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Is the following statement correct?

The Central Limit Theorem (CLT) states that as the sample size tends to infinity, the standardized sample mean distribution approaches the standard normal distribution. Motivated by this theorem, we can say that for a large sample size (greater than 30), the standardized sample mean is approximately normally distributed. However, we cannot use this theorem to state that the sample mean by itself is approximately normally distributed for large sample sizes. This is because, according to the Strong Law of Large Numbers (SLLN), the sample mean converges almost surely to the population mean.

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  • $\begingroup$ Thank you so much for your excellent response. $\endgroup$
    – user385034
    Commented Apr 6, 2023 at 1:59

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There's something wrong with all three parts of this, though the first bit is almost right.

  1. The Central Limit Theorem (CLT) states that as the sample size tends to infinity, the standardized sample mean distribution approaches the standard normal distribution.

    Not quite. As long as the conditions for it to apply are present, sure. They weren't mentioned in your quote.

  2. Motivated by this theorem, we can say that for a large sample size (greater than 30), the standardized sample mean is approximately normally distributed.

    No, there's nothing that makes it work for say n=31. For many reasonable definitions of approximately (including those related to convergence in distribution), counterexamples are not that hard to identify even at n=1000, or n=1000000 or any larger n you like.

    As long as the conditions for the CLT hold, it will work eventually, but it might not happen at any practically possible sample size.

    That said, for a lot of common variables, it will work at pretty moderate sample sizes -- depending on what you start with (and how approximate you need it to be) sometimes n=3 is fine, sometimes 15, sometimes n=25, sometimes n=200 is okay. But sometimes you need much larger n.

    I work with some pretty skewed and heavy-tailed variables, and assuming approximate normality of a standardized sample mean at n=100 would be dangerous.

    Unless you constrain the distributions you're talking about, you really can't make such a general claim.

    Here's an obvious counterexample. Lets say I am not happy to call a (standardized) exponential distribution "approximately normal" in this sense (for some purpose or other). Then if my parent distribution were gamma with shape parameter 1/50 (the random variables are i.i.d Gamma$(0.02,\theta)\,$), then at $n=50$, the sample mean is exponentially distributed, and then the standardized sample mean is the standardized exponential we were not prepared to call approximately standard normal to begin with.

    Clearly, then, $n>30$ doesn't work for this case, even though the CLT applies here.

  3. we cannot use this theorem to state that the sample mean by itself is approximately normally distributed for large sample sizes. This is because, according to the Strong Law of Large Numbers (SLLN), the sample mean converges almost surely to the population mean.

    There's a correct underlying point there (convergence in the standardized $Z_n$ is not convergence in the unstandardized $\bar{X}$; instead the latter obeys the LLN if its conditions hold), but I'd disagree on what is stated here as well.

    The claim we're addressing here is not that of convergence in relation to $\bar{X}$ as $n\to\infty$ but of its cdf being approximately normal at some specific large $n$, in particular, whatever specific $n$ we might have regarded as large enough for the claim about approximation in the previous part in relation to the distribution of standardized means.

    If you believe you can use the CLT to argue for something approximately happening to the distribution of a standardized mean at a finite sample size (like $n=31$ say) then it applies exactly as well to the unstandardized distribution. That is, the usual distance for the standardized distribution $|F_n-\Phi|$ will be exactly the same for the unstandardized equivalents of both. If one is approximately normal in that sense, so must the other be, because the discrepancy from equality is the same.


I would add to all this an additional caution; often this is used to then argue for some other fact (e.g. that a t-statistic converges a normal distribution even when the original sampling distribution is non-normal) -- you must be careful not to leap over the remaining gaps to establish the further claim (in respect of the t-statistic example, you need an additional theorem).

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  • $\begingroup$ Thank you so much for your excellent response. $\endgroup$
    – user385034
    Commented Apr 6, 2023 at 1:59

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