[R] shapiro wilk normality test
(Ted Harding)
Ted.Harding at manchester.ac.uk
Sun Jul 13 22:59:21 CEST 2008
On 13-Jul-08 19:53:47, Johannes Huesing wrote:
> Frank E Harrell Jr <f.harrell at vanderbilt.edu> [Sun, Jul 13, 2008 at
> 08:07:37PM CEST]:
>> (Ted Harding) wrote:
>>> On 13-Jul-08 13:29:13, Frank E Harrell Jr wrote:
>>>> [...]
>>>> A large P-value means nothing more than needing more data. No
>>>> conclusion is possible. Please read the classic paper Absence of
>>>> Evidence is not Evidence for Absence.
>>>
> [...]
>>
>> It's real. Full text is available to all:
>> http://www.bmj.com/cgi/content/full/311/7003/485
>
> The quotation is attributed to the late Carl Sagan who
> seemed to have used it as a strawman argument , see
> http://oyhus.no/AbsenceOfEvidence.html.
This citation of Sagan, and the link therein to Sagan quotes:
http://en.wikiquote.org/wiki/Carl_Sagan
are interesting, as far as they go. However, I disagree with the
proof ("by conditional probability") that absence of evidence is
evidence of absence.
Definition 1 is disputable. But, whether one agrees with it or not,
Definition 2 does not correspond to my interpretation of "absence
of evidence". If A is evidence for B (in terms of P(B|A) etc.),
this means that if we *know* that A is the case, or that not-A
is the case, then we can say something about P(B). But "absence
of evidence", in my interpretation (which I believe is right for
the statistical context of "non-significant P-values"), means that
we do not know about A: we do not have enough information.
That proof needs to be discussed in terms of the available evidence
for A!
The proof is, basically, given in terms of a 2-valued logic where
every term is either TRUE or FALSE. In the real world we have at
least a third possible value: UNKNOWN (or, as R would put it, NA).
Even if you accept (Definition 1) that
"A is evidence for B" == P(B|A) > P(B|not-A)
what can you possibly say about P(B|NA) (other than that it is NA
itself)?
Best wishes to all,
Ted.
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Date: 13-Jul-08 Time: 21:59:16
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