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 95% confidence interval $\bar X \pm 1.96\frac{\sigma}{\sqrt{n}}$ for unknown $\mu$ is correct for normal data when the population standard deviation $\sigma$ is known. It is approximately correct for moderately large $n,$ when $\sigma$ is estimated by the sample standard deviation $S.$
However, you are correct to doubt this so-called 'z-interval' when $\sigma$ is estimated by $S$ and the sample size is small. Then the exact 95% CI for $\mu$ is given by $\bar X \pm t^*\frac{S}{\sqrt{n}},$ where $\pm t^*$ cut probability $0.025 = 2.5\%$ from the upper and lower tails, respectively, of Student's t distribution with $\nu = n-1$ degrees of freedom. [For example, if $n = 10,$ then $t^* = 2.262;$ computation in R.]
At the 95% level in particular, $t^* \approx 2$ when $n \ge 30,$ so the z-interval gives pretty good results for $n \ge 30.$
More generally, for confidence level $(1 - \alpha)\%.$ there are other sample sizes $n$ at which $t^*$ is sufficiently near the $z^*$ that cuts probability $\alpha/2$ from the upper tail of the (symmetrical) standard normal distribution.
[For example, depending on ones degree of fussiness, something like $n=400$ might be large enough for a 98% CI; something like $n=12$ might be large enough for an 80% CI. But it is simpler just to use 't-intervals' whenever $\sigma$ is unknown and estimate by $S.]$
Note: You may sometimes see $n = 30$ given as a large enough sample size to pretend that the z-interval may be used even for non-normal data, and this can be very bad advice depending on the actual population distribution.