Re: Testing two proportions

Yes.


Let us suppose that previously (always previously) a
95% confidence level is chosen. As 97.x*% is larger
we conclude the opposite (reject the null
hypotheses), or expressing in a schematic way: 8/13
<= 3 /13 could not be true.
On the other hand, if 99% was chosen, then I cannot
reject the null Hypotheses and I must admit that 8/13
<= 3 /13
The basic on this *stuff*.
If we are interested (by reasons that are not
STATISTICAL, but economical-scientific) to have a
very conservative position concerning the null we put
the confidence level high:, 99%.
On contrary if we intend to put the null hypotheses
in a somewhat precarious position, a 95% should be
chosen.
This looks like *tolerance* in measurements.
(tolerance: the amount by which the size of a part of
a machine can be different from the standard size
before it prevents the machine from operating
correctly – Macmillan English Dictionary).

It (for ordinary persons, not for Jack) goes without
saying that the expression 8/13 IS NOT A FRACTION but
the statement that it was observed (in a random
experiment) eight successes in the total of thirteen
trials. In this context 8/13 is very different than
80/130 for example.
This point was yet stated in the inaugural post of
this thread.

__________licas (Luis A. Afonso)


Luis, your hypothesis statements are on p1 and p2, which you defined as 8/13 and 3/13, respectively. I happen to be 100% confident that 8/13 is not less than or equal to 3/13. You quoted von Braun, Oppenheimer, and Fermi, but presented no evidence that any of them ever believed that 8/13 <= 3/13.

If 8/13 and 3/13 are sample outcomes rather than p1 and p2, you should give them different names.

Jack
.