Re: Improving Lilliefors Test



In reference to my Sep 7, 2008 3:13 PM post, column k=1, I have to join that after to change n=19 --- > .197 the later adjustment gave
___________y = a * b^n * n^c
(y=critical value, alpha=0.05):
___________a = 0.68517 539
___________b = 0.99725 938
___________c = -.40588 224
The degree of adjust is such that
__________ sigma 0.00025, r = 0.99997 509
I do remember that each critical value was obtained based on 400 000 sized n normal samples.

Luis A. Afonso
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