Question

Let A and B represent two variants (alleles) of the DNA at a certain genome locus (chromosome location). Assume that 40% of all the alleles in a certain population are type A and 30% are type B. The locus is said to be in Hardy-Weinberg equilibrium if the proportion of organisms that are of type AB is (0.40)(0.30) = 0.12 (in other words, alleles A and B are independent). In a sample of 300 organisms, 42 are of type AB. Can you conclude that this locus is not in Hardy-Weinberg equilibrium? 1. Define the parameter of interest and state the hypotheses which should be tested. 2. Compute the test statistic z and the corresponding p-value for your hypotheses. 3. What do you conclude at α = 0.01?

Answer 1

1. Define the parameter of interest and state the hypotheses which should be tested.

Here, we have to use one sample z test for population proportion. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: Locus is in Hardy-Weinberg equilibrium.

Alternative hypothesis: H_a: Locus is not in Hardy-Weinberg equilibrium.

H₀: p = 0.12 versus H_a: p ≠ 0.12

2. Compute the test statistic z and the corresponding p-value for your hypotheses.

We are given

Number of favourable items/obs. = X = 42

Total number of items/obs. = n = 300

Sample proportion = P = X/n = 42/300 = 0.14

Test statistic z is given as below:

Z = (P – p)/sqrt(p*(1 – p)/n)

Z = (0.14 - 0.12)/sqrt(0.12*(1 - 0.12)/300)

Z = (0.14 - 0.12)/ 0.0188

Z = 1.066004

P-value = 0.2864

(by using z-table or excel)

3. What do you conclude at α = 0.01?

We are given

α = 0.01

P-value = 0.2864

P-value > α

So, we do not reject the null hypothesis that Locus is in Hardy-Weinberg equilibrium.

There is insufficient evidence to conclude that Locus is not in Hardy-Weinberg equilibrium.