Chi-Square Test (Goodness of Fit)

First you state the null hypothesis, for example that a cross follows the 3:1 ratio. From it you compute for each phenotype class the expected number E as ratio times total. Then for each class you form (B − E)²/E from the observed number B and expected number E and sum all contributions to the test value χ². You compare it with the critical value from the chi-square table for the degrees of freedom df = classes − 1 and the level α = 5 %. Example: 100 offspring, expected 75:25, observed 80:20 gives χ² = 25/75 + 25/25 = 1.333. At df = 1 the critical value is 3.84. Since 1.333 is smaller, the 3:1 ratio is kept.

In the goodness-of-fit test the degrees of freedom are df = number of classes minus 1. Testing a monohybrid ratio with two phenotype classes gives df = 2 − 1 = 1. For the dihybrid 9:3:3:1 ratio with four classes, df = 4 − 1 = 3. The subtraction of 1 arises because the total is fixed: if all classes but one are known, the last follows automatically. The degrees of freedom determine which row of the chi-square table applies and thus the critical value. A common mistake is to use the total number of individuals as degrees of freedom. If parameters are additionally estimated from the data, you subtract one further degree of freedom per estimated parameter, which does not occur in the pure ratio test of genetics.

The value 3.84 is the critical chi-square value for one degree of freedom and a significance level of 5 %. It marks the boundary from which a deviation is considered meaningful. If the calculated χ² value is below 3.84, the deviation between observed and expected frequencies is compatible with chance, and the null hypothesis is kept. If it is above, the deviation is so large that it would be purely random with a probability under 5 %, and you reject the hypothesis. Other degrees of freedom have other values, such as 5.99 at df = 2 or 7.81 at df = 3. You always read the appropriate value from the chi-square table for your own degrees of freedom and the chosen level.

Dividing by the expected value E weights each deviation relative to its expected size. A deviation of 10 individuals matters strongly for an expected class of 20, but hardly for an expected class of 1000. Without the division, large classes would dominate the test value simply by their size, even though their relative deviation is small. Through (B − E)²/E all classes are made comparable: each contribution measures how strongly the class deviates relative to its expectation. Squaring additionally ensures that positive and negative deviations do not cancel and that larger deviations count disproportionately more. This yields a fair overall measure of goodness of fit across all classes.

A significant result, that is a χ² value above the critical value, means the observed frequencies deviate so strongly from the expected that pure chance is unlikely. In genetics this means the cross probably does not follow the assumed ratio. Possible causes are gene linkage, incomplete dominance, lethal factors or an unfavourable sample. It is important that the test only checks compatibility with a hypothesis and gives no reason; the biological interpretation must be added by you. A non-significant result does not prove the hypothesis, it only shows that the data do not contradict it. The test can therefore reject or keep a hypothesis, but never confirm it definitively.