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Talk:PNAS Response to Letter (view source)
Revision as of 19:54, September 23, 2008
, 19:54, September 23, 2008no edit summary
::::::::: I've been following this discussion for a while and I have to agree with ASchlafly. It hardly seems fair that he should have to, in his spare time, replicate an experiment done by a professional just to "earn" the right to criticize it. I am unfamiliar with statistics, but if some complicated transform goes against common sense, common sense should prevail. After all, there are lies, damned lies, and statistics... [[User:AndyM|AndyM]] 10:57, 23 September 2008 (EDT)
::::::::: I've been following this discussion for a while and I have to agree with ASchlafly. It hardly seems fair that he should have to, in his spare time, replicate an experiment done by a professional just to "earn" the right to criticize it. I am unfamiliar with statistics, but if some complicated transform goes against common sense, common sense should prevail. After all, there are lies, damned lies, and statistics... [[User:AndyM|AndyM]] 10:57, 23 September 2008 (EDT)
(unindent)I'm not asking anyone to write a paper or replicate an experiment. I'm asking ASchlafly to support his statement "The results from a very large sample size would not be weighted equally with the results from a small sample size, '''as you and Lenski have done'''"(bolding mine). I have stated publicly, subject to challenge by others, that the sample sizes (n) of the three replays are four, five, and eight respectively. Furthermore, using n of 4, 5, and 8 in the weighted Z-method DOES NOT weight all the replay experiments equally - it weights replay 3 twice as much as replay 1 and 8/5 as much as replay 2. Tell you what: I'll drop all my questions about Monte Carlo and the Z-transform, and simply ask ASchlafly one question: '''what is the sample size, n, of the second replay experiment?''' He need not even do any calculations - a statement in words that will allow someone else to do the calculation will suffice. This is not a complicated question to answer; the paper states how many replicate cultures there were (340), how many cells there were in each replicate (3.9x10^8), how many replicates gave rise to Cit+ cells (5), and which generations those Cit+ replicates came from (4 from 32,000 and one from 32,500). I will even give ''my'' answer: '''five'''. Furthermore, I will say ''why'' I believe that, using the murderer/age analogy: performing the 340 replicates is the same as interviewing 340 people in order to find out if any of them are convicted murderers. Finding that five replicates gave rise to Cit+ mutants is the same as the survey finding that 5 of those 340 people were convicted murderers. Finding that the Cit+ mutants arose from 4 replicates from generation 32,000 and 1 from generation 32,500 is the same as finding the ages of the murderers. The five data points in the Lenski study allow one to calculate the 'mean generation of clones yielding Cit+': 32,100. This is the same as finding the mean age of the five murderers. If I want to compare this hypothetical murderer age study to some other study of the mean age of murderers, I would weight the studies based on how many murderers were in each study, not on how many non-murderers were included in the initial survey.
Surely ASchlafly can say ''what'' he thinks the n of the second replay is, even if he won't say ''why'' he thinks it. Is it five? 340? The number of replicates times the number of cells per replicate? Something else? No analysis need be performed on the resulting number.--[[User:Brossa|Brossa]] 15:54, 23 September 2008 (EDT)