Confidence intervals — derivation for mean, proportion: Question bank

4.9.19 · D5 · Maths › Probability Theory & Statistics › Confidence intervals — derivation for mean, proportion

Shuru karne se pehle, teen plain-word reminders taaki koi bhi symbol unearned na lage:

Dhyan rakho kaun si cheezein fixed hain ( $μ$ ) aur kaun si wobble karti hain ( $\overset{ˉ}{X}$ , band). Neeche har trap fixed-vs-random confusion ka disguise hai.

True or false — justify

The probability that the true mean lies inside the computed interval

[501.02, 502.98]

is 0.95.

False. Ek baar compute ho jaane ke baad, dono endpoints aur

μ

fixed numbers hain, isliye

μ

ya toh andar hai ya bahar — probability 0 ya 1. 95% method ka long-run success rate describe karta hai, is ek band ka nahi.

A 99% confidence interval is more likely to be "correct" for a single given sample than a 95% one.

True, long-run sense mein: zyada confidence level ka matlab hai zyada fraction of intervals

μ

ko capture karte hain. Iska cost yeh hai ki 99% interval wider hota hai (bada critical value use karta hai), isliye woh kam precise hota hai.

Widening the confidence level from 95% to 99% makes the interval narrower because we are "more sure."

False. Zyada confidence ke liye ek bada critical value chahiye (

z_{0.005} = 2.576 > 1.96

), jo band ko wide karta hai. Certainty aur precision ek dusre ke against trade off karte hain.

For the same data, the

t

-interval is always at least as wide as the

z

-interval.

True. Kisi bhi finite

n - 1

degrees of freedom ke liye,

t_{α /2, n - 1} > z_{α /2}

kyunki

t

-distribution ke tails zyada mote hote hain, isliye uska critical value bada hota hai. Woh sirf

n \to \infty

par converge karte hain — dekho Student's t-distribution.

Doubling the sample size

n

halves the margin of error.

False. Margin

1/ n

ki tarah scale karta hai, isliye

n

ko chaar guna karne par woh half hota hai;

n

ko double karne se woh sirf

1/ 2 \approx 0.71

factor se shrink hota hai. Isliye badi precision gains expensive ho jaati hain — dekho Sample Size Determination.

If a 95% CI for

μ

excludes the value 0, then a two-sided hypothesis test of

μ = 0

at level

α = 0.05

rejects.

True. Yeh CI–test duality hai: CI exactly un null values ka set hai jo reject nahi hote, isliye usse bahar ki value reject ho jaati hai — dekho Hypothesis Testing.

The confidence interval tells us the range in which 95% of individual data values fall.

False. Yeh mean

μ

ko bound karta hai, jiska spread standard error

σ / n

hai, raw data spread

σ

nahi. Individual values ka range (ek prediction/tolerance interval) bahut wider hota hai.

For a proportion, the standard error

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

is largest when

\overset{p}{^} = 0.5

True. Function

p (1 - p)

ka peak

p = 0.5

par hota hai (value

0.25

) aur extremes ki taraf drop karta hai, isliye 50/50 split ko pin down karna sabse mushkil hai — yeh worst-case sample-size planning ko drive karta hai.

The Central Limit Theorem guarantees

\overset{ˉ}{X}

is exactly normal for any sample size.

False. Yeh large

n

ke liye ek approximate normal shape deta hai; ek skewed population se small

n

ke liye approximation poor ho sakti hai. Dekho Central Limit Theorem.

Spot the error

"We're 95% confident, so we look up the 95th percentile

z = 1.645

Error: two-sided intervals

α = 0.05

ko do tails mein equally split karte hain,

0.025

each, isliye hume

z_{0.025} = 1.96

chahiye. Value

1.645

ek one-sided 95% bound hai.

"Since we don't know

σ

, we just use

s

in place of it and keep the

z

critical value, even for

n = 12

Error:

σ

ki jagah random

s

use karne se uncertainty badhti hai, isliye small

n

ke liye hume wider

t_{n - 1}

critical value use karni chahiye,

z

nahi. Sirf large

n

ke liye woh roughly agree karte hain.

"Margin of error

= z_{α /2} σ

Error: missing

n

interval ko absurdly wide bana deta hai. Correct margin

z_{α /2} σ / n

hai kyunki hum mean ko bound karte hain, jiska spread standard error hai — dekho Standard Error.

"With 3 successes in 20 trials,

\overset{p}{^} = 0.15

, we apply the Wald interval directly."

Error: validity check

n \overset{p}{^} = 20 (0.15) = 3 < 5

fail ho jaata hai, isliye normal approximation unreliable hai. Itne kam successes ke saath Wald interval 0 se neeche bhi ja sakta hai; uski jagah exact ya adjusted method use karo.

"The

t

-distribution has

n

degrees of freedom."

Error: iske

n - 1

degrees of freedom hain.

s

compute karna constraint

\sum (X_{i} - \overset{ˉ}{X}) = 0

impose karta hai, ek degree of freedom use kar leta hai aur

n - 1

free deviations chodta hai.

"Because our 95% interval is

[65.7, 74.3]

, if we sampled again we'd get a mean in that range 95% of the time."

Error: interval fixed parameter

μ

ke baare mein hai, future sample means ke baare mein nahi. Ek future

\overset{ˉ}{X}

aasaani se bahar ja sakta hai; 95% yeh refer karta hai ki intervals kitni baar

μ

ko capture karte hain.

"A wider interval means our estimate is better."

Error: wider interval ka matlab kam precision hai — humne

μ

ko ek badi range par pin kiya hai. Narrow intervals (bade

n

ya chhote

σ

se) precise hote hain; width aur quality yahan inverted hain.

Why questions

Why do we split

α

into two equal tails instead of putting it all in one?

Kyunki ek symmetric two-sided interval

μ

ko upar aur neeche dono se bracket karta hai, isliye hum har tail mein

α /2

probability reserve karte hain; central band phir

1 - α

hold karta hai. Ek one-tailed split ek one-sided bound deta hai, jo alag sawaal hai.

Why does the variance of

\overset{ˉ}{X}

shrink as

σ^{2} / n

rather than staying

σ^{2}

n

independent values ko average karne se unke random ups and downs partly cancel ho jaate hain; sum ka variance

n σ^{2}

hai, aur

n

se divide karne par variance

1/ n^{2}

se scale hota hai,

σ^{2} / n

bachta hai.

Why must a proportion be treated as a special case of a mean?

Ek 0/1 (Bernoulli) trial ka mean

p

aur variance

p (1 - p)

hota hai, isliye

\overset{p}{^}

literally in indicators ka sample mean hai; poori mean machinery

σ^{2}

ki jagah

p (1 - p)

ke saath transfer ho jaati hai — dekho Bernoulli & Binomial Distributions.

Why do we plug

\overset{p}{^}

into the standard error when the true SE uses the unknown

p

Hum

p (1 - p) / n

compute nahi kar sakte bina

p

jaane, isliye hum apna best estimate

\overset{p}{^}

substitute karte hain. Yeh "Wald" approximation hai aur isliye interval tabhi reliable hai jab successes aur failures dono plentiful hon.

Why does the

t

-interval get closer to the

z

-interval as

n

grows?

Zyada data ke saath,

s

zyada accurately

σ

estimate karta hai, isliye extra uncertainty jo

t

-tails ko mota karti thi woh khatam ho jaati hai aur

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

Why does "95% confidence" refer to the procedure and not to one interval?

Randomness sampling mein rehta hai,

μ

mein nahi. Sampling se pehle, abhi-banana-wala band ka 95% chance hai ki woh fixed

μ

ko pakad le; data observe karne ke baad, band bhi fixed ho jaata hai, isliye koi probability bolne ke liye nahi bachti.

Edge cases

What happens to the CI as

n \to \infty

(with

σ

fixed)?

Standard error

σ / n \to 0

, isliye margin zero ho jaata hai aur interval true

μ

par collapse ho jaata hai — infinite data mean ko exactly pin kar deta hai.

What if

\overset{p}{^} = 0

(zero successes observed)?

Wald SE

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

, jo degenerate interval

[0, 0]

deta hai, jo clearly galat hai —

p

certainly zero nahi hai. Yahaan exactly approximation break karti hai aur ek adjusted (jaise Wilson) interval required hai.

What if the population is already exactly normal but

n = 2

Shape ke liye CLT approximation ki zaroorat nahi (woh already normal hai), lekin

σ

unknown hai, isliye hum phir bhi

t_{n - 1} = t_{1}

use karte hain, jinke tails extremely fat hote hain — interval bahut wide hota hai, honestly reflect karta hai ki do points hume kitna kam batate hain.

What happens to the margin when the confidence level is 100%?

Critical value

z_{α /2}

infinity ki taraf diverge karta hai (koi finite band normal model mein certain nahi ho sakta), isliye 100% interval

(- \infty, \infty)

hai — technically hamesha correct, practically useless.

What if

σ = 0

(no variability in the population)?

Standard error

0

hai, isliye har sample

\overset{ˉ}{X} = μ

exactly deta hai aur interval single point

{μ}

par collapse ho jaata hai — koi randomness nahi hai toh uncertain hone ki koi wajah nahi.

If two researchers each build a 95% CI from independent samples and the intervals don't overlap, is one of them definitely wrong?

Zaruri nahi — non-overlap means differ karne ka strong evidence hai lekin har interval individually ka abhi bhi 5% miss chance tha, aur non-overlap difference ke liye ek conservative (exact nahi) test hai.

Recall wrap-up

Recall One-line self-checks

Is $μ$ random or fixed? ::: Fixed aur unknown; interval random object hai (data dekhne se pehle). Does higher confidence give a narrower or wider interval? ::: Wider — $μ$ ko zyada baar pakadne ke liye ek bada critical value chahiye. When is the proportion standard error largest? ::: $\overset{p}{^} = 0.5$ par, kyunki $p (1 - p)$ wahaan peak karta hai. Why $n - 1$ and not $n$ degrees of freedom? ::: $s$ estimate karna ek constraint $\sum (X_{i} - \overset{ˉ}{X}) = 0$ impose karta hai, ek free deviation remove karta hai. Non-overlapping 95% CIs — proof of difference? ::: Strong lekin definitive evidence nahi; yeh ek conservative check hai, aur har interval ka apna 5% miss rate tha.

Connections

Confidence intervals — derivation for mean, proportion — parent derivation jise yeh traps stress-test karte hain.
Central Limit Theorem — kyun normal shape (aur small skewed $n$ ke liye uski limits).
Standard Error — woh $σ / n$ jise itne saare errors drop kar dete hain.
Student's t-distribution — fatter tails, $n - 1$ df, $z$ -vs- $t$ traps.
Hypothesis Testing — CI–test duality "excludes 0" item ke peeche.
Bernoulli & Binomial Distributions — proportion-as-mean aur $p (1 - p)$ peak.
Sample Size Determination — "doubling" trap ke peeche $1/ n$ scaling.