MS Module 22 chisq test phenotype equilibrium practice exam questions

(The attached PDF file has better formatting.)

The groups of phenotypes, R, S, and T, are in equilibrium if for some θ:

●    P(R) = p1 = θ2
●    P(S) = p2 = 2θ(1–θ)
●    P(T) = p3 = (1–θ)2

A sample from a population has the following number of observations in each group:

●    Group R: n1 = 101
●    Group S: n2 = 261
●    Group T: n3 = 138

The null hypothesis H0 is that the population is in equilibrium for some parameter θ.


Question 22.1: Maximum likelihood estimate for θ

What is the maximum likelihood estimate for θ?

Answer 22.1: (2 × 101 + 261) / (2 × (101 + 261 + 138) ) = 0.463

(formula derived by maximizing the loglikelihood is θ = (2n1 + n2) / 2(n1 + n2 + n3)


Question 22.2: Expected cell counts

What are the expected cell counts?

Answer 22.2: expected cell counts derived by formulas for p1, p2, and p3

total count = N = 101 + 261 + 138 = 500

●    n1 = N × p1 = N × θ2 = 500 × 0.4632 = 107.1845
●    n2 = N × p2 = N × 2θ(1–θ) = 500 × 2 × 0.463 × (1 – 0.463) = 248.6310
●    n3 = N × p3 = N × (1–θ)2 = 500 × (1 – 0.463)2 = 144.1845


Question 22.3: χ2 statistic

What is the χ2 statistic to test the null hypothesis that the population is in equilibrium?

Answer 22.3:  (observed – expected)2 / expected =

(101 – 107.1845)2 / 107.1845 + (261 – 248.6310)2 / 248.6310 + (138 – 144.1845)2 / 144.1845 = 1.237