Changes

::Oh, and while we're at it, here's some (still voluntary response-based!) [http://en.wikipedia.org/wiki/Category:Moderate_Wikipedians data] for the third category. [[User:John Brown|John Brown]] 22:07, 15 March 2007 (EDT)

::Oh, and while we're at it, here's some (still voluntary response-based!) [http://en.wikipedia.org/wiki/Category:Moderate_Wikipedians data] for the third category. [[User:John Brown|John Brown]] 22:07, 15 March 2007 (EDT)

A bit more can be said when there are moderates. If we accept that WL/AL = ((WL/(WC+WL+WM))/((AL/(AC+AL+AM)) = (2 + 1 + AM/AL)/(1/3 + 1 + WM/WL) = (3 + AM/AL)/(4/3 + WM/WL) = 6, the the lower bound on AM/AL is when WM = 0, so (3 + AM/AL)/(4/3) = 6, so 3 + AM/AL = 8, so AM/AL = 5. The data presented support the conclusion that there are at least 5 times as many moderate Americans as liberal Americans. [[User:Hdelbrook|Hdelbrook]] 22:21, 15 March 2007 (EDT)

 

A bit more can be said when there are moderates '''if we accept as input data''' the assertion that WL/AL = 6 (it does not follow from the other input data). Then WL/AL = ((WL/(WC+WL+WM))/((AL/(AC+AL+AM)) = (2 + 1 + AM/AL)/(1/3 + 1 + WM/WL) = (3 + AM/AL)/(4/3 + WM/WL) = 6, the the lower bound on AM/AL is when WM = 0, so (3 + AM/AL)/(4/3) = 6, so 3 + AM/AL = 8, so AM/AL = 5. The data presented support the conclusion that there are at least 5 times as many moderate Americans as liberal Americans. [[User:Hdelbrook|Hdelbrook]] 22:21, 15 March 2007 (EDT)