Re: Two nit-picks re definition of p-value (Was: goodness of fit ?)


Kevin E. Thorpe wrote:
Reef Fish wrote:
Bruce Weaver wrote:
2. Isn't it really "EQUAL TO or greater than" rather than "greater
than"?

No! That's a nit pick on YOUR nit!

It depends on the STATEMENT of the Alternative Hypothesis.
If Ha is "greater than", the more extreme in p-value is "greater
than":
If Ha is "greater than or equal", the more extreme in p-value is
"greater
than or equal".

It makes no difference whether the test statistic is continuous OR
discrete. It never makes any difference when the test statistic
is continuous. But the STATEMENT and DEFINITION of p-value
follows directly the STATEMENT (and intent) of the Alternative
Hypothesis.


I know that for test statistics that have continuous sampling
distributions, the difference is trivial.

There would be NO difference. Probability at a point for a
continuous random variable (the Test Statistic) is always ZERO.

But not so for those with
discrete sampling distributions. Here's a binomial problem,

See the explanation above. No example is necessary.


for example, with a non-directional alternative hypothesis:

X = number of successes in N = 13 trials
p = p(success)
q = 1-p = p(failure)

H0: p EQ 0.5
H1: p NE 0.5

Observed X = 2

Whoa! What is your TEST STATISTIC?


X p(X|H0)
---------------
0 .0001
1 .0016
2 .0095 <-- observed X
3 .0349
4 .0873
5 .1571
6 .2095
7 .2095
8 .1571
9 .0873
10 .0349
11 .0095
12 .0016
13 .0001
---------------


I was taught to include the 0.0095 when computing the p-value:

But did your professor explain WHY? If he had bothered to, using
the definition of a p-value (as I'll show below), he would have
realized
his own error. That's only one of the troubles of ROTE teaching and
ROTE learning, without understanding the reason(s) WHY.

The Alternative Hypothesis is Ha: p .NE. 0,5

So, the definition of p-value for that problem would be
P(# Succeeses "more extreme" than the observed "2" successes)

= P (# Successes > 2 ) + P( # Failures < 2)

P(# Successes < 2) = .0016 + .0001.
P(# Failures < 2) is also .0016 + .0001,

which is why you multiplied by 2.


p = (0.0095 + 0.0016 + 0.0001)*2 = 0.0224

This is NOT correct. .0095 does NOT belong to the "more extreme"
definition of a p-value.


Do you agree? Thanks for clarifying.

You are welcome. You have to come back with better nits than those!!
:-)

As I recall, in Fisher's Exact Test, the probability of the observed
2x2 table is included in the test statistic. How does that fit with
the foregoing explanation?

You'll have to describe Fisher's Exact Test -- in terms of the
STATEMENT
of Ho and Ha, and how Fisher tests the two-tail alternative, THEN I can

tell you whether Fisher subscribes to the p-value definition or usage.


In Bruce Weaver's nit, the essence of his nit was the inclusion or
exclusion
of the P( exactly 2 successes) in the p-value.

For THAT example, it is unequivocal that the .0095 MUST be excluded,
in order to satisfy the DEFINITION of "more extreme" than the observed
value of the test statistic. "2" is NOT "more extreme" than "2".

-- Reef Fish Bob.

.

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