Re: predicting sex by body height
- From: "Peter / Labo" <peterpc5j@xxxxxxxxxxx>
- Date: Wed, 17 Sep 2008 11:52:30 -0700 (PDT)
Hey there,
To solve this kind of problem, you need Bayes' theorem. It will take
into account different population sizes for men and women. z scores
probably won't help you much.
I assume that the sample means and sds you provide data for are based
on large samples that leave little room for estimation error.
Otherwise you'll have to move to hierarchical Bayes methods.
Let's write length as x, and the event of being a man M and of being a
woman W (or ~M, for "not M").
Bayes theorem states that P(M|x) = P(x|M)*P(M)/P(x). If you have
calculate this so-called posterior probability, you can then of course
deduce P(W|x), as P(~M|x)=1-P(M|x).
So, about the actual calculation... x is in principle a continuous
variable, with a densitity function rather than a discrete probability
function, but that doesn't change the methods. P(x|M), called the
likelihood, can be calculated for any value of x with the probability
density function of the normal distribution. P(M), called the prior
probability, the proportion of males in the population, should be
given or assumed, otherwise you can't get to the chances without a
sample to provide an estimate. P(x) is the probability (or probability
density) of observing the value x, no matter whether it comes from the
male or female subpopulation. Basic probability theory shows this to
equal P(x|M)*P(M)+P(x|W)*P(W).
If you find a x>200, and probabilistic inference leads you to the
conclusion that it's more likely to be a woman than to be a man, then
you will just have to accept that, right? If this is counterintuitive
to you, you will re-evaluate your assumptions about the distribution
of x for each group. Maybe sd=30 for female variation is too large?
Maybe one or both distributions are not normal? And just remember
that, the larger the number of cases you look at, the more likely it
gets that at least one of your observations will be "odd".
.
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