# scipy.stats.bartlett¶

scipy.stats.bartlett(*args)[source]

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
statisticfloat

The test statistic.

pvaluefloat

The p-value of the test.

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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