Given independent and identically distributed (not necessarily normal) random variables $X_1, \dots, X_n$ with unknown mean $\mu$ and unknown variance $\sigma^2$, the Central Limit Theorem states that the sample mean (equiv. sum) approximately follows a normal distribution. In particular, we have that $$ \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1) $$ as the sample size $n$ gets large. Armed with this knowledge, we can create a level-$\alpha$ confidence interval for the mean: $$ \bar{X} \pm z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}} $$ where $z_{1-\alpha/2}$ is the appropriate quantile of the standard normal distribution.

However, this interval depends on the population value $\sigma$ (equiv. $\sigma^2$); when we observe a sample, we don't know this value. What we do observe is the sample variance $S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})^2$, but my understanding is that $S^2$ doesn't follow a chi-square distribution unless the original $X_i$ came from a normal distribution (which is not part of our assumptions or else we don't need the CLT). So, we can't simply replace $\sigma$ with $s$ and $z_{1-\alpha/2}$ with $t_{1-\alpha/2;\ n-1}$ in our confidence interval.

My question is: how do we use the CLT to create a confidence interval if we don't know the population variance?

EDIT: In this question, one of the answers simply replaces $\sigma$ with $s$ and $z$ with $t$ like I mentioned above. Is this inaccurate? Central limit theorem with unknown variance

  • $\begingroup$ stats.stackexchange.com/questions/44189/… In this question, one of the answers simply replaces σ with s and z with t like I mentioned above. Is this inaccurate? $\endgroup$
    – mai
    Mar 11, 2018 at 21:11
  • 1
    $\begingroup$ it does not make a simple substitution: if you read more carefully, you'll notice that 1) Alex applies Slutsky's theorem (or lemma) and 2) he mentions that, because of this, we get a confidence interval which is only accurate in the limit $n\to\infty$, unlike that based on the $t-$test, which is exact for finite sample size when the population is normal. $\endgroup$
    – DeltaIV
    Mar 11, 2018 at 21:21
  • $\begingroup$ I carefully read the post you linked, and I understand that the r.v. with $S$ instead of $\sigma$ converges to $\mathcal{N}(0,1)$ as $n$ goes to infinity. The one I linked is a different post which does make the substitution. $\endgroup$
    – mai
    Mar 11, 2018 at 21:23
  • $\begingroup$ the answer you linked is wrong, because it doesn't specify that to use the $t-$based confidence interval, your $t-$statistics must have the Student's $t-$distribution. A sufficient condition is that the $X_i$ are iid normal. $\endgroup$
    – DeltaIV
    Mar 11, 2018 at 21:50
  • $\begingroup$ Right, that's basically what I said in the OP. Thanks, just wanted to make sure. $\endgroup$
    – mai
    Mar 11, 2018 at 22:49


Browse other questions tagged or ask your own question.