Put your observed counts in a table, build a second table of expected counts (row total × column total ÷ grand total), then type =CHISQ.TEST(observed_range, expected_range). The result is the p-value. If it is below 0.05, the two variables are associated.
When to use it
The chi-square test of independence checks whether two categorical variables are related: course format and pass/fail, region and product choice, age group and yes/no answers. Each person must be counted once, in one cell only.
The example data
190 students took the same course online, in a hybrid format or in person. Did the pass rate depend on the format? Enter the counts like this:
| A | B | C | D | |
|---|---|---|---|---|
| 1 | Passed | Failed | Total | |
| 2 | Online | 38 | 22 | 60 |
| 3 | Hybrid | 51 | 9 | 60 |
| 4 | In person | 55 | 15 | 70 |
| 5 | Total | 144 | 46 | 190 |
Row totals in D2:D4 are =SUM(B2:C2) and so on; column totals in B5:D5 are =SUM(B2:B4).
Step 1: the expected counts
If format and outcome were unrelated, each cell would hold row total × column total ÷ grand total. In F2, type:
=$D2*B$5/$D$5
Fill it across to G2 and down to G4. The dollar signs keep the totals fixed while the formula moves.
| Expected | Passed | Failed |
|---|---|---|
| Online | 45.47 | 14.53 |
| Hybrid | 45.47 | 14.53 |
| In person | 53.05 | 16.95 |
All expected counts are above 5, so the test is valid.
Step 2: the p-value
=CHISQ.TEST(B2:C4, F2:G4)
Result: p = 0.017. The pass rate depends on the course format (at the 0.05 level). Do not include the totals in the ranges.
Step 3: the χ² statistic and degrees of freedom
=SUMPRODUCT((B2:C4-F2:G4)^2/F2:G4)
Result: χ² = 8.14. The degrees of freedom are (rows − 1) × (columns − 1) = 2 × 1 = 2. You can check: =CHISQ.DIST.RT(8.14, 2) returns the same p-value.
Step 4: the effect size (Cramér's V)
=SQRT(8.14/(190*(MIN(3,2)-1)))
Result: V = 0.21, a small-to-medium association. In the formula, 3 and 2 are the number of rows and columns of the observed table.
Select your data, click Run: full results table, assumption checks, a plain-language interpretation and an APA line, free in Google Sheets.
Step 5: which cells drive the result?
A significant test says there is an association, not where. Standardized residuals, (observed − expected) / SQRT(expected), show which cells are far from what was expected. Values close to or beyond ±2 stand out:
| Residual | Passed | Failed |
|---|---|---|
| Online | −1.11 | 1.96 |
| Hybrid | 0.82 | −1.45 |
| In person | 0.27 | −0.47 |
The largest gap is online students who failed: 22 observed against 14.5 expected. Hybrid students failed less often than expected.
Assumptions and alternatives
- Use counts, never percentages or means.
- Each person appears in exactly one cell. For the same people answering twice (before/after), use McNemar's test instead.
- Expected counts should all be at least 5. For a 2 × 2 table with small counts, use Fisher's exact test (ExplainStats Pro includes it).
How to report it (APA style)
A chi-square test of independence showed a significant association between course format and outcome, χ²(2, N = 190) = 8.14, p = .017, Cramér's V = .21. Online students failed more often than expected.
Frequently asked questions
What does CHISQ.TEST return?
The p-value only. It needs two ranges of the same size: the observed counts and the expected counts. To get the χ² value itself, use a SUMPRODUCT formula or CHISQ.INV.RT(p, df).
What is the minimum expected count for a chi-square test?
The usual rule is that all expected counts should be at least 5. For a 2 × 2 table with small counts, use Fisher's exact test instead.
How do I do a chi-square goodness-of-fit test?
Put the observed counts in one column and the expected counts (total × expected proportion) in another, then =CHISQ.TEST(observed, expected). The degrees of freedom are the number of categories minus 1.
Can I use percentages instead of counts?
No. The chi-square test must be run on raw counts. Percentages give a wrong χ² value and a wrong p-value.
Select your data, click Run: full results table, assumption checks, a plain-language interpretation and an APA line, free in Google Sheets.