Sizing the Sample for a Target Width

Question

Daily slippage has standard deviation sigma = 4 basis points. You want a 95% confidence interval for the mean slippage whose total width is at most 1 basis point. How many days of data do you need?

Accepted Answer

The 95% confidence interval is bar x pm z,fracsigmasqrt n with z = 1.96, so its full width (lower bound to upper bound) is W = 2 z fracsigmasqrt n. Requiring W le 1 bp with sigma = 4: sqrt n ge frac2 z sigmaW = frac2(1.96)(4)1 = 15.68 implies n ge 15.68^2 = 245.86, so you need mathbf246 days. (Half-width = 0.5 bp gives the same arithmetic: sqrt n ge 1.96 cdot 4 / 0.5 = 15.68.) The structural lesson is the squared scaling. Width shrinks like 1/sqrt n, so to halve the interval you quadruple the sample. Want the width down to 0.5 bp instead of 1? That is n ge 984, four times as much data for twice the precision. This is the same brutal arithmetic behind every "we need more data" conversation on a quant desk. The trap: confusing half-width with full width and dropping the factor of 2, which makes n come out four times too small (approx 61 instead of 246). State which one the target refers to before you compute. A second trap: using sigma when you only have the sample s, for small n you should use the t critical value, though by n approx 246 the t and z values are indistinguishable. Desk relevance: sizing an experiment, an A/B execution test, a slippage study, a calibration run, is exac

Sizing the Sample for a Target Width

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