Question

Section 4: Sex ratio variation You want to test whether the sex ratio for a species of birds is constant, or if there can be variation among nests in the probability that a chick is female. To do this, you look at nests with 3 chicks in them, counting the number of females in each. You obtain the results in the table below. Number of female chicks in nest Number of nests 32 23 65 33 A) What are the odds that a randomly selected chick is female?(4 points) B) what are the expected counts of nests with each number of females ir the probability that a chick is female is constant among nests? (6 points) C) What test statistic would you calculate to perform the test? Be specific about how you would perform the calculation, not just the name of the statistic. (But you do not need to calculate it). (4 points) D)What threshold would you use for the above statistic if you wanted to test for variation in sex ratio among nests with a false positive rate of 0.01? (3 points)

Answer 1

(a) Total number of families = 32 + 23 + 65 + 33 = 153

Total number of females = 0 * 32 + 1 * 23 + 2 * 65 + 3 * 33 = 252

Total number of chicks = 153 * 3 = 459

Pr(A random chick is a female)= 252/459 = 0.549

in terms of odds = 252 : 459 - 252 = 252 : 207 = 84 : 69

(b) Here that means Pr(A chick is female) = 0.5

here there are 153 families

so expected number of familiies with x number of females = 153 * ³C_x (0.5)^x(0.5)^(3-x)

so here for x = 0, expected number of famility with no females = 153 * (0.5)³ = 19.125

for x = 1, expected number of famility with no females = 153 * 3 * (0.5)³ = 57.375

for x = 2, expected number of famility with no females = 153 * 3 * (0.5)³ = 57.375

for x = 3, expected number of famility with no females = 153 * 3 * (0.5)³ = 19.125

(c) Here we will perform chi-square test..

Here we calculate expected counts so we will calculate chi-square statistic

$\chi^2$ = $\sum (O_i-E_i)^2/E_i$

= (32 - 19.125)²/19.125 + (23 - 57.375)²/19.125 + (65 - 19.125)²/19.125 + (33 - 19.125)²/19.125

= 8.6675 + 20.595 + 1.0133 + 10.066 = 40.34

(d) Here degree of freedom = dF = 4 - 1 = 3

critical test statistic for 0.01 confidence interval = 11.345

so here $\chi^2$ > $\chi^2$(critical) so we reject the null hypothesis.