Cramér's V Calculator

StatisticsLast updated: August 22, 2026

Cramér's V measures the strength of the association between two categorical variables on a 0 to 1 scale. A chi-square test only answers "is there an association?", and as the sample grows almost every table turns out significant; V divides out the effect of sample size and leaves the actual strength of the association behind. That is why contingency table results are reported as the trio χ², p, and V.

This calculator takes the chi-square value, the total number of observations, and the row-by-column size of the table, and returns Cramér's V, the phi coefficient (Cohen's w), Bergsma's bias-corrected V, and the p-value. A separate table shows how the interpretation cutoffs shift with table size: a V of 0.30, merely "moderate" in a 2×2 table, already means a strong association in a 4×5 table.

The chi-square statistic of the test of independence. If you only have the table, compute it first with the chi-square test calculator.
The sum of all the cells in the contingency table — not a row or column total.
How many categories the row variable has.
How many categories the column variable has.
Sets the false-positive risk. In the social sciences 5% is the usual choice.

Cramér's V and Phi Formulas

Phi squared:     φ² = χ² / n
Phi (Cohen's w): φ = √(χ² / n)
Cramér's V:      V = √( χ² / (n · k) ),   k = min(r − 1, c − 1)
In a 2×2 table:  V = φ = √(χ² / n)
Degrees of fr.:  df = (r − 1)·(c − 1)
Corrected V:     φ²ᵈ = max(0; φ² − df/(n−1))
                 r̃ = r − (r−1)²/(n−1),   c̃ = c − (c−1)²/(n−1)
                 Ṽ = √( φ²ᵈ / min(r̃ − 1, c̃ − 1) )
Cutoffs:         small 0.10/√k · moderate 0.30/√k · strong 0.50/√k

Cramér's V squeezes the phi coefficient into the 0–1 range by dividing it by the largest value it could take (√k). That makes association strengths from tables of different sizes comparable — but in exchange, the interpretation cutoffs change with the table size.

How to Calculate

  1. Enter the χ² value from your chi-square test of independence; if you have not computed it yet, use the chi-square test calculator first.
  2. Enter the total number of observations (the sum of all cells) — this is not a row or column total.
  3. Enter the number of rows and columns of the contingency table; the calculator derives the degrees of freedom and k = min(r−1, c−1) itself.
  4. Choose the significance level; the critical chi-square value and the p-value follow from it.
  5. Compare Cramér's V against the cutoffs for your own table size: in a 3×3 table, 0.21 already counts as a moderate association.
  6. Read the "smallest V that would be significant" row; it shows how much your sample size alone eases the way to significance.

Worked Examples

A significant but weak association in a 3×3 table

In a sample of 240, χ²(4) = 18.42 gives p = 0.001, so the association is significant. With φ² = 18.42 / 240 = 0.0768 and k = 2, Cramér's V comes out as 0.1959. Because the cutoffs for a 3×3 table are 0.071 / 0.212 / 0.354, this value sits at the top of the "small" band — significant, but weak. The bias-corrected value is Ṽ = 0.1740.

Cramér's V: 0.1959 · Strength of the association (given the table size): Small · Generic scale label (ignoring df): zayıf

Phi and V coincide in a 2×2 table

In a 2×2 table k = 1, so Cramér's V and phi are the same: V = φ = √(6.35 / 120) = 0.2300. For χ²(1) = 6.35 the p-value is 0.012, so the association is significant at the 5% level, and against the 2×2 cutoffs (0.10 / 0.30 / 0.50) it lies between small and moderate. In this sample the smallest V that would be significant is 0.1789.

Cramér's V: 0.2300 · Strength of the association (given the table size): Small · Generic scale label (ignoring df): zayıf

Significant but negligible in a large sample

In a sample of 5,000, χ²(2) = 9.85 and p = 0.007 make the result significant, yet Cramér's V is only 0.0444: a negligible association. In this sample even an association as small as V = 0.0346 would have been significant — the classic illustration of why reading "strong association" off a p-value is a mistake in large data sets.

Cramér's V: 0.0444 · Strength of the association (given the table size): Negligible · Generic scale label (ignoring df): ihmal edilebilir

Frequently Asked Questions

How large does Cramér's V have to be to count as a strong association?
There is no single cutoff; the boundary depends on the table size. Cohen's w cutoffs (0.10 small, 0.30 moderate, 0.50 large) are divided by √min(r−1, c−1) when moved onto the V scale: 0.10/0.30/0.50 in a 2×2 table, 0.07/0.21/0.35 in a 3×3 table, and 0.05/0.15/0.25 in a 5×5 table. That is why "V = 0.25" cannot be interpreted without stating the table size.
The chi-square test is significant — why is V small?
They answer different questions: p asks "is the association different from zero?", V asks "how strong is it?". Because χ² grows directly with n, in a sample of a few thousand even the tiniest departure turns out significant, while V has the sample size divided out of it. In large samples, associations that are significant but of no practical importance are the rule, which is why the two are always reported together.
What is the difference between phi and Cramér's V?
Phi is computed as √(χ²/n) and stays within 0–1 only in 2×2 tables; in larger tables it can exceed 1, so it is not comparable across tables. Cramér's V divides phi by the largest value it could take (√k), placing it in the 0–1 range for any table size. In a 2×2 table k = 1, so the two are numerically identical.
Can Cramér's V be negative?
No. Because V involves a square root, it always lies between 0 and 1, and it does not show the direction of the association — only its strength. To see the direction, look at the cell percentages, the standardized residuals, or, in a 2×2 table, the odds ratio. The only signed measure here is the phi coefficient computed for a 2×2 table.
How do I get Cramér's V in SPSS and R?
In SPSS, tick the "Phi and Cramer's V" box under Crosstabs → Statistics → Nominal; the output matches the values here. In R, rcompanion::cramerV() or vcd::assocstats() give the same result, and rcompanion applies the Bergsma correction shown here when you pass bias.correct = TRUE. Excel has no direct function: take χ² from CHISQ.TEST and apply the formula by hand.
When do I need the bias-corrected V?
Cramér's V comes out above zero even when the true association is zero, especially in small samples and tables with many cells. Bergsma's (2013) correction removes this inflation by subtracting df/(n−1) from φ². If the average number of observations per cell is below 5, reporting the corrected value is the better choice; in large samples the two values are very close.
What happens if the expected-frequency assumption is violated?
If more than 20% of the cells have an expected frequency below 5, the chi-square approximation breaks down, and the p-value and everything derived from it become unreliable. Use Fisher's exact test for 2×2 tables, and category merging or Monte Carlo exact tests for larger ones. The effect size V can still be computed, but the significance claim should rest on the p-value of the exact test.