Confidence intervals: Question bank — Hinglish deep dive

1.3.21 · D5 · AI-ML › Probability & Statistics › Confidence intervals

In questions mein jo vocabulary hai uski reminder:

$μ$ — true population mean, ek fixed unknown number (kabhi random nahi).
$\overset{x}{ˉ}$ — sample mean, data se compute kiya hua ek number (samples ke across random).
CI — ek interval $\overset{x}{ˉ} \pm (critical value) \cdot \frac{spread}{n}$ .
coverage — repeated intervals ka fraction jo $μ$ ko trap karta.

True ya false — justify karo

Is topic ki sabse badi confusion "probability" word mein chupi hai. Ye items usi ko drill karte hain.

A 95% CI ka matlab hai ki is specific computed interval ke andar

μ

hone ki 95% probability hai

False. Jab numbers plug in ho jaate hain, to interval endpoints aur

μ

dono fixed constants hain, isliye

μ

ya to andar hai ya bahar — 95% procedure ki long-run coverage describe karta hai, is ek interval ki nahi.

"95% confident" aur "kaafi saare repeats mein mere 95% intervals

μ

ko capture karte hain" ek hi baat hai

True. Confidence level exactly method ki long-run coverage hai; "95% confident" shorthand hai "yeh net-throwing procedure

μ

ko 95% time pakadti hai" ke liye.

Agar main same data par 95% CI ko 99% CI mein widen karta hoon, to yeh wider ho jaata hai

True. Higher confidence ko larger critical value chahiye (

2.576

1.96

), aur margin of error us value ke saath scale karta hai, isliye

μ

ko trap karne ke liye interval grow karna hi padega.

4× zyada data collect karne se margin of error roughly half ho jaata hai

True. Margin

1/ n

ke saath scale karta hai, aur

4 = 2

, isliye

n

ko quadruple karne se width half ho jaati hai — diminishing returns square root mein built-in hain.

Wider confidence interval ka matlab hamesha yeh hai ki aapne kuch galat kiya

False. Ek wide interval honestly high uncertainty report karta hai (chhota

n

ya bada spread); asli galti ek tiny sample se "false precision" lena hai, na ki width.

Same data aur confidence level ke liye, t-interval hamesha z-interval se kam se kam utna hi wide hota hai

True. Kyunki

s

khud random hai, t-distribution ke tails heavier hote hain, isliye

t_{α /2, n - 1} \geq z_{α /2}

hamesha, jo equal-or-wider interval deta hai.

Sample size badhne ke saath, t-interval aur z-interval converge karte hain

True. Jab

n \to \infty

s \to σ

aur

t_{n - 1} \to N (0, 1)

, to dono critical values (aur intervals) indistinguishable ho jaate hain — yahi reason hai ki

n \geq 30

par aap safely

z

use kar sakte ho.

Agar do models ke 95% CIs overlap karte hain, to models statistically indistinguishable hain

False. Overlapping CIs suggestive hain lekin decisive nahi; sahi test difference ke liye CI build karta hai — woh 0 ko exclude kar sakta hai (significant) even jab individual intervals overlap karein.

Confidence interval batata hai ki 95% individual data points kahan hain

False. CI mean

μ

ke baare mein hai, jiska spread

σ / n

hai; individual points ke liye prediction/tolerance interval

σ

itself use karta hai aur bahut wider hota hai.

Error dhundho

Har item mein ek real reasoning slip hai jo students actually likhte hain. Batao kya toot raha hai.

n = 8

σ

unknown, isliye maine

z = 1.96

use kiya kyunki yeh standard 95% value hai."

Error:

σ

unknown aur small

n

ke saath, aapko

t_{0.025, 7} = 2.365

use karna chahiye;

z

use karne se margin understate hoti hai aur ek interval banta hai jo bahut narrow hai (aur

μ

ko undercover karta hai).

"Mera 95% CI proportion ke liye low end par negative aaya, to maine report kiya accuracy

- 3%

ho sakti hai."

Error: proportions

[0, 1]

mein bounded hain; normal approximation boundary ke paas break down ho gayi.

[0, \cdot]

par clip karo ya boundary-aware method use karo jaise Wilson ya bootstrap.

"Mujhe achha result mila, isliye maine CI ko resplit data par kai baar rebuild kiya aur sabse narrow wala report kiya."

Error: bahut saare intervals mein se tightest ko cherry-pick karna coverage guarantee destroy kar deta hai — 95% sirf ek single, pre-committed procedure ke liye hold karta hai, attempts par minimum ke liye nahi.

"95% CI

[0.80, 0.94]

hai, isliye 95% future test sets is range mein score karengi."

Error: yeh mean accuracy ke liye CI ko future samples ke prediction interval ke saath confuse karta hai; future single-run scores mean se bahut zyada vary karte hain.

s

bahut huge aaya ek outlier ki wajah se, lekin maine ise formula mein plug in kar diya."

Error: CI assume karta hai ki koi gross outliers nahi hain; ek single point

s

inflate kar sakta hai aur silently near-normal assumption invalidate kar sakta hai. Outlier investigate karo ya robust/bootstrap method use karo.

"Maine 95% CI use kiya, paya ki usne baseline ko exclude kiya, aur conclude kiya ki 5% chance hai ki effect fake hai."

Error: confidence level is result ke baare mein per-conclusion error probability nahi hai; "5%" procedure ki long-run false-coverage refer karta hai, not the posterior probability ki effect null hai.

"Mera test set wahi 100 images hain jinpar maine hyperparameters tune kiye, lekin CI phir bhi apply hoti hai."

Error: CI independent, representative data assume karta hai; tuning data reuse karna "one honest random sample" premise break karta hai (dekho Cross-validation), isliye interval optimistically biased hai.

Why questions

Yeh mechanism ke baare mein poochte hain, fact ke baare mein nahi.

Hum tail probability ko ek side par rakhne ki bajaye

α /2

dono sides par kyun split karte hain?

Kyunki hum ek two-sided interval chahte hain jo

μ

ko too high OR too low hone se miss kar sake; equally split karna interval ko symmetric banata hai aur symmetric distribution ke liye shortest interval deta hai.

Standard error

σ

itself nahi balki

σ / n

kyun use karta hai?

CI sample mean ke spread ko describe karta hai, aur

n

independent values ko average karna variance ko

n

ke factor se shrink karta hai (Central Limit Theorem ke zariye), isliye

\overset{x}{ˉ}

ka standard deviation

σ / n

hai — dekho Standard Error.

Sample variance

s^{2}

compute karte waqt

n

se nahi balki

n - 1

se kyun divide karte hain?

Bessel's correction: humne pehle se ek degree of freedom

\overset{x}{ˉ}

μ

estimate karne mein spend kar diya, isliye

n - 1

se divide karna

s^{2}

σ^{2}

ka unbiased estimate banata hai na ki systematically-too-small wala.

T-distribution ke tails normal se heavier kyun hote hain?

Kyunki yeh

σ

s

se estimate karne ki extra randomness account karta hai; woh added uncertainty extreme values ko zyada likely banati hai, tails ko fatten karti hai taaki hum apna confidence overstate na karein.

CLT hume CI build karne deta hai even jab raw data normal nahi hoti, kyun?

CLT kehta hai sample mean

\overset{x}{ˉ}

large

n

ke liye approximately normal ho jaata hai regardless of population shape, aur CI

\overset{x}{ˉ}

ke baare mein hai — isliye hume sirf mean ki distribution ko normal hona chahiye, data ki nahi.

Do accuracies ke difference par CI rakhna do overlapping CIs ko eyeball karne se zyada informative kyun hai?

Kyunki difference ka apna (smaller-combined) standard error

S E_{A}^{2} + S E_{B}^{2}

hota hai, aur yeh poochna ki kya woh interval 0 ko exclude karta hai mathematically correct significance question hai — yahi logic hai A/B Testing aur Hypothesis Testing ke peeche.

Bayesian credible interval ko "

μ

ke andar hone ki 95% probability" wali meaning kyun milti hai jo frequentist CI ko nahi milti?

Ek credible interval

μ

ko ek random variable treat karta hai posterior distribution ke saath, isliye woh literally woh probability carry kar sakta hai; frequentist CI

μ

ko fixed treat karta hai, aur woh phrasing forbid karta hai.

Edge cases

Har method ki boundaries hoti hain. Yeh formulas ko tab tak push karte hain jab tak woh crack na karein.

n \to \infty

ke saath CI ka kya hota hai?

Margin

z_{α /2} \cdot σ / n \to 0

, isliye interval true

μ

par collapse ho jaata hai — infinite data mean ko exactly pin kar deta hai (halanki yeh kabhi individual spread ke baare mein nahi batata).

Jab

\overset{p}{^} = 0

ho (

n

mein se 0 correct) to proportion CI ka kya hota hai?

Standard error

\overset{p}{^} (1 - \overset{p}{^}) / n = 0

, jo ek zero-width interval

[0, 0]

deta hai — clearly absurd, isliye boundary par aapko Wilson ya bootstrap intervals par switch karna hoga jo degenerate nahi hote.

Agar aap

n = 1

ke saath t-interval compute karein to kya hoga?

Aap nahi kar sakte:

df = n - 1 = 0

s

estimate karne ke liye koi degrees of freedom nahi bachte (

n - 1 = 0

se division), isliye ek one-point sample spread ke baare mein koi information nahi deta aur koi interval exist nahi karta.

Agar population perfectly constant ho (

σ = 0

), to CI kya hai?

Margin

z_{α /2} \cdot 0/ n = 0

, isliye interval single point

\overset{x}{ˉ} = μ

hai — zero variability ke saath, ek sample mean exactly reveal kar deta hai.

Fixed

n

ke liye, confidence

\to 100%

hone par margin of error ka kya hota hai?

Critical value

\to \infty

, isliye margin infinity par explode ho jaata hai —

μ

ko trap karne ka certain hona useless interval

(- \infty, \infty)

require karta hai.

Normal-approximation proportion CI 0 ya 1 ke paas kab unreliable hoti hai?

Jab

n \overset{p}{^}

n (1 - \overset{p}{^})

small ho (roughly

< 5

–

10

); binomial symmetric normal approximation ke liye bahut skewed hai, aur interval

[0, 1]

ke bahar spill kar sakta hai — exact ya Wilson methods use karo.

Agar aapka data heavily skewed hai aur

n

moderate hai, to kya standard CI trustworthy hai?

Poori tarah nahi: CLT ko skewed data ke mean ko normalize karne ke liye larger

n

chahiye, isliye skew ke liye ya to zyada data collect karo ya bootstrap CI prefer karo jo koi normality assumption nahi banata.

Recall Fast self-test

Interval random hai, $μ$ fixed hai :::: Batao "95%" actually kiske randomness ko describe karta hai. Small $n$ , unknown $σ$ :::: Woh exact condition batao jo z ki jagah t-distribution force karta hai. Difference ke liye CI jo 0 ko exclude kare :::: Do models ki accuracies compare karne ka sahi tarika batao.