IB Math AI HL Estimation & Confidence Intervals — Complete Cheatsheet
Everything you need for estimation in IB Mathematics Applications & Interpretation HL — point estimates and unbiased estimators, the standard error, confidence intervals for a mean built on the GDC, the margin of error and interval width, and how to choose the sample size and read a confidence interval correctly. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Estimation is where the Statistics & Probability strand of IB Mathematics Applications & Interpretation HL turns a single sample into a statement about a whole population you can never measure in full — polling, quality control, medical trials all live here. It is applications mathematics at its purest: you collect data, read a point estimate straight off the GDC, then attach a margin of error to produce a confidence interval. A calculator is always allowed, so the interval itself is built by the TInterval / ZInterval routine — the marks live in choosing $t$ or $z$, entering the unbiased sample standard deviation, and interpreting the result in context, not in grinding the arithmetic.
This cheatsheet condenses AHL 4.14–4.15 onto one page — point estimates and unbiased estimators, the standard error $\sigma/\sqrt{n}$, the confidence interval for a mean, the margin of error and interval width, how confidence level and sample size trade off, choosing $n$ for a target accuracy, and the one interpretation examiners actually reward — and flags the traps that quietly cost marks ($s_x$ versus $\sigma_x$, dividing by $\sqrt{n}$ not $n$, $t$ versus $z$, rounding $n$ up). The print-ready PDF is at the bottom, free to download.
§1 — Point estimates & unbiased estimators AHL 4.14
A sample can only estimate a population's parameters. A point estimate is a single best-guess value; the good ones are unbiased — correct on average over all possible samples.
Estimate of the mean:$\hat{\mu}=\bar{x}=\dfrac{\sum x}{n}$ — the sample mean estimates the population mean $\mu$.
Estimate of the variance:$\widehat{\sigma^{2}}=s_{n-1}^{2}=\dfrac{\sum (x-\bar{x})^{2}}{n-1}$ — divide by $n-1$, not $n$.
Unbiased means:$E(\bar{x})=\mu$ and $E(s_{n-1}^{2})=\sigma^{2}$ — right on average, not too big or too small.
TrapDividing by $n$ (the calculator's $\sigma_x$) systematically underestimates the population variance. For estimation IB uses the $n-1$ divisor — the $s_x$ key — so the estimator is unbiased.
NoteThis is the exact opposite of describing a fixed data set (Univariate Statistics), where IB wants $\sigma_x$. Rule of thumb: describing data → $\sigma_x$; estimating a population from a sample → $s_x$.
§2 — The standard error of the mean AHL 4.14
Take another sample and $\bar{x}$ shifts. The standard error measures how much the sample mean itself varies from sample to sample — the smaller it is, the sharper your estimate.
Sampling distribution:$\bar{X}\sim N\!\left(\mu,\dfrac{\sigma^{2}}{n}\right)$ — exact for a normal population, approximate for any population when $n$ is large.
Standard error:$\text{SE}=\dfrac{\sigma}{\sqrt{n}}$ (known $\sigma$) or $\dfrac{s}{\sqrt{n}}$ (unknown $\sigma$) — the standard deviation of $\bar{x}$.
TrickDivide by $\sqrt{n}$, never by $n$. Quadruple the sample and the standard error only halves — precision improves slowly, and that is exactly why big samples are expensive.
NoteCentral Limit Theorem: for $n\ge 30$ the sample mean is approximately normal whatever the population's shape — the reason the interval formulas work so widely in real AI applications.
§3 — Confidence interval for a mean AHL 4.15
A confidence interval wraps the point estimate $\bar{x}$ in a margin of error, giving a range expected to capture $\mu$ a stated proportion of the time.
The interval:$\bar{x}\pm E$ — always centred on the sample mean.
$\sigma$ known ($z$-interval):$\bar{x}\pm z^{*}\dfrac{\sigma}{\sqrt{n}}$.
$\sigma$ unknown ($t$-interval):$\bar{x}\pm t^{*}\dfrac{s}{\sqrt{n}}$, with $t^{*}$ read at $\nu=n-1$ degrees of freedom.
TrapPick the interval by what you are given: a stated population $\sigma$ → $z$; only the sample $s$ → $t$. In AI HL exams $\sigma$ is almost always unknown, so the $t$-interval is the default choice.
Note$t^{*}>z^{*}$ for the same confidence (the $t$-distribution has heavier tails), and $t^{*}\to z^{*}$ as $n$ grows — for large samples the two intervals nearly coincide.
§4 — Building the interval on the GDC AHL 4.15
AI is GDC-first: once you know which interval, the calculator returns the limits directly. There are two input modes.
Data mode:Enter the raw list → choose TInterval (or ZInterval) → set the C-level, e.g. $0.95$.
Stats mode:Type in $\bar{x}$, $s_x$ (or $\sigma$) and $n$ → same routine, same C-level.
Output:The interval $(a,\,b)$, ready to quote, together with $\bar{x}$ and $n$.
TrickFor a $t$-interval, enter the sample standard deviation $s_x$ (the $n-1$ value) in the s.d. field. For a $z$-interval enter the given population $\sigma$. Wrong field, wrong interval.
NoteIf a question wants the margin of error but you built the interval on the GDC, recover it instantly as $E=\dfrac{b-a}{2}$.
§5 — Margin of error & the interval limits AHL 4.15
The margin of error $E$ is the half-width of the interval — the "$\pm$" part — and it fixes both endpoints.
Margin of error:$E=t^{*}\dfrac{s}{\sqrt{n}}$ (or $z^{*}\dfrac{\sigma}{\sqrt{n}}$) — critical value $\times$ standard error.
Limits:lower $=\bar{x}-E$, upper $=\bar{x}+E$.
Critical values $z^{*}$:$1.645,\;1.96,\;2.576$ for $90\%,\;95\%,\;99\%$ confidence.
A confidence interval on a number line: the sample mean $\bar{x}$ sits at the centre, the margin of error $E$ reaches out to each limit, and the full width is $w=2E$. Raising the confidence level lengthens both arrows; increasing $n$ shrinks them.
Trap$E$ multiplies the standard error $s/\sqrt{n}$ — not $s$ on its own. Dropping the $\sqrt{n}$ inflates every limit by a factor of $\sqrt{n}$.
Note$t^{*}$ depends on both the confidence level and $\nu=n-1$; read it from the GDC's inverse-$t$ or straight from the interval — never guess it.
§6 — Width of the interval & reading a reported CI AHL 4.15
The width is the full span of the interval — twice the margin of error — and it lets you reverse-engineer a quoted interval.
Width:$w=2E=2t^{*}\dfrac{s}{\sqrt{n}}$ — upper limit minus lower limit.
Mean from $(a,\,b)$:$\bar{x}=\dfrac{a+b}{2}$ — the midpoint of the interval.
Margin from $(a,\,b)$:$E=\dfrac{b-a}{2}$ — half the gap between the ends.
TrapWidth is the full span, so keep the factor of $2$: $w=2E$. The sample mean sits at the midpoint, never at an end.
TrickGiven only the two limits you can still recover $\bar{x}$ and $E$ with no raw data — handy when a question quotes an interval and asks for the mean or the margin.
§7 — Confidence level vs sample size AHL 4.15
Two levers set the width. Knowing which one moves it — and by how much — is a favourite exam theme.
Change
Effect on the interval
Why
Raise the confidence level (95% → 99%)
wider
larger critical value $z^{*}$/$t^{*}$; the SE is unchanged
Increase the sample size $n$
narrower
standard error $s/\sqrt{n}$ falls
Halve the margin of error
need $\approx 4\times$ the data
width $\propto 1/\sqrt{n}$
More variable data (larger $s$)
wider
standard error rises
Extra margin, 95% → 99%:$(z^{*}_{99}-z^{*}_{95})\dfrac{\sigma}{\sqrt{n}}=(2.576-1.96)\dfrac{\sigma}{\sqrt{n}}$ — same sample, only the critical value changes.
TrapRaising the confidence level widens the interval but does not move its centre $\bar{x}$ — more confidence buys less precision, not a different estimate.
NoteBecause width $\propto 1/\sqrt{n}$, halving the margin of error needs about four times the sample size — a common "how many more?" exam twist.
§8 — Choosing $n$ for a required width AHL 4.15
Turn the margin-of-error formula around to plan a sample before collecting the data.
Solve for $n$:from $E=z^{*}\dfrac{\sigma}{\sqrt{n}}$, $\;n=\left(\dfrac{z^{*}\sigma}{E}\right)^{2}$.
Target a width $w$:put $E=\dfrac{w}{2}$, so $n=\left(\dfrac{2z^{*}\sigma}{w}\right)^{2}$.
TrapAlways round $n$ up to the next whole number — rounding down leaves the interval wider than the target. And $n$ must be a positive integer.
NoteYou need an estimate of $\sigma$ up front (a pilot study or earlier data). Use $z^{*}$, not $t^{*}$: $t^{*}$ depends on $n$, the very quantity you are solving for.
§9 — Interpreting a confidence interval AHL 4.15
One sentence earns the interpretation mark — and one tempting sentence loses it.
Correct wording:"We are $95\%$ confident that the interval contains the population mean $\mu$."
What it means:over many repeated samples, about $95\%$ of the intervals built this way capture $\mu$.
TrapIt is not "there is a $95\%$ probability that $\mu$ lies in this interval". $\mu$ is a fixed number; the interval is what is random. Once computed, this particular interval either contains $\mu$ or it does not.
NoteDual to testing: if a claimed value $\mu_0$ falls outside a $95\%$ interval, that is evidence against $\mu=\mu_0$ at the $5\%$ level — the bridge to hypothesis testing.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
"Find a point estimate of the mean / variance"
Read $\bar{x}$ and $s_x^{2}$ from 1-Var Stats
The variance uses the $n-1$ value $s_x$, not $\sigma_x$
"Construct a 95% confidence interval"
GDC TInterval ($\sigma$ unknown) or ZInterval ($\sigma$ given); set the C-level
Enter $s_x$ in the s.d. field; choose $t$ vs $z$ correctly
"State the margin of error"
$E=\dfrac{b-a}{2}$ from the interval, or $E=t^{*}\dfrac{s}{\sqrt{n}}$
It is the half-width; keep the $\sqrt{n}$ in the SE
"Find the width of the interval"
$w=b-a=2E$
Full span — do not forget the factor of $2$
"The CI is $(a,b)$; find the mean"
$\bar{x}=\dfrac{a+b}{2}$
Mean is the midpoint, not a limit
"Effect of a higher confidence level"
Larger $z^{*}/t^{*}$ → wider; centre unchanged
Only the critical value changes for a fixed sample
"Sample size for a margin below $E$"
$n=\left(\dfrac{z^{*}\sigma}{E}\right)^{2}$, then round up
Round up; use $z^{*}$; need an estimate of $\sigma$
"Interpret the interval"
"95% confident the interval contains $\mu$"
Not a probability statement about $\mu$
Worked Example — A Confidence Interval for a Mean
Question (AI HL Paper 2 style — 9 marks)
A café measures the caffeine content, in mg, of a random sample of 8 espressos from its machine: 118, 122, 125, 127, 130, 131, 134, 137. The caffeine content is assumed to be normally distributed. (a) Find (i) the sample mean $\bar{x}$ and (ii) an unbiased estimate of the population standard deviation. (b) Construct a $95\%$ confidence interval for the mean caffeine content of all espressos from this machine. (c) State the margin of error and interpret your interval in context. (d) The café wants to estimate the mean to within $3$ mg with $95\%$ confidence. Using the sample standard deviation as an estimate of $\sigma$, find the least number of espressos it must test.
Solution
(a) Enter the 8 values into 1-Var Stats. (M1) The mean is $\bar{x}=128$ mg (A1). The unbiased estimate of the population standard deviation is the sample s.d. $s=s_{n-1}=\sqrt{\dfrac{\sum (x-\bar{x})^{2}}{n-1}}=\sqrt{\dfrac{276}{7}}=6.28$ mg (3 s.f.) (A1). (Use $s_x$, not $\sigma_x=5.87$.)
(b) $\sigma$ is unknown, so use a $t$-interval with $\nu=n-1=7$. TInterval with $\bar{x}=128$, $s_x=6.28$, $n=8$, C-level $0.95$ (M1) gives $122.75<\mu<133.25$, i.e. $(122.75,\;133.25)$ mg. (A1)
(c) Margin of error $E=\dfrac{133.25-122.75}{2}=5.25$ mg (A1). Interpretation: we are $95\%$ confident that the interval $(122.75,\,133.25)$ mg contains the true mean caffeine content $\mu$ of all espressos from the machine — a procedure like this captures $\mu$ about $95\%$ of the time. (A1)
(d) Require $E\le 3$ with $z^{*}=1.96$ and $\sigma\approx s=6.28$: $n=\left(\dfrac{z^{*}\sigma}{E}\right)^{2}=\left(\dfrac{1.96\times 6.28}{3}\right)^{2}=(4.10)^{2}=16.8$ (M1). Round up: the café must test at least $n=17$ espressos. (A1)
Examiner's note: three marks are routinely dropped here — quoting $\sigma_x$ instead of $s_x$ for the estimate and the interval; using a $z$-interval when only the sample s.d. is known (it must be $t$); and rounding the sample size down. Keep the units (mg), and remember the interval is a statement about the mean $\mu$, not about individual espressos.
Common Student Questions
Do I use a $z$-interval or a $t$-interval?
It depends on whether you know the population standard deviation $\sigma$. If the question gives you $\sigma$, use a $z$-interval, $\bar{x}\pm z^{*}\sigma/\sqrt{n}$ (GDC: ZInterval). If you only have the sample standard deviation $s$ — the usual case — use a $t$-interval, $\bar{x}\pm t^{*}s/\sqrt{n}$ with $n-1$ degrees of freedom (GDC: TInterval). On AI HL, $\sigma$ is rarely handed to you, so reach for the $t$-interval by default. For large samples the two give almost the same answer because $t^{*}\to z^{*}$.
Why divide by $n-1$ here, when Univariate Statistics told me to use $\sigma_x$?
Because the job is different. In univariate statistics you describe a data set you already hold in full, so the population value $\sigma_x$ (dividing by $n$) is exact. In estimation you have only a sample and are inferring a larger population; dividing by $n$ would systematically underestimate the spread, so the unbiased estimate uses $n-1$ — the $s_x$ key. Rule of thumb: describing → $\sigma_x$; estimating or building a confidence interval → $s_x$.
What does "$95\%$ confident" actually mean?
It describes the method, not this one interval. If you repeated the whole sampling-and-interval process many times, about $95\%$ of the intervals produced would contain the true mean $\mu$. It is wrong to say "there is a $95\%$ probability $\mu$ is in this interval" — $\mu$ is a fixed (if unknown) number, and this particular interval either contains it or it does not. The $95\%$ lives in the long-run success rate of the procedure, and it is why a wider interval means more confidence but less precision.
How do I decide the sample size for a given accuracy?
Set the margin of error $E$ you can tolerate, then solve $E=z^{*}\sigma/\sqrt{n}$ to get $n=\left(z^{*}\sigma/E\right)^{2}$, and round up to the next whole number. You need an estimate of $\sigma$ first (from a pilot sample or a previous study), and you use $z^{*}$ rather than $t^{*}$ because $t^{*}$ would depend on the $n$ you are trying to find. If the target is a full width $w$ rather than a margin, put $E=w/2$ first.
What changes the width of a confidence interval?
Three things. A higher confidence level uses a larger critical value ($z^{*}=1.645,\,1.96,\,2.576$ for $90,95,99\%$), so the interval gets wider. A larger sample size shrinks the standard error $s/\sqrt{n}$, so the interval gets narrower — but only as $1/\sqrt{n}$, so halving the width needs about four times the data. More variable data (a larger $s$) widens it. Changing the confidence level does not move the centre $\bar{x}$ — only the critical value, and hence the width, responds.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every estimator, standard-error and confidence-interval method in full, with the exact GDC TInterval / ZInterval keystrokes and worked interpretations.
Tutorial booklet — 25+ AI HL-style questions from standard-error fluency to full confidence-interval and sample-size applications.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations and the exact GDC screens for every interval.
Predicted-paper questions — the confidence-interval and sample-size question types most likely in the next AI HL session.
