Specifically, two group ANOVA vs unpaired t-test. My understanding was that those would always give the same p-value, but someone just told me that there could be rare cases where the ANOVA and the t-test would lead to different p-values. My questions: (a) is this true? and if yes, (b) under what conditions would this be more likely to occur? Thanks.

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    $\begingroup$ Which t-test, exactly? There are several variants. Many people would automatically use a version that accommodates the possibility of unequal variances in the groups. That is likely to yield a p-value different from ANOVA, which assumes equal variances. $\endgroup$ – whuber Aug 28 '16 at 20:24
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    $\begingroup$ If you compare equal variance t with equal variance anova on 2 groups they should give the same p-values; indeed, $t^2=F$. However if one uses say a Welch-Satterthwaite approach for dealing with unequal variance and the other does not (and either might) then they can differ. For the equal-variance case, I give an outline of the connection in my answer here (in the latter part) both doing some of the algebra and showing a specific numerical example (it doesn't do a complete derivation, however). $\endgroup$ – Glen_b Aug 28 '16 at 22:55

The two should give the same results in all cases. ANOVA is a generalization of t-test.

  • $\begingroup$ Thank you for the prompt response. Do you know of a link to a (ideally simple) mathematical proof of this on line? $\endgroup$ – Sandeep Aug 28 '16 at 18:54
  • $\begingroup$ Not offhand. Probably the thing to do is to examine the formulas for ANOVA and the t test and see the similarities. Others here would be better guides. $\endgroup$ – Peter Flom Aug 28 '16 at 19:10
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    $\begingroup$ ANOVA uses the F distribution, a t test uses the t distribution. The F is a squared t, therefore they will always give the answer to the same question. $\endgroup$ – Repmat Aug 28 '16 at 19:49
  • $\begingroup$ As others have noted, if the pooled sum of squares is used for the denominator of the t then the F is the square of the corresponding t and the p-values are identical. Other forms of the t will differ from the usual F. $\endgroup$ – David Smith Jun 15 '17 at 16:57

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