How many patients does a trial need?
Tonight's idea
A new blood-pressure drug: \(n\) patients get it, \(n\) others a placebo, assigned by coin flip. The coin is the point: randomisation makes the groups alike in everything but the drug and chance. Compare the mean drops, \(\bar x_T - \bar x_C\). Drug, or noise?
Patients vary with standard deviation \(\sigma\), so a mean of \(n\) has spread \(\sigma/\sqrt n\) (the standard error). For a difference of independent means, variances add: \(\text{SE} = \sqrt{\sigma^2/n + \sigma^2/n} = \sigma\sqrt{2/n}\).
Convention says the drug works if the observed difference exceeds \(1.96\,\text{SE}\), the two-sided 5% rule (p-values). If the true effect is \(\delta\), the observed difference is bell-shaped around \(\delta\) with spread \(\text{SE}\), and we want it to clear the bar 80% of the time: the trial's power. So the bar sits at that bell's 20th percentile, \(0.84\,\text{SE}\) below its centre:
\[\delta = (1.96 + 0.84)\,\text{SE} = 2.8\,\sigma\sqrt{\tfrac{2}{n}} \quad\Longrightarrow\quad n = \frac{2 \times 2.8^2\,\sigma^2}{\delta^2}.\]
\(n \approx \dfrac{16\,\sigma^2}{\delta^2}\) per group, for a 5% two-sided test with 80% power. In general \(n = 2(z_{\alpha/2} + z_\beta)^2\sigma^2/\delta^2\).
Read the shape: noise squared on top, effect squared underneath. Halve the effect you are hunting and the trial must be four times larger.
Watch one: a modest blood-pressure drug
Drops in systolic pressure vary with \(\sigma = 12\) mmHg. The drug is hoped to beat placebo by \(\delta = 5\) mmHg. Then \(n \approx \dfrac{16 \times 12^2}{5^2} = \dfrac{16 \times 144}{25} = 92\) per group, so about 184 patients. (The exact \(2 \times 2.8^2 = 15.7\) gives 90; the 16 is a rounding for the head.) If the honest hope were only \(\delta = 2.5\) mmHg, the same arithmetic gives 369 per group, four times as many. Check the logic the other way: with 92 per group, \(\text{SE} = 12\sqrt{2/92} = 1.77\), and \(5/1.77 = 2.83\) standard errors, just over the 2.8 we built in.
Your turn
- \(\sigma = 10\), 50 patients per group. Find the SE of the difference in means. The treated group does 3.0 better. How many SE is that? Does it pass the 1.96 bar? How big a difference would?
- Cholesterol readings have \(\sigma = 1.0\) mmol/L. How many per group to detect a drop of \(\delta = 0.5\) mmol/L with 80% power? And \(\delta = 0.25\)? State the rule that connects the two answers.
- A clinic can recruit only 20 patients per group, with \(\sigma = 10\). Turn the formula round: what is the smallest effect \(\delta\) this trial can detect with 80% power?
- A company trials its drug on 16 patients per group, \(\sigma = 10\), and the true effect is \(\delta = 5\). (a) Find the SE and how many SE \(\delta\) is. (b) The power is the chance the observed difference clears \(1.96\) SE: \(P(Z > 1.96 - \delta/\text{SE})\). Use \(P(Z > 0.55) = 0.29\). (c) The trial finds no significant effect. Has it shown the drug does not work?
Stretch: when the outcome is life or death
If the outcome is "died within a year" (1) or not (0), each patient's value has variance \(p(1-p)\). A heart drug is hoped to cut one-year mortality from 10% to 8%. Take \(\sigma^2 \approx 0.09 \times 0.91\) and \(\delta = 0.02\). How many patients per group? Why do such trials run across dozens of hospitals?
Answers
- \(\text{SE} = 10\sqrt{2/50} = 2.0\). A difference of 3.0 is 1.5 SE, below 1.96, so it does not pass. It would need \(1.96 \times 2.0 = 3.9\) or more.
- \(n = 16 \times 1.0/0.25 = 64\) per group. For \(\delta = 0.25\): \(16 \times 1.0/0.0625 = 256\). Halving the effect quadruples the trial, because \(\delta\) sits squared in the denominator.
- \(\delta = 2.8\,\sigma\sqrt{2/n} = 2.8 \times 10 \times \sqrt{0.1} = 2.8 \times 3.16 = 8.9\). Anything smaller than about 9 units will be missed more than one time in five. The trial can only be trusted to find a large effect.
- (a) \(\text{SE} = 10\sqrt{2/16} = 3.54\), and \(\delta/\text{SE} = 1.41\). (b) Power \(= P(Z > 1.96 - 1.41) = P(Z > 0.55) = 0.29\). (c) No. Even if the drug works exactly as hoped, this trial fails to show it 71% of the time. A non-significant result from an underpowered trial is close to no information. The formula says 64 per group were needed; the company ran a quarter of that.
Stretch. \(\sigma^2 = 0.09 \times 0.91 = 0.0819\) and \(\delta^2 = 0.0004\), so \(n \approx 16 \times 0.0819/0.0004 \approx 3{,}300\) per group, about 6,600 patients. A 2-percentage-point change in a rare event is tiny against the noise of a 0/1 outcome, so cardiac trials enrol thousands and need many hospitals to find them. The same formula decides whether a study of a teaching method can show anything, and most small ones cannot. Compare the screening logic in What a positive test really means: there a false alarm came from a rare disease; here a false reassurance comes from a small trial.