The Hardy-Weinberg law says the following. If gene frequencies are in equilibrium, the genotypes AA, Aa, and

Question:

The Hardy-Weinberg law says the following. If gene frequencies are in equilibrium, the genotypes AA, Aa, and aa occur in a population with frequencies θ2, 2θ(1−θ), and (1−θ)

2. In an i.i.d. sample of size n, with each outcome being an AA, Aa, or aa with the above probabilities, let X1, X2, and X3 be the observed counts. For example, X1 is the number of trials where the observation is AA. Note that X1 + X2 + X3 = n. The joint distribution of (X1, X2, X3) is a trinomial distribution. Hence, Pθ{X1 = x1, X2 = x2, X3 = x3} = n!

x1!x2!x3!

(θ2

)

x1 [2θ(1 − θ)]x2 [(1 − θ)

2

]

x3 for any nonnegative integers x1, x2, and x3 summing to n. Find the MLE and its limiting distribution (suitably normalized). Derive the likelihood ratio and chi-squared tests to test the Hardy-Weinberg law.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question

Testing Statistical Hypotheses

ISBN: 9781441931788

3rd Edition

Authors: Erich L. Lehmann, Joseph P. Romano

Question Posted: