Question

In an experiment using Drosophila melanogaster, white-eyed females were crossed with wild-type (red-eyed) males. The $F_1$ flies were then crossed among themselves. The following data were collected: $224$ white-eyed females, $218$ wild-type females, $243$ white-eyed males, $215$ wild-type males.
According to the chi-square test, what can be concluded about the validity of these results?

Answer 1

As per the date given, the F${}_2$ generation shows 224 white-eyed females, 218 wild-type females, 243 white-eyed males and 215 wild-type males. The expected value (E) is 225.

Hence, chi-square = (O-E)${}^2$ /E

= (224-225)${}^2$/225 + (218-225)${}^2$/225 + (243-225)${}^2$/225 + (215-225)${}^2$/225

= (-1)${}^2$/225 + (-7)${}^2$/225 + 18${}^2$/225 + (-10)${}^2$/225

= 1/ 225 + 49/225 + 324/225 + 100/225

= 0.0044 + 0.22 + 1.44 + 0.44

= 2.1

Hence, the chi-square value is 2.1. It is less than the critical value of 7.81 with 3 degrees of freedom, p=.05. Therefore, it can be concluded that there is no significant difference between observed and expected values.