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

4.9.20 · D4 · Maths › Probability Theory & Statistics › Hypothesis testing — null - alternative, test statistic, p-v

Yeh page ek self-testing ladder hai. Har problem ko solution dekhne se pehle khud solve karo. Levels "pieces pehchano" se lekar "poora test design karo" tak chalte hain. Parent note Hypothesis testing — null - alternative, test statistic, p-value, errors (Type I & II) theory build karta hai; prerequisites Central Limit Theorem, Standard Error, aur Normal Distribution tools provide karte hain. Lekin tumhe yeh page chhodni nahi chahiye — har symbol neeche re-anchor kiya gaya hai pehle use se.

Definition Is page ke symbols, simple words mein

$μ_{0}$ — $H_{0}$ ke under claimed population mean (woh "boring" value jo hum sach maante hain).
$\overset{ˉ}{X}$ — sample mean, un numbers ka average jo humne actually measure kiye.
$σ$ — true population standard deviation (spread), ek known fixed number.
$s$ — sample standard deviation, data se $σ$ ka hamara estimate.
$n$ — sample size, kitne observations collect kiye humne.
$SE$ — standard error $= σ / n$ (ya $s / n$ ): $\overset{ˉ}{X}$ khud kitna wobble karta hai sample se sample tak. Bada $n$ → chhota wobble.
$Z$ — z-statistic $= \frac{X ˉ - μ _{0}}{SE}$ : $\overset{ˉ}{X}$ , $μ_{0}$ se kitne standard errors door hai. $H_{0}$ ke under yeh Normal Distribution follow karta hai mean $0$ , spread $1$ ke saath, likha jaata hai $N (0, 1)$ .
$T$ — t-statistic, $Z$ jaisi shape lekin $σ$ ki jagah $s$ ke saath; yeh Student t-distribution $t_{n - 1}$ follow karta hai jiske tails zyada mote hote hain.
$α$ — significance level, Type I error rate jo hum accept karte hain (usually $0.05$ ).
$P (Z \geq z)$ — right-tail area: probability ki ek standard normal $z$ par ya uske upar land kare.

Standard-normal tail areas ka ek reference table jo hum reuse karte hain (yeh $P (Z \geq z)$ hain, bell curve ke neeche $z$ ke daayein shaded area):

Level 1 — Recognition

Exercise 1.1

Har claim ke liye $H_{0}$ aur $H_{1}$ likho, aur batao ki test two-, left-, ya right-tailed hai. (a) "Ek drug average blood pressure ko known baseline $μ_{0} = 120$ se badalta hai." (b) "Ek naya fertilizer mean yield ko $μ_{0} = 50$ kg se upar badhata hai." (c) "Ek factory ka defect rate old mean $μ_{0} = 0.03$ se neeche hai."

Recall Solution 1.1

Yaad rakho: == $H_{0}$ hamesha equality hold karta hai== (kyun? test ko probabilities compute karne ke liye ek precise value chahiye jisse — ek inequality koi single number nahi deta standardize karne ke liye). Claim mein direction word tail pick karta hai.

(a) "changes" = kisi bhi direction mein ja sakta hai → $H_{0} : μ = 120$ , $H_{1} : μ \neq = 120$ → two-tailed.
(b) "increases / above" → $H_{0} : μ = 50$ , $H_{1} : μ > 50$ → right-tailed.
(c) "below" → $H_{0} : μ = 0.03$ , $H_{1} : μ < 0.03$ → left-tailed.

Exercise 1.2

Ek study report karta hai $p = 0.02$ significance level $α = 0.05$ par. Decision batao aur usse sahi se phrase karo.

Recall Solution 1.2

Rule: $H_{0}$ reject karo jab $p \leq α$ . Yahan $0.02 \leq 0.05$ → reject $H_{0}$ . Yeh rule kyun? $p$ woh chance hai ki data itna extreme ho agar $H_{0}$ sach hota; $α$ woh fluke-rate hai jo hum tolerate karte hain. Jab $p \leq α$ tab data hamare tolerance se zyada rare hote hain, toh pure luck ko blame karna credible nahi rehta. Sahi phrasing: "Data $5%$ level par $H_{0}$ ke saath inconsistent hain; hamare paas $H_{1}$ ke liye evidence hai." Hum yeh nahi kehte " $H_{1}$ proven ho gayi."

Level 2 — Application

Exercise 2.1

Ek coffee machine ko $μ_{0} = 250$ ml pour karni chahiye known $σ = 12$ ml ke saath. $n = 36$ cups ka ek sample $\overset{ˉ}{X} = 245$ ml deta hai. $α = 0.05$ par test karo ki machine mis-set hai ya nahi.

Recall Solution 2.1

$H_{0} : μ = 250$ , $H_{1} : μ \neq = 250$ (two-tailed — "mis-set" = koi bhi direction). Standard error $= σ / n$ . $n$ se kyun divide karte hain? $n$ noisy readings ko average karna bahut sa noise cancel kar deta hai; $\overset{ˉ}{X}$ ka wobble $1/ n$ ki tarah shrink hota hai. Yahan $SE = 12/ 36 = 12/6 = 2$ . Standardize: $z = \frac{X ˉ - μ _{0}}{SE} = \frac{245 - 250}{2} = - 2.5$ . Kyun? Yeh " $5$ ml low" ko " $2.5$ standard errors low" mein convert karta hai, ek unit-free score jo hum look up kar sakte hain. Two-tailed p-value: $p = 2 P (Z \geq 2.5)$ . One-sided tail ko double kyun karte hain? "Mis-set" $250$ se itni door mean ko equally surprising count karta hai chahe upar ho ya neeche. Toh " $z = - 2.5$ jaisa extreme" matlab hai $∣ Z ∣ \geq 2.5$ , yaani tail below $- 2.5$ plus mirror tail above $+ 2.5$ . Kyunki normal curve symmetric hai, un donon tails ka area equal hota hai, toh hum ek tail lete hain aur double karte hain: $p = 2 P (Z \geq 2.5) = 2 (0.00621) = 0.01242.$ $0.01242 \leq 0.05$ → reject $H_{0}$ : machine mis-set hai (low pour kar rahi hai).

Exercise 2.2

Ek teacher claim karta hai ki naya method mean score $μ_{0} = 65$ se upar raise karta hai. Sample: $n = 25$ , $\overset{ˉ}{X} = 68$ , sample SD $s = 10$ . $α = 0.05$ par test karo.

Recall Solution 2.2

$σ$ unknown hai (sirf sample $s$ diya gaya hai) → Student t-distribution statistic $T$ use karo. $H_{0} : μ = 65$ , $H_{1} : μ > 65$ (right-tailed — "raises above"). $SE = s / n = 10/ 25 = 10/5 = 2$ . $T = \frac{68 - 65}{2} = 1.5$ with $df = n - 1 = 24$ . $df = n - 1$ kyun? $s$ estimate karna ek degree of freedom "use up" karta hai ( $\overset{ˉ}{X}$ se deviations zero sum honi chahiye), baaki $n - 1$ free rehte hain. Critical value $t_{0.05, 24} \approx 1.711$ . Kyunki $1.5 < 1.711$ → fail to reject $H_{0}$ . Wording: itna evidence nahi ki method scores raise karta hai.

Level 3 — Analysis

Exercise 3.1

Do students same data test karte hain ( $μ_{0} = 100$ , $σ$ known, $z_{o b s} = 1.8$ ). Student A two-tailed test run karta hai, Student B right-tailed test, dono $α = 0.05$ par. Dono p-values compute karo aur explain karo ki woh decision par kyun disagree karte hain.

Recall Solution 3.1

Table se, $P (Z \geq 1.8) = 0.0359$ .

A (two-tailed): $p = 2 (0.0359) = 0.0718 > 0.05$ → fail to reject.
B (right-tailed): $p = 0.0359 \leq 0.05$ → reject. Same number, opposite verdicts. Kyun? Two-tailed test $α$ dono sides par spend karta hai ( $2.5%$ har side), toh reject karne se pehle zyada extreme statistic maangta hai; one-tailed test saara $5%$ ek side mein daal deta hai. Yahi exact reason hai ki tail data dekhne se pehle choose ki jaani chahiye — warna tum B isliye choose karte kyunki woh tumhe chahiya result deta hai.

Exercise 3.2

Exercise 2.1 mein, Type I aur Type II error ka kya matlab hoga us machine ke liye, words mein describe karo, aur batao kaunsa $α = 0.05$ directly control karta hai.

Recall Solution 3.2

Type I (false alarm, prob $α$ ): machine actually exactly $250$ ml pour kar rahi hai, lekin hum declare karte hain ki mis-set hai aur bilkul sahi machine recalibrate karte hain.
Type II (miss, prob $β$ ): machine sach mein off hai, lekin hamara sample normal lag raha tha, toh hum usse broken chhod dete hain. $α = 0.05$ Type I rate directly control karta hai: $P (reject ∣ H_{0} true) = 0.05$ . Yeh akele $β$ ke baare mein kuch nahi kehta — woh true $μ$ aur $n$ par depend karta hai (dekho Statistical Power & Sample Size).

Level 4 — Synthesis

Exercise 4.1

Ek quality engineer batch reject karegi agar sample mean $μ_{0} = 500$ se door hai. $σ = 8$ , $n = 64$ ke saath (toh $SE = 1$ ), woh $α = 0.05$ par two-tailed test use karti hai, reject karti hai jab $∣ z ∣ \geq 1.96$ . (a) Rejection rule ko $\overset{ˉ}{X}$ par rule mein convert karo (do critical means). (b) Maano machine ka true mean drift karke $μ = 502$ ho gaya. Power find karo = probability ki woh correctly reject karti hai. (Yeh neeche figure mein shaded story hai.)

Figure — ise kaise padhein. Figure do bell curves ek hi $\overset{ˉ}{X}$ -axis par draw karta hai. Violet curve $H_{0}$ hai ( $μ_{0} = 500$ par centred), magenta curve drift ke under $\overset{ˉ}{X}$ ki true distribution hai ( $H_{1}$ , $502$ par centred). Do navy dashed lines critical means $498.04$ aur $501.96$ hain — inke bahar hum reject karte hain. Orange shading magenta ( $H_{1}$ ) curve ka woh slice hai jo rejection region mein pada hai: iska total area power hai. Notice karo ki almost saara orange right tail mein hai, kyunki $502$ upper cutoff ke paas hai aur lower se door.

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

Recall Solution 4.1

(a) $z$ mein rejection boundary: $z = \pm 1.96$ . $\overset{ˉ}{X} = μ_{0} + z \cdot SE$ se convert karo: $\overset{ˉ}{X} = 500 \pm 1.96 (1) = {498.04, 501.96} .$ Toh woh reject karti hai agar $\overset{ˉ}{X} \leq 498.04$ ya $\overset{ˉ}{X} \geq 501.96$ . (b) Agar sach $μ = 502$ hai, toh $\overset{ˉ}{X} \sim N (502, 1^{2})$ — toh ab hum $502$ se standardize karte hain, $μ_{0}$ se nahi. Power $= P (reject ∣ μ = 502)$ = magenta curve ka area har cutoff ke baad:

Upper tail: $P (\overset{ˉ}{X} \geq 501.96) = P (Z \geq \frac{501.96 - 502}{1}) = P (Z \geq - 0.04)$ . $P (Z \geq - 0.04) = 0.5160$ kyun? Standard normal ki symmetry se, $- 0.04$ ke daayein area = ( $0$ ke daayein area) $+$ ( $0$ aur $0.04$ ke beech area) $= 0.5 + 0.0160 = 0.5160$ . Equivalently $P (Z \geq - 0.04) = 1 - P (Z \geq 0.04) = 1 - 0.4840 = 0.5160$ .
Lower tail: $P (\overset{ˉ}{X} \leq 498.04) = P (Z \leq \frac{498.04 - 502}{1}) = P (Z \leq - 3.96) \approx 0.0000$ (essentially itna door koi area nahi).

Power $\approx 0.5160 + 0.0000 = 0.516$ . Interpretation: real $2$ -ml drift ke saath bhi, woh use sirf ~ $52%$ time pakadti hai. Behtar karne ke liye usse $n$ badhana hoga (dono curves narrow hoti hain) — Statistical Power & Sample Size tradeoff, Central Limit Theorem par resting.

Level 5 — Mastery

Exercise 5.1

Tumhe $H_{0} : μ = μ_{0}$ ka two-tailed $z$ -test design karna hai $σ = 10$ aur $α = 0.05$ ke saath. Tumhe power $\geq 0.90$ chahiye $∣ μ - μ_{0} ∣ = 4$ ka true difference detect karne ke liye. Minimum sample size $n$ find karo. (Design formula use karo $n = (\frac{( z _{α /2} + z _{β} ) σ}{μ - μ _{0}})^{2}$ , $z_{α /2} = 1.96$ ke saath $α = 0.05$ ke liye aur $z_{β} = 1.2816$ power $0.90$ ke liye.)

Recall Solution 5.1

$z_{α /2} = 1.96$ , $z_{β} = 1.2816$ , $σ = 10$ , effect $= 4$ plug karo: $n = (\frac{( 1.96 + 1.2816 ) \cdot 10}{4})^{2} = (\frac{3.2416 \cdot 10}{4})^{2} = (\frac{32.416}{4})^{2} = (8.104)^{2} = 65.67.$ Sample size whole number honi chahiye aur hum upar round karte hain (kabhi neeche nahi — neeche round karne se power kho jaati hai): $n = 66.$ Yeh formula kyun? Yeh do requirements ek axis par stitch karta hai: cutoff $μ_{0}$ se $z_{α /2}$ SEs door honi chahiye ( $α$ control karta hai) aur true mean $μ_{1}$ se $z_{β}$ SEs door dosri taraf ( $β$ control karta hai). Means ke beech gap $∣ μ - μ_{0} ∣$ hai; ise $(z_{α /2} + z_{β}) \cdot SE$ ke equal set karna aur $SE = σ / n$ se $n$ ke liye solve karna box deta hai. Dekho Statistical Power & Sample Size aur Standard Error.

Exercise 5.2

Ek clean paragraph mein explain karo kyun $n$ badhana woh ek hi lever hai jo $α$ fixed rakhte hue dono $α$ aur $β$ shrink karta hai, jabki cutoff adjust karna sirf ek ko dosre ke saath trade karta hai.

Recall Solution 5.2

Pehle, do regimes clarify karo taaki hum inhe confuse na karein. Regime 1 — fixed cutoff (raw $\overset{ˉ}{X}$ units mein): cutoff ko $H_{1}$ ki taraf slide karo aur $H_{0}$ tail past it shrink hoti hai ( $α ↓$ ) jabki $H_{1}$ tail galat side par grow hoti hai ( $β ↑$ ): ek pure trade, koi free lunch nahi. Regime 2 — fixed $α$ (normal practice): hum $α = 0.05$ pin karte hain, jo cutoff ko $μ_{0}$ se $z_{α /2}$ standard errors par pin karta hai, toh $α$ ab construction se constant hai — bacha hua sawal sirf yeh hai ki $β$ kitna bada hai. Yahan $n$ badhane ka payoff hai: kyunki $SE = σ / n$ $1/ n$ ki tarah collapse hota hai, dono bell curves narrow hoti hain aur, SE units mein measure karke, apart pull hoti hain. $α$ -cutoff apni fixed $z$ -position par held ke saath, narrower, better-separated $H_{1}$ curve ab almost entirely rejection region ke andar hang karti hai, toh $β$ girta hai jabki $α$ exactly $0.05$ rehta hai. Isi sense mein $n$ "dono" shrink karta hai: yeh $α$ ko wahin hold karta hai jahan tumne set kiya aur $β$ drive karta hai down — kuch aisa jo koi cutoff move nahi kar sakta. Central Limit Theorem guarantee karta hai ki $\overset{ˉ}{X}$ normal rehta hai jabki uski spread collapse hoti hai, toh poori picture kaam karti rehti hai.

Recall Poore ladder par one-line self-check

Kya tumne (i) har tail compute karne se pehle fix ki, (ii) jab bhi $σ$ estimate hua $t$ use kiya, (iii) power $H_{0}$ ke nahi $H_{1}$ ke under standardize ki, (iv) sample sizes upar round kiye, aur (v) kisi bhi cheez par trust karne se pehle normality/CLT assumption check ki? ::: Agar paanchon mein haan, toh tumne classic traps L1–L5 clear kar liye.