Chi Square Test

A chi square test checks whether categorical count data sits too far from what a model would expect by chance alone. It works on counts in categories, never on averages or raw measurements.

The Statistic And Its Symbols

In the most common setup you have an observed count $O$ and an expected count $E$ for each category. The test statistic is

The number grows when observed counts drift farther from expected counts. Squaring makes bigger mismatches matter more, and dividing by $E$ scales each category by how large its expected count is. The core comparison is always the same: what you observed versus what the null hypothesis predicts.

The two common versions differ only in how $E$ is computed. A goodness-of-fit test asks whether one categorical variable follows a claimed distribution; a test of independence asks whether two categorical variables are associated in a contingency table.

Why The Formula Has This Shape

Each term $\frac{(O-E)^2}{E}$ is a squared deviation measured in units of its own expected count. Squaring removes the sign so surpluses and shortfalls both add to the evidence, and dividing by $E$ is what lets you compare categories of very different sizes on the same scale: a gap of $6$ is dramatic when only $20$ were expected but trivial when thousands were. Summing those scaled gaps gives a single number whose distribution, under the null hypothesis, is known, which is what makes a threshold comparison possible.

Worked Example: Goodness Of Fit

A cafe wants to know whether three drink sizes are chosen equally often. Over $60$ orders, the observed counts are Small $26$, Medium $18$, Large $16$.

If the null hypothesis says all three sizes are equally likely, the expected count in each category is

Now compute the statistic:

That is the test statistic, not the conclusion by itself. You compare $\chi^2 = 2.8$ to a chi square distribution with the appropriate degrees of freedom. Here the degrees of freedom are $3 - 1 = 2$, because there are three categories and no parameters were estimated from the data. With $df = 2$, a statistic of $2.8$ is not strong evidence against equal preference at the $5\%$ level. The practical reading: the counts differ from perfect equality, but not by enough to confidently call the true preferences unequal from this sample alone.

Practice And Conditions To Check

Before trusting any chi square result, confirm three conditions hold: the data are counts in categories, the observations are independent (or close enough for your model), and the expected counts are not too small for the approximation you are using. A common introductory rule of thumb is that each expected count should be at least about $5$; that is a useful warning sign, not a universal law.

For your own practice, take a small table of category counts and write down the null hypothesis before doing any arithmetic. That single step prevents the biggest error in chi square problems: using the right formula with the wrong expected counts.

Calculation Pitfalls To Watch

1. Using the test on means, measurements, or percentages instead of category counts.
2. Treating the observed counts as expected counts; the expected counts must come from the null hypothesis.
3. Ignoring small expected counts, which can make the usual chi square approximation unreliable.
4. Reading "statistically significant" as "important in practice"; the test only addresses evidence against the null model.

Chi square tests appear in surveys, genetics, quality control, and market research, anywhere outcomes fall into categories and the real question is whether a pattern is surprising or whether two categorical variables seem related. If the data are numerical rather than categorical, a different tool such as a $t$ test or ANOVA is usually better.