# Independent t-tests. One- or two-tailed? [duplicate]

For an independent t-test, if my hypothesis states there will be no significant difference between the two groups, is that a one- or two-tailed test?

• stats.stackexchange.com/questions/24676/… might be interesting.
– Gala
Jul 10, 2013 at 10:06
• @COOLSerdash Maybe you can make this the answer so that rose can accept it. I don't think much more can be said.
– Gala
Jul 10, 2013 at 10:07
• It depends not on your null, but on your alternative, which you haven't specified. (In general, if you're not certain what you should be using, you should probably be using two-tailed. If you're sure it should be a one tailed test, you might be correct) Jul 10, 2013 at 11:03

As @Glen_b mentiones in the comments: The answer depends on your alternative hypothesis $\text{H}_{1}$. From your question, I assume that your alternative hypothesis is just that the means differ. If your hypothesis is that the two group means are equal vs. that they differ, i.e.: $\text{H}_{0}: \mu_{1}=\mu_{2}$ vs. $\text{H}_{1}: \mu_{1}\neq\mu_{2}$, then you have a two-tailed test. This is because your alternative hypothesis is that the means differ in either direction: the mean of the second group ($\mu_{2}$) could either be higher or smaller than the mean of the first group ($\mu_{1}$). A one-sided hypothesis would for example be: $\text{H}_{0}: \mu_{1}\leq\mu_{2}$ vs. $\text{H}_{1}: \mu_{1}>\mu_{2}$. In this case, the alternative hypothesis is that the mean of the second group is smaller than the mean of the first group. So your alternative hypothesis is one-sided. Note that the null-hypothesis and the alternative hypothesis are complementary: if $\text{H}_{1}: \mu_{1}\neq\mu_{2}$ then $\text{H}_{0}: \mu_{1}=\mu_{2}$, if $\text{H}_{1}: \mu_{1}>\mu_{2}$ then $\text{H}_{0}: \mu_{1}\leq\mu_{2}$ and if $\text{H}_{1}: \mu_{1}<\mu_{2}$ then $\text{H}_{0}: \mu_{1}\geq\mu_{2}$ and so on.
• @what You are correct (I've edited the answer). But it is valid to use, for example the hypotheses: $\text{H}_{1}:\theta<a$ vs. $\text{H}_{0}:\theta=a$, even if the implicit null hypothesis is $\text{H}_{0}:\theta\geq a$. The rationale for using the simplified null hypothesis is that any reasonable decision procedure for deciding between $\text{H}_{0}:\theta=a$ and $\text{H}_{1}:\theta<a$ will also be reasonable for deciding between the claim that $\theta\geq a$ and $\text{H}_{1}$ (see here). Jul 10, 2013 at 13:29
• @what This post deals exactly with your question. To back my claim up further, Young and Smith write (page 65) that possibly, but not necessarily, $\Omega_{0}$ and $\Omega_{1}$ satisfy $\Omega_{0} \cup \Omega_{1} = \Omega$. Jul 11, 2013 at 10:02