Question

Description for Question 7. A multinomial distribution for three nonnegative counts X1, X2, X3 has joint pdf given by 23 P(X1 = X1, X2 = 22, X3 = x3) (21.3.2.) pi? pºp3", X1 X2 X3 where pi + P2 + P3 1. For genotypes AA, Aa, and aa, the Hardy-Weinberg model puts the respective genotype proportions in the population at (1 - 0)?, 20(1 – 0), and 02, where 0 is the gene frequency of gene type "a" (0 < a < 1). A sample of n individuals is drawn, and the counts of AA, Aa, and aa individuals are respectively x1, X2, X3. We know that under a multinomial sampling model for the counts, the maximum likelihood estimate of 0 is (3x2 + x3)/(x1 + x2 + x3). 7. Construct a statistical hypothesis test of whether or not the Hardy-Weinberg model is a plausible description of the data. Give sufficient detail in the formulas for a biologist to be able to calculate the test statistic and look up any critical values for the test.

Answer 1

For genotypes AA, Aa and aa, the hardy-Weinberg model pets the respective genotype proportions in the poppe at p. = (1-02 P2 = 20 (1-0) and P₂ = 82 where o is the gene frequency of gene type "a" 02021. obviously, &, +h2 + p ₂ = 1 A random sample of individuals is drawn and the counts of AA, Aa and aa individuals are respectively xe, de, dz. obviously, tht do tzn. We want to test Ho: Handy-Weinberg model fits the data against t not to. under Ho, 24, Hg, xz follows a multinomial dist with pmf X₂ 24 X₂ X3 Evin 2 X 3 (1-0)***** x da X3 Departure of the data from Ho is measured by the quantity 7(0) = 3 (Xx - n P x)2 (10) 12 $160,X9 ) ( 2 ) (-6)*60)*(-6)*;*** 6, 23) 0**** (1-2 spx K=1 but since pies are unknown as & is unknown; we have to put an estimate of 0. Under a multinomial sampling model for the counts, the mle of d is da ^ 2 + H₂ + X₂ + x3 X + 23 n By invariance property of mestimators, 2 n Ng + H3 ha 12 1 = (1-6)* = (****2) y = 2'ô(-6) - 2 + ť s = 82= (5x+3) n 2

mph 3 kal For testing the hypothesis problem at a level of significance, Test statistic is TÔ) = 3 we - nfhu)" We can use to compute the test statistic. Infact for simplication of calculation, we can write Tê) as & Gak - - nx thanh Xx 3 + [nok -2{xx nh since Exxana 2 = 1 can also use in the above formula to be able to calculate the test statistic For critical values of the test, assuming to be large we note that to) a x² dist 31 but while estimating ô K=1 K Kal K=) 2 de - K2) K=1 KE) We = x under to +(ô) a x 2 =xt, diste under to where & E(0,1), H is rejected if Tlo) 78 on where P(+(6) >c) & as no Gal Thus, c = 32 upper & point of x diste