Here is an attempt at answering the question using numerical experiments: using Monte Carlo estimation it is easy to determine the rate of type I errors for the test with a given distribution of input data. Here I try data from the following distributions:
Normally distributed data: here the t-test is guaranteed to work.
Samples from the uniform distribution on $[-1,1]$: this is a prototype for a distribution with light tail (or rather, the extreme case of no tails).
The double-exponential distribution: this is a distribution with heavier tails than the normal distribution has.
A shifted exponential distribution, $\mathrm{Exp}(1) - 1$: this is a very asymmetric distribution, with a tail only on one side.
The discrete uniform distribution on the set $\{-1,+1\}$: this could be seen as an extreme case of a bi-modal distribution.
The discrete distribution with $P(X=-1) = 0.9$ and $P(X=9)=0.1$: this is very far from a normal distribution because it is both discrete and very asymmetric.
Since we expect the test to get more accurate as $n$ increases, I try only small and moderate values of $n$, namely $n \in \{10, 30, 100\}$. For the significance level I choose the commonly used value $\alpha = 5\%$.
My experiment is performed using the following R script: the script simulates $N=1,000,000$ dataset of size $n$, applies the t-test and counts how often $H_0\colon \mu=0$ is (wrongly) rejected. If the t-test still works, this should be the case in $5\%$ of the cases, any deviation from $5\%$ indicates that for the given distribution and $n$ the t-test did not perform optimally.
Edit: As requested by the OP, I have changed the code to also perform the same experiments for the z-test, so that the performance of both tests can be compared.
set.seed(1)
try.one <- function(gen, n, N=1000000, alpha=0.05) {
crit.t <- qt(1 - alpha/2, n-1)
reject.t <- 0
crit.z <- qnorm(1 - alpha/2)
reject.z <- 0
for (j in 1:N) {
X <- gen(n)
Z <- sqrt(n) * mean(X) / sd(X)
if (abs(Z) > crit.t) {
reject.t <- reject.t + 1
}
if (abs(Z) > crit.z) {
reject.z <- reject.z + 1
}
}
p.t <- reject.t/N
p.z <- reject.z/N
list(prob.t=p.t, sd.t=sqrt(p.t*(1-p.t)/N), prob.z=p.z, sd.z=sqrt(p.z*(1-p.z)/N))
}
distributions <- c("normal", "uniform", "double exponential", "exponential",
"discrete", "asym. discrete")
try.all <- function() {
dist.name <- character(0)
nn <- numeric(0)
fp.rate.t <- numeric(0)
std.err.t <- numeric(0)
fp.rate.z <- numeric(0)
std.err.z <- numeric(0)
for (dist in distributions) {
if (dist == "normal") {
gen <- rnorm
} else if (dist == "uniform") {
gen <- function(n) runif(n, -1, 1)
} else if (dist == "double exponential") {
gen <- function(n) rexp(n) * sample(c(-1,1), n, replace=TRUE)
} else if (dist == "exponential") {
gen <- function(n) rexp(n) - 1
} else if (dist == "discrete") {
gen <- function(n) sample(c(-1,1), n, replace=TRUE)
} else if (dist == "asym. discrete") {
gen <- function(n) sample(c(-1, 9), n, replace=TRUE, prob=c(0.9,0.1))
}
for (n in c(10, 30, 100)) {
row <- try.one(gen, n)
dist.name <- c(dist.name, dist)
nn <- c(nn, n)
fp.rate.t <- c(fp.rate.t, row$prob.t)
std.err.t <- c(std.err.t, row$sd.t)
fp.rate.z <- c(fp.rate.z, row$prob.z)
std.err.z <- c(std.err.z, row$sd.z)
}
}
data.frame(dist.name, n=nn, fp.rate.t, std.err.t, fp.rate.z, std.err.z)
}
print(try.all(), row.names=FALSE)
The output, after some minutes, is
dist.name n fp.rate.t std.err.t fp.rate.z std.err.z
normal 10 0.050029 0.0002180048 0.081694 0.0002738980
normal 30 0.050059 0.0002180667 0.059824 0.0002371605
normal 100 0.049930 0.0002178004 0.052726 0.0002234859
uniform 10 0.054490 0.0002269820 0.084445 0.0002780540
uniform 30 0.050906 0.0002198058 0.060263 0.0002379735
uniform 100 0.050116 0.0002181843 0.053001 0.0002240355
double exponential 10 0.042272 0.0002012090 0.074645 0.0002628177
double exponential 30 0.047506 0.0002127185 0.057410 0.0002326244
double exponential 100 0.049646 0.0002172125 0.052526 0.0002230852
exponential 10 0.099738 0.0002996503 0.130045 0.0003363529
exponential 30 0.072758 0.0002597389 0.082090 0.0002745018
exponential 100 0.058040 0.0002338191 0.060755 0.0002388804
discrete 10 0.021386 0.0001446673 0.109666 0.0003124730
discrete 30 0.042853 0.0002025256 0.042853 0.0002025256
discrete 100 0.056972 0.0002317891 0.056972 0.0002317891
asym. discrete 10 0.350463 0.0004771150 0.350463 0.0004771150
asym. discrete 30 0.044408 0.0002059998 0.191153 0.0003932093
asym. discrete 100 0.067916 0.0002516017 0.067916 0.0002516017
Some observations about these results:
The rate of type I errors is listed in the column fp.rate
. As expected, for the normal distribution this is very close to $5\%$.
In nearly all cases, the rate of type I errors gets closer to $5\%$ as $n$ increases, sometimes from below and sometimes from above. The only exception is the asymmetric discrete distribution.
The weight of the tails seems not to have too much effect: the test performs reasonably well for both uniform and double exponential distributions.
For small sample size ($n=10$) there are notable deviation of the type I error rate from $5\%$, both for the discrete distributions and for the asymmetric distributions.
The worst case is the discrete, asymmetric distribution where the t-test at $5\%$-level shows type I errors in $35\%$ of the cases. Given this huge discrepancy, I would argue that care is required when attempting to use the $t$-test for distributions which are far from normal.
Edit: Using the updated code, we can also compare the performance of the t-test to the performance of a z-test (still using the sample variance):
As expected, for normally distributed data the z-test performs worse that the t-test (because we didn't use the exact variance). The effect is quite noticeable for $n=10$ and nearly disappears for $n=100$. For $n=10$, the t-test seems superior to the $z$-test (using estimated variances) for all examples tested.
The worst case (assymetric+discrete, $n=10$) is equally bad for both tests.
For $n=100$ the results of both tests are very similar, but in some cases the t-test seems to perform slightly better.
This experiment only considers the type I error, but experiments along similar lines could be used to compare type II errors between distributions.