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Module 4 · Statistical Inference · L-INF-2
Maya's reply to the skeptic ended with "a different 300 lands within about ±0.15 h of the truth." Her editor circles the word about: "I can't print 'about'. Give me an exact range — and tell me how sure you are of it." Today Maya builds the sentence she's been chasing since her very first survey: an interval around 6.8 with a stated warranty, good for all 48,000 students.
Live session · join with code OTTO-INF2 · ottolearn.org/join
Same rules as always: vote before I explain, every poll is anonymous, and changing your mind after discussion is the goal.
Today · one demand from the editor, five answers
From a point estimate to an interval with a margin of error
The recipe \(\bar{x} \pm z^* \cdot \sigma/\sqrt{n}\) — and where \(z^*\) comes from
The warranty belongs to the procedure, not to one interval
Three dials: confidence level, \(n\), and \(\sigma\)
Planning ahead: solve for \(n\), always round up
Turn a point estimate into an interval, learn the recipe and where its critical value comes from, pin down what "95% confident" really promises, see what makes intervals wide — then price the editor's precision demand in students. Every step starts with your vote.
Recall · last class · 45 seconds
Warm-up · commit first · 60 seconds · anonymous
Hold your answer — it is today's lesson. By the end of class you'll have the exact range, the exact warranty, and the price list for making the range tighter.
Core concept 1 · C1
The editor's complaint names a real gap. \(\bar{x} = 6.8\) is Maya's single best guess at \(\mu\) — her point estimate — but alone it says nothing about how far off it might be. The fix: attach a margin of error and print a range.
The sample's best single guess at \(\mu\): \(\bar{x} = 6.8\) h. Unbiased (it aims at the truth — last class) but almost surely not exactly right. A different 300 gives a different 6.8-ish number.
\(\bar{x} = 6.8\) hThe new object: \(E\), the half-width of the claim. Built from last class's yardstick — \(E = z^* \times \text{SE}\) — it says how far out from \(\bar{x}\) the claim reaches. Small \(E\) = a precise claim.
\(E = z^* \cdot \text{SE}\)The printable object: \(\bar{x} \pm E\), a range that owns up to its own uncertainty. Instead of "the average is 6.8", the paper prints "the average is between here and here" — with a warranty attached.
\(\bar{x} \pm E\)The warm-up, resolved: the strongest honest sentence is a range with a warranty. Two questions remain — how wide is the range (the recipe), and what exactly does the warranty promise (the day's summit). First, where the range comes from:
See it · C1 → C2 · which settings fit?
Last class's cereal machine: \(\sigma = 12\) g, \(n = 36\), so \(\text{SE} = 2\) g, and the middle 95% of sample means land within \(1.96 \times 2 = 3.92\) g of the setting. Now the setting is hidden and one sample averages \(\bar{x} = 498.6\) g. Predict first: which settings could have produced it? Name a lowest and a highest.
This set of plausible settings is the 95% confidence interval for μ.
Setting 505 g: its middle 95% of sample means runs 501.08 to 508.92 g, so 498.6 g would be unusual. Does not fit.
Core concept 2 · C2
The settings that fit the cereal sample ran from \(498.6 - 3.92\) to \(498.6 + 3.92\): the sample mean, plus or minus \(1.96\) standard errors. Every confidence interval for a mean is that same assembly: start at the point estimate, step out \(z^*\) standard errors in both directions. Two ingredients are already on Maya's desk; only one is new.
The sample mean \(\bar{x}\) — the interval is built around it. Maya's: 6.8 h from her 300 responses. This is the only ingredient the data hands you directly.
\(\bar{x} = 6.8\) hThe standard error \(\text{SE} = \sigma/\sqrt{n}\) — last class's object, unchanged. It sets the scale of the reach: how far averages typically wander from \(\mu\). Maya's: \(\approx 0.075\) h.
\(\text{SE} = \dfrac{\sigma}{\sqrt{n}}\)The critical value \(z^*\) — how many SEs to reach out. You choose it by choosing a confidence level: more confidence, bigger \(z^*\), wider interval. Where its numbers come from is next.
\(z^*\)The whole recipe: \(\bar{x} \pm z^* \cdot \dfrac{\sigma}{\sqrt{n}}\) — center, dial, yardstick. License conditions first, though: a random sample, and \(n \geq 30\) (or a normal population) so the CLT vouches for the bell. One ingredient still needs unpacking: \(z^*\).
I do · watch a full build
Watch me decide every step out loud — check the license, choose the warranty, compute the margin, assemble the claim and say it in words.
Random sample? Yes — the 300 were drawn by a plan, not by whoever answered. Size? \(n = 300 \geq 30\), so the CLT vouches for the bell. Spread? The survey's \(s = 1.3\) h stands in for \(\sigma\) — fine at this \(n\), and I'm noting the IOU out loud: paying it properly is next class.
The paper's standard is 95%. Middle 95% of the bell means 2.5% left out in each tail — so \(z^*\) is the value with left-tail area 0.975: \(z^* = 1.96\). I chose this number; the data didn't hand it to me.
Yardstick first: \(\text{SE} = 1.3/\sqrt{300}\) \(\approx 0.075\) h. Then the reach: \(E = 1.96 \times 0.075\) \(\approx 0.147\) h — about 9 minutes. Before I assemble, a sanity check: two yardsticks wide, so a tenth-of-an-hour-ish margin. It is.
\(6.8 \pm 0.147\) \(\Rightarrow (6.65,\ 6.95)\) h. In words: "We are 95% confident the campus average sits between 6.65 and 6.95 hours." Notice what the range does: even its top is more than an hour below the 8-hour recommendation.
The habit: check the license, choose the warranty (\(z^*\)), compute \(E = z^* \times \text{SE}\), assemble \(\bar{x} \pm E\) — then say the claim in words.
We do · you finish it
Same survey, bigger warranty: you turn the dial to 99% and supply the new interval — and watch what the extra confidence costs.
Before printing, the paper's lawyer asks what the range would be at 99% confidence. Same survey, same numbers: \(\bar{x} = 6.8\), \(\text{SE} \approx 0.075\) h. Re-run the build.
Finish it: \(E = ?\) — and the 99% interval is… wider or narrower than (6.65, 6.95)?
Practice · build the interval, solo · fresh numbers on demand
Now you build whole intervals solo — fresh numbers every re-roll, the three critical values on screen, support at its minimum.
Core concept 3 · C3
The warranty dial isn't magic — every \(z^*\) is an ordinary inverse table lookup, the skill from the normal-distribution class. The only subtlety is which area you look up.
\(z^*\) is the z-score that fences in the middle C% of the standard normal. For 95%: leave \(2.5\%\) in each tail, so look up the value whose left-tail area is \(0.975\) — the table answers \(1.96\).
\(\Phi(z^*) = 1 - \alpha/2\)90% \(\rightarrow 1.645\) · 95% \(\rightarrow 1.96\) · 99% \(\rightarrow 2.576\). These three run every inference problem in this module — and any other level is just another inverse lookup away.
\(1.645,\ 1.96,\ 2.576\)For a 95% interval, do not look up the z with left-tail 0.95 — that's 1.645, the 90% dial. The confidence level is the area in the middle; the table wants the area to the left. Convert first: \(0.95 + 0.025 = 0.975\).
\(z^* \neq 0.95\)One picture holds all of it: the level is the tinted middle, the tails split what's left, and \(z^*\) is the fence post. See the anatomy:
Quick check · the lookup trap
See it · C3
Predict first: as I slide the confidence level from 90% up to 99%, which way do the fence posts move? And where would they sit for 100% confidence?
Commit first · today's summit
Core concept 4 · C4
The trap has one root: treating \(\mu\) as if it moves. It doesn't. The true campus average is some fixed number — unknown, but not random. What's random is the interval, because it's built from a random sample.
The true mean is a constant. Once (6.65, 6.95) is computed, \(\mu\) is either inside it or not — probability 1 or 0, we just can't see which. Probability language about this interval has nothing left to attach to.
\(\mu\) is a constantNot the chance that \(\mu\) is in this interval. Not the fraction of students inside the range. Not the chance a repeat survey's average lands inside it. Each one misplaces the randomness: on \(\mu\), on the students, or on only one of two surveys, instead of on the sampling that built this interval.
"We are 95% confident the campus average is between 6.65 and 6.95 h." — the accepted shorthand, read as: we used a method that captures the truth in 95% of surveys like this one. Same words as her draft, one crucial repair: the 95% credits the method.
One picture settles it: run the survey many times, build many intervals, and watch — the intervals jump around, \(\mu\) stands still, and about 95 in 100 of them catch it. Watch it live:
See it · C4 · the anchor, resolved
Predict first: at 95%, roughly how many of 100 intervals will miss \(\mu\) entirely? \(\mu\) stays hidden until I show it. Then: when I drop the level to 90%, do the same intervals get wider or narrower — and do more or fewer miss?
Show μ to score the intervals.
Did your interval capture μ? With μ hidden, there is no way to tell.
See it · C4 · the replication bet
One study, one 95% interval. A colleague repeats it exactly, same method, same \(n\). Predict first: out of 100 repeats, how many of their averages land inside the first study's interval — about 95, more, or fewer?
Predict: how many of 100 replications will land inside? Then press Replicate.
Practice · what may the report claim? · fresh numbers on demand
Core concept 5 · C5
The 99% re-build already showed it: width is not fate, it's arithmetic. \(\text{width} = 2E = 2\,z^* \sigma/\sqrt{n}\) — three quantities, three dials, but they are not equally available.
Raise the level and \(z^*\) grows: Maya's margin runs 0.123 h at 90%, 0.147 at 95%, 0.193 at 99%. More confidence is more width — a safer claim is a vaguer claim, mechanically.
\(z^* \uparrow\ \Rightarrow\ E \uparrow\)The one dial you truly own — and it fights back with a square root: to halve the width you must quadruple \(n\). Precision is bought in bulk, at √-scale prices (last class's diminishing returns, now with a price tag).
\(4n\ \Rightarrow\ E/2\)The population's own spread. More variable populations give wider intervals — but \(\sigma\) is a fact about the students, not a setting on the survey. You can't ask a campus to sleep more consistently.
\(\sigma\)So the honest routes to a tighter printed range are exactly two: accept a weaker warranty, or collect more data. Watch the three dials move one interval:
Quick check · the price of half
See it · C5
Predict first: press \(\times 4\ n\) — how much narrower does the interval get? Then: can any setting of \(n\) undo a jump from 95% to 99%? And what does a bigger \(\sigma\) do?
Same data: the 99% interval is always 56.6% wider than the 90% one (2.576 ÷ 1.645).
Change one knob: the old interval stays above as “before”.
Values are rounded to two decimals; the bounds and the × ratios use the unrounded E.
Core concept 6 · C6
So far the data came first and the width fell out. Real surveys run backwards: the editor names a margin of error first, and Maya must find the smallest \(n\) that delivers it — before collecting anything.
Set \(E = z^* \sigma/\sqrt{n}\) and rearrange: \(n = (z^* \sigma / E)^2\), rounded up. Every symbol on the right is known before the survey: the level picks \(z^*\), history supplies a \(\sigma\), the demand sets \(E\).
\(n = \left\lceil \left(\dfrac{z^* \sigma}{E}\right)^{\!2} \right\rceil\)If the formula says 61.47, the answer is 62 — never 61. Rounding down leaves \(n\) short and the margin over spec: the ceiling isn't a convention, it's the difference between meeting the demand and missing it.
\(\lceil 61.47 \rceil = 62\)\(E\) and the confidence level are decisions, made before any student is surveyed — the formula just prices them. That's the planning move: negotiate the spec while it's still cheap to change.
The editor's demand is about to get a price tag. First, see the whole trade-off as one curve:
I do · watch a full pricing
Watch me decide every step out loud — name the demand (\(E\), level), rearrange the margin formula for \(n\), compute, and round UP to the price.
The editor wants the printed range tight: a margin of \(\pm 3\) minutes \(= 0.05\) h, at the paper's standard 95%. So the spec is \(E = 0.05\), \(z^* = 1.96\) — and history supplies the spread: \(\sigma \approx 1.3\) h from the survey. All three knowns, no data needed.
The margin formula \(E = z^*\sigma/\sqrt{n}\), solved for the one unknown: \(n = \left(\dfrac{z^*\sigma}{E}\right)^{\!2}\). I say the units out loud as a check: SEs per demanded margin, squared — a count of students. It's the width machine, run in reverse.
Inside first: \(\dfrac{1.96 \times 1.3}{0.05}\) \(= \dfrac{2.548}{0.05}\) \(= 50.96\). Then square: \(50.96^2\) \(\approx 2596.9\). Expectation check before I round: 3 minutes is a third of her current 9-minute margin, so the bill should be about \(3^2 = 9\) times her 300. It is.
\(n = \lceil 2596.9 \rceil = \mathbf{2{,}597}\) students. Not 2,596 — that one falls just short of spec. In words: "±3 minutes costs nearly nine surveys' worth of students." Now the editor can decide if the tightness is worth the field work.
The habit: name the demand (\(E\), level, \(\sigma\)), rearrange \(n = (z^*\sigma/E)^2\), compute, round UP — then say the price in words.
We do · you finish it
Same formula, softer demand: you re-run it at ±6 minutes and supply the new \(n\) — and watch what relaxing the spec refunds.
Twenty-six hundred students is too many. The editor relaxes the demand to \(\pm 6\) minutes \(= 0.1\) h, same 95%. Re-run the pricing: \(z^* = 1.96\), \(\sigma = 1.3\), \(E = 0.1\).
Finish it: \(n = ?\) — and check it against the quadruple rule: the demand doubled, so the bill should…
See it · C6
Predict first: as the demanded margin \(E\) shrinks toward zero, does the required \(n\) grow steadily — or explode? Find where "a little tighter" starts costing hundreds of students.
Which tool applies? · mixes earlier classes + today
Practice · price the spec, solo · fresh numbers on demand
Now you price whole specs solo — fresh numbers every re-roll, the formula on screen, support at its minimum.
Production task · the sentence she files
Exit ticket · 60 seconds · anonymous
What Maya files
The sentence the editor wanted exists, and it's exact: "We are 95% confident the campus average is between 6.65 and 6.95 hours." Built as \(\bar{x} \pm z^* \cdot \sigma/\sqrt{n}\) — center from the data, yardstick from last class, warranty dial chosen and owned.
The 95% is a warranty on the method: across many surveys like hers, 95 in 100 such intervals capture \(\mu\). This one either did or didn't — \(\mu\) is a constant, and no probability attaches to it. That one repair is what makes the sentence printable.
And width has a price list: 99% would stretch the range to (6.61, 6.99); halving the width costs \(4\times\) the students; the editor's ±3-minute wish priced out at 2,597 students and was renegotiated to ±6 minutes for 650. Precision is a budget line now.
One IOU is still open: Maya used her survey's \(s = 1.3\) as if it were the true \(\sigma\) — a stand-in the large sample forgives. Next class pays that debt properly: what happens when \(\sigma\) is honestly unknown and \(n\) isn't large — the t-distribution, and intervals that own up to one more layer of uncertainty.
Vote counts on today's slides are simulated for rehearsal; live sessions show the room's real votes.