Question

In a certain population of newts, being poisonous (P) is dominant over not being poisonous (p). In the parent population, the genotype frequencies are:

PP: .60

Pp: .34

pp: .06

In a sample of 175 newts washed downstream after a storm, you find that 91 are PP, 71 are Pp and 13 are pp. You would like to test whether the new population of newts follows the same genotype frequency distribution as the parent population at a significance level of 5%.

Answer 1

We are given the observed frequencies here as:
O(PP) = 91,
O(Pp) = 71,
O(pp) = 13

Using the given proportions, the expected frequencies are computed here as:
E(PP) = 175*0.6 = 105
E(Pp) = 175*0.34 = 59.5
E(pp) = 175*0.06 = 10.5

Therefore the chi square test statistic here is computed as:

Now for n - 1 = 2 degrees of freedom, the p-value here is computed from the chi square distribution tables as:

As the required p-value here is 0.1 > 0.05 which is the level of significance, therefore the test is not significant and we cannot reject the null hypothesis here. Therefore we dont have sufficient evidence here to reject the hypothesis that the new population of newts follows the same genotype frequency distribution as the parent population