I have problems to understand when a non-centrality parameter is relevant. As far as I understood, if the t distribution is not centered on 0, a non-central t distribution is used. For example, in power analysis, it is assumed that an effect is present, therefore, a non-central t is used to calculate the probabilities. Now to my application: If I carry out a t test where the null hypothesis is not that the difference is 0, eg:

H0: μ1 − μ2 ≤ 3

H1: μ1 − μ2 > 3

Do I need a noncentral t distribution here to calculate the pvalue for H0? Is this non-centrality parameter = 3?

Does anyone have an good source for the relevance of non-central distributions in social sciences?

Thanks for your answers! (and patience!)

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    $\begingroup$ Under the null hypothesis $\mu_1 - \mu_2 = 3$, what's the expectation of $\bar{X}_1 - \bar{X}_2 - 3$ (where $\bar{X}_j$ is the mean of the $j$th sample)? $\endgroup$ – Scortchi Jan 16 '17 at 11:35
  • $\begingroup$ I suppose 3? Do you need more background info? $\endgroup$ – 00schneider Jan 16 '17 at 11:37
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    $\begingroup$ Just a hint - I think your question's clear enough. (The answer's not 3 though.) $\endgroup$ – Scortchi Jan 16 '17 at 11:54
  • $\begingroup$ Then I suppose it is 0 and I do not need a noncentral t? Is my understanding of noncentral t, as posted in the original post, correct? $\endgroup$ – 00schneider Jan 16 '17 at 14:17

Apart from not being centred around zero, a non-central T is also non-symmetric about its mean. As @Scortchi mentions in the comments, the null hypothesis can be expressed in another way to make it zero-centred.

Under the null hypothesis: u1 - u2 - 3 = 0

E(X1 - X2 -3) = E(X1) - E(X2) - 3 = u1 - u2 -3 = 0

Using the zero centred T distribution here would test the hypothesis assuming that X1 - X2 -3 is symmetric about zero. If that's not the hypothesis, then you should go for non-central T test.

  • $\begingroup$ Thank you. Could you please elaborate why the t distribution is not symmetrical over non-zero values? The plots seem to show that the higher the df, the more symmetrical they are, is that correct? $\endgroup$ – 00schneider Jan 16 '17 at 14:18
  • $\begingroup$ See the distributions here: en.wikipedia.org/wiki/Noncentral_t-distribution . For non-zero mean, the distribution is not symmetric about the mean/mode. $\endgroup$ – Ujjwal Kumar Jan 16 '17 at 15:59

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