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Luis A. Afonso
Posts:
4,743
From:
LIsbon (Portugal)
Registered:
2/16/05
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Neyman-Pearson problem ?
Posted:
Oct 19, 2012 1:02 PM
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Neyman-Pearson problem ?
I have no intention to improve at this matter; however, I had devise a strategy based on the outputs of multi-tests on a single sample.
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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), p77-79.
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Table 1- Cut-off 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.
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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 cut-off quantities.
Luís Amaral Afonso
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