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:
- re: theoretical distribution modification. However the distribution tested against is still t - which still seems parametric to me
- 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.