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Module « scipy.stats »

Fonction poisson_means_test - module scipy.stats

Signature de la fonction poisson_means_test

def poisson_means_test(k1, n1, k2, n2, *, diff=0, alternative='two-sided') 

Description

help(scipy.stats.poisson_means_test)

Performs the Poisson means test, AKA the "E-test".

This is a test of the null hypothesis that the difference between means of
two Poisson distributions is `diff`. The samples are provided as the
number of events `k1` and `k2` observed within measurement intervals
(e.g. of time, space, number of observations) of sizes `n1` and `n2`.

Parameters
----------
k1 : int
    Number of events observed from distribution 1.
n1: float
    Size of sample from distribution 1.
k2 : int
    Number of events observed from distribution 2.
n2 : float
    Size of sample from distribution 2.
diff : float, default=0
    The hypothesized difference in means between the distributions
    underlying the samples.
alternative : {'two-sided', 'less', 'greater'}, optional
    Defines the alternative hypothesis.
    The following options are available (default is 'two-sided'):

      * 'two-sided': the difference between distribution means is not
        equal to `diff`
      * 'less': the difference between distribution means is less than
        `diff`
      * 'greater': the difference between distribution means is greater
        than `diff`

Returns
-------
statistic : float
    The test statistic (see [1]_ equation 3.3).
pvalue : float
    The probability of achieving such an extreme value of the test
    statistic under the null hypothesis.

Notes
-----

Let:

.. math:: X_1 \sim \mbox{Poisson}(\mathtt{n1}\lambda_1)

be a random variable independent of

.. math:: X_2  \sim \mbox{Poisson}(\mathtt{n2}\lambda_2)

and let ``k1`` and ``k2`` be the observed values of :math:`X_1`
and :math:`X_2`, respectively. Then `poisson_means_test` uses the number
of observed events ``k1`` and ``k2`` from samples of size ``n1`` and
``n2``, respectively, to test the null hypothesis that

.. math::
   H_0: \lambda_1 - \lambda_2 = \mathtt{diff}

A benefit of the E-test is that it has good power for small sample sizes,
which can reduce sampling costs [1]_. It has been evaluated and determined
to be more powerful than the comparable C-test, sometimes referred to as
the Poisson exact test.

References
----------
.. [1]  Krishnamoorthy, K., & Thomson, J. (2004). A more powerful test for
   comparing two Poisson means. Journal of Statistical Planning and
   Inference, 119(1), 23-35.

.. [2]  Przyborowski, J., & Wilenski, H. (1940). Homogeneity of results in
   testing samples from Poisson series: With an application to testing
   clover seed for dodder. Biometrika, 31(3/4), 313-323.

Examples
--------

Suppose that a gardener wishes to test the number of dodder (weed) seeds
in a sack of clover seeds that they buy from a seed company. It has
previously been established that the number of dodder seeds in clover
follows the Poisson distribution.

A 100 gram sample is drawn from the sack before being shipped to the
gardener. The sample is analyzed, and it is found to contain no dodder
seeds; that is, `k1` is 0. However, upon arrival, the gardener draws
another 100 gram sample from the sack. This time, three dodder seeds are
found in the sample; that is, `k2` is 3. The gardener would like to
know if the difference is significant and not due to chance. The
null hypothesis is that the difference between the two samples is merely
due to chance, or that :math:`\lambda_1 - \lambda_2 = \mathtt{diff}`
where :math:`\mathtt{diff} = 0`. The alternative hypothesis is that the
difference is not due to chance, or :math:`\lambda_1 - \lambda_2 \ne 0`.
The gardener selects a significance level of 5% to reject the null
hypothesis in favor of the alternative [2]_.

>>> import scipy.stats as stats
>>> res = stats.poisson_means_test(0, 100, 3, 100)
>>> res.statistic, res.pvalue
(-1.7320508075688772, 0.08837900929018157)

The p-value is .088, indicating a near 9% chance of observing a value of
the test statistic under the null hypothesis. This exceeds 5%, so the
gardener does not reject the null hypothesis as the difference cannot be
regarded as significant at this level.

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