How Bad Is My Estimator?
A standalone mini-lesson in a small, growing set on estimator properties. The running example is still the fair die, \(\theta = 3.5\), \(\sigma \approx 1.708\) — the same die from unbiased, but never consistent, now asking a sharper question: not just "does \(\bar X_n\) tighten as n grows," but "by how much, exactly, and how far off should I expect to be right now?"
New vocabulary is underlined like this — hover or tap any term for a one-line explanation that appears right below the line you are reading.
- Think · 2 minYou roll a die n times and get \(\bar X_n = 4.1\), when the true θ is 3.5. Is that a bad estimate? Would your answer change between n = 5 and n = 500 — and what single number would you need to know to answer precisely, instead of by gut feel?
I've done the Think step — reveal Pair & Share
- Pair · 3 minCompare what each of you would need to know before calling 0.6 "big" or "small."
- Share · 2 minAgree as a table on a rule of thumb, then check it against the widget below.
How far should you expect to be off?
The standard error of the mean answers exactly this question. It's the standard deviation of \(\bar X_n\) itself, across hypothetical repeated samples:
$$SE(\bar X_n) = \frac{\sigma}{\sqrt n}$$
Not a bound and not a guarantee — a typical scale. Being off from θ by about one SE is ordinary; being off by three or four SEs would be a surprise. Larger n shrinks SE like \(1/\sqrt n\), which is the precise version of "more data means a better estimate."
Compute SE at n = 5 using \(\sigma \approx 1.708\): \(SE = 1.708/\sqrt5\). Then find the n at which SE first drops below 0.2.
Show the arithmetic
At n = 5: \(SE = 1.708/2.236 \approx 0.764\) — check it against the widget's readout. For \(SE < 0.2\): \(1.708/\sqrt n < 0.2 \Rightarrow \sqrt n > 8.54 \Rightarrow n > 72.9\), so n = 73 is the first integer sample size where the typical error drops below 0.2 dots.
FAQ
The widget uses the true σ to compute SE. In practice we don't know σ — now what?
You plug in an estimate of σ instead — the sample standard deviation s, computed with Bessel's correction so it isn't itself biased low. \(SE \approx s/\sqrt n\) is the version you'd actually compute from real data, and it's what most software reports by default.
Does "within 2 SE" mean a 95% guarantee?
No — that's a rule of thumb from the normal approximation (the Central Limit Theorem), and it's only accurate for large enough n. At small n, especially for a bounded, non-normal population like a die, the actual percentage can differ noticeably from 95% — which is exactly what the widget's live percentages let you check, rather than assume.