Worked examples — Hypothesis testing — null - alternative, test statistic, p-value, errors (Type I & II)

4.9.20 · D3 · Maths › Probability Theory & Statistics › Hypothesis testing — null - alternative, test statistic, p-v

Yeh page ek shooting range hai: is topic mein jo bhi tarah ke hypothesis-test questions aa sakte hain, ek ke baad ek, poori tarah se solve kiye gaye hain. Hum pehle saare cases ka map banate hain, phir har cell ko picture-first, guess-first method se walk karte hain. Yahan sab kuch parent Hypothesis testing — null - alternative, test statistic, p-value, errors (Type I & II) pe build hota hai aur Central Limit Theorem, Standard Error, Normal Distribution aur Student t-distribution pe rely karta hai.

Koi bhi symbol aane se pehle, yahan plain-words dictionary hai jo hum baar baar use karte rahenge, taaki kuch bhi "assumed" na rahe:

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The scenario matrix

Is topic ka har question is table ki ek row hai. Neeche ke examples har cell ko tick off karte hain.

#	Case class	Kya badalta hai	Example jo isko hit karta hai
A	Two-tailed, known $\sigma$ → z	direction $\ne$	Ex 1
B	Right-tailed, known $\sigma$ → z	direction $>$, statistic positive	Ex 2
C	Left-tailed, unknown $\sigma$ → t	direction $<$, statistic negative	Ex 3
D	Boundary / degenerate: $p=\alpha$ exactly, aur $\bar{X}={\mu }_0$ toh $z=0$	zero input	Ex 4
E	Non-rejection sahi se padha gaya (evidence ki absence)	large p-value	Ex 5
F	Limiting behaviour: same effect, growing $n$ → power badhti hai	$n\to \infty$	Ex 6
G	Type-II / power computation	$\beta$ compute karo	Ex 7
H	Real-world word problem (proportions)	alag data type	Ex 8
I	Exam twist: woh p-value dete hain aur ek trap question poochte hain	conceptual	Ex 9

Test statistic ke teen signs — positive, negative, zero — B, C, D se cover hote hain. Teen tail directions — two-tailed, right, left — A, B, C hain. Acchi coverage ka matlab hai: koi bhi student us case se na takraye jo humne skip kiya ho.

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Ex 1 — Case A: two-tailed z-test (dono signs matter karte hain)

Forecast: pehle andaza lagao — 64 bottles mein $3$ ml neeche jaana surprising hai, ya normal wobble? Aage padhne se pehle yes/no likhlo.

Step 1. $H_0:\mu =500$, $H_1:\mu \ne 500$ likhte hain. Yeh step kyun? "Miscalibrated" direction ke baare mein kuch nahi kehta — zyada full hona utna hi bura hai jitna zyada khaali hona — toh dono tails count hote hain. Yahi hai $\ne$ ka matlab.

Step 2. Standard error $\text{SE}=\frac{\sigma }{\sqrt{n}}=\frac{8}{\sqrt{64}}=\frac{8}{8}=1$ ml. Yeh step kyun? Sample mean ek akele bottle se kam wobble karta hai kyunki averaging noise ko cancel karti hai; Central Limit Theorem kehta hai woh wobble $\sigma /\sqrt{n}$ ke barabar hai. Dekho Standard Error.

Step 3. $z=\frac{\bar{X}-{\mu }_0}{\text{SE}}=\frac{497-500}{1}=-3$. Yeh step kyun? Hum "3 ml neeche" ko "3 wobbles neeche" mein convert karte hain, jo ek unit-free count hai. Figure s01 dekho: observed mean centre se 3 tick-marks baayein hai.

Step 4. Two-tailed p-value $=2P(Z\ge |-3|)=2P(Z\ge 3)=2(0.001350)=0.0027$. Yeh step kyun? $\ne$ ke liye "as extreme or more" ka matlab hai ya toh far-left ya far-right, isliye hum ek tail ka area lete hain aur double karte hain (Normal Distribution symmetric hai).

Step 5. $0.0027<0.05\Rightarrow$ reject $H_0$. Machine off hai.

Verify: SE ki units ml hain ($8\text{ml}/8$), $z$ unitless hai (ml/ml) ✓. $-3$ ka $z$ famously rare hai ("0.13%" tail), aur double karne se $0.27\%$ milta hai — intuition se match karta hai ki 3-wobble miss ek red flag hai.

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Ex 2 — Case B: right-tailed z-test (positive statistic)

Forecast: 100 logon mein 2 minutes zyada. Reject ya nahi? Andaza lagao.

Step 1. $H_0:\mu =30$, $H_1:\mu >30$. Yeh step kyun? Shak ki ek direction hai ("zyada") → sirf right tail. Hum left tail ko bilkul ignore karte hain.

Step 2. $\text{SE}=\frac{10}{\sqrt{100}}=\frac{10}{10}=1$ min.

Step 3. $z=\frac{32-30}{1}=+2$. Yeh step kyun? Positive isliye kyunki $\bar{X}$ claim ke upar hai — $z$ ka sign literally record karta hai "kaun sa side."

Step 4. Right-tailed p-value $=P(Z\ge 2)=0.0228$. Yeh step kyun? Sirf right tail "as extreme or more" hai jab $H_1$ right point karta hai. Koi doubling nahi yahan (Ex 1 se contrast karo).

Step 5. $0.0228<0.05\Rightarrow$ reject $H_0$: evidence hai ki commutes zyada lambi hain.

Verify: Wahi $|z|=2$ ek two-tailed p of $0.0456$ deta — still significant, lekin bada, dikhata hai ki one-tailed test pre-chosen direction mein zyada powerful hota hai. Units: min/min = unitless ✓.

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Ex 3 — Case C: left-tailed t-test (negative statistic, σ unknown)

Forecast: chhota sample, 1.5 kg kam. Reject karein?

Step 1. $H_0:\mu =50$, $H_1:\mu <50$ (left tail — "underweight" left point karta hai).

Step 2. $t$ kyun aur $z$ kyun nahi? Hum true spread $\sigma$ nahi jaante; hamare paas sirf sample estimate $s$ hai. Spread estimate karne se extra uncertainty add hoti hai, isliye hum fatter-tailed Student t-distribution use karte hain jisme $df=n-1=8$ hai.

Step 3. $\text{SE}=\frac{s}{\sqrt{n}}=\frac{2}{\sqrt{9}}=\frac{2}{3}\approx 0.6667$ kg.

Step 4. $t=\frac{48.5-50}{0.6667}=\frac{-1.5}{0.6667}=-2.25$. Yeh step kyun? Negative = claim se neeche, left-tail suspicion se match karta hai.

Step 5. Critical value $t_{0.05,8}\approx 1.860$ (right side); left tail ke liye hum $-1.860$ se compare karte hain. Kyunki $-2.25<-1.860$, hum cutoff se zyada baahir hain $\Rightarrow$ $H_0$ reject karo.

Verify: $df=8$ ke saath, $t=-2.25$ ki one-sided p-value lagbhag $0.0273<0.05$ hai ✓ — critical-value decision se agree karta hai. Do tarike, ek hi verdict.

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Ex 4 — Case D: the degenerate boundary ($z=0$ aur $p=\alpha$)

Forecast: test ko kya kehna chahiye jab aapka data bilkul claim pe land kare?

Step 1 (a). $z=\frac{500-500}{1}=0$. Yeh step kyun? Zero distance → zero wobbles. Data sabse kam surprising possible hai.

Step 2 (a). Two-tailed $p=2P(Z\ge 0)=2(0.5)=1$. Yeh step kyun? Har possible dataset exact centre se "as extreme or more" hai, isliye poora probability mass qualify karta hai → $p=1$. Hum fail to reject karte hain — jaisa hona chahiye.

Step 3 (b). $\alpha =0.05$ two-tailed ke liye hume chahiye $2P(Z\ge |z|)=0.05$, yaani $P(Z\ge |z|)=0.025$, jo famous $|z|=1.96$ deta hai. Yeh step kyun? Yeh razor's edge hai. Convention hai: exactly $p=\alpha$ pe hum reject karte hain (rule hai $p\le \alpha$). Toh $|z|=1.96$ sabse chhota $|z|$ hai jo phir bhi reject karta hai.

Verify: (a) $2\times 0.5=1$ ✓. (b) $P(Z\ge 1.96)\approx 0.025$ toh doubled $=0.05=\alpha$ ✓ — critical value $1.96$ exactly recover hota hai.

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Ex 5 — Case E: ek sahi non-rejection (evidence ki absence)

Forecast: kya 1 point upar prove karta hai ki method kaam karti hai?

Step 1. $H_0:\mu =70$, $H_1:\mu >70$ (right tail).

Step 2. $\text{SE}=\frac{6}{\sqrt{16}}=\frac{6}{4}=1.5$.

Step 3. $t=\frac{71-70}{1.5}=\frac{1}{1.5}=0.6667$, $df=15$.

Step 4. Critical $t_{0.05,15}\approx 1.753$. Kyunki $0.6667<1.753\Rightarrow$ fail to reject $H_0$.

Step 5 — woh wording jo marks dilati hai. Humne method ko useless prove nahi kiya. Sirf 16 students pe 1-point rise luck ki reach ke andar hai; ek bada $n$ shayad phir bhi ek real effect reveal kare. Yeh evidence ki absence hai, absence ki evidence nahi.

Verify: $t=0.667$, $df=15$ ke liye one-sided p-value lagbhag $0.258$ hai — bahut bada, $0.05$ ke paas kahin nahi ✓. "$H_0$ pe believe karte raho" se consistent hai.

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Ex 6 — Case F: limiting behaviour, $n$ ko infinity tak badhao

Forecast: jab data pile up hota hai toh kya ek fixed 2-minute gap zyada ya kam significant ho jaata hai?

Step 1. $z=\frac{2}{10/\sqrt{n}}=\frac{2\sqrt{n}}{10}=0.2\sqrt{n}$. Yeh step kyun? Numerator fix karo; sirf SE shrink hoti hai. $z$ ko $n$ ke function ke roop mein likhne se trend samajh aata hai.

Step 2. Plug in: $z(25)=0.2\cdot 5=1.0,z(100)=0.2\cdot 10=2.0,z(400)=0.2\cdot 20=\mathrm{4.0.}$

Step 3 — the limit. Jab $n\to \infty$, $z=0.2\sqrt{n}\to \infty$, toh $p\to 0$. Yeh step kyun? Kisi bhi size ka real gap detectable ho jaata hai jab aap enough data gather karo — SE $\to 0$ null curve ko ek spike mein narrow kar deti hai. Figure s03 dekho: null bell $n$ badhne ke saath tight hoti jaati hai.

Verify: $z(25)=1<1.96$ (not significant, two-tailed), $z(100)=2>1.96$ (significant), $z(400)=4$ (wildly significant) ✓. Confirm karta hai: ek fixed effect ke liye statistical significance sample size se kharidi jaati hai — dekho Statistical Power & Sample Size.

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Ex 7 — Case G: Type II error $\beta$ aur power compute karna

Forecast: agar true mean 6 upar hai, toh hum isko miss karne ki kitni sambhavna hai?

Step 1 — reject region real units mein. $\text{SE}=\frac{15}{\sqrt{25}}=3$. Hum reject karte hain jab $z\ge 1.645$, yaani jab $\bar{X}\ge {\mu }_0+1.645\cdot \text{SE}=100+1.645(3)=104.935$. Yeh step kyun? $\beta$ real-unit cutoff ke baare mein hai, isliye hum $z$-cutoff ko $\bar{X}$ threshold mein translate karte hain. Yahan $1.645$ one-tailed 5% critical z hai.

Step 2 — truth ke neeche evaluate karo. Agar sach mein ${\mu }_1=106$ hai, toh mean $\bar{X}$, $106$ pe centred Normal hai same $\text{SE}=3$ ke saath. Ek miss (Type II) tab hota hai jab $\bar{X}<104.935$: $\beta =P\left(Z<\frac{104.935-106}{3}\right)=P(Z<-0.355)\approx \mathrm{0.361.}$ Yeh step kyun? $\beta =P(\text{fail to reject}\mid H_0\text{ false})$ — true distribution ke neeche measure hota hai, null ke neeche nahi. Figure s04 dekho: amber "true" bell accept region ke saath overlap karti hai.

Step 3. Power $=1-\beta \approx 1-0.361=0.639$.

Verify: $z$-cutoff back-check: $(104.935-100)/3=1.645$ ✓. Miss z: $(104.935-106)/3=-0.355$, aur $P(Z<-0.355)\approx 0.361$ ✓. Power $\approx 64\%$ — modest, jaisi 2-SE-sized true effect ke liye expect ki jaati hai. Zyada $n$ ya effect isko raise karega (Statistical Power & Sample Size, Confidence Intervals).

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Ex 8 — Case H: real-world word problem (ek proportion, coin)

Forecast: 100 mein se 62 heads — fluke ya bias?

Step 1. $H_0:p=0.5$, $H_1:p>0.5$.

Step 2 — proportion ke liye SE. $H_0$ ke neeche, har flip ka variance $p_0(1-p_0)$ hai, isliye $\text{SE}=\sqrt{\frac{p_0(1-p_0)}{n}}=\sqrt{\frac{0.5\cdot 0.5}{100}}=\sqrt{0.0025}=0.05$. Yeh step kyun? Hum SE null value $p_0=0.5$ use karke compute karte hain, kyunki poora test "agar $H_0$ sach hoti" wali duniya mein rehta hai.

Step 3. $z=\frac{\hat{p}-p_0}{\text{SE}}=\frac{0.62-0.5}{0.05}=\frac{0.12}{0.05}=2.4$.

Step 4. $p=P(Z\ge 2.4)\approx 0.0082<0.05\Rightarrow$ $H_0$ reject karo: heads ki taraf bias ka evidence hai.

Verify: SE $=0.05$, $z=2.4$, tail $P(Z\ge 2.4)\approx 0.0082$ ✓. Central Limit Theorem $\hat{p}$ ko approximately Normal treat karna justify karta hai kyunki $n{p}_0=50$ aur $n(1-p_0)=50$ dono bade hain.

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Ex 9 — Case I: exam twist (woh p-value dete hain)

Forecast: kya $p=0.03$ null ke sach hone ki probability measure karta hai? Padhne se pehle decide karo.

Step 1 — trap ka naam lo. Yeh claim $P(\text{data}\mid H_0)$ ko $P(H_0\mid \text{data})$ se confuse karti hai. Yeh prosecutor's fallacy hai. Yeh step kyun? $p$ $H_0$ sach maanke compute kiya jaata hai — aap phir isko $H_0$ ki apni probability ke roop mein nahi padh sakte; uske liye ek prior chahiye hoga, jo Bayesian Inference mein rehta hai.

Step 2 — isko sahi se kaho. "$p=0.03$" ka matlab hai: agar $H_0$ sach hoti, toh itna extreme ya zyada extreme data sirf 3% time hota. Yeh directly $P(H_0)$ ke baare mein kuch nahi kehta.

Step 3 — sahi decision. Kyunki $0.03\le 0.05$, hum $H_0$ reject karte hain. Sahi verdict, lekin classmate ki reasoning galat hai.

Verify: Sirf logic check — numeric comparison $0.03\le 0.05$ sach hai, isliye "reject" sahi action hai; interpretation mein error hai. ✓

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Recall Self-test: yeh kaun sa cell hai?

"Testing whether a mean is different (either way) with known $\sigma$." ::: Case A — two-tailed z (Ex 1). "Data lands exactly on ${\mu }_0$; two-tailed p-value kya hai?" ::: Case D — $p=1$, fail to reject (Ex 4). "Same effect size, ever-larger $n$: $p$ ka kya hota hai?" ::: Case F — $p\to 0$; ek fixed effect hamesha significant ho jaata hai (Ex 6). "True mean given ho, toh usse miss karne ka chance find karo." ::: Case G — $\beta$ compute karo, power $=1-\beta$ (Ex 7). "Koi kehta hai p, $H_0$ ke sach hone ki probability hai." ::: Case I — prosecutor's fallacy; $p=P(\text{data}\mid H_0)$ (Ex 9).