Understanding Uncertainty · Mini-Lesson · pairs with Class 04 & Class 10

One Sample, One Point — Estimating Variance

Same collapse, a different target — and a connector that has to bridge two different scales

Understanding Uncertainty · in-class group work

A third version of one sample, one point, back on the familiar die, \(\theta = 3.5\), \(\sigma^2 = 35/12 \approx 2.92\) — but this time each sample collapses to its own variance estimate, \(S^2_N\), not its mean. That single change breaks something the first two versions could take for granted.

A companion exploration built to accompany the sample-mean and sample-variance material in Class 04, and the sampling-distribution material in Class 10 — no external dataset, the widget generates its own die rolls in the browser.

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 · Pair · Share · can the connector still be a straight line?

Think · 2 minIn the mean version of this widget, a die roll of "4" and the collapsed estimate \(\bar X_N = 4\) sit at the same spot on the number line — both are just "die-face units." Now the collapse is to \(S^2_N = \frac1N\sum(x_i-\bar x)^2\) instead. Is a value like \(x_i = 4\) (a die roll) measured in the same units as \(S^2_N = 4\) (a variance estimate)? What would go wrong if you tried to plot both on the exact same axis?

I've done the Think step — reveal Pair & Share

Pair · 3 minCompare: what are the actual units of \(S^2_N\), and why is squaring the deviations responsible for the units changing?
Share · 2 minAgree on why the widget below needs two separately-scaled axes instead of one shared number line.

The top histogram is still one sample's N raw die rolls, in ordinary die-face units, 1 through 6. But the collapse this time is \(S^2_N\), the sample variance — and squaring deviations changes the units, so the bottom strip lives on a completely different scale (squared-deviation units, not die-face units). The curved connector threads the currently selected sample's own histogram down to its variance estimate, crossing from one scale to the other instead of dropping straight down.

Top: histogram of one sample's N raw rolls (die-face units). Bottom: every sample's variance estimate collapsed to one dot (squared-deviation units), plus σ² (true, dashed) and σ̂² (running average, solid). Toggle Bessel's correction and watch every dot shift at once.

Toggling the correction moves the whole swarm, not just one number. Every stored sample keeps its raw sum of squared deviations; flipping the checkbox only changes the divisor applied to that same stored sum. Watch σ̂² (the solid line) slide from noticeably left of σ² — undershooting, as in the Bessel's-correction mini-lesson — to sitting right on top of it, the instant you check the box.

Check your intuition

At N = 5, draw 30 samples with the correction off, and note where σ̂² lands relative to σ² ≈ 2.92. Turn the correction on without resetting. Does σ̂² jump to exactly 2.92, or does it land somewhere else nearby?

Show answer

With the correction off, σ̂² should sit noticeably below 2.92 — the familiar undershoot. Turning the correction on shifts every dot's divisor from n to n − 1, which multiplies every stored estimate by \(n/(n-1)\); at n = 5 that's a factor of 1.25. σ̂² moves to somewhere close to 2.92, not exactly on it — "unbiased" is a statement about the expectation over infinitely many samples, and 30 is still a finite, noisy batch.

FAQ

Why does the connector curve instead of dropping straight down like the mean versions?

In the mean versions, a raw event and its collapsed average share the same units — die faces, or minutes — so both panels could share one x-axis and the connector is just a vertical line. Variance breaks that: squaring deviations changes the units entirely, so the two panels are genuinely different scales, and a straight vertical line would silently misrepresent that as if they were the same number line.

Why can't N go down to 1 here, like the mean versions allow?

With N = 1, there's only one roll, so \(\bar x\) equals that roll exactly and every deviation is zero — the sum of squares is 0 regardless of what you rolled, and Bessel's correction would need to divide by \(n - 1 = 0\). Variance needs at least two points to say anything at all about spread.

More estimator-properties mini-lessons: One sample, one point (mean, die) One sample, one point (mean, skewed) Unbiased, but never consistent What the sample mean minimizes Why the sample variance undershoots How bad is my estimator?