[issue21184] statistics.pvariance with known mean does not work as expected

[Steven D'Aprano]

New submission from Steven D'Aprano:

If you know the population mean mu, you should calculate the sample variance by passing mu as an explicit argument to statistics.pvariance. Unfortunately, it doesn't work as designed:

py> data = [1, 2, 2, 2, 3, 4]  # sample from a population with mu=2.5
py> statistics.pvariance(data)  # uses the sample mean 2.3333...
0.8888888888888888
py> statistics.pvariance(data, 2.5)  # using known population mean
0.8888888888888888

The second calculation ought to be 0.91666... not 0.88888...

The problem lies with the _ss private function which calculates the sum of square deviations. Unfortunately it is too clever: it includes an error adjustment term

ss -= _sum((x-c) for x in data)**2/len(data)

which mathematically is expected to be zero when c is calculated as the mean of data, but due to rounding may not be quite zero. But when c is given explicitly, as happens if the caller provides an explicit mu argument to pvariance, then the error adjustment has the effect of neutralizing the explicit mu.

The obvious fix is to just skip the adjustment in _ss when c is explicitly given, but I'm not sure if that's the best approach.

----------
assignee: stevenjd
components: Library (Lib)
messages: 215802
nosy: stevenjd
priority: normal
severity: normal
stage: needs patch
status: open
title: statistics.pvariance with known mean does not work as expected
type: behavior
versions: Python 3.4, Python 3.5

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Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue21184>
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