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According to my understanding, to test whether the population mean is equal to a specific value when the variance is unknown, a one-sample t test could be used. This is always valid when the population follows a normal distribution, since the t test statistic will follow a t distribution.

When the population is non-normal in distribution: the t test should be valid if achieving a sufficient sample size. According to central limit theorem, the numerator would follow an approximately normal distribution. But what about the denominator? I'm not aware of any limit theorem suggesting that it goes in distribution to $\chi^2$.

How can the t test be applicable under this situation?

AdamO
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Stephen Ma
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  • This is a good question. I think it has been addressed in other threads. See https://stats.stackexchange.com/questions/69898 for a discussion of a situation where it obviously is *not* applicable, for instance. – whuber Apr 16 '18 at 12:52
  • @whuber thanks for the response! From the website that you have given, does it mean that even under sufficient sample, t test is not always applicable for every non normal population? – Stephen Ma Apr 17 '18 at 00:16
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    It depends on what you mean by "sufficient sample." For populations with finite nonzero variances and independent random sampling, in the *limit* of large samples the t test is a good one--for essentially the same reasons the Z test (Normal approximation) will be good. However, for any given sample size $n$ there exist populations for which the distribution of the means of independent random samples of size $n$ will be far from Normal (and therefore a t test may be misleading). – whuber Apr 17 '18 at 12:56

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If you would like to consider the problem theoretically, just use law of large number to deal with the denominator and use Slutsky's theorem to obtain the asymptotic distribution.

If you would like to be know how to use t-test in real data, just use permutation method (or randomization method) to determine the critical value, which can produce exact level tests for very general distributions.

kfeng123
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  • Thanks for the response! I would look to know more about the details in theory. I have searched thru google and found the following link: http://econometricsense.blogspot.hk/2011/03/students-t-and-assumptions-of-normality.html?m=1. Not certain about the proof but from my point of view on the conclusion, together with CLT and Slutsky’s theorem, the t statistics would converge in probability to normal distribution. In case this is true, why z test is not used instead? – Stephen Ma Apr 16 '18 at 13:58
  • @StephenMa Because such argument only holds asymptotically. For small sample case, t-test may be more accurate. – kfeng123 Apr 16 '18 at 14:11
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    "may be more accurate" is not very compelling. One might as easily claim that it may be less accurate. (Why should we use a t-distribution rather than the z without some kind of argument that it actually will better -- and if so, when that will occur) ... The point in your answer -- that you can do a permutation test on the t-statistic avoids any such small sample issue of course, and is a point well made. – Glen_b Apr 17 '18 at 11:18