Visual walkthrough — Confidence intervals — derivation for mean, proportion

4.9.19 · D2 · Maths › Probability Theory & Statistics › Confidence intervals — derivation for mean, proportion

Yeh page parent note ke andar jaata hai. Iske neeche jo engine chal raha hai woh hai Central Limit Theorem, aur jo "wobble" hum baar baar measure karte hain woh hai Standard Error.

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Step 1 — Ek sample ek fuzzy guess hai

KYA. Hamare paas ek bahut badi population hai jiska koi true average value hai jise hum $\mu$ kehte hain (Greek letter "mu", bas ek naam us true mean ke liye jo hum dekh nahi sakte). Hum $n$ items uthate hain, measure karte hain, aur average nikalte hain: $\bar{X}=\frac{1}{n}{\sum }_{i=1}^n{X}_i.$ Yahan $X_i$ $i$-va measurement hai, $\sum$ ka matlab hai "sab ko jodo", $n$ hai kitne liye, aur $\bar{X}$ (padho "X-bar") unka average hai — hamara point estimate.

KYUN. $\bar{X}$ $\mu$ ke liye hamara best single guess hai. Lekin agar kisi dost ne alag handful liya hota, toh unhe thoda alag $\bar{X}$ milta. Toh $\bar{X}$ ek fixed number nahi hai — yeh sample se sample par kaampti hai.

PICTURE. Neeche, true mean $\mu$ yellow line hai. Har blue dot ek possible sample se ek possible $\bar{X}$ hai. Woh $\mu$ ke around scatter hote hain — kabhi bilkul uss par nahi lagte.

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Step 2 — Kaampan ko collect karo: sampling distribution

KYA. Agar hum sochein ki hamesha ke liye sample pe sample draw karte rahein aur saare $\bar{X}$ values ko pile karte rahein, woh ek smooth hill banate hain — sampling distribution of the mean. Iska center $\mu$ par baithta hai, aur iska spread ek number se measure hota hai jise standard error kehte hain: $\text{SE}=\frac{\sigma }{\sqrt{n}}.$ $\sigma$ ("sigma") ek single measurement ka spread hai; $\sqrt{n}$ sample size ka square-root hai.

$\sqrt{n}$ se kyun divide karo? Averaging flukes ko cancel karti hai. Zyada items ($n$ bada) → high wale aur low wale zyada cancel hote hain → $\bar{X}$ ki hill narrow hoti jaati hai. Maths, aur yahan bilkul wahi jagah hai jahan Step 1 ka i.i.d. assumption apna kaam karta hai: independent variables ke variances add hote hain. Kyunki $X_i$ independent hain, $\mathrm{Var}\left({\sum }_{i=1}^n{X}_i\right)={\sum }_{i=1}^n\mathrm{Var}(X_i)=n{\sigma }^2$ (yeh additivity independence ke bina galat hai — correlated terms extra cross-terms laate). Sum ko $n$ se divide karne par variance $1/n^2$ scale hota hai, jisse ${\sigma }^2/n$ milta hai; square root lo aur $\sigma /\sqrt{n}$ milta hai.

PICTURE. Dekho hill kaisi $n$ ke 4 se 100 tak badhne par tall aur skinny hoti jaati hai. Same center $\mu$, shrinking width.

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Step 3 — Hill bell kyun hai (CLT)

KYA. Chahe individual measurements ki shape koi bhi ho (skewed, lumpy, uniform), averages ki hill $n$ badhne par ek normal bell curve approach karti hai. Yahi hai Central Limit Theorem.

KYUN matter karta hai. Ek bell curve sirf do numbers se puri tarah describe hoti hai — uska center aur uski width. Toh jab hum believe karte hain ki $\bar{X}$ approximately bell-shaped hai, hum compute kar sakte hain ki $\bar{X}$ typically $\mu$ se kitna door jaata hai. Yahi ek fact poore interval ko possible banata hai.

PICTURE. Ek lopsided population se shuru karo (left). Size $n$ ke samples average karo. Un averages ki distribution (right) ek clean symmetric bell hai jo $\mu$ par center hai.

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Step 4 — Universal ruler par rescale karo: $Z$ score

KYA. Har bell curve ek master bell — standard normal, center $0$, width $1$ — ki stretched/shifted copy hoti hai. Hum uspar jaate hain center subtract karke aur width se divide karke: $Z=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}.$ Term by term: $\bar{X}-\mu$ hai humara estimate truth se kitna door landa; $\sigma /\sqrt{n}$ se divide karna us doori ko standard-error units mein measure karta hai. Toh $Z$ ka jawab hai: "main $\mu$ se kitne SE's door landa?"

KYUN. Alag-alag problems mein alag-alag $\mu$, $\sigma$, $n$ hote hain. $Z$ mein convert karna yeh sab erase kar deta hai — ab ek fixed picture har problem ke liye kaam karti hai.

PICTURE. Raw $\bar{X}$ bell (blue, messy units) standard $Z$ bell (green, centered at 0) par map hoti hai. Wahi point "center se itne steps" ban jaata hai.

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Step 5 — Central 95% karo carve ($z_{\alpha /2}$)

KYA. Ek confidence level $1-\alpha$ choose karo. 95% ke liye, $\alpha =0.05$. Hum woh number $z_{\alpha /2}$ dhundhte hain jisse central band $-z_{\alpha /2}$ se $+z_{\alpha /2}$ tak probability $1-\alpha$ hold kare: $P\left(-z_{\alpha /2}\le Z\le z_{\alpha /2}\right)=1-\alpha .$

$\alpha$ ko half mein kyun split karo? Hum extreme 5% total ko exclude karna chahte hain, lekin bell ke do tails hain. Har tail mein $\alpha /2=2.5\%$ chodne se middle $95\%$ rehta hai. Poora $5\%$ ek tail mein use karna ek one-sided bound hoga — ek alag sawaal.

PICTURE. Green bell ek shaded central band ($95\%$) ke saath, aur do chhote red tails ($2.5\%$ each). Band ke edges $\pm z_{\alpha /2}=\pm 1.96$ label hain.

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Step 6 — $\mu$ ko trap karne ke liye inequality flip karo

KYA. Hum ab probability statement ke andar plain algebra karte hain, ek line at a time. $Z=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}$ substitute karke shuru karo: $-z_{\alpha /2}\le \frac{\bar{X}-\mu }{\sigma /\sqrt{n}}\le z_{\alpha /2}.$ Line 1 — har part ko $\sigma /\sqrt{n}$ se multiply karo. Yeh quantity positive hai (ek positive root par standard deviation), toh isse multiply karne par har $\le$ same direction mein rehta hai: $-z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\le \bar{X}-\mu \le z_{\alpha /2}\frac{\sigma }{\sqrt{n}}.$ Line 2 — har part se $\bar{X}$ subtract karo. Subtraction teeno pieces ko equally shift karti hai; directions unchanged: $-\bar{X}-z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\le -\mu \le -\bar{X}+z_{\alpha /2}\frac{\sigma }{\sqrt{n}}.$ Line 3 — har part ko $-1$ se multiply karo. Ek negative number se multiply karne par inequality ki direction reverse ho jaati hai (agar $2\le 4$, toh $-2\ge -4$: chhota number bada ban jaata hai jab signs flip hote hain). Toh dono $\le$ ban jaate hain $\ge$, aur left/right ends roles swap karte hain: $\bar{X}+z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\ge \mu \ge \bar{X}-z_{\alpha /2}\frac{\sigma }{\sqrt{n}}.$ Line 4 — left-to-right padho (usi chain ko smallest-first mein rewrite karo): $\boxed{\bar{X}-z_{\alpha /2}\frac{\sigma }{\sqrt{n}}\le \mu \le \bar{X}+z_{\alpha /2}\frac{\sigma }{\sqrt{n}}}$

KYUN. Step 5 ek statement tha ki $\bar{X}$ relative to $\mu$ kahan girta hai. Lekin $\mu$ woh hai jo hum jaante nahi! $\mu$ ko middle mein isolate karna usi fact ko hamare estimate ke around ek band mein badal deta hai — half-width $E=z_{\alpha /2}\frac{\sigma }{\sqrt{n}}$ margin of error hai.

PICTURE. Number line par: $\bar{X}$ center mein baitha hai; length $E$ ke arrows left aur right tak jaate hain; shaded band interval hai. True $\mu$ (yellow) andar baitha hai — is baar.

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Step 7 — "95% confidence" kaisa dikhta hai (crucial subtlety)

KYA. Interval random hai kyunki $\bar{X}$ random hai. Kuch intervals aise lagte hain ki $\mu$ cover hota hai; kuch (lagbhag 5%) miss karte hain.

KYUN yahi poora point hai. "95% confidence" ek statement hai procedure ke baare mein over many samples, ek computed interval ke baare mein nahi. Jab aapke paas $[501.02,502.98]$ aa jaata hai, woh fixed range ya toh fixed $\mu$ contain karti hai ya nahi — probability $0$ ya $1$.

PICTURE. Bees intervals vertically stacked hain, har ek apne sample se. Yellow vertical line $\mu$ hai. Green intervals use cross karte hain (hit); red wale miss karte hain. Lagbhag 1 in 20 red hain.

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Step 8 — Degenerate & edge cases

KYA & KYUN, case by case:

• $\sigma$ unknown, small $n$ → $t$ use karo. Hum $\sigma$ ko abhi defined sample SD $s$ se replace karte hain. Kyunki $s$ khud ek jittery guess hai (yeh chhote $n$ ke liye $\sigma$ underestimate karta hai), true statistic ke fatter tails hote hain — Student's t-distribution $n-1$ degrees of freedom ke saath. Phir $\bar{X}\pm t_{\alpha /2,n-1}s/\sqrt{n}$. Scope caveat: $t$-interval exact coverage tab deta hai jab underlying population khud (kam se kam approximately) normally distributed ho — woh normality woh assumption hai jo $T=(\bar{X}-\mu )/(s/\sqrt{n})$ ko exact $t_{n-1}$ law follow karaati hai. Heavily non-normal populations ke liye, dono $t$-interval aur uski coverage sirf approximate hain aur large $n$ (CLT ke zariye) par depend karti hain. $n\to \infty$ par, $t\to$ normal, toh dono methods agree karte hain.
• $n\to \infty$. SE $=\sigma /\sqrt{n}\to 0$: band $\bar{X}$ par collapse ho jaata hai, jo khud $\mu$ par home in karta hai. Limit mein certainty.
• $n=1$. SE $=\sigma$: koi averaging nahi hua, toh interval ek raw measurement jitna wide hai — formula abhi bhi bolta hai, bas uselessly wide (aur $s$ yahan undefined hai, kyunki $n-1=0$).
• Proportion (0/1 data). Ek success/failure trial (ek Bernoulli variable) ka mean $p$ aur variance $p(1-p)$ hota hai, toh $\hat{p}$ sirf 0's aur 1's ka ek sample mean hai. Upar sab kuch transfer hota hai: $\hat{p}\pm z_{\alpha /2}\sqrt{\hat{p}(1-\hat{p})/n}$, valid jab $n\hat{p}\ge 5$ aur $n(1-\hat{p})\ge 5$ (bell appear hone ke liye dono taraf se kaafi).

PICTURE. Left: $t$ bell (red) normal (green) ke upar fatter-tailed baitha hai — wider critical marks. Right: $t$ band $n$ badhne par $z$ band ke saath converge ho raha hai.

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Ek-picture summary

Sab kuch ek canvas par: population → sample → $\bar{X}$ ki bell $\mu$ par centered width $\sigma /\sqrt{n}$ ke saath → $\pm z_{\alpha /2}$ se carved central $95\%$ → ek band mein flip jo $\bar{X}$ ke around $\mu$ ko trap kare.

Recall Feynman retelling — poori walkthrough simple words mein

Tum ek bade soup pot ki true saltiness $\mu$ guess karne ke liye kuch chamche taste karte ho, pehle hilaake taki har chamcha ek independent, fair sample ho. Chamchon ki ek handful ka average $\bar{X}$ deta hai — kareeb lekin exact nahi (Step 1). Agar kai dost apne-apne chamche taste karein, unke saare averages true saltiness ke center par ek hill mein pile ho jaate hain, aur — kyunki chamche independent hain — hill utni skinnier hogi jitne zyada chamche har ek taste kare, woh skinniness $\sigma /\sqrt{n}$ hoti hai (Step 2). Kaafi badi handful ke liye hill bell-shaped ho jaati hai (Step 3 — sirf approximately, aur sirf jab handful kaafi badi ho). Hum "main kitna door hoon" ko us skinniness ki units mein measure karte hain, isse $Z$ kehte hain (Step 4). Master bell par hum beech ka 95% mark karte hain, har side se 2.5% katke — edge marks $\pm 1.96$ hain (Step 5). "$\bar{X}$ $\mu$ ke 1.96 skinny-units ke andar hai" ko "$\mu$ $\bar{X}$ ke 1.96 skinny-units ke andar hai" mein rearrange karna — $-1$ se multiply karte waqt inequality ko carefully flip karte hue — hamare guess ke around ek band deta hai (Step 6). Bees doston ke bands draw karo: lagbhag unnees true line cross karte hain, ek miss karta hai — yahi hai 95% confidence ka matlab, method ki ek property, tumhare ek band ki nahi (Step 7). Agar spread nahi pata aur thode chamche liye, toh apne hi chamchon ki spread $s$ se measure karo aur fatter-tailed $t$ ruler use karo (trust karo ki pot ki saltiness roughly bell-shaped hai); yes/no sawaalon ke liye bilkul wahi recipe $\hat{p}$ use karti hai (Step 8).

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Connections

• Central Limit Theorem — kyun averages ki hill ek bell hai (Step 3).
• Standard Error — shrinking width $\sigma /\sqrt{n}$ (Step 2).
• Student's t-distribution — fatter-tailed ruler jab $\sigma$ estimate kiya gaya ho (Step 8).
• Hypothesis Testing — ${\mu }_0$ CI ke bahar ⇔ level $\alpha$ par reject karo.
• Bernoulli & Binomial Distributions — proportions ke peeche 0/1 variance $p(1-p)$.
• Sample Size Determination — $n$ ke liye solve karne ke liye $E=z_{\alpha /2}\sigma /\sqrt{n}$ invert karo.