0

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?

rose
  • 1
  • 1
  • 1
  • 1
  • 1
    http://stats.stackexchange.com/questions/24676/difference-between-one-tailed-and-two-tailed-testing might be interesting. – Gala Jul 10 '13 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 '13 at 10:07
  • 2
    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) – Glen_b Jul 10 '13 at 11:03

1 Answers1

2

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.

COOLSerdash
  • 25,317
  • 8
  • 73
  • 123
  • -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:μ1μ2. 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 –  Jul 10 '13 at 12:11
  • @what You are correct (I've edited the answer). But it is valid to use, for example the hypotheses: $\text{H}_{1}:\theta – COOLSerdash Jul 10 '13 at 13:29
  • 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. –  Jul 11 '13 at 09:57
  • 1
    @what [This post](http://stats.stackexchange.com/questions/18988/do-null-and-alternative-hypotheses-have-be-to-exhaustive-or-not) deals exactly with your question. To back my claim up further, [Young and Smith](http://www.amazon.com/Essentials-Statistical-Inference-Probabilistic-Mathematics/dp/0521548667/ref=sr_1_1?ie=UTF8&qid=1373536792&sr=8-1&keywords=young+smith+essentials+inference) write (page 65) that possibly, *but not necessarily,* $\Omega_{0}$ and $\Omega_{1}$ satisfy $\Omega_{0} \cup \Omega_{1} = \Omega$. – COOLSerdash Jul 11 '13 at 10:02
  • 1
    @what: see also [this comment](https://stat.ethz.ch/pipermail/r-help/2011-September/289202.html). 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. – COOLSerdash Jul 11 '13 at 10:10
  • Thank you, I understand. I'm not statistically advanced enough to agree or disagree, but currently this would be my stance, quoted from the linked post: "You rule out the possibility of being surprised, and learning something interesting." –  Jul 11 '13 at 10:30