Biology Reference
In-Depth Information
extensive MLE package (maxLik, see
http://www.maxlik.org/
). To use MLE we must first
write the log-likelihood function, an element of which is what the probabilities are that brows
would have been scored as “M,” “?,” or “F” given known sex. For the current example we
will simply use our data from the reference sample. Thus, for a male the probability of an
“M” score is 51/60, the probability of a “?” score is 4/60, and the probability of an “F” score
is 5/60. For a female the probabilities of the three scores are 16/54, 6/54, and 32/54, respec-
tively. In matrix form the log-likelihood function is
2
4
3
5
;
51=60
35
11
54
4=60
5=60
ln LK ¼ ln
p
M
1 p
M
(11.1)
16=54
6=54
32=54
which after the matrix multiplications is
17p
m
20
p
m
þ
8ð1 p
m
Þ
27
15
þ
1 p
m
ln LK ¼35 ln
þ 11 ln
9
p
m
(11.2)
12
þ
16ð1 p
m
Þ
þ 54 ln
:
27
Since we harbor no particular affection for or facility with algebra, we obtained Equation
11.2
from Equation
11.1
using wxMaxima (see
http://andrejv.github.com/wxmaxima/
), a free-
ware symbolic algebra and calculus package that can be downloaded for Windows, Mac
OS X, or as source code. Equation
11.2
has a single parameter (p
m
), so we can have the max.-
Lik object in “R” search across the function to find the maximum likelihood, which occurs at
a value of 0.0982. This means that we estimate that about 10% of our sample of 100 crania is
male and 90% is female, based on our previous scores of 35 crania with a “male” brow, 11
“undetermined,” and 54 with a “female” brow.
Having a sample that is only 10% male seems like a substantial deviation from the
expected 50% (based on the human sex ratio of approximately 1:1), so it would be useful
to have some indication of how well the parameter is estimated. While skeletal biology arti-
cles are replete with analyses and reported percentages that do not have standard errors, this
should and can be avoided. We calculate the variance of the estimate by taking the reciprocal
of the negative of the second derivative of the log-likelihood. The second derivative of the
log-likelihood can be found using the maxLik function “hessian.” From this we obtain a stan-
dard error (the square root of the variance of the estimate) equal to 0.0855. Assuming a normal
distribution, we expect about 95% of the estimates to fall within 1.96 standard errors of the
true population value for the proportion of males. Consequently, we have a 95% confidence
interval of
0:0982 1:96 0:0855;
or from
0.06938 to 0.26578. In other words, 95% of the
estimates place the percentage of males in our sample between
6.95% and 26.58%, which
removes a 1:1 (50%) sex ratio from the realm of possibility. Note that the percentage of males
cannot be below 0%, or conversely that the percentage of females cannot be above 100%. We
obtained a negative percentage because the percentage of males is low in this example and
our estimate has a large standard error. It would be more correct in this setting to give
a “one-tailed” 95% confidence interval (
0:0982 þ 1:64 0:0855
) of from 0 to 23.84%, which
again does not include a 1:1 sex ratio.
