wwwpub.utdallas.edu/~herve/Abdi-Lillie2007-pretty.pdf

UUUUUUUUUUUUUUUUUUUUUUUUUUUUU

MY VALUES




size___alpha=5%_1% ___Connover_____Abdi.
_10___0.264___0.305___.258_.294__.2616_.3037
_15___0.220___0.255___.220_.257__.2196_.2545
_20___0.192___0.224___.190_.231__.1920_.2226
_25___0.174___0.202___.173_.200__.1726_.2010
_30___0.159___0.185___.161_.187__.1590_.1848
_35___0.148___0.172_____________.1478_.1720
_40___0.139___0.161_____________.1386_.1616
_45___0.131___0.152_____________.1309_.1525
_50___0.124___0.145_____________.1246_.1457

(for each sample size, 500´000 samples were
simulated
by my work, 100´000 by Abdi & Molin).


JACK TOMSKY is so unlearned and shameless that
deserves to be exposed every time he posts an opinion
on Hypotheses Tests. The less experience people
should take in attention that HE IS A CLOWN.

Readers. Do appreciate what I found out at WEB.

*** Lilliefors/Van Soest´s test of normality ***
_____Hervé Abdi & Paul Molin

1. OVERVIEW

The normality assumption is at the core of the
majority of standard statistical procedures, and it
is important to be able to test this assumption. In
addition, showing that a sample does not come from a
normally distributed population is sometimes of
importance per se. Among the many procedures used to
test this assumption, one of the most well-known is a
modification of the Kolmogorov-Smirnov test of
goodness of fit, generally referred to as the
Lilliefors test for normality (or Lilliefors test for
short).This test was developed independently by
Lilliefors (1967) and by Van Soest (1967). The null
hypotheses for this test is that the error is
normally distributed (i.e. there is no difference
between the observed distribution f and the normal
distribution). The alternative hypotheses is that the
error is not normally distributed.
Like most statistical tests, this test of normality
defines a criterion and gives its sampling
distribution. When the probability associated with
the criterion s smaller than a given [alpha]-level,
the alternative hypotheses is accepted (i.e. we
conclude that the sample does not come from a normal
distribution). An interesting peculiarity of the
Lilliefors test is the technique used to derive the
sampling distribution of the criterion. In general
mathematical statisticians derive the sampling
distribution of the criterion using analytical
techniques. However in this case, this approach fails
and consequently Lilliefors decided to calculate an
approximation of the sampling distribution by using
the Monte Carlo technique.
Essentially the procedure consists of extracting a
large number of samples from a Normal Population and
computing the value of the criterion for each of
these samples. The empirical distribution of the
values o the criterion gives an approximation of the
sampling distribution of the criterion under the null
hypotheses.
Specifically, both Lilliefors and Van Soest used, for
each sample size chosen, 1000 random samples derived
from a standardized normal distribution to
approximate the sampling distribution of a
Kolmogorov-Smirnov criterion of goodness of fit. He
critical values given by Lilliefors and Van Soest are
quite similar, the relative error being of the order
of 10^ (-2).
According to Lilliefors (1967) this test of normality
is more powerful than others procedures for a wide
range of nonnormal conditions. Dagnelie (1968)
indicated, in addition, that the critical values
reported by Lilliefors can be approximated by an
analytical formula. Such a formula facilitates
writing computer routines because it eliminates the
risk of creating errors when keying in the values of
the table. Recently, Molin and Abdi (1998), refined
the approximation given by Dagnelie and computed new
tables using a larger number o runs (i.e. K=100,000)
in their simulations. ***
(End of citation).

____TOMSKY´s ABSOLUTELY K.O.!!!!!

____licas (Luis A. Afonso)

Significance level, alpha, by CDF



Acceptance (no-rejection) interval,
right bounded: interval (-infinity, b] such that
________1- alpha = F(b)

The rejection is (b, infinity) defined by
________ alpha = 1- F(b) - p(X=b)

This way to define them is the same the density f(X)
be continuous, i.e. p(X=b)=0 or has at *b* a
discontinuity.

Jack Tomsky wrote:

*** What happens if there is no b such that 1- alpha
= F(b)? Then your acceptance region is undefined.
Jack ***

My response

It seems to me *homework* *master* Jack. You must
tell what you got yet in this matter. Ask your
teacher to direct you in the right way.

________licas (Luis A. Afonso)

***Jack wrote:

You obviously don't know that the answer is that your
definition is wrong. Jack ***

My response

Let be X= Bin (n=4, p=0.5)
___p(X=0) = 1/16
______ F(0) = 1/16
___p(X=1) = 4*1/16
___ F(1) = 5/16
___p(X=2) = 6*1/16
___1- F(2) = 1-11/16 = 5/16
_________ = 0.3125
___p(X=3) = 4*1/16
___1- F(3) = 1-15/16 =1/16
__________= 0.0625
___p(X=4) = 1*1/16
______F(4) = 16/16

if we try to solve X such that
_________ 1- F(X) = alpha = 0.05
_____(I)
_________ you get a value between
X=3 and X=4.
THERE IS SUCH A VALUE for X.
________Equation (I) has not solution.

When you Jack Tomsky start to *mettre en marche* your
little grey cells (Hercule Poirot)?

_______licas (Luis A. Afonso)

Jack Tomsky wrote

*** According to your own calculations, there is no x
for which 1-F(x) = 0.05.

For 3 <= x < 4, 1-F(x) = 0.0625.
For x = 4, 1-F(x) = 0. Jack
***

My response

*** And according of yours, Tomsky, what is the x
value that makes F(x)=0.95 ? ***

________licas (Luis A. Afonso)

I MUST CONCLUDE that your comment:

*** You obviously don't know that the answer is that
your definition is wrong.***

It is one more STUPID and UNLEARNED statement of
yours.

The Significance Level ,alpha, defined by means of
the Cumulative Distribution Function is correct,
exactly the opposite you claimed.

______licas (Luis A. Afonso)