An even fainter edge and the data it demands

Question

A weaker signal has a per-observation signal-to-noise ratio of $0.05$ (half the size of a stronger benchmark signal). Its z-statistic grows as $z = 0.05\sqrt{n}$ .

About how many observations are needed for this edge to just reach two-sided significance at 5%, and what does the size of that number illustrate?

Accepted Answer

The computation. Two-sided significance needs z \geq 1.96. Solving 0.05\sqrt{n} = 1.96: \sqrt{n} = \frac{1.96}{0.05} = 39.2, \qquad n = 39.2^2 \approx 1537. The lesson in the number. Halving the per-observation edge (from 0.1 to 0.05) did not double the required sample, it roughly quadrupled it (from about 385 to about 1537). That is because z grows only as \sqrt{n}, so to compensate for a smaller effect you need the square of the shortfall in data. Faint effects are extraordinarily data-hungry. Why it matters. This is also why a non-significant result on a small dataset proves little: the ed…

An even fainter edge and the data it demands

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