Find Critical χ² Value
- Goodness-of-fit test: df = (number of categories) - 1
- Test of independence: df = (rows - 1) × (columns - 1)
- Test of homogeneity: df = (rows - 1) × (columns - 1)
What the critical value means
Each printed entry is an upper-tail cutoff c such that P(χ²_df > c) = α. For categorical goodness-of-fit and independence tests, reject the null for a sufficiently large statistic: χ²_observed > c.
These tests use the upper tail even though observed counts can depart in either direction. The statistic squares departures, so do not halve α or double the upper-tail p-value as though χ² were a signed t statistic.
Upper-tail critical values
| df | α = 0.10 | α = 0.05 | α = 0.025 | α = 0.01 | α = 0.001 |
|---|---|---|---|---|---|
| 1 | 2.7055 | 3.8415 | 5.0239 | 6.6349 | 10.8276 |
| 2 | 4.6052 | 5.9915 | 7.3778 | 9.2103 | 13.8155 |
| 3 | 6.2514 | 7.8147 | 9.3484 | 11.3449 | 16.2662 |
| 4 | 7.7794 | 9.4877 | 11.1433 | 13.2767 | 18.4668 |
| 5 | 9.2364 | 11.0705 | 12.8325 | 15.0863 | 20.5150 |
| 6 | 10.6446 | 12.5916 | 14.4494 | 16.8119 | 22.4577 |
| 7 | 12.0170 | 14.0671 | 16.0128 | 18.4753 | 24.3219 |
| 8 | 13.3616 | 15.5073 | 17.5345 | 20.0902 | 26.1245 |
| 9 | 14.6837 | 16.9190 | 19.0228 | 21.6660 | 27.8772 |
| 10 | 15.9872 | 18.3070 | 20.4832 | 23.2093 | 29.5883 |
| 12 | 18.5493 | 21.0261 | 23.3367 | 26.2170 | 32.9095 |
| 15 | 22.3071 | 24.9958 | 27.4884 | 30.5779 | 37.6973 |
| 20 | 28.4120 | 31.4104 | 34.1696 | 37.5662 | 45.3147 |
| 24 | 33.1962 | 36.4150 | 39.3641 | 42.9798 | 51.1786 |
| 25 | 34.3816 | 37.6525 | 40.6465 | 44.3141 | 52.6197 |
| 30 | 40.2560 | 43.7730 | 46.9792 | 50.8922 | 59.7031 |
| 40 | 51.8051 | 55.7585 | 59.3417 | 63.6907 | 73.4020 |
| 60 | 74.3970 | 79.0819 | 83.2977 | 88.3794 | 99.6072 |
| 100 | 118.4980 | 124.3421 | 129.5612 | 135.8067 | 149.4493 |
The interactive lookup supports other df and probabilities. Printed values are rounded to four decimals.
Get degrees of freedom from the design
| Procedure | Usual df | Qualification |
|---|---|---|
| Goodness of fit, k categories with fixed probabilities | k − 1 | Probabilities specified independently of these counts |
| Goodness of fit with m fitted parameters | k − 1 − m | Under appropriate fitting and binning conditions |
| Independence in an r×c table | (r − 1)(c − 1) | Independent units and suitable expected counts |
| Normal-population variance inference | n − 1 | A different statistic and tail choice |
The same chi-square distribution appears in several procedures. The table cannot decide which procedure your data require.
Worked independence example
Observe counts 30, 20 in the first row and 20, 30 in the second. Both row totals and both column totals are 50; total n = 100. Under independence, every expected cell count is 50 × 50/100 = 25.
For a 2×2 table, df = 1. At α = 0.05, c = 3.8415. The uncorrected Pearson statistic exceeds it, with upper-tail p ≈ 0.0455.
This example uses no Yates continuity correction. A corrected or exact test can give a different answer near the threshold. State the method and report the actual proportions alongside the test. An association does not establish causation.
Check expected counts, not just the total
For independence, E = row total × column total / total. For goodness of fit, E = n × specified category probability. Expected counts can be fractional and should retain precision.
Requiring every E ≥ 5 is a conservative introductory check of the approximation. Sparse tables may require exact or Monte Carlo methods; a large overall sample does not prevent some categories from having tiny expectations. Paired categorical outcomes need a paired analysis.
The chi-square lesson explains a complete goodness-of-fit calculation, Cramér’s V, and residual interpretation.
A variance interval uses both tails
For independent normal observations, (n − 1)s²/σ² has a chi-square distribution. A two-sided interval for σ² uses lower and upper quantiles:
Here q denotes a cumulative quantile. In an upper-tail table, q₍₁₋α/₂₎ is found at tail α/2, while q₍α/₂₎ is found at tail 1 − α/2. The chi-square distribution is asymmetric, so do not use ± one cutoff.
This variance procedure is sensitive to departures from normality. It is not the categorical-count test described above.
Practice
A 3×4 independence table has adequate expected counts. What df and 5% critical value do you use?
Show the solution
df = (3 − 1)(4 − 1) = 6. The upper-tail 5% cutoff is 12.5916. The statistic must exceed that value to reject at this level under the approximation.
References and next steps
NIST’s contingency-table reference describes the categorical independence statistic. Review the degrees of freedom guide and chi-square lesson. The site’s Bartlett calculator compares numerical variances; it is not a categorical chi-square calculator.