The frequency of A allele may be estimated using exoression
	P(A) = [2 N(AA) + N(AT)]/[2 N] = [2 27 + 219]/[2 660] = 0.206818181818
Then, if Hardy-Weinberg equilibrium is true, expected frequencies 
of genotypes are
	P(AA) = P(A) P(A) = 0.0427737603306
	P(AT) = 2 P(A) P(T) = 2 P(A) (1-P(A)) = 0.328088842975
	P(TT) = P(T) P(T) = (1-P(A)) (1-P(A)) = 0.629137396694
and the expected number of people with these genotypes are
	28.2306818182, 216.538636364, 415.230681818 
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 = 
		(27 - 28.2306818182)^2/28.2306818182 +
			(219 - 216.538636364)^2/216.538636364 +
				(414 - 415.230681818)^2/415.230681818 = 
		0.085275583703
This CHI2 has one degree of freedom, consequently, if CHI2>3.84, 
the null hypothesis of HWE is to be rejected.
In this case 0.085275583703 < 3.84 and, consequently, the null hypothesis holds.