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Luis A. Afonso
Posts:
4,743
From:
LIsbon (Portugal)
Registered:
2/16/05


Re: Common sense and NHST
Posted:
Jun 20, 2013 12:10 PM


As promised I will show that a sequence of two nosignificant tests could very well lead to a significant over all result: two not sufficient evidence to reject H0  > reject this hypothesis. It is used the fact that pvalues, p(i), do follow a Uniform Distribution *** and k of them are such that (Ronald Fisher) : ________H = 2 * Sum (1<= j <= k) Log(p(j)) H is distributed like a Chisquare, k degrees of freedom. *** The conditions: ___1___i.i.d. data, ___2___Testing H0: miu= true Population value (not shifted)
Then for k=2, H = 2*Log(p(1))  2*Log(p(2)) 5% significance Critical Value H = 5.99146. Simulating 1000´000 p(2), random numbers, when fixed p(1) successively from 0.05(0.05)0.95 the following Table was calculated: ____________p(1)___prob. to 5% significant ____________0.05________1.000__ ____________0.10________0.500__ ____________0.15________0.332__ ____________0.20________0.251__ ____________0.25________0.200__ ____________0.30________0.167__ ____________0.35________0.143__ ____________0.40________0.125__ ____________0.45________0.110__ ____________0.50________0.098__ ____________0.55________0.091__ ____________0.60________0.078__ ____________0.65________0.076__ ____________0.70________0.067__ ____________0.75________0.067__ ____________0.80________0.064__ ____________0.85________0.060__ ____________0.90________0.055__ ____________0.95________0.051__ The 2nd column is the frequencies to which H was exceeded owing to p(2)=RND. Note that for p(1)=0.95 we almost attain the 5% level, as it must be from construction when p(1)=1.00000.
Luis A. Afonso



