HookWhat the ±3 in every opinion poll actually promises
Every election poll comes with the same small print: 'margin of error ±3 percentage points'. Almost nobody who reads it could say what it promises. It does not mean the true figure is within 3 points of the headline number — sometimes it isn't, and the pollster has done nothing wrong. It is a 95% confidence statement: the polling procedure, run over and over on fresh random samples of about a thousand people, produces an interval that captures the truth 19 times in 20. The twentieth poll misses — cleanly, honestly, by design. The machinery behind that small print was formalised by Jerzy Neyman in a 1937 paper that scandalised statisticians of the day precisely because of that twentieth poll: Neyman insisted, correctly, that once a specific interval is computed you cannot say there is a 95% probability the truth lies inside it. The truth is not random. The interval is.
That inversion — randomness lives in the procedure, not the parameter — is the whole intellectual content of this section, and the calculations are deliberately simple by comparison. A point estimate \(\bar{x}\) on its own is almost dishonest: it will essentially never equal \(\mu\), and it carries no confession of how far off it might be. A confidence interval is the estimate plus the confession, and it comes in three builds — \(\sigma\) known, \(\sigma\) estimated from a large sample, and \(\sigma\) estimated from a small one, where Student's \(t\) from the previous section returns to charge its fat-tailed premium.
ModelThe machine — known variance
Take a random sample of size \(n\) from \(\mathrm{N}(\mu,\sigma^2)\) with \(\sigma\) known. Then \(\bar{X}\sim\mathrm{N}\!\left(\mu,\frac{\sigma^2}{n}\right)\), so \(\bar{X}\) lands within 1.96 standard errors of \(\mu\) with probability 0.95. Flip that sentence around — if \(\bar{X}\) is within 1.96 standard errors of \(\mu\), then \(\mu\) is within 1.96 standard errors of \(\bar{X}\) — and you have the symmetric 95% confidence interval
\[\bar{x}\pm 1.96\frac{\sigma}{\sqrt{n}}.\]
The general recipe is \(\bar{x}\pm z\frac{\sigma}{\sqrt{n}}\), where \(z\) cuts off \(\frac{\alpha}{2}\) in each tail of the standard normal: 1.645 for 90%, 1.96 for 95%, 2.576 for 99%. Anatomy matters more than arithmetic here. The centre is the data's estimate \(\bar{x}\); the half-width \(z\frac{\sigma}{\sqrt{n}}\) is the confession, and it obeys two laws worth stating in every evaluation answer. Confidence costs width: pushing 95% to 99% swaps 1.96 for 2.576 and widens the interval by a third — certainty is bought with vagueness. And precision costs sample size at a brutal exchange rate: the \(\sqrt{n}\) in the denominator means halving the width requires quadrupling the sample.
A filling machine dispenses juice with standard deviation \(\sigma=2.5\) ml, known from long experience and unaffected by the volume setting. A random sample of 50 bottles has mean \(\bar{x}=330.6\) ml. The standard error is \(\frac{2.5}{\sqrt{50}}=0.3536\) ml, so the 95% confidence interval for the mean fill is \(330.6\pm 1.96\times 0.3536=330.6\pm 0.693\), i.e. \((329.9,\ 331.3)\) ml to 4 s.f. At 99% the interval becomes \(330.6\pm 2.576\times 0.3536=330.6\pm 0.911\), i.e. \((329.7,\ 331.5)\) — wider, exactly as the confidence-costs-width law demands. Suppose the manufacturer wants a 95% interval of total width at most 0.7 ml: solve \(2\times 1.96\times\frac{2.5}{\sqrt{n}}\leq 0.7\Rightarrow\sqrt{n}\geq\frac{9.8}{0.7}=14\Rightarrow n\geq 196\). Four times the precision of \(n=50\)? Nearly four times the bottles.
ModelLarge samples — when \(s\) can stand in for \(\sigma\)
'\(\sigma\) known' is a textbook fiction; in practice the sample must supply its own spread via \(s\), the divisor-\((n-1)\) estimate. When the sample is large — as a working convention, \(n\geq 30\), and comfortably so at the sample sizes AQA uses in this setting — two graces are extended at once. First, \(s\) is computed from enough data to be a reliable stand-in for \(\sigma\), so the extra wobble it introduces is negligible and normal \(z\)-values remain honest. Second, the Central Limit Theorem makes the distribution of \(\bar{X}\) approximately normal whatever the population's own shape, so the interval
\[\bar{x}\pm z\frac{s}{\sqrt{n}}\]
is approximately valid even when the underlying data are skewed — journey times, component lifetimes, incomes. That second grace is the one to cite when a question asks why no normality assumption was stated: with \(n=100\) daily journey times, the population can be as lopsided as it likes and \(\bar{X}\) is still approximately normal. Both statements are approximations, and the exam wants that word: the large-sample interval is an approximate 95% interval, and saying so is a mark, not a weakness.
A courier firm logs a random sample of 80 delivery times, finding \(\bar{x}=42.3\) minutes and \(s=11.6\) minutes (divisor 79). Delivery times are right-skewed — a few disasters drag the tail — but \(n=80\) is ample for the CLT. The approximate 95% confidence interval for the mean delivery time is \(42.3\pm 1.96\times\frac{11.6}{\sqrt{80}}=42.3\pm 1.96\times 1.297=42.3\pm 2.54\), i.e. \((39.8,\ 44.8)\) minutes. Note what was not assumed: normality of individual delivery times. The CLT carried the whole argument, and one sentence saying so is the interpretation mark.
ModelSmall samples — the \(t\)-interval and its premium
Shrink the sample and both graces are withdrawn. With \(n=10\), \(s\) is a shaky estimate of \(\sigma\), and there is no CLT rescue — so the population itself must be assumed normal, and the standard normal must give way to Student's \(t\) with \(\nu=n-1\) degrees of freedom. The interval becomes
\[\bar{x}\pm t_{n-1}\,\frac{s}{\sqrt{n}},\]
where \(t_{n-1}\) is the critical value cutting \(\frac{\alpha}{2}\) into each tail of the \(t\)-distribution — 2.262 at \(\nu=9\) for 95%, against the normal's 1.96. The structure is identical to the known-\(\sigma\) machine; only the multiplier changes, and it changes in one direction: \(t\)-intervals are always wider. That widening is not pessimism but honesty — the interval must cover its 95% of cases despite the denominator itself being a noisy estimate, and the fat tails of \(t\) price that noise exactly. As \(n\) grows the premium melts away (\(t_{29}=2.045\), \(t_{\infty}=1.96\)), which is the small-sample and large-sample stories agreeing at the boundary, as they must.
Ten specimens of a new alloy are tested for tensile strength, giving \(\bar{x}=4.38\) kN and \(s=0.60\) kN (divisor 9); strengths may be assumed normally distributed. With \(\nu=9\), the 95% critical value is \(t_9=2.262\). Standard error: \(\frac{0.60}{\sqrt{10}}=0.1897\) kN. Interval: \(4.38\pm 2.262\times 0.1897=4.38\pm 0.429\), i.e. \((3.95,\ 4.81)\) kN. Had we wrongly used \(z=1.96\), the interval would be \(4.38\pm 0.372=(4.01,\ 4.75)\) — about 13% narrower than ten observations can honestly justify. The narrower interval looks better and is worth less: its true capture rate is below the 95% on its label.
MechanismReading an interval — inference without a hypothesis test
The construction marks are routine; the inference marks are where candidates separate. Rule one, Neyman's rule: once computed, an interval like \((329.9,\ 331.3)\) either contains \(\mu\) or it does not — the correct sentence is 'we are 95% confident that the mean lies between 329.9 and 331.3 ml', where the 95% describes the long-run capture rate of the method. The forbidden sentence is 'there is a probability of 0.95 that \(\mu\) lies in this interval': \(\mu\) is a fixed number, not a random variable, and mark schemes police the distinction explicitly.
Rule two: an interval is a claim-testing device. If a manufacturer claims \(\mu=330\) and the 95% interval is \((329.9,\ 331.3)\), the claimed value sits inside, so the sample is consistent with the claim at this level — no ground to dispute it. If the claim were \(\mu=329\), it falls outside, and the data cast significant doubt on the claim: the same verdict a two-tailed 5% test would return, because a value of \(\mu_0\) lies outside the 95% interval exactly when the corresponding test rejects \(H_0:\mu=\mu_0\) at 5%. Phrase verdicts in the language of evidence and consistency, not proof — the interval no more proves \(\mu\neq 329\) than a test does. Rule three: interrogate the width before trusting any conclusion. Width scales with the confidence level (up), with \(\sigma\) or \(s\) (up), and with \(\frac{1}{\sqrt{n}}\) (down) — and a question that asks 'how could the estimate be made more precise?' is fishing for exactly two answers: a larger sample, or a lower confidence level, each with its stated cost.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Announce the build before the arithmetic: which case you are in (\(\sigma\) known / large sample / small sample with \(t\)), the distribution and critical value you will use, and — for the small-sample case — the normality assumption stated in words. Then the fixed sequence: standard error, half-width, interval quoted as a pair of values to sensible accuracy with units. The three habitual slips are using a one-tailed table value (a 95% interval needs the 97.5% point — 1.96, not 1.645), computing \(s\) with divisor \(n\) instead of \(n-1\), and forgetting that \(t\) needs \(\nu=n-1\), not \(n\). If the answer will feed a comparison ('is the claim plausible?'), keep an extra decimal place until the final line — borderline verdicts die by premature rounding.
The interpretation marks have set phrasings, and it pays to use them verbatim. For the meaning of the level: 'if many samples were taken, about 95% of the intervals constructed would contain the true mean'. For a claim check: state whether the claimed value lies inside, then conclude 'the sample is consistent with the claim' or 'there is significant evidence against the claim at this level' — never 'proves' either way. For a precision question: 'increase \(n\)' (with the \(\sqrt{n}\) cost quoted) or 'accept a lower confidence level'. And when a question ends 'comment on the validity of the interval', audit the assumptions like a checklist — random sample? normal population, or \(n\) large enough for the CLT? \(\sigma\) genuinely known, or should \(t\) have been used? — because one named, context-tested assumption outscores a paragraph of generalities.