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I'm still confused as to why it is often recommended to use an independent samples T-Test when comparing two means rather than a One-way Anova.

If t^2 = F, and the p values are the same, why use a T-Test? What is the difference?

I've read that its because a T-Test 'is more flexible', but why is this the case? Is it just because the T distribution is two-tailed, whereas the F distribution is one-tailed?

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With questions like this there is always a bit of ambiguity, since there are lots of different tests that use the T-distribution as the null distribution, and hence, are referred to as "T-tests". Comparing to the ANOVA test requires consideration of which particular T-test you are talking about.

Gossett's T-test (the two-sample t-test with assumed equal variances) is formally equivalent to a one-way two-group ANOVA test from Gaussian linear regression. For these tests it can be shown that $T^2 = F$ and so the p-values of both tests are the same. Since these are the same test, it makes no difference which you use. Where the T-test becomes "more flexible" is when you remove the assumption of equal variances for the groups and use Welch's T-test or some similar variant. With this variation you now have a T-test that is not equivalent to the one-way two-group ANOVA test from Gaussian linear regression. The greater "flexibility" occurs because the test accomodates cases where the variances of the two groups are substantially different.

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  • $\begingroup$ Thanks this is really helpful. Does a one-way two-group ANOVA have any way of accounting for unequal variances? Also, when I look at the equations behind the scenes for Gossett's T-Test and the one-way ANOVA, I don't see how the calculations end up producing the relationship T^2 = F. Do you know of any resources to help me understand how this relationship comes about? $\endgroup$ – thymeandspace Sep 4 '18 at 0:37
  • $\begingroup$ ANOVA is a technique built on an underlying model, so it depends on the assumptions of the underlying model. It is possible to generalise a two-group Gaussian linear regression model to allow heteroscedasticity (i.e., different variance for the two groups), and in this case you might be able to get an ANOVA that accounts for differences in variance of the groups. Whether or not this would be equivalent to Welch's test depends on how you set it up. $\endgroup$ – Reinstate Monica Sep 4 '18 at 3:38
  • $\begingroup$ If you want to understand the formal equivalence of Gossett's T-test with a one-way two-group ANOVA test from Gaussian linear regression, you could have a look at some simple proofs like this, or some more general questions like this. $\endgroup$ – Reinstate Monica Sep 4 '18 at 3:42
  • $\begingroup$ So something like a Greenhouse-Geisser correction isn't really correcting heteroscedasticity? $\endgroup$ – thymeandspace Sep 4 '18 at 22:02
  • $\begingroup$ As I understand it, that method adds an additional parameter to represent non-sphericity of the model, and then estimates that parameter from the data. My understanding is that it is used to allow correlation between a subject factor and the treatment in an experiment. I'm not aware of it being used to correct heteroscedasticity, but maybe I'm just not aware of it. In any case, I'm not really clear on why you think that something I've said implies anything for this test. $\endgroup$ – Reinstate Monica Sep 5 '18 at 2:36

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