Two desks, very different latency spreads: Welch's t-test

Question

Two trading desks report order latencies in milliseconds. Desk A: 30 orders, mean 48 ms, sample standard deviation 6 ms. Desk B: 40 orders, mean 55 ms, sample standard deviation 15 ms. The spreads are clearly unequal, so a pooled test is inappropriate.

Run Welch's two-sample t-test at the 5% level (two-sided): do the mean latencies differ? Report the t-statistic and its approximate degrees of freedom.

Accepted Answer

Welch's test does not assume equal variances. Hypotheses: H0: \muA = \muB versus H1: \muA 
eq \muB. Standard error of the difference: 	ext{SE} = \sqrt{\frac{sA^2}{nA} + \frac{sB^2}{nB}} = \sqrt{\frac{36}{30} + \frac{225}{40}} = \sqrt{1.20 + 5.625} = \sqrt{6.825} = 2.613 	ext{ ms}. Test statistic: t = \frac{\bar xA - \bar xB}{	ext{SE}} = \frac{48 - 55}{2.613} = \frac{-7}{2.613} = -2.68. Welch-Satterthwaite degrees of freedom: 	ext{df} = \frac{(sA^2/nA + sB^2/nB)^2}{\dfrac{(sA^2/nA)^2}{nA-1} + \dfrac{(sB^2/nB)^2}{nB-1}} = \frac{6.825^2}{\dfrac{1.20^2}{29} + \dfrac{5.625^2}{39}} = \frac{46.…

Two desks, very different latency spreads: Welch's t-test

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