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Re: st: Re: single sample pre/post comparison of proportions
From |
José Maria Pacheco de Souza <[email protected]> |
To |
<[email protected]> |
Subject |
Re: st: Re: single sample pre/post comparison of proportions |
Date |
Thu, 11 Jun 2009 09:03:15 -0300 |
bound p(known), test a hypothesis Ha: p(new)>p(known) vs H0:
p(new)=p(known), using the at risk of improving.
Or as Svend presented, just estimate the proportion of new, among those
at risk. In this case, aftward it will difficult to resist the temptation to
compare this result with the p(known).
Cheers,
José Maria
Jose Maria Pacheco de Souza, Professor Titular (aposentado)
Departamento de Epidemiologia/Faculdade de Saude Publica, USP
Av. Dr. Arnaldo, 715
01246-904 - S. Paulo/SP - Brasil
fones (11)3061-7747; (11)3768-8612;(11)3714-2403
www.fsp.usp.br/~jmpsouza
-----
Michael asked and Joseph responded (not shown) - and Michael then wrote:
The suggestion of a one-sample test restricted to pre-intervention
ADOPT=NO crowd makes sense. I think you are also sneakily suggesting that
the most obvious null hypothesis -- "H0: p = 0" is not a good choice;
there would probably be some adoption even in the absence of the
intervention, and the intervention probably cannot be called a success
unless the proportion of adopters exceeds a minimum cost/benefit
threshold. Instead, I could choose, e.g., "H0: p < .25" (a one-tailed
test). That seems reasonable.
===============================================================
I wonder whether a P-value related to a somewhat arbitrary null hypothesis
is useful. I think the following is more informative:
Assume that you had 90 participants, 40 of whom already had the good
habit, leaving 50 "at risk" for improvement. 20 (40%) of these improved.
The 95% CI for this estimate is 26%-55%:
. cii 50 20 , binomial
-- Binomial
Exact --
Variable | Obs Mean Std. Err. [95% Conf.
Interval]
-------------+---------------------------------------------------------------
| 50 .4 .069282 .2640784
.548206
Hope this helps
Svend
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