What is continuity correction in statistics

Question

Can someone please explain to me the idea behind continuity correction and when is it necessary to add or subtract $\dfrac{1}{2}$ from the desired number (how do we tell whether we need to add or subtract), how do we tell when we need to use continuity correction?

Answer 1

The continuity correction comes up most often when we are using the normal approximation to the binomial. It comes up sometimes when we are approximating a Poisson distribution with large $\lambda$ by a normal.

Let $X$ be a binomially distributed random variable that represents the number of successes in $n$ independent trials, where the probability of success on ay trial is $p$. Let $Y$ be a normal random variable with the same mean and the same variance as $X$.

Suppose that $npq$ is not too small. Then if $k$ is an integer, $\Pr(X\le k)$ is reasonably well-approximated by $\Pr(Y\le k)$. It is ordinarily better approximated by $\Pr(Y\le k+\frac{1}{2})$. The difference can be significant when $n$ is not large. When $np(1-p)$ is big, say bigger than $100$, the continuity correction makes little practical difference.

The continuity correction is less important than it used to be. For with modern software, we can compute $\Pr(X\le k)$ essentially exactly.

It is easy to get confused when using the continuity correction. In particular, the question that you asked comes up: when do we add $\frac{1}{2}$, and when do we subtract? I deal with that by remembering only one rule. To repeat,

Rule: If $k$ is an integer, then $\Pr(X\le k)\approx \Pr(Y\le k+\frac{1}{2})$, where $Y$ is a normal with the same mean and variance as $X$.

Let us look at a couple of examples. Let $X$ have binomial distribution. Approximate the probability that $X\lt k$, where $k$ is an integer. This doesn't quite look like our Rule. Note we have $\lt k$, not $\le k$. But $X\lt k$ if and only if $X\le k-1$. Now we are of the right shape. The answer is, approximately, $\Pr(Y\le (k-1+\frac{1}{2}$, where $Y$ is the appropriate normal. This is $\Pr(Y\le k-\frac{1}{2}$, so in a sense we sutracted. But it all came from the one Rule, where we always add, but pay close attention to the difference between $\lt$ and $\le$.

What is the probability that $X\gt k$? This is $1-\Pr(X\le k)$. Thus we get that the result is approximately $1-\Pr(Y\le k+\frac{1}{2})$.

A numerical example: Toss a fair coin $100$ times. Approximate the probability that the number of heads is $\le 55$.

By working directly with the binomial, and software, I get this is, to $6$ figures, $0.864373$. That's the "right" answer.

Using $\Pr(Y\le 55)$, where $Y$ is normal mean $50$, standard deviation $5$, no continuity correction, I get the approximation $0.8413$.

Using the continuity correction, I get the approximation $0.8643$. I should really do a few other examples, the continuity correction is too good here!