I'm working with binomial response data with wildly varying sample sizes and was trying to use the de Moivre-Laplace approximation for some $t$-tests. However, I've seen two different definitions of the standard deviation:
$$ \sigma = \sqrt{np(1-p)}, \\ \sigma = \sqrt{\frac{p(1-p)}{n}}. $$
The second is suggested when $n$ is "large", but I've not seen any rules of thumb for what is considered large, or related and underlying assumptions for those definitions. Is there a good crossover point or is it dependent on context?
(The definitions are provided on the Wikipedia pages for the Binomial Distribution and the de Moivre-Laplace theorem.)
Edit: I misunderstood the source of the second formula, which refers to the standard deviation of repeated samples of a binomial-random population. The answers given regarding sufficient sample size are still relevant.
We (usually) use central limit theorem when $n\geq 30$, and so for $n\geq 30$, $n$ is considered to be large. There are some books out there that says that $n\geq 20$ suffices as being large (and so you can apply the second standard deviation given above in this situation), but I go by the rule that for $\color{blue}{n\geq 30}$, it is considered to be $\color{blue}{\text{large enough}}$.
Here are some additional information which you may find useful:
$\textbf{Addendum}.$ I've contacted my academy colleagues who are experts in operations research and probability and statistics. Here are their (summarized) comments, which I hope are helpful.
To answer the general question "what constitutes $n$ to be large enough to use the Central Limit Theorem and related theorems?", it heavily depends on the underlying distribution of where the data came from as well as how close is "close enough". For many distributions, an $n$ of $10$ or fewer is sufficient (in fact if the underlying distribution is normal, then $n=1$ is sufficient as you are not relying on the asymptotic properties of the central limit theorem in that case). For any underlying distribution other than normal, the distribution of $(\overline{X} - \mu)/(\sigma/\sqrt{n})$ is only going to be approximately normal for $\textit{any}$ finite $n$. But in many cases that approximation becomes very good very quickly. However, $n=30$ is often suggested in textbooks as a rule of thumb and for the vast majority of distributions, $n=30$ is more than sufficient.
With respect to the binomial distribution, the answer becomes a little more complicated. I am guessing that mattpolicastro is trying to apply the $t$-tests (more on this in a minute) to the mean of the binomial distribution, i.e., $np$. It could also be that he is trying to test $p$ directly. This is a nuanced difference, almost certainly above the level of an introductory student, but in the second case you aren’t actually applying the central limit theorem directly. And which question you are actually answering makes a difference with respect to Robert Israel’s comment. Additionally, with binomial distributions, there are two possible “$n$”s running around with binomial random variables. Do you have a single binomial random variable with “$n$” trials (that could also be expressed as $n$ Bernoulli random variables), or do you have, let’s say “$m$” binomial random variables, each with “$n$” trials? Since he mentions the De Moivre-LaPlace theorem, I think it’s safe to assume that mattpolicastro is in the first situation, i.e., he wants to estimate $n p$ where $n$ represents either the number of trials in a single binomial random variable or, equivalently, $n$ independent Bernoulli trials.
Then to address the specific comment about “a binomial distribution with $n=30$ and $p=1/30$, say, is not very well approximated by a normal distribution,” namely, that for $\textit{very small}$, or $\textit{very }$ $\textit{large}$, $p$, $n= 30$ is not “large enough” to ensure a “good” approximation. The general "rule of thumb" with a $\color{green}{\text{binomial distribution}}$ is that
$$ \color{green}{\text{both $n p$ and $n (1-p)$ should be $\geq 10$.}} $$ So, the example $p = 1/30$ and $n=300$ should provide a $\textit{good}$ approximation. And if $p = 0$ or $1$, then you are completely out of luck.
Finally, as a side note, mattpolicastro says in the original question that he wants to do $t$-tests. While this is done somewhat routinely in practice without too many people raising a fuss about it, it is not actually correct. The CLT (or De Moivre-LaPlace) says absolutely $\underline{\textbf{nothing}}$ about the $t$-distribution. In order to correctly use this distribution, the distribution of the underlying data $\underline{\textbf{must}}$ be normal (here, it clearly is not). The reason that most people don’t raise too much of a fuss is that the $t$-distribution is also asymptotically normal and for finite $n$, it is more conservative, i.e., it would yield higher $p$-values and wider confidence intervals.