The frequency of A allele may be estimated using exoression
	P(A) = [2 N(AA) + N(AT)]/[2 N] = [2 131 + 335]/[2 843] = 0.35409252669
Then, if Hardy-Weinberg equilibrium is true, expected frequencies 
of genotypes are
	P(AA) = P(A) P(A) = 0.125381517458
	P(AT) = 2 P(A) P(T) = 2 P(A) (1-P(A)) = 0.457422018465
	P(TT) = P(T) P(T) = (1-P(A)) (1-P(A)) = 0.417196464077
and the expected number of people with these genotypes are
	105.696619217, 385.606761566, 351.696619217 
in AA, AT and TT groups, respectively
The chi-square statistics is calculated to test is actual data fit 
HWE expectation:
	CHI2 = SUM (OBS-EXP)^2/EXP = 
		(131 - 105.696619217)^2/105.696619217 +
			(335 - 385.606761566)^2/385.606761566 +
				(377 - 351.696619217)^2/351.696619217 = 
		14.5196245667
This CHI2 has one degree of freedom, consequently, if CHI2>3.84, 
the null hypothesis of HWE is to be rejected.
In this case 14.5196245667 > 3.84 and, consequently, the null hypothesis is to be rejected.