scipy.stats.median_test#

scipy.stats.median_test(*samples, ties='below', correction=True, lambda_=1, nan_policy='propagate')[source]#

Perform a Mood’s median test.

Test that two or more samples come from populations with the same median.

Let n = len(samples) be the number of samples. The “grand median” of all the data is computed, and a contingency table is formed by classifying the values in each sample as being above or below the grand median. The contingency table, along with correction and lambda_, are passed to scipy.stats.chi2_contingency to compute the test statistic and p-value.

Parameters
sample1, sample2, …array_like

The set of samples. There must be at least two samples. Each sample must be a one-dimensional sequence containing at least one value. The samples are not required to have the same length.

tiesstr, optional

Determines how values equal to the grand median are classified in the contingency table. The string must be one of:

"below":
    Values equal to the grand median are counted as "below".
"above":
    Values equal to the grand median are counted as "above".
"ignore":
    Values equal to the grand median are not counted.

The default is “below”.

correctionbool, optional

If True, and there are just two samples, apply Yates’ correction for continuity when computing the test statistic associated with the contingency table. Default is True.

lambda_float or str, optional

By default, the statistic computed in this test is Pearson’s chi-squared statistic. lambda_ allows a statistic from the Cressie-Read power divergence family to be used instead. See power_divergence for details. Default is 1 (Pearson’s chi-squared statistic).

nan_policy{‘propagate’, ‘raise’, ‘omit’}, optional

Defines how to handle when input contains nan. ‘propagate’ returns nan, ‘raise’ throws an error, ‘omit’ performs the calculations ignoring nan values. Default is ‘propagate’.

Returns
statfloat

The test statistic. The statistic that is returned is determined by lambda_. The default is Pearson’s chi-squared statistic.

pfloat

The p-value of the test.

mfloat

The grand median.

tablendarray

The contingency table. The shape of the table is (2, n), where n is the number of samples. The first row holds the counts of the values above the grand median, and the second row holds the counts of the values below the grand median. The table allows further analysis with, for example, scipy.stats.chi2_contingency, or with scipy.stats.fisher_exact if there are two samples, without having to recompute the table. If nan_policy is “propagate” and there are nans in the input, the return value for table is None.

See also

kruskal

Compute the Kruskal-Wallis H-test for independent samples.

mannwhitneyu

Computes the Mann-Whitney rank test on samples x and y.

Notes

New in version 0.15.0.

References

1

Mood, A. M., Introduction to the Theory of Statistics. McGraw-Hill (1950), pp. 394-399.

2

Zar, J. H., Biostatistical Analysis, 5th ed. Prentice Hall (2010). See Sections 8.12 and 10.15.

Examples

A biologist runs an experiment in which there are three groups of plants. Group 1 has 16 plants, group 2 has 15 plants, and group 3 has 17 plants. Each plant produces a number of seeds. The seed counts for each group are:

Group 1: 10 14 14 18 20 22 24 25 31 31 32 39 43 43 48 49
Group 2: 28 30 31 33 34 35 36 40 44 55 57 61 91 92 99
Group 3:  0  3  9 22 23 25 25 33 34 34 40 45 46 48 62 67 84

The following code applies Mood’s median test to these samples.

>>> g1 = [10, 14, 14, 18, 20, 22, 24, 25, 31, 31, 32, 39, 43, 43, 48, 49]
>>> g2 = [28, 30, 31, 33, 34, 35, 36, 40, 44, 55, 57, 61, 91, 92, 99]
>>> g3 = [0, 3, 9, 22, 23, 25, 25, 33, 34, 34, 40, 45, 46, 48, 62, 67, 84]
>>> from scipy.stats import median_test
>>> stat, p, med, tbl = median_test(g1, g2, g3)

The median is

>>> med
34.0

and the contingency table is

>>> tbl
array([[ 5, 10,  7],
       [11,  5, 10]])

p is too large to conclude that the medians are not the same:

>>> p
0.12609082774093244

The “G-test” can be performed by passing lambda_="log-likelihood" to median_test.

>>> g, p, med, tbl = median_test(g1, g2, g3, lambda_="log-likelihood")
>>> p
0.12224779737117837

The median occurs several times in the data, so we’ll get a different result if, for example, ties="above" is used:

>>> stat, p, med, tbl = median_test(g1, g2, g3, ties="above")
>>> p
0.063873276069553273
>>> tbl
array([[ 5, 11,  9],
       [11,  4,  8]])

This example demonstrates that if the data set is not large and there are values equal to the median, the p-value can be sensitive to the choice of ties.