Recipe
Compute a standard error, and how it changes with n
Use when: you know a population standard deviation \sigma and a sample size n, and need SE itself, or need to know how SE changes after n changes
Goal: the standard error itself, or a correctly scaled new SE after a change in n
Taught in Standard error
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1
Apply SE=\sigma/\sqrt n directly, using the population standard deviation \sigma and the sample size n.
Example
\sigma=12, n=36: SE = 12/\sqrt{36} = 12/6 = 2.
Why: SE depends only on these two quantities; nothing else about the population's shape enters the formula.
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2
Figure out whether the problem gives an actual new sample size to plug in, or only describes how n changes by some factor, such as doubling or quadrupling.
Example
From Step 1's SE=2 at n=36: quadrupling n to 144 means k=4, so the shortcut gives SE_{\text{new}} = 2/\sqrt4 = 1, matching a direct recomputation of 12/\sqrt{144}=1.
- an actual new n is given — recompute SE from scratch with Step 1's formula, using the new n in place of the old one.
- only a multiplicative factor k is given for how n changes, with no specific new n stated — skip recomputing and multiply the old SE by 1/\sqrt k instead.
Why: because SE scales as 1/\sqrt n, a multiplicative change in n produces a predictable multiplicative change in SE, so recomputing from \sigma every time is unnecessary once the scaling factor is known.
Common mistakes
- Assuming SE scales linearly with n, such as expecting quadrupling n to quarter SE instead of halving it.
- Dropping the square root and treating SE as proportional to n itself.
- Applying the scaling shortcut with the wrong factor, such as using k=2 when n actually quadrupled (k=4).
- Noting only that a larger sample shrinks SE, without pinning down by how much — the square-root relationship this recipe is built around.