How do you use a chi-squared test to judge whether two species are associated?

A chi-squared test of independence checks whether two species occur together more or less often than expected if their distributions were unrelated. Start by stating the null hypothesis, the claim you will test: in the sampled habitat, the presence of species A is independent of the presence of species B.

Here is an invented survey of 120 independently selected quadrats:

Under the null hypothesis, the expected proportion of quadrats containing B is the same in each row. Calculate each expected count using

$$E=\frac{\text{row total}\times\text{column total}}{\text{grand total}}.$$

The first expected count is $60\times50/120=25$. The four expected counts, read across the rows, are therefore $25,35,25,35$. Their row and column totals match the observations.

Each cell contributes $(O-E)^2/E$. If the bottom-right contribution is missing, calculate it as

$$\frac{(46-35)^2}{35}=\frac{121}{35}\approx3.457.$$

Now add all four contributions:

$$\chi^2=\frac{121}{25}+\frac{121}{35}+\frac{121}{25}+\frac{121}{35}\approx16.59.$$

There are two rows and two columns, excluding totals, so

$$\mathrm{df}=(2-1)(2-1)=1.$$

At the 5% significance level, the critical value is approximately 3.84. Since $16.59>3.84$, reject the null hypothesis. The counts provide evidence of an association. A and B occur together more often than expected here: 36 quadrats rather than 25.

That conclusion does not establish that A helps B. Both might prefer the same soil moisture, for example. Likewise, a negative association would not by itself prove competition. To investigate a causal interaction, measure environmental variables and consider a controlled removal experiment.

Before using the test, check that each observation is counted in just one category, the quadrats were sampled appropriately and the expected counts are large enough. All four expected counts exceed five here. Nearby quadrats that repeatedly sample the same patch may not be independent observations. A non-significant result would mean that the evidence is not strong enough to reject independence; it would not prove that the species never interact.

Try the beach strawberry and yellow sand verbena investigation in the Biology SL mock exam. Work through the missing contribution and total chi-squared value, then decide what the result can tell you about association and what further evidence you would need for competition.