Module « scipy.stats »
Signature de la fonction bartlett
def bartlett(*args)
Description
bartlett.__doc__
Perform Bartlett's test for equal variances.
Bartlett's test tests the null hypothesis that all input samples
are from populations with equal variances. For samples
from significantly non-normal populations, Levene's test
`levene` is more robust.
Parameters
----------
sample1, sample2,... : array_like
arrays of sample data. Only 1d arrays are accepted, they may have
different lengths.
Returns
-------
statistic : float
The test statistic.
pvalue : float
The p-value of the test.
See Also
--------
fligner : A non-parametric test for the equality of k variances
levene : A robust parametric test for equality of k variances
Notes
-----
Conover et al. (1981) examine many of the existing parametric and
nonparametric tests by extensive simulations and they conclude that the
tests proposed by Fligner and Killeen (1976) and Levene (1960) appear to be
superior in terms of robustness of departures from normality and power
([3]_).
References
----------
.. [1] https://www.itl.nist.gov/div898/handbook/eda/section3/eda357.htm
.. [2] Snedecor, George W. and Cochran, William G. (1989), Statistical
Methods, Eighth Edition, Iowa State University Press.
.. [3] Park, C. and Lindsay, B. G. (1999). Robust Scale Estimation and
Hypothesis Testing based on Quadratic Inference Function. Technical
Report #99-03, Center for Likelihood Studies, Pennsylvania State
University.
.. [4] Bartlett, M. S. (1937). Properties of Sufficiency and Statistical
Tests. Proceedings of the Royal Society of London. Series A,
Mathematical and Physical Sciences, Vol. 160, No.901, pp. 268-282.
Examples
--------
Test whether or not the lists `a`, `b` and `c` come from populations
with equal variances.
>>> from scipy.stats import bartlett
>>> 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 = bartlett(a, b, c)
>>> p
1.1254782518834628e-05
The very 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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