Author: Andrew Wagner
Date: 14:47:06 05/25/04
Go up one level in this thread
On May 25, 2004 at 17:02:01, Bas Hamstra wrote: >On May 25, 2004 at 16:39:38, Andrew Wagner wrote: > >>On May 25, 2004 at 15:49:22, Bas Hamstra wrote: >> >>>Supposed you played a huge automated tournament, and ratings were automatically >>>calculated (like in Arena) and displayed for all participants. Then, would it be >>>possible to derive/calculate the sigma, 2*sigma, 3*sigma rating >>>rating-intervals? If so how? >>> >>> >>> >>>Ciao, >>> >>>Bas. >> >>Hi. I'm not sure what you mean by 2*sigma and 3*sigma. The are not statistically >>significant, so far as I know. Variance is sigma^2 (i.e. sigma * sigma). The >>formula for this is: sum((x - average)^2)/N. So, let's take a few ratings for an >>example: >>2400 >>2432 >>2414 >>2443 >> >>The average is 2422.25. Therefore, we have (22.25^2) + (9.75^2) + (8.25^2) + >>(19.75^2) in the numerator. That's 1048.25. And our N here is 4 because we have >>4 ratings, so our answer is 1048.25/4 or 262. That's the variance. Taking the >>square root, we get 16.2, which is sigma (standard deviation). Hope that helps. >>Andrew > >What I mean is in a "normal" distribution you can expect values to lie in the >interval [Mean-sigma, Mean+sigma] with 65% confidence. For +- 2*sigma borders >the confidence is 95% and for 3*sigma it's 99.7%. > >What I like to calculate is after a tournament with n participants, over a >certain number of rounds, with 95% confidence: what is the rating interval of >program A, program B, etc. Ok, so for each program, calculate the mean and sigma (standard deviation) as I did above. Then just multiply by 2 or 3 and you have your window around the average.
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