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Is it possible to reject

$$H_0\!: \mu_1 = \mu_2 = \mu_3,$$

for example, with an F test in the context of an ANOVA table, but not reject any of the following: \begin{align} H_0\!: \mu_1 &= \mu_2 \\ H_0\!: \mu_3 &= \mu_2 \\ H_0\!: \mu_1 &= \mu_3 \end{align} for example, using a Fisher's LSD test?

It seems like one of those cases where at least one member of a set of twins or triplets is guilty, but it cannot be determined which one is guilty.

If so, like how?

If not, please prove it.

gung - Reinstate Monica
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BCLC
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    My first reaction is that it might be possible to reject one, but not the other primarily due to the problem of multiple testing. The first test is testing just one hypothesis, and so an ANOVA under the assumptions will be exact. However, the second setup will require 3 tests, and hence require correcting for multiple testing. To the best of my knowledge, there are no known exact corrections for multiple testing, and thus it may be possible to draw different conclusions in each of the two setups. Although such cases might often be borderline. – Greenparker Mar 17 '16 at 17:40
  • @Greenparker Some multiple testing is conditioned on $H_0: u_i \ \text{all equal}$ being rejected right? Does that make a difference? If not conditioned, it's possible. If conditioned...still possible? Or impossible by construction? – BCLC Mar 17 '16 at 17:45
  • No there is no multiple testing required in $H_0: \mu_i$ all equal, since only one hypothesis is tested. – Greenparker Mar 17 '16 at 17:46
  • @Greenparker I mean sometimes testing $H_0: \mu_1 = \mu_2$ can be conditioned on rejection of $H_0: \mu_1 = \mu_2 = \mu_3$? – BCLC Mar 17 '16 at 17:48
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    I see, yes, typically in Tukey HSD test, one tests for all pairwise differences conditioned on the overall hypothesis being rejected. In such cases, I believe, you will notice at least one of the hypothesis being rejected. However, it is possible for one of the pairwise hypothesis to be rejected conditioned on the overall not rejected. I remember running into such an example at some point. I believe it is impossible by construction for none of the 3 hypothesis to be rejected conditioned on $H_0: \mu_i$ all equal. However, I don't have a reference, and I might be wrong. – Greenparker Mar 17 '16 at 17:54
  • @Greenparker Might such impossibility depend on the test/s used? – BCLC Mar 17 '16 at 17:59
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    Yes, I believe it might depend on the test. Sorry, I realize I am not much help here. I only speak from intuition. – Greenparker Mar 17 '16 at 18:09
  • @Greenparker We have comments and not only answers for reasons like these I believe. No apologies needed. Thank you for your comments – BCLC Mar 17 '16 at 18:11
  • @Greenparker you can get this even without changing p-values for multiple testing. (There are examples on site.) – Glen_b Mar 17 '16 at 23:07
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    It looks to me that it is the same question than http://stats.stackexchange.com/questions/83030/can-anova-be-significant-when-none-of-the-pairwise-t-tests-is/83083#83083 ? – peuhp Mar 18 '16 at 11:34
  • @peuhp thanks! May you please close this as a duplicate now? ^-^ – BCLC Mar 18 '16 at 11:37

1 Answers1

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In mathematical terms, this amounts to check whether or not the region made of the intersection of

  1. (x-y)²<8
  2. (x-z)²<8
  3. (z-y)²<8
  4. (x-z)²+(x-y)²-(x-y)(x-z)>9

is empty. The first three constraints correspond to testing whether $X-Y\sim\mathcal{N}(0,2)$ is not too unlikely (and the same for $X-Z$ and $Y-Z$), while the last constraint is derived from the distribution $$(x-y\quad x-z)\left[\begin{matrix}2 &1\\ 1&2\end{matrix}\right]^{-1}\left(\begin{matrix}x-y\\ x-z\end{matrix} \right)\sim\chi^2_2$$under the null.

Xi'an
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