A 1:2:1 phenotype ratio
Of 93 chickens, 23 are extremely frizzled, 50 slightly frizzled and 20 normal. Test the genetic prediction that the three types occur in the ratio 1:2:1, at level 0.05. Use \chi^2_{2;0.95}=5.991.
Worked answer (click to show)
The expected counts are 93/4=23.25, 93/2=46.5 and 23.25. Then
T=\frac{(23-23.25)^2}{23.25}+\frac{(50-46.5)^2}{46.5}+\frac{(20-23.25)^2}{23.25}\approx 0.720.
There are 3-1=2 degrees of freedom. Since 0.720<5.991, the test does not reject the predicted proportions at level 0.05.