tag:blogger.com,1999:blog-14974164.post3821780414955354488..comments2023-10-20T21:46:49.945+08:00Comments on Philippine Commentary: How To Measure the Accuracy of a Counting MachineDeany Bocobohttp://www.blogger.com/profile/01443168826029321831noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-14974164.post-24102137587495990232009-05-25T06:18:13.349+08:002009-05-25T06:18:13.349+08:00GabbyD,
You have a deeper understanding of the mar...GabbyD,<br />You have a deeper understanding of the margin of error than most, but the simple point is that at 20,000 points, your statistical precision is plus or minus 0.7 percent. That means you can only "bin" the results into boxes of about that width. But if you want to tell the difference between a machine accuracy of 99.995% (acceptable level) and 99.990% (no acceptable accuracy level) then your statistical test better have precision of plus or minus 0.0025% (half the difference you are trying to detect) -- in order for your claim to be valid that you could reject any machine that falls below the spec. <br /><br />Yet at 0.7% precision the marks on the ruler are "too fat".Deany Bocobohttps://www.blogger.com/profile/01443168826029321831noreply@blogger.comtag:blogger.com,1999:blog-14974164.post-5067426893610051362009-05-25T05:54:00.818+08:002009-05-25T05:54:00.818+08:00two comments:
1) you may have to be more careful i...two comments:<br />1) you may have to be more careful in calculating the standard error at the 95% CI<br /><br />the standard error of the average is affected by the population standard error and the sample size. <br /><br />effectively, SWS is assuming that the pop standard error is (root(.5*.5))=.5, from the assumption that its a binomial probability. [variance of binomial is root[p*(1-p)]<br /><br />2. the interpretation of CI is:<br /><br />if i run 100 sets of 100 ballots on a machine, 95 of the sets will have an average success rate of 99.995 for 95 of 100 sets (containing 100 ballots).<br /><br />3. lets say that jimenez is correct: the point estimate is 99.995%. note that ANY standard error would mean the min bound of the CI is going to be less than 99.995%. <br /><br />i wonder therefore, if jimenez means that the point estimate is in fact 99.995%.GabbyDnoreply@blogger.comtag:blogger.com,1999:blog-14974164.post-31194661196461702022009-05-25T00:41:35.307+08:002009-05-25T00:41:35.307+08:00A simple general rule of thumb relates the so call...A simple general rule of thumb relates the so called Statistical Margin of Error to the sample size of respondents in a survey or the number of test points in a test of election counting machines. The Margin of Error is equal to the reciprocal of the square root of the number of respondents or test points. For 1200 respondents in SWS surveys, the margin of error is plus or minus 2.8 percent (often rounded up to 3 percent in news reporting). For 20,000 proposed test points, the margin of error or statistical precision of the measurement of accuracy is 0.7 percent. This is is much too coarse, it can only distinguish a counting machine with accuracy equal to 99.995% and one with 99.295%. Way too coarse!Deany Bocobohttps://www.blogger.com/profile/01443168826029321831noreply@blogger.comtag:blogger.com,1999:blog-14974164.post-67962785025187045732009-05-25T00:22:10.759+08:002009-05-25T00:22:10.759+08:00Gabby 0.7% or 0.007 is the reciprocal of the squar...Gabby 0.7% or 0.007 is the reciprocal of the square root of 20,000. This is the general formula that I use to estimate "Margin of Error" based on sample size.Deany Bocobohttps://www.blogger.com/profile/01443168826029321831noreply@blogger.comtag:blogger.com,1999:blog-14974164.post-78317619895864603992009-05-24T11:50:13.458+08:002009-05-24T11:50:13.458+08:00how did you calculate.7%?how did you calculate.7%?GabbyDnoreply@blogger.com