Sample size for a chi-square test on a 2 × 2 table

When to use it

A 2 × 2 comparison that will be analysed with Pearson’s chi-square test.

Not for

Unequal groups, or when the analysis is a two-proportion z test with a continuity correction. Use the two-proportion test.
Larger tables. Use the test of independence.

Notation

$N$: Total participants, split equally between the groups
$p_1,\ p_2$: Expected proportions in groups 1 and 2
$w$: Cohen’s effect size for a contingency table
$P_{ij},\ E_{ij}$: Expected cell probabilities under the alternative and under independence
$\alpha$: Significance level: the chance of a false positive
$1-\beta$: Power: the chance of detecting the effect if it is real
$\rho$: Intracluster correlation coefficient (ICC)
$m$: Average number of participants per cluster
$\mathrm{CV}$: Coefficient of variation of cluster size
$C$: Clusters per group
$\ell$: Expected loss of whole clusters
$q$: Expected dropout

Formulas

Cohen’s w for the table

w = \sqrt{\sum_{i,j} \frac{(P_{ij} - E_{ij})^{2}}{E_{ij}}}

With equal groups each row holds half the participants, so P = ½(p₁, 1 − p₁, p₂, 1 − p₂), and E replaces p₁ and p₂ by their average [1].

The smallest N with enough power

\Pr\!\left(\chi'^{2}_{1,\lambda} > \chi^{2}_{1,\,1-\alpha}\right) \ge 1-\beta, \qquad \lambda = N w^{2}

When the proportions differ, the statistic follows a noncentral chi-square distribution with one degree of freedom [1].

Clustering

\mathrm{DE} = 1 + \left[(\mathrm{CV}^{2} + 1)\,m - 1\right]\rho

When participants are recruited or randomised in clusters, observations within a cluster repeat information, and the sample is multiplied by the design effect. With equal clusters (CV = 0) this is 1 + (m − 1)ρ [2, 3].

Few clusters

F_t = \left(\frac{t_{2(C-1),\,1-\alpha/\tau} + t_{2(C-1),\,1-\beta}}{z_{1-\alpha/\tau} + z_{1-\beta}}\right)^{2}

Optional. A cluster trial is analysed on cluster-level summaries, so its critical values come from t on 2(C − 1) degrees of freedom rather than the normal distribution. With few clusters this costs participants; because C depends on the sample, it is iterated until it settles [4].

All adjustments together

n_{\text{adj}} = \left\lceil n \mathrm{DE}\;F_t \right\rceil

The adjustments multiply, in this order, and the result is rounded up once. Any adjustment not used contributes a factor of 1.

Dropout

n_{\text{recruit}} = \left\lceil \frac{n_{\text{adj}}}{1 - q} \right\rceil

Applied last, to the participants actually recruited.

Clusters to recruit

C = \left\lceil \frac{n_{\text{recruit}}}{m} \right\rceil, \qquad C_{\text{recruit}} = \left\lceil \frac{C}{1 - \ell} \right\rceil

Per group. Losing a whole cluster loses every participant in it, so an expected loss of clusters inflates the clusters recruited rather than the participants.

Parameters

Expected proportion in group 1 $p_1$

Values: Between 0 and 1

From routine data or previous studies.

Expected proportion in group 2 $p_2$

Values: Between 0 and 1, different from p₁

The proportion that would make the difference worth detecting.

Significance level $\alpha$

Values: Between 0 and 1; usually 0.05

The false-positive rate you accept. A smaller α, for example to allow for several primary outcomes, needs a larger sample.

Power $1-\beta$

Values: Between 0 and 1; usually 0.80 or 0.90

The chance of a significant result if the assumed effect is real. Funders and ethics committees commonly expect at least 0.80.

Expected dropout $q$ Optional

Values: 0% up to, but not including, 100%

The share of participants expected to provide no outcome, from comparable studies. Recruiting more restores the numbers, not the comparability of the groups: if dropout is related to the outcome, the estimate can still be biased.

Intracluster correlation (ICC) $\rho$ Optional

Values: From 0 to 1

For clustered designs only. Published ICCs for health outcomes in primary-care and community clusters are mostly small, often below 0.05 [5]; prefer an estimate from a comparable trial.

Average cluster size $m$ Optional

Values: At least 1

The average number of participants per cluster. Given with the ICC.

Variation in cluster size $\mathrm{CV}$ Optional

Values: 0 or more; 0 for equal clusters

The SD of cluster sizes divided by their mean. Unequal clusters cost power on their own account [3].

Expected loss of whole clusters $\ell$ Optional

Values: 0% up to, but not including, 100%

Clusters, such as practices or schools, expected to withdraw entirely.

Correction for few clusters $F_t$ Optional

Values: On or off

Worth applying when there are few clusters per group, roughly fewer than 15–20.

Assumptions

The two groups are equal in size and independent; participants are independent apart from any clustering specified.
Every cell is expected to hold at least 5 participants, so the chi-square approximation holds.
With clustering: clusters are independent, with the same ICC throughout, and the analysis accounts for the clustering, for example with a mixed model or cluster-level summaries [2].
Dropout is unrelated to the outcome; recruiting more participants restores the numbers but cannot remove bias.

Worked examples

Thirty against forty-five per cent

p₁ = 0.3 and p₂ = 0.45 give w = 0.155; α = 0.05, 80% power: N = 328, 328 / 2 = 164 per group, close to the 163 of the two-proportion test.

Sample size: 328

Protocol statement
To detect a difference between proportions of 30% and 45% with 80% power at a significance level of 0.05, using a chi-square test on a 2 × 2 table with two equal groups, the required sample size is 328 participants. Calculated with StatWisely (statwisely.com) using: N: solve P[χ²(df=1, λ=Nw²) > χ²_(1-α,1)] = 1-β. These are planning estimates; confirm the final design with a qualified statistician.

Open this example in the calculator

Allowing for dropout

Expecting 10% to provide no outcome: 328 / 0.9 = 364.4, so 365 are recruited.

Needed for analysis: 328 · To recruit: 365

Protocol statement
To detect a difference between proportions of 30% and 45% with 80% power at a significance level of 0.05, using a chi-square test on a 2 × 2 table with two equal groups, the required sample size is 328 participants. Allowing for 10% dropout, 365 participants should be recruited. Calculated with StatWisely (statwisely.com) using: N: solve P[χ²(df=1, λ=Nw²) > χ²_(1-α,1)] = 1-β. These are planning estimates; confirm the final design with a qualified statistician.

Open this example in the calculator

Validation

Every worked example on this page is recomputed by the StatWisely engine whenever the software is built. If a number or formula here stopped matching the calculator, the release would be stopped until this document was corrected.
An independent implementation in R recomputes this test every night and before each production release, and the results are compared with the engine’s.

How to cite

StatWisely. Sample size for a chi-square test on a 2 × 2 table. Method document, engine version 1.0.0, reviewed 2026-10-01. https://statwisely.com/methods/proportions-chi-square/

Cite the original references for the formulas your design uses as well.

References

1.Cohen J. Statistical Power Analysis for the Behavioral Sciences. 2nd ed. Hillsdale, NJ: Lawrence Erlbaum Associates; 1988.
2.Donner A, Klar N. Design and Analysis of Cluster Randomization Trials in Health Research. London: Arnold; 2000.
3.Eldridge SM, Ashby D, Kerry S. Sample size for cluster randomized trials: effect of coefficient of variation of cluster size and analysis method. International Journal of Epidemiology. 2006;35(5):1292–1300. doi:10.1093/ije/dyl129
4.Hayes RJ, Moulton LH. Cluster Randomised Trials. 2nd ed. Boca Raton: CRC Press; 2017.
5.Adams G, Gulliford MC, Ukoumunne OC, Eldridge S, Chinn S, Campbell MJ. Patterns of intra-cluster correlation from primary care research to inform study design and analysis. Journal of Clinical Epidemiology. 2004;57(8):785–794. doi:10.1016/j.jclinepi.2003.12.013