning and not-winning. Doesn't matter if >>not-winning consists of draws and losses. >> >>On January 17, 2006 at 05:07:22, Joseph Ciarrochi wrote: >> >>>thank you. that's great!. I assume that this formula does not require >>>dichotomous data,e g., it is approrate for data with losses wins, and draws? >>> >>>best >>>Joseph >>> >>>On January 17, 2006 at 04:01:57, Renze Steenhuisen wrote: >>> >>>>On January 17, 2006 at 03:52:21, Joseph Ciarrochi wrote: >>>> >>>>>Does anybody have the formula for calculating standard error bars for win >>>>>percentages? e.g., given a a sample of N, and an observed win rate of x %, I 'd >>>>>like to construct a 95% confidence interval around x. >>>>> >>>>>It would also be cool if you had the formula for calculating confidence >>>>>intervals for ratings? >>>>> >>>>>best >>>>>Joseph >>>> >>>>If you have M positive occurences of a certain event (i.e., a win), from N >>>>samples. Then, the %-interval is given by: >>>> >>>>(M + (L^2)/2 \pm L*SQRT(M*(1-(M/N))+(L^2)/4)) / (N + (L^2)) >>>> >>>>with \pm being minus for the lowerbound, and plus for the upperbound. >>>> >>>>L depends on the %-confidence you like. 95% confidence needs L=1.96. >>>> >>>>Cheers >>>> >>>>Renze Steenhuisen
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