Chi-Square Test Calculator

How to Use the Chi-Square Test Calculator

1. 1. Pick the test. Choose test of independence when you have two categorical variables cross-tabulated, or goodness of fit when you have one variable and a claimed distribution to compare it with.
2. 2. Enter the counts. For independence, type each cell's count and rename the rows and columns; add or remove them to fit anything from 2 × 2 to 8 × 8. For goodness of fit, list the observed counts and, if the categories are not expected to be equal, the expected proportions, percentages, or counts.
3. 3. Check the expected counts. The results show every expected count. If any falls below 5, a warning explains the problem and, for a 2 × 2 table, points you to Fisher's exact test.
4. 4. Read and report. You get χ², df, the p-value, an effect size, adjusted residuals that show which cells drive an association, the working, and a copy-ready APA line.

Independence or Goodness of Fit?

How the Statistic Is Built

Each cell gets an expected count — what you would see if the null hypothesis were exactly true. In a test of independence that is E = (row total × column total) / N; in a goodness-of-fit test it is E = N × p for each category's expected proportion p. The statistic adds up how far each observed count O sits from its expectation, scaled by the expectation: χ² = Σ (O − E)² / E. Large values mean the table is unlikely under H₀, and the p-value is the upper-tail area of the chi-square distribution with the test's degrees of freedom. To look up critical values instead, use the chi-square distribution table.

• Cramér's V = √(χ² / (N(min(r, c) − 1))) runs from 0 (no association) to 1 (perfect association) and lets you compare tables of different sizes.
• φ (phi) = √(χ² / N) is the 2 × 2 special case, where it equals V.
• Cohen's w = √(χ² / N) measures goodness-of-fit departures. Cohen's benchmarks are .10 small, .30 medium, and .50 large; for V, divide those benchmarks by √(min(r, c) − 1).

Worked Example: Study Mode and Pass Rates

Click Try an example to load results for 160 students: online, 45 passed and 15 failed; in person, 30 and 30; hybrid, 25 and 15. Overall 100 of 160 passed, so independence predicts 37.5 passes among the 60 online students (60 × 100 / 160).

Reported in APA style: χ²(2, N = 160) = 8.00, p = .018, V = .22. The adjusted residuals locate the association: online students passed more often than independence predicts (+2.53) and in-person students less often (−2.53), while hybrid students matched expectation exactly. Because the table has two columns, min(r, c) − 1 = 1 and Cohen's benchmarks apply to V unchanged, so .22 sits between small (.10) and medium (.30).

When the Approximation Breaks Down

The p-value comes from a large-sample approximation. A common rule (Cochran, 1954) asks that no expected count fall below 1 and no more than 20% fall below 5; the calculator flags any expected count under 5. For a 2 × 2 table with small counts, use Fisher's exact test. Yates's continuity correction is the older fix for 2 × 2 tables: it shrinks χ² and makes the test conservative, so many statisticians now skip it when counts are adequate. R's chisq.test and SciPy's chi2_contingency apply it to 2 × 2 tables by default, which is why their χ² can differ from SPSS's Pearson row — tick or untick the box to match the one you are checking.

• Entering percentages instead of counts. χ² depends on N; percentages throw the sample size away and give a meaningless result.
• Counting one person in two cells. Every observation must land in exactly one cell. Repeated measurements on the same people need McNemar's test or another paired method.
• Reading a small p as a strong association. With a large N, trivial associations become significant — report V or w alongside p.

Frequently Asked Questions

What does the chi-square test of independence tell you?

It tests whether two categorical variables are associated in the population — whether the distribution of one changes across the levels of the other. A significant result says the observed table is unlikely if the variables were independent. It does not say how strong the association is (use Cramér's V for that) or which cells are responsible (look at the adjusted residuals).

How do you calculate expected frequencies for a chi-square test?

For a contingency table, multiply the cell's row total by its column total and divide by the grand total N. For a goodness-of-fit test, multiply N by each category's expected proportion. Expected counts do not have to be whole numbers — 37.5 is a perfectly good expected count.

What is a good Cramér's V?

There is no universal threshold, but Cohen's conventions, adjusted for table size, are the usual reference: for a 2 × 2 table (where V equals φ), about .10 is small, .30 medium, and .50 large. For larger tables divide those values by √(min(r, c) − 1) — in a 3 × 3 table the medium benchmark becomes about .21.

Should I use Yates's continuity correction?

Only if your course, journal, or software comparison expects it. The correction applies to 2 × 2 tables only and makes the test conservative — p-values come out too large. With adequate expected counts, the uncorrected Pearson χ² is accurate; with small counts, Fisher's exact test is the better remedy.

Can I run a chi-square test on percentages or means?

No. The test needs raw frequencies — the number of cases in each category. Convert percentages back to counts using the group sizes before you enter them; a table of means or scores needs a different test, such as a t-test or ANOVA.

How do I report a chi-square test in APA style?

Write χ² with the degrees of freedom and the sample size in parentheses, then the value and p: χ²(2, N = 160) = 8.00, p = .018. Add an effect size such as V = .22. In APA style the Greek χ is not italicised, but N, p, and V are; p has no leading zero, and anything below .001 is written p < .001.