scipy.stats.barnard_exact(table, alternative='two-sided', pooled=True, n=32)[source]

Perform a Barnard exact test on a 2x2 contingency table.

tablearray_like of ints

A 2x2 contingency table. Elements should be non-negative integers.

alternative{‘two-sided’, ‘less’, ‘greater’}, optional

Defines the null and alternative hypotheses. Default is ‘two-sided’. Please see explanations in the Notes section below.

pooledbool, optional

Whether to compute score statistic with pooled variance (as in Student’s t-test, for example) or unpooled variance (as in Welch’s t-test). Default is True.

nint, optional

Number of sampling points used in the construction of the sampling method. Note that this argument will automatically be converted to the next higher power of 2 since scipy.stats.qmc.Sobol is used to select sample points. Default is 32. Must be positive. In most cases, 32 points is enough to reach good precision. More points comes at performance cost.


A result object with the following attributes.


The Wald statistic with pooled or unpooled variance, depending on the user choice of pooled.


P-value, the probability of obtaining a distribution at least as extreme as the one that was actually observed, assuming that the null hypothesis is true.

See also


Chi-square test of independence of variables in a contingency table.


Fisher exact test on a 2x2 contingency table.


Boschloo’s exact test on a 2x2 contingency table, which is an uniformly more powerful alternative to Fisher’s exact test.


Barnard’s test is an exact test used in the analysis of contingency tables. It examines the association of two categorical variables, and is a more powerful alternative than Fisher’s exact test for 2x2 contingency tables.

Let’s define \(X_0\) a 2x2 matrix representing the observed sample, where each column stores the binomial experiment, as in the example below. Let’s also define \(p_1, p_2\) the theoretical binomial probabilities for \(x_{11}\) and \(x_{12}\). When using Barnard exact test, we can assert three different null hypotheses :

  • \(H_0 : p_1 \geq p_2\) versus \(H_1 : p_1 < p_2\), with alternative = “less”

  • \(H_0 : p_1 \leq p_2\) versus \(H_1 : p_1 > p_2\), with alternative = “greater”

  • \(H_0 : p_1 = p_2\) versus \(H_1 : p_1 \neq p_2\), with alternative = “two-sided” (default one)

In order to compute Barnard’s exact test, we are using the Wald statistic [3] with pooled or unpooled variance. Under the default assumption that both variances are equal (pooled = True), the statistic is computed as:

\[T(X) = \frac{ \hat{p}_1 - \hat{p}_2 }{ \sqrt{ \hat{p}(1 - \hat{p}) (\frac{1}{c_1} + \frac{1}{c_2}) } }\]

with \(\hat{p}_1, \hat{p}_2\) and \(\hat{p}\) the estimator of \(p_1, p_2\) and \(p\), the latter being the combined probability, given the assumption that \(p_1 = p_2\).

If this assumption is invalid (pooled = False), the statistic is:

\[T(X) = \frac{ \hat{p}_1 - \hat{p}_2 }{ \sqrt{ \frac{\hat{p}_1 (1 - \hat{p}_1)}{c_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{c_2} } }\]

The p-value is then computed as:

\[\sum \binom{c_1}{x_{11}} \binom{c_2}{x_{12}} \pi^{x_{11} + x_{12}} (1 - \pi)^{t - x_{11} - x_{12}}\]

where the sum is over all 2x2 contingency tables \(X\) such that: * \(T(X) \leq T(X_0)\) when alternative = “less”, * \(T(X) \geq T(X_0)\) when alternative = “greater”, or * \(T(X) \geq |T(X_0)|\) when alternative = “two-sided”. Above, \(c_1, c_2\) are the sum of the columns 1 and 2, and \(t\) the total (sum of the 4 sample’s element).

The returned p-value is the maximum p-value taken over the nuisance parameter \(\pi\), where \(0 \leq \pi \leq 1\).

This function’s complexity is \(O(n c_1 c_2)\), where n is the number of sample points.



Barnard, G. A. “Significance Tests for 2x2 Tables”. Biometrika. 34.1/2 (1947): 123-138. DOI:dpgkg3


Mehta, Cyrus R., and Pralay Senchaudhuri. “Conditional versus unconditional exact tests for comparing two binomials.” Cytel Software Corporation 675 (2003): 1-5.


“Wald Test”. Wikipedia.


An example use of Barnard’s test is presented in [2].

Consider the following example of a vaccine efficacy study (Chan, 1998). In a randomized clinical trial of 30 subjects, 15 were inoculated with a recombinant DNA influenza vaccine and the 15 were inoculated with a placebo. Twelve of the 15 subjects in the placebo group (80%) eventually became infected with influenza whereas for the vaccine group, only 7 of the 15 subjects (47%) became infected. The data are tabulated as a 2 x 2 table:

    Vaccine  Placebo
Yes     7        12
No      8        3

When working with statistical hypothesis testing, we usually use a threshold probability or significance level upon which we decide to reject the null hypothesis \(H_0\). Suppose we choose the common significance level of 5%.

Our alternative hypothesis is that the vaccine will lower the chance of becoming infected with the virus; that is, the probability \(p_1\) of catching the virus with the vaccine will be less than the probability \(p_2\) of catching the virus without the vaccine. Therefore, we call barnard_exact with the alternative="less" option:

>>> import scipy.stats as stats
>>> res = stats.barnard_exact([[7, 12], [8, 3]], alternative="less")
>>> res.statistic
>>> res.pvalue

Under the null hypothesis that the vaccine will not lower the chance of becoming infected, the probability of obtaining test results at least as extreme as the observed data is approximately 3.4%. Since this p-value is less than our chosen significance level, we have evidence to reject \(H_0\) in favor of the alternative.

Suppose we had used Fisher’s exact test instead:

>>> _, pvalue = stats.fisher_exact([[7, 12], [8, 3]], alternative="less")
>>> pvalue

With the same threshold significance of 5%, we would not have been able to reject the null hypothesis in favor of the alternative. As stated in [2], Barnard’s test is uniformly more powerful than Fisher’s exact test because Barnard’s test does not condition on any margin. Fisher’s test should only be used when both sets of marginals are fixed.