Sample size for a chi-square goodness-of-fit test

When to use it

Testing whether categories occur in stated proportions: equal use of four clinics, or a population’s known distribution of blood groups.

Not for

Two variables against each other. Use the test of independence.

Notation

$N$: Number of participants
$k$: Number of categories
$e_i,\ a_i$: Category probabilities under the null and the alternative
$w$: Cohen’s effect size
$\alpha$: 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 from the two distributions

w = \sqrt{\sum_{i=1}^{k} \frac{(a_i - e_i)^{2}}{e_i}}

The chi-square discrepancy per participant between the expected and true distributions [1].

The smallest N with enough power

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

On k − 1 degrees of freedom. More categories at the same w need more participants.

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 distribution $e_1, \dots, e_k$

Values: One probability per category, each above 0, adding to 1

The distribution the test is against: the null hypothesis.

True distribution to detect $a_1, \dots, a_k$

Values: One probability per category, each above 0, adding to 1

The smallest departure from the expected distribution that would matter.

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

Participants are independent apart from any clustering specified, each in exactly one category.
Every category expects 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

Four categories

Expected 25% in each of four categories; true 35%, 25%, 25% and 15%: w = √(0.04 + 0.04) = 0.283 on 3 degrees of freedom; α = 0.05, 80% power: N = 137.

Sample size: 137

Protocol statement
To detect a departure from the expected distribution of w = 0.283 with 80% power at a significance level of 0.05, using a chi-square goodness-of-fit test across 4 categories, the required sample size is 137 participants. Calculated with StatWisely (statwisely.com) using: N: solve the noncentral chi-square power at λ=Nw² on k-1 df. 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: 137 / 0.9 = 152.2, so 153 are recruited.

Needed for analysis: 137 · To recruit: 153

Protocol statement
To detect a departure from the expected distribution of w = 0.283 with 80% power at a significance level of 0.05, using a chi-square goodness-of-fit test across 4 categories, the required sample size is 137 participants. Allowing for 10% dropout, 153 participants should be recruited. Calculated with StatWisely (statwisely.com) using: N: solve the noncentral chi-square power at λ=Nw² on k-1 df. 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 goodness-of-fit test. Method document, engine version 1.0.0, reviewed 2026-10-01. https://statwisely.com/methods/proportions-goodness-of-fit/

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