On 3/5/2020 2:45 AM, Bruce Kellett wrote:
Now sequences with small departures from equal numbers will still give probabilities within the confidence interval of p = 0.5. But this confidence interval also shrinks as 1/sqrt(N) as N increases, so these additional sequences do not contribute a growing number of cases giving p ~ 0.5 as N increases. So, again within factors of order unity, the proportion of sequences consistent with p = 0.5 decreases without limit as N increases. So it is not the case that a very large proportion of the binary strings will report p = 0.5. The proportion lying outside the confidence interval of p = 0.5 is not vanishingly small -- it grows with N.
I agree with you argument about unequal probabilities, in which all the binomial sequences occur anyway leading to inference of p=0.5. But in the above paragraph you are wrong about the how the probability density function of the observed value changes as N->oo. For any given interval around the true value, p=0.5, the fraction of observed values within that interval increases as N->oo. For example in N=100 trials, the proportion of observers who calculate an estimate of p in the interval (0.45 0.55) is 0.68. For N=500 it's 0.975. For N=1000 it's 0.998.
Confidence intervals are constructed to include the true value with some fixed probability. But that interval becomes narrower as 1/sqrt(N). So the proportion lying inside and outside the interval is relatively constant, but the interval gets narrower.
Brent -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/f0a35d39-6287-2955-bcf5-3f6bc01c1317%40verizon.net.
