Module 09 · Categorical data
Compare observed counts with the counts independence would predict. Keep conditional percentages, effect size, and evidence in view.
A two-way table records two categorical variables for each observation. In an illustrative study, a retailer records a customer's shopping channel and whether the customer returns an item. Each customer appears once. The four cells are online/return, online/no return, store/return, and store/no return.
Raw counts can be misleading when the groups have different sizes. Compare the return count with the total in its own channel. That gives a conditional proportion: the return rate among online customers versus the return rate among store customers. Keep the denominator explicit when describing a percentage.
A table of counts also differs from a table of percentages. The chi-square calculation uses actual frequencies. Entering percentages as though they were counts silently changes the sample size and therefore changes the evidence.
Edit the four illustrative customer counts. Compare the rates and inspect how each cell departs from independence.
The pale bars are expected counts under independence, using the current margins. The test is Pearson chi-square without a continuity correction.
At α = 0.05, reject independence. Online and store return rates are 30.0% and 50.0%. This describes association; it does not establish causation.
N = 200. Online return rate 30.0%; store return rate 50.0%. Expected counts, in displayed order: 40.00, 60.00, 40.00, 60.00. Chi-square 8.333 with 1 degree of freedom; p 0.0039.
The null hypothesis says shopping channel and return status are independent in the population. Under independence, both channels share a common return probability. Estimate that probability using the overall return proportion, then apply it to each row total.
The expected count in a cell is E = row total × column total / grand total. Expected counts preserve the observed margins, and they need not be whole numbers. The alternative says that the two variables are associated; it does not specify which cells differ or why.
Pearson's statistic adds (O − E)² / E across the cells. Large discrepancies relative to expected counts produce large statistics. With r rows and c columns, the approximate reference distribution has (r − 1)(c − 1) degrees of freedom. A 2 × 2 table therefore has one degree of freedom.
Suppose the illustrative counts are 30 returns and 70 nonreturns online, versus 50 returns and 50 nonreturns in store. The return rates are 30% and 50%, a difference of −20 percentage points for online minus store.
There are 200 customers and 80 returns. Each channel has 100 customers, so the expected return count is 100 × 80 / 200 = 40. The expected nonreturn count is 60 in each channel. Thus χ² = 100/40 + 100/60 + 100/40 + 100/60 = 8.333. With one degree of freedom, p ≈ 0.00389.
At a prespecified 5% significance level, this is evidence against independence if the sampling assumptions hold. Cramér's V is √(χ²/N) = 0.204 for this table. Report the actual rates as well: V summarizes association without showing its direction.
A small p-value indicates that the observed discrepancy is unusual under the independence model. It does not give the probability that the null hypothesis is true, nor does it measure the commercial importance of the difference. A very large sample can detect an association too small to matter operationally.
Expected counts matter for the approximation. This lesson flags any expected count below 5. For a sparse 2 × 2 table, an appropriate exact method such as Fisher's exact test may be preferable. That method is outside this calculator. Do not combine categories only to obtain a desired result.
Finally, an observational association between channel and returns does not show that the channel caused the difference. Customer mix, product type, and other variables may explain it. Aggregation can even reverse the apparent direction of a relationship, so examine important subgroups and the data-collection process before making a policy recommendation.
Try each question before revealing the solution.
E = 80 × 50 / 200 = 20. This uses the marginal totals; it does not require knowing the observed count in the cell.
The rates remain 30% and 50%. Each squared discrepancy divided by its expected count doubles, so χ² becomes 16.667. Both χ² and N double, leaving V = √(χ²/N) unchanged. More data strengthen the evidence without changing the observed effect size.
No. A sparse expected count makes the asymptotic approximation questionable. Review the sampling design and use an appropriate exact or simulation-based procedure. The small approximate p-value does not override the method’s conditions.
Further reading