z-test, t-test, chi-squared goodness of fit, F-test: Worked examples

4.9.21 · D3 · Maths › Probability Theory & Statistics › z-test, t-test, chi-squared goodness of fit, F-test

Yeh page ek drill hai. Parent note ne machinery build ki thi — yahan hum har case class ke through walk karte hain jo yeh charon tests aap par throw kar sakte hain: known vs unknown spread, one-sided vs two-sided, tiny samples, degenerate/limiting inputs, aur exam traps.

Symbols aane se pehle, ek plain-words reminder taaki pehli line samajh aaye:

Definition Teen symbols jo hum neeche baar baar use karenge

$H_{0}$ = null hypothesis, yaani "boring guess" ya status-quo claim jise hum giraaney ki koshish karte hain — jaise "mean exactly $μ_{0}$ hai" ya "die fair hai." Har test luck curve compute karte waqt $H_{0}$ ko true maanta hai, phir poochhta hai ki data abhi bhi fit baitta hai ya nahi. Hum ya toh reject $H_{0}$ karte hain (data luck ke liye bahut weird hai) ya fail to reject $H_{0}$ (luck ek achhi explanation hai). Hum kabhi " $H_{0}$ prove true" nahi karte.
$k$ = chi-squared goodness-of-fit test mein categories (bins) ki sankhya — jaise die ke $6$ faces $k = 6$ dete hain.
Critical-value subscript convention: hum $c_{p, ν}$ likhte hain jahan subscript $p$ right ki taraf tail area hai (far tail mein cut-off area) aur $ν$ degrees of freedom hai. Toh $z_{0.025} = 1.96$ matlab "woh point jiske upar standard Normal ka $2.5%$ hai," aur $t_{0.025, 5} = 2.571$ matlab $ν = 5$ wale $t$ ke liye wahi idea. Two-sided test ke liye level $α$ par hum $α$ ko do tails mein split karte hain aur tail area $α /2$ use karte hain; one-sided test ke liye poora $α$ ek tail mein rakhte hain aur tail area $α$ use karte hain.

Definition p-value — critical values ka alternative

p-value woh probability hai, $H_{0}$ true maankar computed, ki test statistic utna hi extreme ya aur zyada extreme mile jitna humne actually dekha. Chhota p-value = hamaara data $H_{0}$ ke under ek rare fluke hota = $H_{0}$ ke against evidence. Do equivalent verdict rules hain, aur neeche har example dono batata hai:

Critical-value rule: $H_{0}$ reject karo agar statistic fence $c_{p, ν}$ ko cross kare.
p-value rule: $H_{0}$ reject karo agar p-value $< α$ ho. Yeh hamesha agree karte hain kyunki fence defined hai us point ke roop mein jiska tail area $α$ (ya $α /2$ per side) ke barabar hai: "statistic fence ke baar" aur "tail area $α$ se neeche" ek hi baat hai. "Extreme" ka matlab two-sided test ke liye dono tails aur one-sided test ke liye ek tail (predicted direction) hai. Dekho p-value.

The scenario matrix

Har row ek case class hai jise aapko handle karna aana chahiye. Right column us worked example ka naam deta hai jo uspe land karta hai.

#	Case class	Tricky kyun hai	Example
A	Mean, $σ$ known, two-sided	denominator ek fixed constant hai	Ex 1
B	Mean, $σ$ known, one-sided	tail sirf ek taraf rehti hai (cutoff ka sign)	Ex 2
C	Mean, $σ$ unknown, small $n$	$s$ use karna padta hai → fatter-tailed $t$	Ex 3
D	Limiting case: $n \to \infty$	$t$ collapse hokar $z$ ban jaata hai	Ex 4
E	GOF, uniform expected	saare $E_{i}$ equal, $ν = k - 1$	Ex 5
F	GOF, estimated parameter	ek extra d.f. lose hota hai, $ν = k - 1 - m$	Ex 6
G	Degenerate GOF: $O_{i} = E_{i}$	zero surprise, $χ^{2} = 0$	Ex 7
H	F-test, two variances, two-sided	bada $s^{2}$ upar, $F \geq 1$	Ex 8
I	Word problem end-to-end	khud sahi test choose karna	Ex 9
J	Exam twist: same data, wrong test	verdict kaise flip hota hai	Ex 10

Hum poore mein Central Limit Theorem, Student's t-distribution, Chi-squared Distribution, F-distribution, Degrees of Freedom aur Bessel's Correction ka sahara lete hain — parent note ne inhe define kiya; yahan hum use karte hain.

Figure — z-test, t-test, chi-squared goodness of fit, F-test

Figure kaise padhein (chaar chhote panels):

Top-left mein standard Normal (blue, $z$ curve) aur $ν = 3$ wala $t$ (orange) overlay hai. $\pm 1.9$ ke baar shaded orange slivers woh extra tail area hai jo $t$ carry karta hai — woh shaded region precisely isliye hai ki $t$ critical value (Ex 3 mein $2.571$ ) $z$ ke $1.96$ se farther baithta hai.
Top-right mein do $χ^{2}$ curves hain (Ex 5–7 mein use); red dashed line $7.82$ par $ν = 3$ upper-tail fence hai — note karo $χ^{2}$ sirf positive axis par rehta hai, toh koi lower tail nahi hai.
Bottom-left mein $F_{7, 10}$ (Ex 8) apni upper critical value $4.03$ ke saath hai.
Bottom-right mein $ν = 1$ phir $ν = 8$ wale $t$ curves Normal ke saath stack hain, taaki aap tails ko thin hote dekh sakein jab $ν$ badhta hai — Ex 4 mein " $t \to z$ " ka visual matlab.

Yeh figure saamne rakho: neeche har example iske chaar panels mein se kisi ek ko point karta hai.

Cell A — Mean, $σ$ known, two-sided

Worked example Ex 1 · Cereal box weight

Ek machine ko boxes $μ_{0} = 500$ g fill karne chahiye. Lambi factory history se known $σ = 6$ g milta hai. $n = 36$ boxes ka sample average $\overset{x}{ˉ} = 502.4$ g hai. $α = 0.05$ par, kya machine off-target hai (kisi bhi taraf)?

Forecast: abhi andaaza lagao — kya $∣ Z ∣$ $1.96$ beat karega, ya nahi? Ek yes/no note karo.

Standard error $SE = \frac{σ}{n} = \frac{6}{36} = \frac{6}{6} = 1$ . Yeh step kyun? $36$ boxes ka average lena wobble ko $36 = 6$ se shrink karta hai; mean ek box se $6 \times$ zyada steady hai.
Z-statistic $Z = \frac{x ˉ - μ _{0}}{SE} = \frac{502.4 - 500}{1} = 2.4$ . Yeh step kyun? Yeh count karta hai ki sample target se kitne standard errors upar baitha hai — ek pure, unit-free distance.
Critical-value rule: two-sided tail area $α /2 = 0.025$ se $z_{0.025} = 1.96$ milta hai; $∣2.4∣ > 1.96$ . Yeh step kyun? "Kisi bhi taraf" off-target matlab dono tails count hoti hain, toh hum $α$ split karte hain aur tail area $0.025$ per side use karte hain (figure ka top-left panel).
p-value rule: two-sided $z = 2.4$ ke liye, $\pm 2.4$ ke baar tail area $2 \times P (Z > 2.4) \approx 2 \times 0.0082 = 0.0164$ hai. Kyunki $0.0164 < 0.05$ , same verdict. Yeh step kyun? p-value quantify karta hai kitna rare — $H_{0}$ ke under hum itna bada deviation sirf lagbhag $1.6%$ time hi dekhenge.
Dono rules $\Rightarrow$ reject $H_{0}$ . Machine significantly off-target hai. Yeh step kyun? Statistic fence ke baar aur p-value $α$ se neeche — yeh ek hi fact hai do taraf se kaha gaya.

Verify: $σ$ known tha, toh denominator ek constant hai — sirf $\overset{x}{ˉ}$ random hai, aur Central Limit Theorem se yeh Normal hai. $z$ (na ki $t$ ) use karna sahi hai. Units: g $/$ g $=$ dimensionless ✓.

Cell B — Mean, $σ$ known, one-sided (cutoff ke sign ka dhyan rakho)

Worked example Ex 2 · Kya drug

faster hai? Ek standard painkiller $μ_{0} = 30$ min mein kaam karta hai, known $σ = 8$ min. Nayi goli $n = 64$ patients par average $\overset{x}{ˉ} = 28$ min karti hai. Kya yeh faster kaam karta hai ( $H_{1} : μ < 30$ )? $α = 0.05$ use karo.

Forecast: one-sided cutoff $1.96$ se closer to zero hai — kya isse verdict badlega?

$SE = \frac{8}{64} = \frac{8}{8} = 1$ . Yeh step kyun? Same shrink-by- $n$ logic; $n = 64 \Rightarrow$ factor $8$ .
$Z = \frac{28 - 30}{1} = - 2.0$ . Yeh step kyun? Negative isliye ki sample neeche claim se hai — exactly woh direction jo "faster" predict karta hai.
Cutoff ka sign. Hamara subscript rule $z_{0.05} = 1.645$ deta hai, jo woh point hai jiske upar Normal ka $5%$ hai. Lekin $H_{1} : μ < 30$ rejection region ko lower tail mein push karta hai, toh actual fence mirror-image point $- z_{0.05} = - 1.645$ hai. Critical-value rule: reject karo agar $Z < - 1.645$ . Yeh step kyun? Subscript $p$ hamesha ek upper tail area naam karta hai, toh lower-tail test sign flip karta hai: fence $- z_{0.05}$ hai, $+ z_{0.05}$ nahi. Minus bhoolna ek classic error hai.
$- 2.0 < - 1.645 \Rightarrow$ statistic lower fence cross karta hai. Yeh step kyun? Result lower tail mein cutoff se farther baitha hai jo akele luck allow karta hai.
p-value rule: one-sided, lower tail, $p = P (Z < - 2.0) \approx 0.0228$ . Kyunki $0.0228 < 0.05$ , same verdict. Yeh step kyun? Hum sirf predicted (lower) direction count karte hain, toh hum ise double nahi karte — ek one-sided p-value same $∣ Z ∣$ ke two-sided ka exactly aadha hai.
Dono rules $\Rightarrow$ reject $H_{0}$ . Evidence hai ki goli faster hai.

Verify: same $∣ Z ∣ = 2.0$ two-sided p $\approx 0.0455$ aur one-sided p $\approx 0.0228$ deta hai — exactly aadha. One-sided test predicted direction mein reject karna easier hai. Consistent ✓.

Cell C — Mean, $σ$ unknown, small $n$

Worked example Ex 3 · Handmade nails ki length

Target $μ_{0} = 50$ mm. Hum $n = 6$ nails measure karte hain: $51, 49, 52, 50, 48, 50$ . Humare paas koi known $σ$ nahi hai. Kya length $α = 0.05$ par off-target hai (two-sided)?

Forecast: sirf $6$ nails aur unknown spread ke saath, kya aap expect karte ho ki critical value $1.96$ ke paas hoga ya noticeably bada?

Mean $\overset{x}{ˉ} = \frac{51 + 49 + 52 + 50 + 48 + 50}{6} = \frac{300}{6} = 50$ . Yeh step kyun? Data ka centre — $μ$ ka hamara best guess.
Deviations $1, - 1, 2, 0, - 2, 0$ ; squares $1, 1, 4, 0, 4, 0$ ; sum $= 10$ . Yeh step kyun? Squaring signs remove karta hai taaki plusses aur minuses cancel na karein; bade misses zyada count hote hain.
Sample variance with Bessel's Correction: $s^{2} = \frac{10}{n - 1} = \frac{10}{5} = 2$ , toh $s = 2 \approx 1.414$ . $n - 1$ se kyun divide karte hain? Deviations $\overset{x}{ˉ}$ se hain, true $μ$ se nahi; data apne khud ke mean se chipakta hai, toh $n$ use karke hum spread underestimate kar dete. Ek d.f. constraint $\sum (x_{i} - \overset{x}{ˉ}) = 0$ par spend hoti hai.
$SE = \frac{s}{n} = \frac{1.414}{6} \approx 0.577$ . Yeh step kyun? Single nails ki spread ko unke average ki spread mein convert karo $n$ se divide karke.
$t = \frac{x ˉ - μ _{0}}{SE} = \frac{50 - 50}{0.577} = 0$ , with $ν = n - 1 = 5$ . Yeh step kyun? Same "SE mein distance" idea, lekin estimated $s$ ke saath → hum ise Student's t-distribution ke against padhte hain (top-left/bottom-right panels).
Critical-value rule: two-sided tail area $α /2 = 0.025$ se $t_{0.025, 5} = 2.571$ milta hai; $∣0∣ < 2.571$ . p-value rule: $p = 2 \times P (t_{5} > 0) = 2 \times 0.5 = 1.0 > 0.05$ . Dono $\Rightarrow$ fail to reject. Yeh step kyun? Exactly $0$ ka statistic sabse kam surprising possible outcome hai; iska p-value maximum, $1.0$ , hai.

Verify: $\overset{x}{ˉ}$ exactly $μ_{0}$ par land kiya, toh numerator $0$ hai aur $t = 0$ regardless of $s$ — maximally boring outcome. Off-target ka koi evidence nahi ✓. Note karo $2.571 > 1.96$ : $t$ zyada demanding hai, jaise uski fatter tails require karti hain.

Cell D — Limiting case $n \to \infty$

Worked example Ex 4 · Jab

$t$ becomes $z$ Ex 3 jaisa style lekin huge sample ke saath: $n = 2001$ , $\overset{x}{ˉ} = 50.2$ , $s = 1.414$ , target $μ_{0} = 50$ . Two-sided, $α = 0.05$ .

Forecast: $ν = 2000$ ke saath, kya critical value $1.96$ ke closer hoga ya $2.571$ ke?

$SE = \frac{1.414}{2001} \approx \frac{1.414}{44.73} \approx 0.03162$ . Yeh step kyun? Enormous $n$ mean ko extremely steady bana deta hai.
$t = \frac{50.2 - 50}{0.03162} \approx 6.32$ , $ν = 2000$ . Yeh step kyun? Ex 3 jaisa hi standardized-distance formula; tiny $SE$ ek modest $0.2$ mm gap ko standard errors ki badi sankhya mein badal deta hai.
Critical-value rule: two-sided tail area $0.025$ se $t_{0.025, 2000} \approx 1.961$ milta hai — essentially $z$ value $1.96$ ; $6.32 > 1.961$ . p-value rule: $p = 2 \times P (t_{2000} > 6.32) \approx 3 \times 1 0^{- 10}$ , $0.05$ se bahut neeche. Dono $\Rightarrow$ reject $H_{0}$ . Yeh step kyun? Jaise $n \to \infty$ , $s \to σ$ , extra denominator-noise gayab ho jaata hai, aur $t \to N (0, 1)$ ; fat tails thin ho jaati hain (bottom-right panel).

Verify: ratio $t_{0.025, 2000} /1.96 \approx 1.0005$ — Normal se $0.1%$ ke andar. Yeh " $t \to z$ " ka concrete matlab hai. Yahan $z$ use karna same verdict deta ✓.

Cell E — GOF, uniform expected counts

Worked example Ex 5 · Kya spinner fair hai?

Ek $4$ -colour spinner $n = 100$ baar spun kiya jaata hai: $O = (30, 20, 22, 28)$ . Ek fair spinner predict karta hai ki har colour equally aayegi. Yahan categories ki sankhya $k = 4$ hai. $α = 0.05$ par fairness test karo.

Forecast: counts $25$ ke aaspaas wobble karte hain — kya aap sochte ho yeh wobble "just luck" hai?

Expected har cell $E_{i} = \frac{100}{4} = 25$ . Yeh step kyun? "Fair" total ko $k = 4$ categories mein evenly spread karta hai.
$E_{i} \geq 5$ rule check karo. Saare $E_{i} = 25 \geq 5$ , toh $χ^{2}$ approximation valid hai — aage badho. Yeh step kyun? Chi-squared curve count-distribution ko tabhi approximate karta hai jab expected cells tiny na hon; fence pe trust karne se pehle confirm karo.
Cell contributions $\frac{( O _{i} - E _{i} ) ^{2}}{E _{i}}$ : deviations $5, - 5, - 3, 3$ ; squares $25, 25, 9, 9$ ; har ek $25$ se divide: $1, 1, 0.36, 0.36$ . $E_{i}$ se kyun divide karte hain? Ek cell mein count roughly Poisson behave karta hai variance $\approx E_{i}$ ke saath, toh $\frac{O _{i} - E _{i}}{E _{i}}$ ek standardized $z$ -jaisi quantity hai; squaring aur summing ek Chi-squared Distribution deta hai.
$χ^{2} = 1 + 1 + 0.36 + 0.36 = 2.72$ . Yeh step kyun? Per-cell surprises jodne se ek total "model kitna badly fit karta hai" number milta hai.
Degrees of freedom $ν = k - 1 = 4 - 1 = 3$ (total $100$ par fixed hai, ek d.f. cost karta hai). Yeh step kyun? Counts $n$ mein sum hone chahiye, toh sirf $k - 1$ freely wander karte hain. Dekho Degrees of Freedom.
Critical-value rule: tail area $α = 0.05$ se $χ_{0.05, 3}^{2} = 7.815$ milta hai; $2.72 < 7.815$ . p-value rule: $p = P (χ_{3}^{2} > 2.72) \approx 0.437 > 0.05$ . Dono $\Rightarrow$ spinner fair lagta hai. Yeh step kyun? Chi-squared mein sirf upper tail hai (surprise negative nahi ho sakta), toh "extreme" ka matlab "large" hai aur poora $α$ right par baithta hai.

Verify: saare $E_{i} = 25 \geq 5$ ✓. Hamara surprise ( $2.72$ ) bahut kam hai jo luck routinely produce karta hai ( $\sim 7.8$ tak); p-value $\approx 0.44$ kehta hai aisa mismatch $44%$ time chance se hota hai ✓.

Cell F — GOF with an estimated parameter

Worked example Ex 6 · Typos per page count karna (Poisson fit)

Hum $n = 100$ pages par typos count karte hain aur unhe $k = 4$ categories (0, 1, 2, 3+) mein bin karte hain. Humne Poisson rate $λ$ same data se estimate kiya (yeh $m = 1$ fitted parameter hai). Expected counts $E = (37, 37, 18, 8)$ aur observed $O = (40, 34, 20, 6)$ aate hain. Kya Poisson model $α = 0.05$ par fit karta hai?

Forecast: $λ$ estimate karna ek d.f. cost karta hai — kya $ν$ $3$ hoga ya $2$ ?

$E_{i} \geq 5$ rule check karo. Expecteds $37, 37, 18, 8$ hain — sabse chhota $8 \geq 5$ hai, toh $χ^{2}$ approximation hold karta hai aur koi pooling zaroorat nahi. Yeh step kyun? Agar koi $E_{i}$ $5$ se neeche girta (jaise "3+" bin mein $E = 2$ aata), toh hum use ek neighbouring bin ke saath merge karte (jaise "2" aur "3+" combine), recompute karte, aur $k$ ek se reduce karte — tiny expecteds statistic ko inflate karte hain aur approximation todte hain.
Cell contributions $\frac{( O _{i} - E _{i} ) ^{2}}{E _{i}}$ :
- $\frac{( 40 - 37 ) ^{2}}{37} = \frac{9}{37} \approx 0.2432$
- $\frac{( 34 - 37 ) ^{2}}{37} = \frac{9}{37} \approx 0.2432$
- $\frac{( 20 - 18 ) ^{2}}{18} = \frac{4}{18} \approx 0.2222$
- $\frac{( 6 - 8 ) ^{2}}{8} = \frac{4}{8} = 0.5$ Yeh step kyun? Har cell ki squared miss ko wahan expected count se scale kiya jaata hai, raw errors ko standardized, comparable surprises mein badal kar.
$χ^{2} \approx 0.2432 + 0.2432 + 0.2222 + 0.5 = 1.2087$ . Yeh step kyun? Charon standardized surprises jodne se woh single total milta hai jo hum Chi-squared Distribution ke against test karte hain.
Degrees of freedom $ν = k - 1 - m = 4 - 1 - 1 = 2$ . Yeh step kyun? Ab do constraints hain: total fixed hai (−1) aur humne $λ$ ko data match karne ke liye force kiya (−1 aur). Har fitted parameter ek extra constraint khareedta hai.
Critical-value rule: tail area $α = 0.05$ se $χ_{0.05, 2}^{2} = 5.991$ milta hai; $1.21 < 5.991$ . p-value rule: $p = P (χ_{2}^{2} > 1.2087) \approx 0.546 > 0.05$ . Dono $\Rightarrow$ Poisson model fit karta hai. Yeh step kyun? Total surprise upper-tail fence ke samne tiny hai, aur itna chhota mismatch chance se aadhe se zyada time hota hai.

Verify: $ν = 3$ use karne ke trap se compare karo: critical $7.815$ hota aur $p = P (χ_{3}^{2} > 1.2087) \approx 0.751$ — yahan verdict dono mein survive karta, lekin borderline statistic ke saath wrong $ν$ ise flip kar deta. Hamesha fitted parameters $m$ subtract karo ✓.

Cell G — Degenerate GOF: perfect match

Worked example Ex 7 · Suspiciously perfect die

Ek die $60$ baar roll kiya jaata hai aur $O = (10, 10, 10, 10, 10, 10)$ aata hai — exactly fair expectation $E_{i} = 10$ . Yahan categories ki sankhya $k = 6$ hai.

Forecast: ek $χ^{2}$ kabhi se se chhota ho sakta hai?

$E_{i} \geq 5$ rule check karo. Har $E_{i} = 10 \geq 5$ ✓ — approximation valid hai. Yeh step kyun? Ex 5–6 jaisi hi guard; degenerate case ko bhi fence ka matlab hone ke liye itne bade expecteds chahiye.
Har cell $\frac{( O _{i} - E _{i} ) ^{2}}{E _{i}} = \frac{( 10 - 10 ) ^{2}}{10} = 0$ . Yeh step kyun? Har jagah zero deviation matlab zero squared surprise.
$χ^{2} = 0$ . Yeh step kyun? Zeros ka sum zero hota hai — model ne har category exactly match kiya.
$ν = k - 1 = 5$ ke saath: critical-value rule — koi bhi fence $> 0$ hai, toh hum kabhi reject nahi karte. p-value rule — $p = P (χ_{5}^{2} > 0) = 1.0$ , maximum. Dono $\Rightarrow$ kabhi reject mat karo; fit arithmetically possible se utna hi achha hai. Yeh step kyun? $χ^{2} \geq 0$ hamesha (squares over positives ka sum); $0$ floor hai, toh kuch bhi ise rejection tail mein push nahi karta, aur iska p-value $1$ hai.

Verify: $χ^{2}$ kabhi negative nahi ho sakta aur $0$ hai tabhi jab har $O_{i} = E_{i}$ ; phir p-value $= 1$ . Real life mein perfect match khud hi odd hai, lekin GOF test sirf bad fits upper tail mein flag karta hai, "too good" fits nahi — toh ek suspiciously tidy dataset ko ek alag check chahiye, yeh wala nahi. ✓

Cell H — F-test, two variances, two-sided

Yeh figure kaise padhein: blue curve do sample variances ke ratio ke liye $F_{7, 10}$ luck curve hai. $4.03$ par red dashed line upper-tail fence hai; uske baar red-shaded region rejection zone hai. Orange line hamaara observed $F = 3.0$ mark karta hai — fence ke left baitha, "could be luck" region ke andar.

Worked example Ex 8 · Kaun sa lathe zyada consistent hai?

Lathe A: $s_{1}^{2} = 45$ , $n_{1} = 8$ . Lathe B: $s_{2}^{2} = 15$ , $n_{2} = 11$ . Kya unke variances $α = 0.05$ par significantly different hain (two-sided)?

Forecast: ratio $3$ hai — critical $F$ beat karne ke liye kaafi bada?

Bada $s^{2}$ upar rakho: $F = \frac{s _{1}^{2}}{s _{2}^{2}} = \frac{45}{15} = 3.0$ . Yeh step kyun? $F \geq 1$ force karne se hum ek single upper-tail critical value use kar sakte hain, lower tail chase karne ki jagah. Dekho F-distribution.
Degrees of freedom $ν_{1} = n_{1} - 1 = 7$ (numerator), $ν_{2} = n_{2} - 1 = 10$ (denominator). Yeh step kyun? Har variance estimate ek $χ^{2} / ν$ hai; $F$ do aisi ka ratio hai, har estimate ke liye ek d.f.
Critical-value rule: $α = 0.05$ par two-sided $2.5%$ har tail mein dalta hai; bade variance ko upar rakh kar hum upper value padhte hain tail area $α /2 = 0.025$ ke saath: $F_{0.025, 7, 10} \approx 4.03$ ; $3.0 < 4.03$ . Yeh step kyun? "Different" allow karta hai ki koi bhi lathe noisier ho, toh hum $5%$ split karte hain; upar bada rakhna matlab sirf upper $2.5%$ tail hi test hoti hai.
p-value rule: $p = 2 \times P (F_{7, 10} > 3.0) \approx 2 \times 0.0503 \approx 0.101 > 0.05$ . Same verdict. Yeh step kyun? Hum one-tail area double karte hain kyunki two-sided variance test kisi bhi taraf extreme ratio allow karta hai; $0.101$ kehta hai equal variances ke under itna dur ratio $1$ se lagbhag $10%$ time hota hai.
Dono rules $\Rightarrow$ fail to reject. Variances significantly different nahi hain.

Verify: figure $F = 3.0$ (orange) ko $4.03$ (red dashed) ke left dikhata hai — "could be luck" region ke andar, p $\approx 0.10 > 0.05$ match karta hai. Order-of-magnitude $F \approx 3$ with $\sim 10$ d.f. $H_{0}$ ke under common hai ✓.

Cell I — Word problem, test khud choose karo

Worked example Ex 9 · Café owner

"Meri espresso machine ko har shot mein $30$ ml dispense karna chahiye. Mujhe factory ki spread par trust nahi, toh maine khud $10$ shots measure kiye: $\overset{x}{ˉ} = 31.2$ ml, $s = 2.4$ ml. Kya main $α = 0.05$ par over-dispensing kar raha hoon?"

Forecast: charon tests mein se kaun sa, aur kyun? Step 1 padhne se pehle decide karo.

Test chunno. Hum ek mean ko ek target se compare kar rahe hain, $σ$ unknown hai (owner factory value par trust nahi karta) aur $n = 10$ small hai $\Rightarrow$ one-sample $t$ -test, one-sided ( $H_{1} : μ > 30$ ). Yeh step kyun? Unknown $σ$ + estimated $s$ = exact signature jo $z$ ki jagah $t$ force karta hai (Hypothesis Testing).
$SE = \frac{s}{n} = \frac{2.4}{10} \approx 0.7589$ . Yeh step kyun? Single shots ki spread ko unke average ki spread mein badlo $n$ se divide karke — woh quantity ki wobble jo hum actually test karte hain.
$t = \frac{x ˉ - μ _{0}}{SE} = \frac{31.2 - 30}{0.7589} \approx 1.581$ , $ν = 9$ . Yeh step kyun? Standardized distance: measured average target se kitne standard errors upar baitha hai, $ν = n - 1 = 9$ wale $t$ ke against padha jaata hai.
Critical-value rule. "Over-dispensing" ek one-direction upper claim hai, toh poora $α = 0.05$ upper tail mein baithta hai: $t_{0.05, 9} = 1.833$ ; reject karo agar $t > 1.833$ . Yahan $1.581 < 1.833$ — koi sign flip zaroorat nahi kyunki claim upper-tail hai, toh fence $+ t_{0.05, 9}$ hai. Yeh step kyun? Ex 2 se contrast karo: wahan claim lower-tail tha toh humne $- z_{0.05}$ use kiya; yahan yeh upper-tail hai toh hum plus sign rakhte hain. $H_{1}$ ki direction cutoff ka sign dictate karti hai.
p-value rule. $p = P (t_{9} > 1.581) \approx 0.074 > 0.05$ . Same verdict. Yeh step kyun? One-sided, upper direction sirf, toh doubling nahi; $0.074$ narrowly $α$ se zyada hai.
Dono rules $\Rightarrow$ fail to reject. Over-dispensing ka kaafi evidence nahi.

Verify: p-value $\approx 0.074 > 0.05$ "fail to reject" ke consistent hai, aur yeh $α$ ke close hai — thoda bada $\overset{x}{ˉ}$ ise tip over kar deta. Units ml $/$ ml cancel ✓.

Cell J — Exam twist: same data, wrong test verdict flip karta hai

Worked example Ex 10 · Tempting shortcut

Ex 9 ke numbers lo ( $\overset{x}{ˉ} = 31.2$ , $s = 2.4$ , $n = 10$ , target $30$ ). Ek student kehta hai: "CLT sab kuch Normal bana deta hai, toh main sirf $z$ -test $1.645$ ke saath use karunga." Kya verdict badlega?

Forecast: same statistic value $1.581$ — lekin ek alag reference curve. Reject ya nahi?

Student same ratio compute karta hai $z = \frac{31.2 - 30}{2.4/ 10} \approx 1.581$ (woh $s$ wahan plug karta hai jahan $σ$ hona chahiye). Yeh step kyun? Ex 9 ke $t$ se numerically identical; sirf reference curve alag hai, toh arithmetic galti reveal nahi kar sakta.
Wrong critical value. Woh one-sided $z_{0.05} = 1.645$ se compare karta hai $t_{0.05, 9} = 1.833$ ki jagah. Yeh step kyun? $z$ -test Normal padhta hai, jiska one-sided $5%$ point $1.645$ hai; $ν = 9$ wale sahi $t$ curve ka farther fence $1.833$ par hai.
Wrong p-value bhi. Normal p-value $P (Z > 1.581) \approx 0.057$ sahi $t$ p-value $P (t_{9} > 1.581) \approx 0.074$ se chhota hai, kyunki $t$ ki fatter tail $1.581$ ke baar zyada probability pile karta hai. Yeh step kyun? Fence aur p-value dono distorted ho jaate hain; shortcut result ko actual se zyada significant dikhata hai.
$1.581 < 1.645 \Rightarrow$ yahan abhi bhi fail to reject — lekin barely. Agar $\overset{x}{ˉ}$ $31.3$ hota, toh wrong-test $z$ ( $z$ p $\approx 0.041 < 0.05$ ) reject karta jabki sahi $t$ nahi karta. Yeh step kyun? Kyunki $1.645 < 1.833$ , $z$ rejection region wider hai: shortcut bahut aasaani se reject karta hai, Type I error rate ko promised $5%$ se upar inflate karta hai.

Verify: unknown $σ$ aur $n = 10$ ke saath, sahi critical value $1.833$ hai, $1.645$ nahi. Gap $1.833 - 1.645 = 0.188$ $s$ ko $σ$ samajhne ki exact cost hai ✓.

Recall Quick self-check

Kaun se cell mein $χ^{2}$ exactly $0$ hai, aur kab? ::: Cell G — degenerate GOF, jab bhi har $O_{i} = E_{i}$ ; $χ^{2}$ squares ka sum hai toh $0$ iska floor hai aur iska p-value $1$ hai. Ex 2 mein fence $- 1.645$ kyun hai $+ 1.645$ nahi? ::: Subscript $p$ ek upper tail area naam karta hai, lekin $H_{1} : μ < 30$ rejection region lower tail mein dalta hai, toh hum $- z_{0.05}$ par mirror karte hain. Ex 8 mein bada variance upar kyun hai? ::: Taaki $F \geq 1$ ho aur hum sirf ek single upper-tail critical value padh sakein. Ex 6 mein $ν = 2$ kyun hai $3$ nahi? ::: Humne ek parameter ( $λ$ ) data se estimate kiya, ek extra degree of freedom cost karke: $ν = k - 1 - m = 4 - 1 - 1$ . Dono verdict rules batao. ::: Critical-value rule: reject karo agar statistic $c_{p, ν}$ ke baar ho. p-value rule: reject karo agar p-value $< α$ ho. Yeh hamesha agree karte hain.