Confidence intervals — derivation for mean, proportion: Worked examples

Before starting, one reminder of the whole toolkit in one place:

Every symbol above is defined in the parent — if any looks unfamiliar, read the parent note first, then come back.

The scenario matrix

Every confidence-interval problem is one cell of this grid. The point of this page is to leave no empty cell.

#	Cell class	What makes it special	Example
1	Mean, $σ$ known, large $n$	textbook $z$ case	(a)
2	Mean, $σ$ unknown, small $n$	must use $t$ , fatter tails	(b)
3	Mean, $σ$ unknown, large $n$	$t \approx z$ , they agree	(c)
4	One-sided bound	only $α$ in one tail, not $α /2$	(d)
5	Proportion, $\overset{p}{^}$ near $0.5$	widest SE, easy	(e)
6	Proportion, $\overset{p}{^}$ near an edge	small $\overset{p}{^}$ , check validity	(f)
7	Degenerate proportion $\overset{p}{^} = 0$	Wald formula breaks	(g)
8	Limiting behaviour: $n \to \infty$	interval collapses to a point	(h)
9	Word problem / exam twist	choose the right recipe under disguise	(i)

We now walk each cell. Confidence level is 95% unless stated otherwise, so $z_{0.025} = 1.96$ .

(a) Cell 1 — Mean, σ known, large n

Worked example Cell 1

A soda machine has a known fill standard deviation $σ = 4$ ml. A sample of $n = 64$ bottles averages $\overset{ˉ}{X} = 500$ ml. Build a 95% CI for the true mean fill $μ$ .

Forecast: Guess the half-width before computing — is it closer to $1$ ml or $10$ ml?

Standard error $= \frac{σ}{n} = \frac{4}{64} = \frac{4}{8} = 0.5$ ml. Why this step? The interval bounds the mean's spread, which is $σ / n$ — see Standard Error — not the single-bottle spread $σ$ .
Critical value $z_{0.025} = 1.96$ . Why this step? $σ$ is known and $n$ is large, so $\overset{ˉ}{X}$ is normal by the Central Limit Theorem; 95% leaves $2.5%$ in each tail.
Margin $E = 1.96 \times 0.5 = 0.98$ ml. Why this step? Margin = how-sure-number $\times$ how-wobbly.
CI $= 500 \pm 0.98 = [499.02, 500.98]$ ml.

Verify: Units are ml (correct). Half-width $0.98$ ml is small because $n$ is fairly large — matches the forecast "closer to 1 ml." ✓

(b) Cell 2 — Mean, σ unknown, small n

(c) Cell 3 — Mean, σ unknown, large n (t ≈ z)

(d) Cell 4 — One-sided bound (don't halve α!)

Worked example Cell 4

A safety engineer only cares about an upper bound on mean impurity. Known $σ = 2$ ppm, $n = 25$ , $\overset{ˉ}{X} = 10$ ppm. Give a 95% one-sided upper confidence bound.

Forecast: Should the critical value be $1.96$ or $1.645$ ?

One tail only. We want $P (μ \leq bound) = 0.95$ , so all $5%$ sits in one tail. Why this step? A one-sided bound doesn't split $α$ ; that halving was only for a two-sided band.
Critical value $z_{0.05} = 1.645$ (not $1.96$ ).
Standard error $= \frac{2}{25} = \frac{2}{5} = 0.4$ ppm.
Upper bound $= \overset{ˉ}{X} + z_{0.05} \cdot SE = 10 + 1.645 \times 0.4 = 10 + 0.658 = 10.658$ ppm.
CI $= (- \infty, 10.658]$ ppm.

Verify: The one-sided critical value $1.645 < 1.96$ , so a one-sided bound is tighter on its single side — exactly why the parent's "forgetting to halve α" mistake matters in reverse. This links to Hypothesis Testing (a one-sided test at $α = 0.05$ ). ✓

(e) Cell 5 — Proportion near p = 0.5

Figure — Confidence intervals — derivation for mean, proportion

(f) Cell 6 — Proportion near an edge

(g) Cell 7 — Degenerate: p̂ = 0 (formula breaks)

Worked example Cell 7

A drug trial sees $0$ severe side-effects in $n = 50$ patients. What is the Wald 95% CI for $p$ ?

Forecast: The formula will give a width of zero. Is that believable?

Point estimate $\overset{p}{^} = \frac{0}{50} = 0$ .
Validity check FAILS: $n \overset{p}{^} = 0 < 5$ . The normal approximation is invalid.
Wald margin $E = 1.96 \frac{0 \times 1}{50} = 1.96 \times 0 = 0$ . Why this step? We compute it only to expose the breakdown: the formula returns the useless interval $[0, 0]$ .
The honest fix — the "rule of three": with $0$ events in $n$ trials, an approximate 95% upper bound is $\frac{3}{n} = \frac{3}{50} = 0.06$ , giving $[0, 0.06]$ . Why this step? Zero observed doesn't mean zero risk — it means "small," and $3/ n$ is the standard patch.

Verify: Wald gives width $0$ (absurd — links to Wald's failure at the boundary). The rule-of-three upper bound $3/50 = 0.06$ is positive and sensible. ✓

(h) Cell 8 — Limiting behaviour as n → ∞

(i) Cell 9 — Word-problem / exam twist

Worked example Cell 9 (disguised recipe choice)

"A quality lab reports that a batch's mean tensile strength is $\overset{ˉ}{X} = 220$ MPa with sample SD $s = 12$ MPa from $n = 9$ specimens, and asks: are we 95% sure the mean exceeds $210$ MPa?"

Forecast: Which recipe — $z$ or $t$ ? One-sided or two-sided?

Choose the tool. $σ$ unknown, $n = 9$ small → $t$ with df $= 8$ . The question "exceeds $210$ " is one-sided. Why this step? Small $n$ + estimated spread forces $t$ ; "exceeds" means a lower bound, one tail.
Critical value $t_{0.05, 8} = 1.860$ (one tail at $5%$ ).
Standard error $= \frac{12}{9} = \frac{12}{3} = 4$ MPa.
Lower bound $= \overset{ˉ}{X} - t_{0.05, 8} \cdot SE = 220 - 1.860 \times 4 = 220 - 7.44 = 212.56$ MPa.
Answer the actual question: since the 95% lower bound $212.56 > 210$ , yes — we're 95% confident the mean exceeds $210$ MPa.

Verify: Lower bound $212.56$ MPa is above the target $210$ , so the conclusion "exceeds" holds. Units MPa. This is CI ↔ test duality in action (Hypothesis Testing). ✓

Recall

Recall Which cell, which tool?

Match each clue to its recipe. $σ$ known, big $n$ ::: $z$ , two-sided $z_{α /2}$ $σ$ unknown, small $n$ ::: $t_{n - 1}$ "at least / at most / exceeds" ::: one-sided, use $z_{α}$ or $t_{α, n - 1}$ (no halving) $0$ successes observed ::: Wald fails; rule of three $3/ n$ Want narrower interval ::: raise $n$ (width $\propto 1/ n$ )

Connections

Central Limit Theorem — why $z$ works once $n$ is large.
Standard Error — the $σ / n$ that shrinks in cell 8.
Student's t-distribution — the fat tails of cells 2 and 9.
Hypothesis Testing — cells 4 and 9 are one-sided tests in disguise.
Bernoulli & Binomial Distributions — proportion cells 5–7.
Sample Size Determination — invert cell 8's width law for $n$ .
Parent: CI derivations