To side-step dependencies arising when we consider the sample variance, we write
$$(n-1)s^2 = \sum_{i=1}^n\Big((X_i-\mu) -(\bar x-\mu)\Big)^2$$
$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2-2\sum_{i=1}^n\Big((X_i-\mu)(\bar x-\mu)\Big)+\sum_{i=1}^n\Big(\bar x-\mu\Big)^2$$
and after a little manipualtion,
$$=\sum_{i=1}^n\Big(X_i-\mu\Big)^2 - n\Big(\bar x-\mu\Big)^2$$
Therefore
$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \sigma^2- \frac {\sqrt n}{n-1}n\Big(\bar x-\mu\Big)^2 $$
Manipulating,
$$\sqrt n(s^2 - \sigma^2) = \frac {\sqrt n}{n-1}\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2 $$
$$=\frac {n\sqrt n}{n-1}\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sqrt n \frac {n-1}{n-1}\sigma^2- \frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$
$$=\frac {n}{n-1}\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right] + \frac {\sqrt n}{n-1}\sigma^2 -\frac {n}{n-1}\sqrt n\Big(\bar x-\mu\Big)^2$$
The term $n/(n-1)$ becomes unity asymptotically. The term $\frac {\sqrt n}{n-1}\sigma^2$ is determinsitic and goes to zero as $n \rightarrow \infty$.
We also have $\sqrt n\Big(\bar x-\mu\Big)^2 = \left[\sqrt n\Big(\bar x-\mu\Big)\right]\cdot \Big(\bar x-\mu\Big)$. The first component converges in distribution to a Normal, the second convergres in probability to zero. Then by Slutsky's theorem the product converges in probability to zero,
$$\sqrt n\Big(\bar x-\mu\Big)^2\xrightarrow{p} 0$$
We are left with the term
$$\left[\sqrt n\left(\frac 1n\sum_{i=1}^n\Big(X_i-\mu\Big)^2 -\sigma^2\right)\right]$$
Alerted by a lethal example offered by @whuber in a comment to this answer, we want to make certain that $(X_i-\mu)^2$ is not constant. Whuber pointed out that if $X_i$ is a Bernoulli $(1/2)$ then this quantity is a constant. So excluding variables for which this happens (perhaps other dichotomous, not just $0/1$ binary?), for the rest we have
$$\mathrm{E}\Big(X_i-\mu\Big)^2 = \sigma^2,\;\; \operatorname {Var}\left[\Big(X_i-\mu\Big)^2\right] = \mu_4 - \sigma^4$$
and so the term under investigation is a usual subject matter of the classical Central Limit Theorem, and
$$\sqrt n(s^2 - \sigma^2) \xrightarrow{d} N\left(0,\mu_4 - \sigma^4\right)$$
Note: the above result of course holds also for normally distributed samples -but in this last case we have also available a finite-sample chi-square distributional result.
1. Normal data, variance known: If you have observations $X_1, X_2, \dots, X_n$ sampled at random from
a normal population with unknown mean $\mu$ and known standard deviation $\sigma,$ then a 95% confidence interval (CI) for $\mu$ is $\bar X \pm 1.95 \sigma/\sqrt{n}.$ This is the only situation in which the z interval is exactly correct.
2. Nonnormal data, variance known: If the population distribution is not normal and the sample is 'large enough', then $\bar X$ is approximately normal and the same formula provides an approximate 95% CI. The rule that $n \ge 30$ is 'large enough' is unreliable here. If the population distribution is heavy-tailed, then $\bar X$ may not have a distribution that is close to normal (even if $n \ge 30).$ The 'Central Limit Theorem', often provides
reasonable approximations for moderate values of $n,$ but it is a limit theorem,
with guaranteed results only as $n \rightarrow \infty.$
3. Normal data, variance unknown. If you have observations $X_1, X_2, \dots, X_n$ sampled at random from
a normal population with unknown mean $\mu$ and standard deviation $\sigma,$ with $\mu$ estimated by the sample mean $\bar X$ and $\sigma$ estimated by the sample standard deviation $S.$ Then a 95% confidence interval (CI) for $\mu$ is $\bar X \pm t^* S/\sqrt{n},$ where $S$ is the sample standard deviation and
where $t^*$ cuts probability $0.025$ from the upper tail of Student's t distribution with $n - 1$ degrees of freedom. This is the only situation in which the t interval is exactly correct.
Examples: If $n=10$, then $t^* = 2.262$
and if $n = 30,$ then $t^* = 2.045.$ (Computations from R below; you could also use a printed 't table'.)
qt(.975, 9); qt(.975, 29)
[1] 2.262157 # for n = 10
[1] 2.04523 # for n = 30
Notice that 2.045 and 1.96 (from Part 1 above) both round to 2.0. If $n \ge 30$ then $t^*$ rounds to 2.0. That is the basis for
the 'rule of 30', often mindlessly parroted in other contexts where it is not relevant.
There is no similar coincidental rounding for CIs with confidence levels other than 95%. For example, in Part 1 above
a 99% CI for $\mu$ is obtained as $\bar X \pm 2.58 \sigma/\sqrt{n}.$ However,
$t^*=2.76$ for $n = 30$ and $t^* = 2.65$ for $n = 70.$
qnorm(.995)
[1] 2.575829
qt(.995, 29)
[1] 2.756386
qt(.995, 69)
[1] 2.648977
4. Nonnormal data, variance unknown: Confidence intervals based on the t distribution (as in Part 3 above) are known to be 'robust' against moderate departures from normality.
(If $n$ is very small, there should be no far outliers or evidence of severe skewness.) Then, to a degree that is difficult to predict, a t CI may provide a useful CI for $\mu.$
By contrast, if the type of distribution is known, it may be possible
to find an exact form of CI.
For example, if $n = 30$ observations from a (distinctly nonnormal)
exponential distribution with unknown mean $\mu$ have $\bar X = 17.24,\,
S = 15.33,$ then the (approximate) 95% t CI is $(11.33, 23.15).$
t.test(x)
One Sample t-test
data: x
t = 5.9654, df = 29, p-value = 1.752e-06
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
11.32947 23.15118
sample estimates:
mean of x
17.24033
However,
$$\frac{\bar X}{\mu} \sim \mathsf{Gamma}(\text{shape}=n,\text{rate}=n),$$
so that $$P(L \le \bar X/\mu < U) = P(\bar X/U < \mu < \bar X/L)=0.95$$
and an exact 95% CI for $\mu$ is $(\bar X/U,\, \bar X/L) = (12.42, 25.16).$
qgamma(c(.025,.975), 30, 30)
[1] 0.6746958 1.3882946
mean(x)/qgamma(c(.975,.025), 30, 30)
[1] 12.41835 25.55274
Addendum on bootstrap CI: If data seem non-normal, but the actual population
distribution is unknown, then a 95% nonparametric bootstrap CI may be the best
choice. Suppose we have $n=20$ observations from an unknown distribution, with $\bar X$ = 13.54$ and values shown in the stripchart below.
The observations seem distinctly right-skewed and fail a Shapio-Wilk normality test with P-value 0.001. If we assume the data are exponential and use the method in Part 4, the 95% CI is $(9.13, 22.17),$ but we have no way to know whether the data are exponential.
Accordingly, we find a 95% nonparametric bootstrap
in order to approximate $L^*$ and $U^*$ such that
$P(L^* < D = \bar X/\mu < U^*) \approx 0.95.$ In the R code below
the suffixes .re indicate random 're-sampled' quantities based on
$B$ samples of size $n$ randomly chosen without replacement from among the
$n = 20$ observations. The resulting 95% CI is $(9.17, 22.71).$ [There are
many styles of bootstrap CIs. This one treats $\mu$ as if it is a scale
parameter. Other choices are possible.]
B = 10^5; a.obs = 13.54
d.re = replicate(B, mean(sample(x, 20, rep=T))/a.obs)
UL.re = quantile(d.re, c(.975,.025))
a.obs/UL.re
97.5% 2.5%
9.172171 22.714980
Best Answer
The third case is given by Hahn and Meeker in their handbook Statistical Intervals (2nd ed., Wiley 2017):
Here, $z_{(q)}$ denotes $\Phi^{-1}(q)$, the $q$ quantile of the standard normal distribution.
For example, let's assume we drew $n=20$ samples from a normal distribution with unknown mean and standard deviation. The sample mean was $\bar{x}=10.5$ and the sample standard deviation was $s=3.19$. Then, the two-sided $95\%$ confidence interval for the $q=0.25$ quantile $x_{0.25}$ would be given by $(6.42; 9.76)$.
Here is some
Rcode and a small simulation to check the coverage. By changing the parameters, you can run your own simulations: