I can't find what is the distribution corresponding to the default option in:

DistributionFitTest[data, Automatic, "HypothesisTestData"];

Is it N(0,1), uniform?

  • 2
    $\begingroup$ Does this answer your question? $\endgroup$
    – Szabolcs
    Commented Mar 20, 2014 at 15:53

1 Answer 1


By default DistributionFitTest attempts to test fit against the family of normal distributions. To test if the data is standard normal (i.e. N(0,1)) you would use.

DistributionFitTest[data, NormalDistribution[]]

There are examples that show this to be the case in the Properties & Relations section of the documentation for DistributionFitTest.

  • $\begingroup$ When you have time, could you take a look whether this looks reasonable? I have an actual practical need for this. My distribution looks like a histogram, but it comes from probabilistic bisection (see pages 13-15) and not from counting data points. $\endgroup$
    – Szabolcs
    Commented May 12, 2014 at 20:37
  • $\begingroup$ @Szabolcs I don't see anything wrong with your approach. I don't have access to the source so I can't promise that the Infinity is safe across the board. You know the typical caveat about undocumented internals being subject to change. $\endgroup$
    – Andy Ross
    Commented May 14, 2014 at 13:45
  • $\begingroup$ Suggest this may be a Beta distribution, not a normal one. Before testing the fit to any distribution, you should identify which one it is. I do this 100 times and find the best one looping FD = FindDistribution[data, 5, "BIC", "RandomSeed" -> foundi - 1, PerformanceGoal -> "Quality"] in Mathematica 10.3 The functionality may be different in other versions. $\endgroup$
    – Carl
    Commented Aug 25, 2016 at 21:26
  • $\begingroup$ @Andy Ross No, first of all ND is bounded on $(-\infty ,\infty )$ and your results as well as BD are bounded on $(0,1)$. Such that you data may be BD. Best to let Mathematica tell you which distribution it most likely is see FindDistribution. That will go through a lot of distributions and select the best ones. $\endgroup$
    – Carl
    Commented Aug 26, 2016 at 21:53
  • $\begingroup$ @Andy Ross Sorry, my bad. The results pertain to probabilistic bisection (see pages 13-15) and Szabolcs's comment. $\endgroup$
    – Carl
    Commented Aug 26, 2016 at 22:34

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