Chi-Square Test of Independence

The chi-square test of independence brings two categorical variables together in a contingency table and compares how many people fall in each cell with how many would be expected there if the two variables were entirely unrelated. Expected counts come from multiplying the row and column totals and dividing by the grand total; the test statistic adds up the squared departures between observed and expected counts, each scaled by the expected value. A large chi-square means the pattern in the table is hard to attribute to chance once the row and column totals are held fixed. The result points to the presence of an association, not to its direction or shape: even when the categories are ordered, the test ignores that ordering and looks only at how the frequency pattern departs from expectation. YouReply Analyze reports the chi-square value, the degrees of freedom and the p value, computes Cramer's V for the strength of the association, and prints the observed and expected tables side by side.

• Is the preferred contact channel related to the respondent's age bracket?
• Does survey completion vary with the channel through which the invitation was sent?
• Are brand preference and type of settlement independent of one another?
• Does awareness of a campaign differ across levels of education?

• When both variables are categorical and each participant falls into exactly one category of each.
• When the question is whether two classifications are independent, rather than a comparison of means or ranks.
• When your data are counts: the cells must hold numbers of people, not percentages or averages.
• When the table is 2x2 or larger, since all that is required is at least two categories in each of the two variables.
• When the observations are independent; two categories ticked by the same person before and after cannot be compared this way.

• First variable: a categorical column, whether stored as text labels or as numeric category codes.
• Second variable: again a categorical column. The test treats the two symmetrically, so which one you consider the outcome changes nothing in the statistic, only whether it appears in rows or columns.
• One row per participant. If you upload a table of pre-aggregated frequencies the cells will not be read as people and the result will be wrong.
• Bringing a continuous variable into this test means collapsing it into categories, which discards information and should be done deliberately through the derived column feature on the Variable tab.

Independence of observations

Each participant contributes once, to a single cell. Answers given by the same person at two points in time or under two conditions are not independent, and that design calls for paired categorical tests.

Count data

The cells have to contain frequencies. Computing chi-square over percentages, proportions or averages inflates or deflates the statistic in ways that carry no meaning.

At least two categories and a sufficient total

Building the table requires at least two categories in each variable and at least five valid observations overall.

Adequate expected cell counts

The chi-square approximation holds when the expected counts are not too small. The familiar rule asks for an expected count of at least 5 in every cell, while more permissive readings allow up to a fifth of the cells to fall below that.

• Independence of observations: The panel does not check this assumption automatically; the researcher evaluates it.
• Count data: The panel builds the contingency table itself from row-level data, so in correct use the cells are already counts; there is no check that would notice an aggregated table had been uploaded.
• At least two categories and a sufficient total: Both conditions are enforced during the run and the analysis stops with an error when they are not met. Since missing values are dropped row by row first, the number of categories can shrink in the cleaned data.
• Adequate expected cell counts: THE PANEL DOES NOT CHECK THIS. For the test of independence no warning or assumption flag is produced for the smallest expected count, and there is no automatic switch to Fisher's exact test. The check is yours to make from the expected frequency table printed in the result: find the smallest cell, and if it fails the rule, consider merging categories or running Fisher's exact test as a separate method. (The goodness-of-fit version of the chi-square test does report a minimum expected count; the test of independence described here does not.)

1. 1Drop the data file onto the panel. The upload area accepts drag and drop and reads XLSX files as well.
2. 2In the grid on the Data tab, check how the categories are spelled; the same category written with different capitalisation counts twice and enlarges the table for no reason.
3. 3On the Variable tab mark both columns as nominal, declare the missing-value codes, and build a derived column if you want to merge sparse categories.
4. 4On the Analysis tab pick this method from the categorical data group; the data requirements box shows which columns meet the two-category condition.
5. 5In the parameter form name the two columns that will form the table and decide whether to leave the Yates continuity correction switched on.
6. 6Run the analysis. The stacked bar chart of the contingency table appears above the collapsible sections holding the tables and statistics.
7. 7Before reading the statistics, open the expected frequency table and inspect its smallest cell, then export the result to Excel and the chart to PNG.

Chi-square statistic and degrees of freedom

The degrees of freedom are the number of rows minus one times the number of columns minus one. In a 2x2 table with the Yates correction requested, the reported value is the corrected statistic.

Significance value

A p value based on the chi-square distribution. No exact probability is computed, and with small expected counts this approximation loosens and an exact test is preferable.

Observed frequency table

The contingency table built from the cleaned data, with a count of people in each cell.

Expected frequency table

The value expected in each cell under independence, listed with decimals. This table is where you are expected to carry out the assumption check, since the panel raises no separate warning about the smallest value.

Cramer's V with a verbal label

The strength of the association, derived from the reported chi-square value, the total number of observations and the smaller dimension of the table, accompanied by a strength label.

Total observations and correction status

The number of valid observations entering the analysis and whether the Yates continuity correction was requested are both stated in the result.

Stacked bar chart

The contingency table is drawn as a stacked bar chart that makes the category distributions comparable at a glance, and can be saved as a PNG.

Cramer's V

Ranges from 0 to 1 independently of the size of the table and expresses the strength of the association. The panel's labels treat values below 0.1 as negligible, below 0.2 as weak, below 0.4 as moderate, below 0.6 as relatively strong and anything above as strong. The labels only ease reading; in large samples a small V alongside a very small p value is entirely ordinary, and it is V that then settles how much the association weighs in practice.

No confidence interval is returned for this method. The chi-square value, the p value and Cramer's V arrive as point estimates, and no interval is computed for differences between proportions, relative risk or an odds ratio. An interval estimate for cell proportions has to be produced outside the panel.

The numbers below are a representative example, not data from a real study or a real user.

Research question

In an illustrative public services study, is the use of e-government services independent of gender?

Variables

First variable: gender (women, men) · Second variable: use of e-government services (uses, does not use)

Example result

The table was built from 380 valid responses. Of 196 women, 118 use the services and 78 do not; of 184 men, 133 use them and 51 do not. Expected counts range from 62.45 to 129.45, so even the smallest cell sits far above the threshold of 5. The Yates continuity correction was left on: chi-square(1) = 5.63, p = 0.018, Cramer's V = 0.12 (weak).

Interpretation

Usage stands at 60.2 per cent among women and 72.3 per cent among men, and the 12-point gap is statistically significant while Cramer's V describes the association as weak. Significance says that a departure of this size would be uncommon under independence in a sample of 380; how strong the association is comes from V, and where it originates from a comparison of the observed and expected tables. Since the test carries no direction, nothing here licenses the claim that gender determines usage, and variables such as age or education may well sit behind the gap. The figures are illustrative.

Research question: Does the purchase decision differ by education level?

First variable

education

Second variable

purchased

Correction applied

true

Computation credits: scipy 1.18.0 · statsmodels 0.14.6 · scikit-learn 1.9.0 · numpy 2.5.1 · pandas 3.0.5 · semopy 2.3.11 · 89fc29a · Data seed: 20260914

Use of e-government services was significantly associated with gender, chi-square(1, N = 380) = 5.63, p = .018, Cramer's V = .12, with the Yates continuity correction applied.

An example sentence close to APA style; the numbers are representative.

• Because the adequacy of expected counts is not checked for this test, the p value can be optimistic in sparse tables; do not report a result without reading the expected frequency table.
• No standardized residuals are returned to show which cell drives the association, so cell-level reading has to be done by hand from the observed and expected tables.
• It does not apply to designs in which the same people are measured twice; paired categorical data need a different test.
• Ordering among the categories is ignored even when it exists, so a rank-based measure is more sensitive when the question concerns a trend between two ordered variables.
• The statistic grows with sample size, which is why a p value on its own says nothing about how much an association matters.
• Merging categories and rerunning the test undermines the finding if it turns into a hunt for significance; any merge should rest on a conceptual rationale.

• Fisher's Exact TestWhy: When expected counts stay small; it computes the exact probability and is more trustworthy in sparse tables, and it is chosen as a separate method.
• Cross-TabulationWhy: When row and column percentages and a description of the pattern matter more than a significance test.
• Chi-Square Goodness of FitWhy: When a single categorical variable is compared against an expected distribution rather than against another variable.
• Logistic RegressionWhy: When a binary outcome needs to be explained by several variables at once, holding other effects constant.

• Logistic Regression
• Spearman Rank Correlation

Some cells have expected counts below 5. Will the panel warn me?

Not in the test of independence. This method does not report the minimum expected count as a field of its own, raises no assumption flag, and never switches to Fisher's exact test by itself. The check is yours: open the expected frequency table in the result, find the smallest cell, and if it fails the rule either merge categories on defensible conceptual grounds or run Fisher's exact test separately. You may have seen the chi-square goodness-of-fit method report a minimum expected count; that is a different method and does not change the behaviour here.

Should I leave the Yates continuity correction on?

The correction is on by default and takes effect only where the degrees of freedom equal one, that is, in 2x2 tables; in larger tables it does not alter the computation. It shifts the p value in the conservative direction. Opinion is divided between treating it as useful in small samples and as overly cautious, so what matters is reporting your choice, and the result card states whether the correction was requested.

How do I find out which category creates the association?

No cell-level follow-up is returned and standardized residuals are not computed. The practical route is to compare the observed and expected tables, mark the cells clearly above or below expectation, and read that alongside the stacked bar chart. If you need row and column percentages, run the cross-tabulation method as well.

The p value is tiny. Does that mean a strong association?

Those are two separate questions. The chi-square statistic grows with sample size, so in large samples even a faint pattern can produce a very small p value. Judge strength from Cramer's V and report the percentages. A significant result also carries no causal direction: the table treats both variables symmetrically, and which one acts on the other is settled by the study design rather than by the statistic.

• Agresti, A. (2018). An Introduction to Categorical Data Analysis
• Field, A. (2018). Discovering Statistics Using IBM SPSS Statistics
• Yates, F. (1934). Contingency Tables Involving Small Numbers and the Chi-Square Test
• scipy.stats.chi2_contingency documentation