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The equation for this test is the following (that's how it is stated in most texts). I got a proof that it indeed has a t-distribution:

$$T=\frac{(\bar{X}-\bar{Y})-(\mu_x-\mu_y)}{S_p\sqrt{\frac{1}{n}+\frac{1}{n}}}$$

However, I've also seen it in the following form:

$$T=\frac{(\bar{X}-\bar{Y})}{S_p\sqrt{\frac{1}{n}+\frac{1}{n}}}$$

How can we prove that also the second test has a t-distribution? Why some books describe the test with the second equation?

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Under the null hypothesis, $H_o: \mu_x = \mu_y$, so the equations are identical.

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  • $\begingroup$ So it works like: we take the first formula, assume the means are identical in both populations and see what's the probability of them being equal based on the distribution of T (second equation)? If the probablity is very low, we deduce that it's not very likely the means are equal. $\endgroup$ – user4205580 Aug 13 '15 at 16:39
  • $\begingroup$ yes. that is basically it. $\endgroup$ – mandata Aug 13 '15 at 17:06
  • $\begingroup$ @user4205580 & mandata Without context it's hard to say, but my guess is the first formula is meant to test the null hypothesis $E(X) - E(Y) = \mu_X - \mu_Y$. Still, in both cases the statistic is t-distributed under the null hypothesis. $\endgroup$ – A. Donda Aug 13 '15 at 18:18

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