There's still a problem with posting comments to the blog, apparently, so I'll say a bit more here, and then try to post a link on there [Edit: this was successful].
In his article, when discussing possible explanations for Bem's data, in the section entitled "Selective Reporting", Prof. Schimmack estimated the total number of 100-trial experiments that would have been needed to produce (on average) 9 that were significant at p<0.05. As p<0.05 occurs by chance one time in 20, the number of experiments would be 20x9=180, and the total number of trials would be 18,000. (Schimmack doesn't explicitly rule this possibility out, but I think the tenor of his comments suggests he finds it implausible. It seems very implausible to me, as it would involve almost 7 trials a day on every weekday for a decade.)
In his reply to my question, he discusses a scenario in which pilot experiments, each with 10 trials, would be run, and if they were significant at p<0.05 they would be continued with a further 90 trials, but otherwise they would be discarded. To end up with 9 significant 100-trial experiments, Schimmack's argument would again imply 20x9=180 experiments - but now pilots of only 10 trials each - and 9 continuations of 90 trials each, giving a total of 2610 trials. (The figure he gave in his comment was different from that, because apparently there was an arithmetic slip, and the calculation was for 10 significant trials, whereas Bem had 9.)
The problem with this estimate is that it implicitly assumes that if a pilot experiment of 10 trials is significant at p<0.05, then it will remain significant when 90 continuation trials are added to make a total of 100. Obviously that's not the case. Only a certain proportion of the selected pilot experiments would remain significant. The figure of 2610 is therefore an underestimate of the total number of trials required in this scenario.
If we assume all the trials are just binary choices with equal chances of success and failure (as most of Bem's were), then we can use the binomial distribution to allow for the loss of significance during the continuation phase, and work out the correct number of trials required. So that anyone interested can check my calculation, the details are given below. In doing the calculation, I have slightly relaxed the criterion of significance for the pilot phase, to p<0.055. (Unless this is done, the criterion requires 9 or more successes out of 10, which occurs in only about 1% of cases, and that pushes the total number of trials required up above 50,000.)
The answer is that in total about 18,400 trials are needed for the two-phase selection process involving pilot experiments with N=10.
Although this is similar to the figure in Schimmack's article for the scenario in which full experiments of 100 trials are run and then selected for significance, it should really be compared with the result of an exact binomial calculation for that case (because significance requires 59 or more successes out of 100, which happens only about 4.4% of the time). That gives a larger figure - about 20,300 trials. So the two-phase selection process is slightly more economical in terms of trials, but by only about 10%.
It might be asked whether a different choice for the size of the pilot experiments would reduce the number of trials required. But a corresponding calculation for pilots with 20 trials instead of 10 (with a similar tweak of the significance criterion in the pilot phase, to p<0.058), gives a similar figure of about 20,400 trials.
Therefore it seems that Schimmack's scenario, involving the selection of pilot experiments, would involve no significant saving in terms of the total number of trials, and therefore - to my mind - does not offer a plausible explanation for Bem's results.
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[Binomial probabilities from
http://stattrek.com/online-calculator/binomial.aspx]
For pilot experiments of 10 trials, with significance criterion p<0.055, there are three possible outcomes for significance:
(a) 10 out of 10, with probability 0.000977.
(b) 9 out of 10, with probability 0.009766.
(c) 8 out of 10, with probability 0.043945.
The total probability of significance for the pilot experiment is therefore 0.05469.
For significance at p<0.05 in a full experiment of 100, we require 59 or more successful trials (the probability of 59 or more successes is 0.04431).
So in these three cases, for the continuations of 90 trials, the possible outcomes for significance are:
(a) 49 or more out of 90, with probability 0.230396.
(b) 50 or more out of 90, with probability 0.171417.
(c) 51 or more out of 90, with probability 0.123053.
Forming the products of these with the probabilities above and adding the products, we find the total probability of significance in both phases is 0.007307.
Scaling these numbers up, the number of experiments needed for the expected number of significant results to be 1, is 136.9 pilots of 10 trials and 7.485 continuations of 90 trials, totalling 2043 trials.
Therefore for an expectation of 9 significant results, the total number of trials required would be about 18,400.