Suppose I have iid observations $X_i$ (empirical mean $\bar X_n$), drawn from a distribution with unknown mean $\mu$ and known variance $\sigma^2$. To build a confidence interval for $\mu$ I can use the central limit theorem that states:

\begin{equation} \begin{aligned} \frac{\sqrt{n}(\bar X_n - \mu)}{\sigma} \approx \mathcal{N}(0,1) \end{aligned} \end{equation}

and get the following approximation (if I am not mistaken), with $\phi$ being the quantile function of the standard normal distribution: \begin{equation} \begin{aligned} \mathbb{P}(\mu \in [\bar X_n - \frac{\sigma}{\sqrt{n}}\phi_{1-\frac{\alpha}{2}}; \bar X_n + \frac{\sigma}{\sqrt{n}}\phi_{1-\frac{\alpha}{2}} ]) \approx 1-\alpha \end{aligned} \end{equation}

I've always been told to just provide this as an answer for an interval with $1-\alpha$ confidence level. But what about the real confidence level? It must be something like $1-\alpha -\epsilon _n$, right? What about $\epsilon_n$?

  • 1
    $\begingroup$ You are correct. Actually, for large $n$, i.e. large samples, it does not really matter whether you give an exact confidence interval or whether you use CLT. They are considered equally good. They are of course, not mathematically equivalent, but you can forget about $\epsilon_n$. Very good for this observation! $\endgroup$ – Landon Carter Jan 2 '17 at 13:38

@LandonCarter's Comment is correct when the CLT applies. However, you should use this method only when you're sure that the CLT applies. It is usually OK if the population distribution is nearly symmetrical and the sample size is fairly large. I am writing this because I have misgivings about the unrestricted language in your Question, which may lead some to believe that this is a "general-purpose CI", which it is not.

For example, if the population distribution is nearly exponential, then it is best to observe that the sample mean has nearly a gamma distribution (chi-squared with a little re-scaling) and to use quantiles of that distribution to make a CI. (Also, in the case of an exponential population, if you know $\sigma$ then you know $\mu$ exactly, and there is no need for a CI for $\mu.$)

In the real world, even for nearly normal data, there are few cases in which $\sigma$ is known and $\mu$ is not. If $\sigma$ needs to be estimated by the sample SD $S$, then $\frac{\sqrt{n}(\bar X_n - \mu)}{S}$ has approximately Student's t distribution with $\nu = n-1$ degrees of freedom. Then you use Student's t distribution to make the CI. For a 95% CI, results will be about the same as with normal if the sample size is above 30. (But there are different boundaries on the sample size than 30 for confidence levels other than 95%.)

When the population distribution is clearly not normal and of an unknown shape, it might be better to use a bootstrap CI than to try to use your normal approximation, even if the sample size is moderately large.


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