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When I google the central limit theorem it says the following: The central limit theorem states that if you have a population with mean $μ$ and standard deviation $σ$ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

My question is the following: does the central limit theorem only apply to sampling distributions of the sample mean? Or can it apply to any sampling distribution of any statistic? Such as sample variance, sample proportion, or difference between two sample means, etc..

Thank You For Your Time

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    $\begingroup$ The general philosophy of CLT is that, if you have random variables $X_1, X_2, \dots, X_n$ which are (1) "small" in certain cense, and (2) are "almost independent", then their sum $S_n=X_1+\cdots+X_n$ will behave like a normal distribution. So anything that can be viewed in this sense may have chances to be a subject of CLT. $\endgroup$ Jul 31, 2021 at 18:02
  • $\begingroup$ So something like sample variance will also have a sampling distribution that is normally distributed if n is large correct? So CLT applies to sampling distributions of ANY statistic in general. Not just sampling distributions of the sample mean yes? $\endgroup$
    – Laskas
    Jul 31, 2021 at 19:37
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    $\begingroup$ I have little acquaintance with statistic used in statistics, so let me not make any general comment about it. However, it is indeed true that CLT also holds for sample variance assuming finite 4th moment. $\endgroup$ Jul 31, 2021 at 20:18

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The Central Limit Theorem does not apply to the sampling distribution of all statistics, not even those involving the sample means.

Here is a counterexample. Let $X_1,\dots ,X_n \sim_{i.i.d.} F_X$ and let $Y_1,\dots, Y_m \sim_{i.i.d.} F_Y$, independent of the $X_i$. Let the statistic be $\hat \theta = \sum_{i=1}^n X_i/n - \sum_{j=1}^m Y_j/m$. Suppose $n\rightarrow \infty$, but $m$ is fixed. Then the distribution of $$\frac{\hat \theta -E(\hat \theta)}{Var^{1/2}(\hat \theta)}$$ does not converge to $N(0,1)$ or any other normal distribution.

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  • $\begingroup$ Why the downvote? $\endgroup$ Aug 1, 2021 at 21:06

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