By what factor does a bigger sample sharpen the average?

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An A/B test starts with n = 25 users per arm and later grows to n = 225 users per arm. By what factor does the standard error of the estimated average shrink?

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This is the same \sqrt{n} rule, run in the forward direction. The standard error is 	ext{SE} = \frac{\sigma}{\sqrt{n}}, so the SE at two sample sizes changes by the ratio of the square roots. Going from 25 to 225: \frac{	ext{SE}{	ext{old}}}{	ext{SE}{	ext{new}}} = \frac{\sqrt{225}}{\sqrt{25}} = \frac{15}{5} = 3. The sample grew ninefold (25 	o 225), but precision only improved by \sqrt{9} = 3. The standard error is now one-third of what it was. Takeaway. Multiplying the sample by 9 improves precision by only \sqrt{9} = 3: gains in precision always lag the growth in sample size, because S…

By what factor does a bigger sample sharpen the average?

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