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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?

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    $\begingroup$ stats.stackexchange.com/questions/24676/… might be interesting. $\endgroup$
    – Gala
    Jul 10, 2013 at 10:06
  • $\begingroup$ @COOLSerdash Maybe you can make this the answer so that rose can accept it. I don't think much more can be said. $\endgroup$
    – Gala
    Jul 10, 2013 at 10:07
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    $\begingroup$ 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) $\endgroup$
    – Glen_b
    Jul 10, 2013 at 11:03

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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.

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  • $\begingroup$ -1: Part of your answer is wrong. H0 is always H̅ 1 (not-H1, the complement in set theory). There must not be a rest. If H0:μ1=μ2, then H1:μ1≠μ2. If H0:μ1≥μ2, then H1:μ1<μ2. If H0:μ1≤μ2, then H1:μ1>μ2. These are the three possible hypothesis pairs. In your set of H0:μ1=μ2 vs. H1:μ1>μ2 there is a third that you leave uncovered (H2:μ1<μ2), so your two hypotheses cover only a part of the field you want to model, and that must not be. $\endgroup$
    – user14650
    Jul 10, 2013 at 12:11
  • $\begingroup$ @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). $\endgroup$ Jul 10, 2013 at 13:29
  • $\begingroup$ Unfortunately that book is not available to me. Maybe you could explain to me how you would deal with the situation where you establish the hypotheses H0: θ = a and H1: θ < a, but your true parameter is in fact θ > a. You establish your hypotheses before you design your study and collect data. You have a hypothesis about the outcome, but you do not know the outcome. Therefore you must be able to catch all possible outcomes, because probabilities P(H0) + P(H1) = 1. If you have P(H0) + P(H1) + P(H2) = 1, how are you even going to calculate anything?!? Please enlighten me. $\endgroup$
    – user14650
    Jul 11, 2013 at 9:57
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    $\begingroup$ @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 $\Omega_{0} \cup \Omega_{1} = \Omega$. $\endgroup$ Jul 11, 2013 at 10:02
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    $\begingroup$ @what: see also this comment. I want to stress that you are correct conceptually, that's why I edited my answer. But in pratices, you'll often encouter non-exhaustive hypotheses because the tests seem not to differ. $\endgroup$ Jul 11, 2013 at 10:10

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