3.1 Prevalence
Definition 3.1.1. The point prevalence of a disease at time \(t\) is the proportion of a population that has the disease at \(t\). The period prevalence over an interval is the proportion that has the disease at any time during the interval.
Prevalence is a proportion, without units. It counts every existing case, however long ago it began, so it measures the burden of a disease on a population, the quantity needed for planning services, rather than how often the disease begins.
In a cross-sectional survey of \(n\) people drawn at random, of whom \(X\) have the disease, \(X \sim \text {Bin}(n, p)\), and the prevalence \(p\) is estimated by \(\hat p = X/n\), unbiased with variance \(p(1 - p)/n\). Three confidence intervals are in use.
The Wald interval
The normal approximation \(\hat p \approx N(p, p(1 - p)/n)\) with the variance estimated by \(\hat p(1 - \hat p)/n\) gives \[ \hat p \pm z \sqrt {\hat p(1 - \hat p)/n} , \] where \(z = 1.96\) for \(95\%\). It is simple, but its coverage is poor when \(np\) is small: it can include negative values, and when \(X = 0\) it collapses to the single point \(0\).
The Wilson (score) interval
Instead of estimating the variance, keep \(p\) in it and find all \(p\) for which \(|\hat p - p| \leq z\sqrt {p(1 - p)/n}\). Squaring gives a quadratic inequality in \(p\), \[ (1 + z^2/n)\,p^2 - (2\hat p + z^2/n)\,p + \hat p^2 \leq 0 , \] whose roots are the limits \begin {equation} \label {eq:wilson} \frac {\hat p + \dfrac {z^2}{2n} \pm z\sqrt {\dfrac {\hat p(1 - \hat p)}{n} + \dfrac {z^2}{4n^2}}}{1 + z^2/n} . \end {equation} The interval always lies inside \([0, 1]\) and is centred not at \(\hat p\) but at a point pulled towards \(\tfrac 12\). Its coverage is close to the nominal level even for small \(n\), and it is the interval to use by default.
The exact (Clopper–Pearson) interval
The exact interval inverts two one-sided binomial tests: its lower limit \(p_L\) is the value of \(p\) for which \(P(X \geq x) = 0.025\), and its upper limit \(p_U\) the value for which \(P(X \leq x) = 0.025\), where \(x\) is the observed count. These are found from the beta distribution.
Lemma 3.1.2. If \(X \sim \text {Bin}(n, p)\) and \(1 \leq x \leq n\), then \(P(X \geq x) = P(B \leq p)\) where \(B \sim \text {Beta}(x, n - x + 1)\).
Proof. Let \(U_1, \dots , U_n\) be independent uniform on \((0, 1)\) and \(X\) the number of them that are at most \(p\); then \(X \sim \text {Bin}(n, p)\). The event \(X \geq x\) is the event that the \(x\)-th smallest of the \(U_i\), \(U_{(x)}\), is at most \(p\). The \(x\)-th order statistic of \(n\) uniforms has the \(\text {Beta}(x, n - x + 1)\) distribution. □
Hence \(p_L\) is the \(0.025\) quantile of \(\text {Beta}(x, n - x + 1)\) and, applying the lemma to \(n - X\), \(p_U\) is the \(0.975\) quantile of \(\text {Beta}(x + 1, n - x)\); by convention \(p_L = 0\) when \(x = 0\) and \(p_U = 1\) when \(x = n\). The coverage of this interval is never below \(95\%\), so it is conservative: usually somewhat wider than it need be.
Example 3.1.3. In a survey \(46\) of \(400\) adults have hypertension, so \(\hat p = 0.115\). The \(95\%\) intervals are:
| Method | Interval |
| Wald | \((0.084,\ 0.146)\) |
| Wilson | \((0.087,\ 0.150)\) |
| Clopper–Pearson | \((0.085,\ 0.150)\) |
With this many cases the three agree closely. With \(2\) cases in \(40\) (\(\hat p = 0.05\)) they do not: Wald gives \((-0.018,\ 0.118)\), an impossible lower limit, while Wilson gives \((0.014,\ 0.165)\) and Clopper–Pearson \((0.006,\ 0.169)\).
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