Confidence intervals — derivation for mean, proportion: Worked examples

4.9.19 · D3 · Maths › Probability Theory & Statistics › Confidence intervals — derivation for mean, proportion

Shuru karne se pehle, poori toolkit ki ek reminder ek jagah:

Upar har symbol parent mein define hai — agar koi unfamiliar lage, pehle parent note padho, phir wapas aao.

Scenario matrix

Har confidence-interval problem is grid ka ek cell hai. Is page ka point yeh hai ki koi bhi cell khaali na rahe.

#	Cell class	Kya khaas hai	Example
1	Mean, $σ$ known, large $n$	textbook $z$ case	(a)
2	Mean, $σ$ unknown, small $n$	$t$ use karna zaroori hai, fatter tails	(b)
3	Mean, $σ$ unknown, large $n$	$t \approx z$ , dono agree karte hain	(c)
4	One-sided bound	sirf ek tail mein $α$ , $α /2$ nahi	(d)
5	Proportion, $\overset{p}{^}$ near $0.5$	widest SE, easy	(e)
6	Proportion, $\overset{p}{^}$ near an edge	small $\overset{p}{^}$ , validity check karo	(f)
7	Degenerate proportion $\overset{p}{^} = 0$	Wald formula break ho jaati hai	(g)
8	Limiting behaviour: $n \to \infty$	interval ek point pe collapse ho jaata hai	(h)
9	Word problem / exam twist	disguise mein sahi recipe choose karo	(i)

Ab hum har cell chalte hain. Confidence level 95% hai jab tak kuch aur nahi kaha, toh $z_{0.025} = 1.96$ .

(a) Cell 1 — Mean, σ known, large n

Worked example Cell 1

Ek soda machine ka known fill standard deviation $σ = 4$ ml hai. $n = 64$ bottles ka sample average $\overset{ˉ}{X} = 500$ ml hai. True mean fill $μ$ ke liye 95% CI banao.

Forecast: Computing se pehle half-width ka guess karo — kya yeh $1$ ml ke karib hoga ya $10$ ml ke?

Standard error $= \frac{σ}{n} = \frac{4}{64} = \frac{4}{8} = 0.5$ ml. Yeh step kyun? Interval mean ki spread ko bound karta hai, jo $σ / n$ hai — dekho Standard Error — single-bottle spread $σ$ nahi.
Critical value $z_{0.025} = 1.96$ . Yeh step kyun? $σ$ known hai aur $n$ large hai, toh $\overset{ˉ}{X}$ Central Limit Theorem se normal hai; 95% har tail mein $2.5%$ chhodta hai.
Margin $E = 1.96 \times 0.5 = 0.98$ ml. Yeh step kyun? Margin = how-sure-number $\times$ how-wobbly.
CI $= 500 \pm 0.98 = [499.02, 500.98]$ ml.

Verify: Units ml hain (sahi). Half-width $0.98$ ml chhota hai kyunki $n$ kaafi large hai — forecast "closer to 1 ml" se match karta hai. ✓

(b) Cell 2 — Mean, σ unknown, small n

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

(d) Cell 4 — One-sided bound (α ko half mat karo!)

Worked example Cell 4

Ek safety engineer sirf mean impurity ka upper bound chahta hai. Known $σ = 2$ ppm, $n = 25$ , $\overset{ˉ}{X} = 10$ ppm. 95% one-sided upper confidence bound do.

Forecast: Critical value $1.96$ hona chahiye ya $1.645$ ?

Sirf ek tail. Hum chahte hain $P (μ \leq bound) = 0.95$ , toh poora $5%$ ek tail mein hai. Yeh step kyun? One-sided bound $α$ ko split nahi karta; woh halving sirf two-sided band ke liye thi.
Critical value $z_{0.05} = 1.645$ ( $1.96$ nahi).
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: One-sided critical value $1.645 < 1.96$ , toh ek one-sided bound apni single side par tighter hai — exactly isliye parent ki "α ko half karna bhoolna" wali galti ulta matter karti hai. Yeh Hypothesis Testing se connect hota hai (ek 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

Ek drug trial mein $n = 50$ patients mein $0$ severe side-effects dikhte hain. $p$ ke liye Wald 95% CI kya hai?

Forecast: Formula width zero dega. Kya yeh believable hai?

Point estimate $\overset{p}{^} = \frac{0}{50} = 0$ .
Validity check FAILS: $n \overset{p}{^} = 0 < 5$ . Normal approximation invalid hai.
Wald margin $E = 1.96 \frac{0 \times 1}{50} = 1.96 \times 0 = 0$ . Yeh step kyun? Hum ise sirf breakdown expose karne ke liye compute karte hain: formula useless interval $[0, 0]$ return karta hai.
Honest fix — "rule of three": $n$ trials mein $0$ events ke saath, approximate 95% upper bound hai $\frac{3}{n} = \frac{3}{50} = 0.06$ , jo $[0, 0.06]$ deta hai. Yeh step kyun? Zero observed ka matlab zero risk nahi hai — matlab hai "small," aur $3/ n$ standard patch hai.

Verify: Wald width $0$ deta hai (bakwaas — Wald ki boundary par failure se link hai). Rule-of-three upper bound $3/50 = 0.06$ positive aur sensible hai. ✓

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

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

Worked example Cell 9 (disguised recipe choice)

"Ek quality lab report karta hai ki ek batch ki mean tensile strength $\overset{ˉ}{X} = 220$ MPa hai, sample SD $s = 12$ MPa $n = 9$ specimens se, aur poochha jaata hai: kya hum 95% sure hain ki mean $210$ MPa se zyada hai?"

Forecast: Kaunsa recipe — $z$ ya $t$ ? One-sided ya two-sided?

Tool choose karo. $σ$ unknown, $n = 9$ small → $t$ with df $= 8$ . Sawaal "exceeds $210$ " one-sided hai. Yeh step kyun? Small $n$ + estimated spread $t$ force karta hai; "exceeds" matlab lower bound, ek 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.
Actual sawaal ka jawab: kyunki 95% lower bound $212.56 > 210$ hai, haan — hum 95% confident hain ki mean $210$ MPa se zyada hai.

Verify: Lower bound $212.56$ MPa target $210$ se upar hai, toh "exceeds" ka conclusion hold karta hai. Units MPa. Yeh CI ↔ test duality in action hai (Hypothesis Testing). ✓

Recall

Recall Kaunsa cell, kaunsa tool?

Har clue ko uski recipe se match karo. $σ$ known, big $n$ ::: $z$ , two-sided $z_{α /2}$ $σ$ unknown, small $n$ ::: $t_{n - 1}$ "at least / at most / exceeds" ::: one-sided, use karo $z_{α}$ ya $t_{α, n - 1}$ (halving nahi) $0$ successes observed ::: Wald fail; rule of three $3/ n$ Narrower interval chahiye ::: $n$ badhao (width $\propto 1/ n$ )

Connections

Central Limit Theorem — kyun $z$ kaam karta hai jab $n$ large hota hai.
Standard Error — woh $σ / n$ jo cell 8 mein shrink hota hai.
Student's t-distribution — cells 2 aur 9 ki fat tails.
Hypothesis Testing — cells 4 aur 9 one-sided tests hain disguise mein.
Bernoulli & Binomial Distributions — proportion cells 5–7.
Sample Size Determination — cell 8 ka width law $n$ ke liye invert karo.
Parent: CI derivations