9.1.11, 4.2.3, D.1.8, D.1.13, D.2.2, D.2.3 - Chi-squared decisions

Choosing chi-squared for categorical frequencies

Chi-squared compares the frequencies observed in categories with the frequencies expected under a null model.

In a learning study, different participants view either a helping model who is praised or a neutral demonstration, and then choose whether to help in a harmless task. The outcome is nominal: “helped” and “did not help” are categories, not scores with meaningful equal intervals. We count people in each outcome category for each group. This is a chi-squared test of difference in the outcome pattern between independent groups, mathematically a test of independence in a contingency table. It is not a test of the difference between mean helpfulness ratings.

The data must be frequencies, not percentages or averages. Categories must be mutually exclusive and cover the outcomes being analysed, so each participant contributes to one cell only. Observations must be independent: counting repeated acts from the same person as separate people, or placing the same participant in both groups, breaks that assumption. Wilcoxon or another appropriate paired procedure is needed for related score data; this chi-squared procedure does not handle before-and-after pairs.

This lesson teaches the specified frequency test. It does not replace thematic analysis, which identifies patterns of meaning in qualitative material rather than comparing counts against expected frequencies.

Calculating the expected frequencies

Expected frequencies describe the pattern predicted if group membership and the categorical outcome are independent.

The following illustrative counts were created to demonstrate the calculation; no learner or actual study is claimed to have produced them.

The null hypothesis is that the outcome distribution does not differ between the two demonstration groups. Under that model, each group would have the same overall helping proportion. Across everyone, 28 of 48 helped, so a group containing 24 people would be expected to include $24(28/48)=14$ helpers. This is an expected count under the null, not a predicted score for an individual person.

For “praised model, did not help”, $E=24(20)/48=10$. Both row totals are 24, so the expected table is:

Expected counts preserve the observed row totals, column totals and grand total. Equal expected counts in corresponding rows arise here because the groups have equal sizes; unequal group sizes would give different expected counts. It would be wrong to divide 48 equally among four cells and use 12 everywhere, because that ignores the outcome totals.

The usual simple approximation rule for these classroom tables is that every expected cell frequency is at least 5. Check expected, rather than observed, counts: an observed count below 5 does not by itself establish failure. Sparse expected counts make the approximation less dependable. Use a suitable larger sample or an appropriate alternative analysis; do not selectively merge categories after seeing which change makes the result significant.

Calculating chi-squared and degrees of freedom

Chi-squared adds the weighted discrepancy between the observed and expected count in every cell.

The supplied Pearson formula uses O for the observed count and E for its expected count: