What if an expected count is less than 5?

The chi-square approximation becomes unreliable. You must either combine adjacent categories to increase the expected counts or use an exact test, such as Fisher's Exact Test, which does not rely on large-sample approximations.

Do I need to apply Yates' continuity correction?

For $2 imes 2$ tables, Yates' correction subtracts 0.5 from the absolute difference between observed and expected counts before squaring. It is traditionally used to prevent overestimation of statistical significance for small datasets, though many modern statisticians prefer Fisher's Exact Test instead. It is not used for tables larger than $2 imes 2$.

Can I run a chi-square test on continuous data?

No. The chi-square test of independence strictly requires categorical variables. To use it with continuous data, you must first bin the continuous data into discrete categories (e.g., converting age into age brackets).

How to run a chi-square test of independence

The chi-square test of independence evaluates whether two categorical variables are associated in a population. It applies when you have counts of independent observations organized in a contingency table, provided that the expected cell counts are sufficiently large.

The setup

Define the null hypothesis $H_0$ as the two variables being independent, and the alternative hypothesis $H_a$ as the two variables being dependent. Organize your raw frequency data into an $r imes c$ contingency table, where $r$ is the number of rows (categories of the first variable) and $c$ is the number of columns (categories of the second variable). Let $O_{ij}$ represent the observed count in row $i$ and column $j$ .

The steps

Calculate the sum of each row, the sum of each column, and the grand total $n$ . 2. Calculate the expected count for each cell using the formula $E_{ij} = \frac{( ext{Row } i ext{ Total}) imes ( ext{Column } j ext{ Total})}{n}$ . 3. Compute the chi-square test statistic using $\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$ summing over all cells. 4. Determine the degrees of freedom using $df = (r - 1)(c - 1)$ . 5. Compare the calculated $\chi^2$ value to the critical value from the chi-square distribution table for the chosen significance level $\alpha$ and calculated $df$ , or find the p-value. Reject $H_0$ if $\chi^2$ is greater than the critical value.

Checking the result

Verify that the sum of all expected counts exactly equals the sum of all observed counts (the grand total $n$ ). The test statistic $\chi^2$ must always be non-negative. If $\chi^2 = 0$ , the observed counts perfectly match the expected counts.

Common errors

A frequent error is using percentages, proportions, or fractions instead of raw frequency counts; the chi-square formula strictly requires raw counts. Another common error is proceeding with the test when one or more expected cell counts are less than 5, which invalidates the chi-square distribution approximation.

Worked example

A researcher wants to know if preferred exercise type (Cardio vs. Weights) is independent of shift worked (Day vs. Night). The observed counts are: Day/Cardio = 30, Day/Weights = 10, Night/Cardio = 15, Night/Weights = 45. Test at $\alpha = 0.05$ .

Hypotheses: $H_0$ : Exercise type and shift are independent. $H_a$ : Exercise type and shift are dependent. Row totals: Day = 30 + 10 = 40. Night = 15 + 45 = 60. Column totals: Cardio = 30 + 15 = 45. Weights = 10 + 45 = 55. Grand total $n$ = 100. Expected counts: $E_{ ext{Day,Cardio}} = (40 imes 45) / 100 = 18$ . $E_{ ext{Day,Weights}} = (40 imes 55) / 100 = 22$ . $E_{ ext{Night,Cardio}} = (60 imes 45) / 100 = 27$ . $E_{ ext{Night,Weights}} = (60 imes 55) / 100 = 33$ . Test statistic: $\chi^2 = \frac{(30-18)^2}{18} + \frac{(10-22)^2}{22} + \frac{(15-27)^2}{27} + \frac{(45-33)^2}{33} = \frac{144}{18} + \frac{144}{22} + \frac{144}{27} + \frac{144}{33} = 8.000 + 6.545 + 5.333 + 4.364 = 24.242$ . Degrees of freedom: $df = (2 - 1)(2 - 1) = 1$ . Critical value for $\alpha = 0.05$ with $df = 1$ is 3.841. Since 24.242 > 3.841, reject $H_0$ . Exercise type and shift are dependent.

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References: OpenStax Introductory Statistics, Chapter 11 · Khan Academy: Chi-square tests for categorical data