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The Chi-Square Test
	The chi-square test is a non-parametric technique which has a distribution similar
to the binomial distribution without being limited to two variables. The test is flexible
and easy to carry out. The test gives the probability that the differences between two
samples are not due to random variation. It effectively consists of a comparison between
the difference of observed and expected values squared divided by the expected value.

X2 = (Oi - E i)2 * (1/ E i)

Dividing by E i standardizes the statistic by weighting the contribution of each class so
that the largest classes do not necessarily produce the largest statistic (Thomas, 1976).
	The null hypothesis for calculating biological distance using the chi-squared
statistic is that the samples being compared are from the same original population. The
statistic is then compared with the chi-squared distribution tables to determine how close 
the statistic is to that obtained randomly. A result that would be expected to occur
randomly less than 5% of the time, is considered significant enough to reject the
hypothesis that there is no significant difference between the populations.
	An outcome that is not significant at the 5% level does not necessarily mean that
the two samples are from the same population, but may also mean that there is a close
affinity between the populations and the differences are slight. The results from the
chi-square test correlate well with those of other more sophisticated methods and is
generally considered sensitive enough given the nature of the data (Constandse-
Westermann, 1972). These methods all treat the traits as though they were independent
from each other which is not neccessarily true
	The chi-squared statistic for biological distance is calculated with the following
formula from Constandse-Westermann (1972): 

Dk2 = {(p1jk - p2jk)2/ pjkw}

To judge the significance of Dk2:

T2 = {(2n1 * 2n2) / (2n1 + 2n2)} * Dk2 

D = distance
p = the % of individuals possessing the trait
i = designates population
j = designates trait
k = class of trait
so, pijk = the frequency of k class of j trait in i population
pjkw = the weighted expected value calculated by: 

(2n1j* pijk + 2n2j* p2jk) / (2n1j + 2n2j)

(n = sample size)

	This X2 value is then multiplied by 100 to standardize sample sizes and
calculated for each trait between each population. All of the X2 values are added
together to get the X2 statistic. The degrees of freedom are calculated by k-1 and the
statistic with the given number of degrees of freedom is compared with the value from
the distribution table for a 5% significance level. If the X2 statistic is greater than would
be expected by chance, the null hypothesis is rejected.

Simple Example of Chi-Square test:
		p1jk 		 p2jk
		n=301		n=93
trait 1		0.220		0.218
trait 2		0.155		0.188

pjkw = (2n1j* pijk + 2n2j* p2jk) / (2n1j + 2n2j)

(602*0.220 + 186*0.218) / 788 = 0.222 weighted mean for trait #1
(602*0.155 + 186*0.188) / 788 = 0.163 weighted mean for trait #2

Dk2 = {(p1jk - p2jk)2/ pjkw}

{(0.220-0.218)2 / 0.222} + {(0.155-0.188)2 / 0.163} = 0.006699

T2 = {(2n1 * 2n2) / (2n1 + 2n2)} * Dk2

{(602*186) / (602*186)}* 0.006699 = 0.9519041, degrees of freedom= 1


The distribution table has a value of 3.84146 for 5% significance with one degree of
freedom. The differences between the two populations are not significant at the 5% level
for these two traits. 

Squared Difference
	To get an overall idea of the resemblance between populations, the coefficient of
relationship R can be calculated with the following formula:

R2 = d2jk / nj     d= (p1jk - p2jk)2 / n

variance R = (variance of p1jk - variance of p2jk) / nj

Pooled Dispersion Matrix
	In order to separate out the relative contributions of each trait to the difference
between populations, the pooled dispersion matrix cjkl  is calculated by this formula:

cjkl = {(n21ij* aijkl) +...+ (n22ij* aijkl)}/ (n1ij +...+ nij )

aijkl = [ {pijk (1-pijk) /n1ij ] in the form of a matrix, ie.
[ x   -x ]
[ -x   x ]

An example of the pooled dispersion matrix (Constandse-Westermann, 1972):
Relative contributions to the difference between two populations of the MN and ABO
systems:
		Pop1		Pop2		Pop3
gene freq.	n=100		n=200		n=200
M		0.20		0.50		0.30
N		0.80		0.50		0.70
p		0.40		0.30		0.50
q		0.10		0.30		0.20
r		0.50		0.40		0.30

aijkl for MN system:
pop1= [(0.20*0.80)/100] = 0.0016
pop2 = [(0.50*0.50)/100] = 0.00125
pop3 = [(0.30*0.70)/100] = 0.00105

cijkl = {(n21ij* aijkl) +...+ (n22ij* aijkl)}/ (n1ij +...+ nij )
{(100*100*0.0016) + (200*200*0.00125) + (200*200*0.00105)} / 500
cijkl = 108/500 = 0.216

contribution of MN system to djk of pop1 and 2=

[1/cijkl] (freq. n in p1 - freq. n in p2)2 = (1/0.216) (-0.30)2 = 0.4167 = 41.67%

calculating cijkl for the ABO system (remembering that aijkl is a matrix reduced above
by dropping one row and one column), gives a percent contribution of 65.77
The relative contribution can then be calculated for the differences between the other
sample combinations.

Angular Transformation
	For multinomial qualitative traits, a simple angular transformation can be used to
calculate biological distance. This method is based on the geometric formula a2 + b2 =
c2, with the square root of c equaling a linear distance value (Constandse-
Westermann, 1972). Using the symbol key above:

djk = p1jk - p2jk

The result is then expressed as an angle rather than a frequency through the linear
transformation, which also standardizes the variances:

pijk> = sin-1 * (square root of) pijk

the other angle, 0ijk = sin-1 * (1-2pijk)1

the variance pijk= 1/(4nij)     variance 0ijk = 1/ nij 

Using the weighted mean for trait 1 (0.222) from the above example:

pijk> = sin-1 * (square root of) pijk
sin-1* 0.471
28.11 degrees
variance pijk= 1/(4nij)
1/8
0.125
0ijk > = sin-1 * (1-2pijk)1
sin-1 * (0.556)
33.78 degrees
variance 0ijk = 1/ nij
0.125
The distance, a chord, between the two groups can be measured by subtracting pijk - 0ijk
and the 2 standardized variances can be added. 


Mean Measure of Divergence
	The linear transformation is also used in the Mean Measure of Divergence
formula (Berry and Berry, 1967). 
 
(01 - 02)2 - (1/n1 + 1/n2)

Where again the 0 = the angular transformation of the incidence frequency per
population into radians: 

0 = sin -1 (1-2p)  with variance = 1/n

D = 01 - 02 has variance (1/n1) + (1/n2)
and (01 - 02 ) / Var will have an approximate X2 distribution (df = 1)

The variance of D2 = 4D2 (1/n1 + 1/n2) for a pair of traits
The variance of D2 = 4 (1/n1 + 1/n2) sum D2 for a pair of populations

The variance for the MMD between two populations =
4 (1/n1 + 1/n2)[(01 - 02)2 - (1/n1 + 1/n2)] / n of traits

Freeman and Tukey (1950) define 0 as

0 = sin -1 [root of) r / (n+1)] + sin -1 [root of) (r+1)(n+1)] 
r = radians

The difference is apparently in the variance which is calculated incorrectly and
underestimated with the original MMD formula (Souza and Houghton, 1977).
This linear transformation however, is most neccessary when the sample sizes are under
100 and alpha is between 0.05- 0.95. For this study neither of those conditions
apply which leads me to believe that the formula from Berry and Berry (1967) will suffice.