Author: Joseph Ciarrochi
Date: 02:23:52 01/17/06
Go up one level in this thread
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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