Module « scipy.stats »
Signature de la fonction levene
def levene(*args, center='median', proportiontocut=0.05)
Description
levene.__doc__
Perform Levene test for equal variances.
The Levene test tests the null hypothesis that all input samples
are from populations with equal variances. Levene's test is an
alternative to Bartlett's test `bartlett` in the case where
there are significant deviations from normality.
Parameters
----------
sample1, sample2, ... : array_like
The sample data, possibly with different lengths. Only one-dimensional
samples are accepted.
center : {'mean', 'median', 'trimmed'}, optional
Which function of the data to use in the test. The default
is 'median'.
proportiontocut : float, optional
When `center` is 'trimmed', this gives the proportion of data points
to cut from each end. (See `scipy.stats.trim_mean`.)
Default is 0.05.
Returns
-------
statistic : float
The test statistic.
pvalue : float
The p-value for the test.
Notes
-----
Three variations of Levene's test are possible. The possibilities
and their recommended usages are:
* 'median' : Recommended for skewed (non-normal) distributions>
* 'mean' : Recommended for symmetric, moderate-tailed distributions.
* 'trimmed' : Recommended for heavy-tailed distributions.
The test version using the mean was proposed in the original article
of Levene ([2]_) while the median and trimmed mean have been studied by
Brown and Forsythe ([3]_), sometimes also referred to as Brown-Forsythe
test.
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda35a.htm
.. [2] Levene, H. (1960). In Contributions to Probability and Statistics:
Essays in Honor of Harold Hotelling, I. Olkin et al. eds.,
Stanford University Press, pp. 278-292.
.. [3] Brown, M. B. and Forsythe, A. B. (1974), Journal of the American
Statistical Association, 69, 364-367
Examples
--------
Test whether or not the lists `a`, `b` and `c` come from populations
with equal variances.
>>> from scipy.stats import levene
>>> a = [8.88, 9.12, 9.04, 8.98, 9.00, 9.08, 9.01, 8.85, 9.06, 8.99]
>>> b = [8.88, 8.95, 9.29, 9.44, 9.15, 9.58, 8.36, 9.18, 8.67, 9.05]
>>> c = [8.95, 9.12, 8.95, 8.85, 9.03, 8.84, 9.07, 8.98, 8.86, 8.98]
>>> stat, p = levene(a, b, c)
>>> p
0.002431505967249681
The small p-value suggests that the populations do not have equal
variances.
This is not surprising, given that the sample variance of `b` is much
larger than that of `a` and `c`:
>>> [np.var(x, ddof=1) for x in [a, b, c]]
[0.007054444444444413, 0.13073888888888888, 0.008890000000000002]
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