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In an answer to another question it is asserted that the Welch correction to the t-test makes it non-parametric, and this assertion isn't challenged by a frequent contributor with a deep level of statistical knowledge.

In what way does the Welch correction make the test non-parametric? Is it simply that the change in the criterion df means that the t-distribution is no longer strictly being used, or is something more profound happening at a conceptual level?

If something more profound is happening at a conceptual level, and/or the answer can contain a description of what the Welsch's correction is accomplishing without resorting to a pile of equations - that would be great.

The answer author describes their reasoning as follows:

The reason I don't recommend the Welch correction is that it doesn't just change the degrees of freedom and subsequent theoretical distribution from which the p-value is drawn. It makes the test non-parametric. To perform a Welch corrected t-test one still pools variance as if equal variance can be assumed but then changes the final testing procedure implying either that equal variance cannot be assumed, or that you only care about the sample variances. This makes it a non-parametric test because the pooled variance is considered non-representative of the population and you conceded that you're just testing your observed values.

There appear to be at least two arguments being made:

  1. re: theoretical distribution modification. However the distribution tested against is still t - which still seems parametric to me
  2. That pooled variance is considered non-representative and subjected to a modification to make it representative and that makes the test non-parametric - which still seems parametric to me; after all don't we do something similar when we modify the sample variance to come up with the sample estimate of the population variance?

Neither quite makes sense to me just yet - so I'm looking for additional clarification in a response.

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    $\begingroup$ It has been asserted (I'm afraid I cannot remember the source, but recall it's a respectable one) that the Student t-test itself can be considered non-parametric, too. I believe part of the argument is that it approximates a permutation test. "Parametric" does not refer to the sample distribution of the test statistic, as your post seems to imply: it refers to the underlying distribution of the data. Please note that the answer to which you are reacting has (in the week since it was posted) collected no upvotes. $\endgroup$ – whuber Jul 5 '17 at 13:22
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    $\begingroup$ Have you seen a follow-up question by John with an answer by whuber? stats.stackexchange.com/questions/287747 $\endgroup$ – amoeba says Reinstate Monica Jul 5 '17 at 13:25
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So I'll follow the Occam's razor approach and give the simplest answer against the assertion that Welch's t-test is non parametric. Starting with the generally accepted definition of non parametric:

Non parametric: not involving any assumptions as to the form or parameters of a frequency distribution.

Here's a publication that starts with the view that "non-parametric" does not have an established accepted definition, but still concludes to the above stated definition as appropriate.

t-test for testing equal means of 2 distributions requires:

  1. Normality assumption for distribution (relates to form of data)
  2. Homoscedasticity assumption for variances (related to parameter of data)

Welch's t-test disregards the homoscedasticity assumption but still needs the normality assumption. Now this is where we draw the distinction between the sample being considered and the underlying population. This warrants looking at the specific use case and the data collected eg.- sample of heights where the true population is normally distributed. So there might be cases where there is compelling evidence that t-test could be treated as non parametric, but I will refrain from concretely recommending it.

If non parametric test is required, Wilcoxon - Mann - Whitney would be better.

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    $\begingroup$ Does the normality assumption for distribution apply to the source distribution of the data instead of the distribution of samplable means? Isn't the ability to abstract from the underlying distributional shape to one with known characteristics why we envoke CLT in the first place? $\endgroup$ – russellpierce Jul 5 '17 at 16:39
  • $\begingroup$ For sanity I will use the terms sample and true population. One would expect that the sample is representative of the true population for all good sampling practices. In the definition I cite, the assumptions apply to "frequency distribution", I in everyday practice use it on sample. Does that answer what you were getting to? $\endgroup$ – DivyaJyoti Rajdev Jul 5 '17 at 16:46
  • $\begingroup$ @Divya.Jyoti Rajdev rpierce is talking about a different thing. $\endgroup$ – Subhash C. Davar Jul 8 '17 at 10:14

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