On Tue, 5 Oct 2010, "Ingo Althöfer" wrote:
Hello Jonas,
The scores were
A vs C 3-1
B vs C 0-4
With respect to the 5 % level this refutes the hypothesis that
A is not stronger than B.
To prove this you have to find upper bounds for
(1-p)^3 * [1-p + 4p] * p^4, where p is in the interval [0,1].
This maximum is clearly below 0.05.
I do not completely agree with this calculation.
What you have done could be seen as multiplying p-values. Forbidden.
Why?
Probabilities of independent events can be multiplied.
Because multiplying p-values does not yield a p-value. Maybe there was
some misunderstanding: that's what I describe as forbidden, not your
calculation of probabilities of single events, or their summation. The
fact that you agree further on that I am correct after that shows that you
understand it.
By the way, for non-statisticians: to compute a p-value, you have to put
a total order on all possible events. Even if they are naturally
multidimensional. Ideally this order should be: the greater the observed
event, the less ``likely'' it is that H0 is true. Then your p-value is the
probability of seeing the observed event or any greater event for the
order you have used.
Jonas
In total I did not only multiply, but built an expression
which had both multiplication and summation.
Specifically, you have computed the probability under p of having both
A-C 3-1 or better for A
B-C 0-4 or worse for B
The problem is that you are comparing A to B. So you have to take all
the cases where the apparent difference in strength is at least as big.
This notably includes the case:
A-C 4-0
B-C 1-3
You are correct.
In fact in this case that's the only part to be added.
Right. And I had included that term in my computation
a year (or so) ago.
We still get a p-value less than 0.04, it's still significant.
Correct.
Ingo.
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