Let W = X/n, where X is a binomial variable with parameters n and p.

**NORMAL APPROXIMATION TO THE BINOMIAL**

Let *W = X*/*n*, where *X* is a binomial variable with parameters *n* and *p*. Then, since *W* is just a constant times *X*, *E*(*W*) = *p*
and Var(*W*) = *p*(1 – *p*)/*n*. *W*
represents the pro-portion of successes when *X* is the number of successes. Because often we wish to estimate the
proportion *p*, we are interested in
the mean and variance of *W* (the
sam-ple estimate for the proportion p). In the example where *n* = 3 and *p* = 0.5, *E*(*W*) = 0.5 and Var(*W*) = 0.5(0.5)/3 = 0.25/3 = 0.0833.

The central limit theorem applied to the sample
mean of *n* Bernoulli trials tells us
that for large *n* the random variable *W*, which is the sample mean of the *n* Bernoulli trials, has a distribution
that is approximately normal, with mean *p*
and variance *p*(1 - *p*)/*n*.
As *p* is unknown, the common way to
normalize to obtain a statistic that has an approximate standard normal
distribution for a hypothesis test would be *Z*
= (*W* - *p*_{0})/ √[*p*_{0}(1 – *p*_{0})/*n*], where *p*_{0} is the hypothesized value of *p* under the null hypoth-esis. Sometimes *W* itself is used in place of *p*_{0}
in the denominator, since *W*(1 – *W*) is a consistent estimate of the
Bernoulli variance *p*(1 – *p*) for a single trial. Multiplying both
the numerator and denominator by *n* we
see that algebraically *Z* is also
equal to (*X* – *np*_{0})/√{n[*p*_{0}(1
– *p*_{0})}].

Because the binomial distribution is discrete and
the normal distribution is con-tinuous, the approximation can be improved by
using what is called the continuity correction. We simply make *Z* = (*X*
– *np*_{0} – 1/2)/√{n[*p*_{0}(1 – *p*_{0})]}. The normal ap-proximation to the binomial works
fairly well with the continuity correction when *n **≥** *30, provided that 0.3 < *p*
< 0.7. However, in clinical trials we are often interested in *p* > 0.90; these cases require *n* to be several hundred before the *Z* approximation works well. For this
reason and because of the computational speed of modern com puters, exact
binomial methods commonly are used now, even for fairly large sam-ple sizes
such as *n* = 1000

To express *Z*
in terms of *W* in the continuity
corrected version, we divide both the numerator and denominator by *n*. The result is *Z* = (*W* – *p*_{0} – 1/{2*n*})/ √[*p*_{0}(1 – *p*_{0})/*n*].

We use this form for *Z* as it provides a better approximation to expressions such as *P*(*W*
≤ *a*) or *P*(*W* > *a*). On the other hand, if we consider *P*(*W* < *a*) or *P*(*W* ≥ *a*), then
we should use *Z* = (*X* – *np*_{0}
+ 1/2)/√{n[*p*_{0}(1 – *p*_{0})]} or, equivalently, *Z* = (*W*
– *p*_{0} + 1/{2*n*})/ {*p*_{0}(1 – *p*_{0})/*n*}.

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