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Luis A. Afonso
Posts:
4,699
From:
LIsbon (Portugal)
Registered:
2/16/05


NeymanPearson problem ?
Posted:
Oct 19, 2012 1:02 PM


NeymanPearson problem ?
I have no intention to improve at this matter; however, I had devise a strategy based on the outputs of multitests on a single sample.
_____________ The Tests below, for Normality, were T1= Lilliefors, T3= Cramer  von Mises, the 2nd , testing Uniformity, is: L. Amaral Afonso, Pedro Duarte, Revue de Statistique Appliquée, tome 40, nº1, (1992), p7779. ________________
Table 1 Cutoff values and right tail probability, Normal Data, when a triple test is performed
___N=60______0.170____0.5075____0.476___ _____________(5.15%)__ (5.25)____ (5.53)__ ___N=70______0.165____0.505_____0.535___ _____________ (5.01)___ (5.37)____ (5.27)__ ___N=80______0.160____0.503____ 0.590___ _____________ (4.93)___ (5.18)____(5.05)___ ___N=90______0.156____0.502____ 0.645___ _____________ (5.25)___ (4.94)____(4.93)___ ___N=100_____0.153___ 0.501_____0.700___ _____________(5.18)____(4.72)____(4.97)___
For example suppose a Neyman  Pearson test as this: H0: Normal against H1: Uniform. Given that, with N=60, the output frequencies are: __Normal Data: [1 0 0] = 0.526, [1 0 1] = 0.475 __Uniform : [1 0 0] = 1.000 The latter case: the 1st test always rejects normality, the 2nd and 3rd fails to reject.
We dare if we, testing a sample, observe [1 0 0] we can say that alpha, the probability to reject normality when testing normal data is as high as, 1.000  0.526 = 0.474.
________________
Let N=100, this time: __Normal Data: [1 0 0] = 0.950, [1 0 1] = 0.050 __Uniform : [1 0 0] = 1.000
and alpha= 5%. This value is exactly the one we ascribe to the cutoff quantities.
Luís Amaral Afonso



