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

4.9.21 · D4 · Maths › Probability Theory & Statistics › z-test, t-test, chi-squared goodness of fit, F-test

Quick symbol refresher (kuch naya assume nahi kiya ja raha):

Rejection rule har jagah: ek test statistic compute karo, uska size ek table se liye gaye critical value se compare karo. Critical se bada ⇒ ==reject $H_{0}$ ; chhota ⇒ fail to reject==. Dekho Hypothesis Testing aur p-value.

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

Upar ki picture poora game hai: bell/hump "luck alone $H_{0}$ ke under kya karta hai" hai; shaded ends critical region hain. Agar tumhara statistic shade mein land kare, toh luck-alone implausible ho jaata hai.

Level 1 — Recognition

L1·Q1 — Test pehchano

Har situation ke liye z, t, χ², ya F naam batao. (a) Ek coin ka true mean bias $0.5$ ke barabar hai ya nahi, test karna hai, population standard deviation known hai aur $n = 200$ . (b) $n = 6$ bottles se mean fill volume $250$ ml hai ya nahi, $σ$ unknown. (c) Kya ek class mein eye colours ki tally ek claimed ratio se match karti hai. (d) Kya do machines equal variance wale parts produce karti hain.

Recall Solution

(a) $σ$ known, large $n$ ⇒ z-test. (b) $σ$ unknown, small $n$ , a mean ⇒ t-test. (c) Categories mein observed counts ko expected se compare karna ⇒ chi-squared goodness of fit. (d) A ratio of two variances ⇒ F-test. Parent se mnemonic: Z Sure, T Tries, Chi Counts, F Fights.

L1·Q2 — Critical value padhna

$α = 0.05$ , two-sided par, critical cutoff batao: (a) ek z-test ke liye, (b) $ν = 4$ ke saath t-test ke liye, (c) $ν = 5$ ke saath χ² GOF ke liye (one-sided upper).

Recall Solution

(a) $z_{0.025} = 1.96$ . (b) $t_{0.025, 4} = 2.78$ (z se fatter tails kyunki $s$ uncertainty add karta hai — Student's t-distribution). (c) $χ_{0.05, 5}^{2} = 11.07$ (χ² upper end par one-tailed hai kyunki "surprise" sirf positive sum ki tarah accumulate hoti hai — Chi-squared Distribution).

Level 2 — Application

L2·Q1 — One-sample z

Ek factory claim karti hai ki wire $μ_{0} = 120$ N par toot ti hai. Known $σ = 15$ N. Sample $n = 25$ , $\overset{x}{ˉ} = 114$ . $α = 0.05$ , two-sided par test karo.

Recall Solution

KYA: claim se standardized distance compute karo. $SE = \frac{σ}{n} = \frac{15}{25} = \frac{15}{5} = 3$ . $25$ kyun? 25 values ko average karna wobble ko $25 = 5$ se shrink karta hai. $Z = \frac{x ˉ - μ _{0}}{SE} = \frac{114 - 120}{3} = \frac{- 6}{3} = - 2.0$ . Compare: $∣ - 2.0∣ = 2.0 > 1.96$ ⇒ ==reject $H_{0}$ ==. Wire claimed se significantly zyada weak hai.

L2·Q2 — One-sample t

$H_{0} : μ = 100$ . Data: $103, 98, 101, 99, 104$ ( $n = 5$ ), $σ$ unknown. $α = 0.05$ , two-sided par test karo.

Recall Solution

Mean: $\overset{x}{ˉ} = \frac{103 + 98 + 101 + 99 + 104}{5} = \frac{505}{5} = 101$ . $\overset{x}{ˉ}$ se deviations: $2, - 3, 0, - 2, 3$ . Squares: $4, 9, 0, 4, 9$ , sum $= 26$ . Variance (Bessel's $n - 1$ ): $s^{2} = \frac{26}{5 - 1} = 6.5$ , toh $s = 6.5 \approx 2.55$ . $n - 1$ kyun? deviations $\overset{x}{ˉ}$ se hain, jo apne data ke sabse paas hoti hai, toh hum constraint $\sum (x_{i} - \overset{x}{ˉ}) = 0$ ki wajah se 1 d.f. khote hain — dekho Bessel's Correction. SE: $s / 5 = 2.55/2.236 \approx 1.140$ . Statistic: $t = \frac{101 - 100}{1.140} \approx 0.877$ , $ν = 4$ . Compare: $∣0.877∣ < 2.78$ ⇒ fail to reject. Koi evidence nahi ki mean 100 se different hai.

L2·Q3 — χ² goodness of fit

Ek bag mein red:blue:green $2 : 1 : 1$ ratio mein hone chahiye. $n = 100$ ka sample deta hai $O = (46, 28, 26)$ . $α = 0.05$ par ratio test karo.

Recall Solution

Expected counts claimed ratio se (total $2 + 1 + 1 = 4$ parts): $E_{red} = 100 \cdot \frac{2}{4} = 50$ , $E_{blue} = 100 \cdot \frac{1}{4} = 25$ , $E_{green} = 25$ . Har cell ki surprise $(O - E)^{2} / E$ : red $\frac{( 46 - 50 ) ^{2}}{50} = \frac{16}{50} = 0.32$ ; blue $\frac{( 28 - 25 ) ^{2}}{25} = \frac{9}{25} = 0.36$ ; green $\frac{( 26 - 25 ) ^{2}}{25} = \frac{1}{25} = 0.04$ . Sum: $χ^{2} = 0.32 + 0.36 + 0.04 = 0.72$ . d.f.: $k = 3$ categories, koi parameter estimate nahi ( $m = 0$ ), toh $ν = k - 1 - m = 2$ . Critical $χ_{0.05, 2}^{2} = 5.99$ . Compare: $0.72 < 5.99$ ⇒ fail to reject. Data $2 : 1 : 1$ ke saath consistent hai.

Level 3 — Analysis

L3·Q1 — Same data z se reject kyun karta hai lekin t se nahi?

$n = 5$ ka ek sample deta hai $\overset{x}{ˉ} = 101$ , $s = 2.55$ , $H_{0} : μ = 100$ test kar rahe hain (yeh L2·Q2 hai). Maano ek colleague galti se $s$ ko known $σ$ treat karta hai aur z-test run karta hai. Dono conclusions dikhao aur mechanism explain karo.

Recall Solution

Galat (z) path: $Z = \frac{101 - 100}{2.55/ 5} = 0.877$ . Wahi number $t$ statistic ke jitna — kyunki arithmetic identical hai! Fark yardstick mein hai, statistic mein nahi.

z cutoff $1.96$ use karta hai: $0.877 < 1.96$ ⇒ yahan bhi fail to reject. Isse bite karte hain: $\overset{x}{ˉ}$ ko $103$ kar do. Tab $t = \frac{3}{1.140} = 2.63$ .
t (sahi): cutoff $2.78$ ; $2.63 < 2.78$ ⇒ fail to reject.
z (galat): cutoff $1.96$ ; $2.63 > 1.96$ ⇒ reject. Mechanism: chhote $n$ ke saath estimate $s$ khud noisy hoti hai. Kabhi kabhi $s$ accidentally chhoti nikal aati hai, ratio ko luck se bada dikha deti hai. $t$ distribution ke fatter tails cutoff ko widen karte hain exactly is extra luck ko absorb karne ke liye. $1.96$ use karna ise ignore karta hai aur ek false "significant" result manufacture karta hai. Jab $n \to \infty$ , toh $s \to σ$ aur dono cutoffs merge ho jaate hain.

L3·Q2 — Degrees of freedom bookkeeping

Tum $k = 6$ bins mein counts ke liye ek Poisson model fit karte ho, lekin tumne Poisson rate $λ$ same data se estimate ki. Tumhe $χ^{2} = 9.5$ milta hai. $α = 0.05$ par fit test karo.

Recall Solution

d.f. accounting: $k = 6$ cells se shuru karo. Fixed total ke liye 1 ghataao ( $\sum O = \sum E = n$ ). Ek aur ghataao kyunki $λ$ data se estimate ki gayi ( $m = 1$ ). $ν = k - 1 - m = 6 - 1 - 1 = 4.$ Critical $χ_{0.05, 4}^{2} = 9.49$ . Compare: $9.5 > 9.49$ (barely hi) ⇒ reject the fit. Kyun matter karta hai: agar tum estimated parameter bhool jaate aur $ν = 5$ use karte, cutoff $11.07$ hoti, aur $9.5 < 11.07$ galat kehta "fit theek hai." Har woh parameter jo tum $E_{i}$ build karne ke liye data se "churaate" ho ek degree of freedom cost karta hai, cutoff tight karta hai.

Level 4 — Synthesis

L4·Q1 — F-test as a gatekeeper

Machine A: $s_{1}^{2} = 45$ , $n_{1} = 8$ . Machine B: $s_{2}^{2} = 15$ , $n_{2} = 10$ . Pehle decide karo ki unke variances equal hain ya nahi ( $α = 0.05$ , two-sided). Phir batao ki unke means compare karne ke liye kaun sa test use karoge.

Recall Solution

Step 1 — F-test. Bada variance upar rakho taki $F \geq 1$ rahe aur hum ek upper tail use karein: $F = \frac{s _{1}^{2}}{s _{2}^{2}} = \frac{45}{15} = 3.0, ν_{1} = n_{1} - 1 = 7, ν_{2} = n_{2} - 1 = 9.$ Two-sided $α = 0.05$ par $F_{0.025, 7, 9} \approx 4.20$ use karta hai. $3.0 < 4.20$ ⇒ variances ki equality ko fail to reject. F kyun? Har $s^{2}$ (scale tak) ek $χ^{2} / ν$ hai, aur dono ka ratio by definition $F$ hota hai — dekho F-distribution aur Chi-squared Distribution. Step 2 — agla test. Kyunki hum equal variances reject nahi kar sake, means ke liye two-sample t-test with pooled variance (equal-variance assumption satisfied) appropriate follow-up hai. Agar F reject karta, toh hum unequal-variance (Welch) t-test use karte. Yeh chaining exactly ANOVA ke peeche ka logic hai: F-ratios decide karte hain ki group spreads/means different hain ya nahi.

L4·Q2 — p-value se decision, carefully

Ek two-sided z-test deta hai $Z = 1.7$ . (a) Kya yeh $α = 0.05$ par significant hai? (b) $α = 0.10$ par? (c) Ek student conclude karta hai "p-value bada sa hai, toh $H_{0}$ sach hai." Unhe correct karo.

Recall Solution

(a) Cutoff $z_{0.025} = 1.96$ . $1.7 < 1.96$ ⇒ $5%$ par fail to reject. (b) Cutoff $z_{0.05} = 1.645$ . $1.7 > 1.645$ ⇒ $10%$ par reject. Bilkul wahi data jab $α$ change hota hai conclusion flip kar deta hai — significance ek chosen threshold hai, nature ka fact nahi. (c) Fail to reject matlab hum $H_{0}$ ke against insufficient evidence paaye, yeh proof nahi hai uska. Absence of evidence ≠ evidence of absence. Ek bada sample shayad line cross kar leta. Dekho p-value aur Type I and Type II Errors.

Level 5 — Mastery

L5·Q1 — Design karo aur trap pakdo

Tumhare paas $μ_{0} = 10$ claim ki gayi quantity ke $n = 4$ measurements hain $12, 15, 11, 14$ , $σ$ unknown. Ek colleague report karta hai ek z-test "reject at 5% since $Z = 2.5 > 1.96$ ." Unka number reproduce karo, phir analysis correct karo.

Recall Solution

Reproduce karo (unka flawed path). $\overset{x}{ˉ} = \frac{12 + 15 + 11 + 14}{4} = \frac{52}{4} = 13$ . Deviations $- 1, 2, - 2, 1$ ; squares $1, 4, 4, 1$ , sum $= 10$ . $s^{2} = \frac{10}{4 - 1} = 3.333$ , $s = 1.826$ . $SE = 1.826/ 4 = 0.913$ . Statistic $= \frac{13 - 10}{0.913} = 3.286$ . (Colleague ne shayad $n - 1$ ki jagah $n$ se divide kiya: $s^{2} = 10/4 = 2.5$ ke saath, $s = 1.581$ , SE $= 0.79$ , stat $= 3.80$ — ya simply "2.5" round kiya. Dono tarah arithmetic crime nahi hai.) Asli error: $σ$ unknown aur $n = 4$ ke saath, yeh ek t-test hai, $ν = n - 1 = 3$ , cutoff $t_{0.025, 3} = 3.18$ . Sahi comparison: statistic $3.286 > 3.18$ ⇒ ==reject $H_{0}$ == — lekin barely, aur honest, wider cutoff $3.18$ use karke, dishonest $1.96$ nahi. z cutoff bahut weak evidence par bhi reject kar deta. Mastery point: wahi data genuinely significant ho sakta hai, phir bhi colleague galat isliye hai kyunki unhone uncertainty understate ki. Galat reason se sahi conclusion phir bhi ek methodological error hai — thode weaker data ke saath ( $\overset{x}{ˉ} = 12.5$ , statistic $\approx 2.74$ ) flawed z-test reject karta hai jabki correct t-test nahi karta.

L5·Q2 — Full χ² with estimated proportion

Ek quality line $k = 5$ severity classes mein defects per shift record karta hai: $O = (20, 15, 10, 8, 7)$ , total $n = 60$ . Ek geometric-type model jismein ek estimated parameter hai predict karta hai $E = (18, 14, 11, 9, 8)$ . $α = 0.05$ par fit test karo.

Recall Solution

Cell surprises $(O - E)^{2} / E$ : $\frac{( 20 - 18 ) ^{2}}{18} = \frac{4}{18} = 0.2222$ ; $\frac{( 15 - 14 ) ^{2}}{14} = \frac{1}{14} = 0.0714$ ; $\frac{( 10 - 11 ) ^{2}}{11} = \frac{1}{11} = 0.0909$ ; $\frac{( 8 - 9 ) ^{2}}{9} = \frac{1}{9} = 0.1111$ ; $\frac{( 7 - 8 ) ^{2}}{8} = \frac{1}{8} = 0.1250$ . Sum: $χ^{2} = 0.2222 + 0.0714 + 0.0909 + 0.1111 + 0.1250 = 0.6206$ . d.f.: $k = 5$ , total-count constraint $- 1$ , ek estimated parameter $m = 1$ , toh $ν = 5 - 1 - 1 = 3$ . Critical $χ_{0.05, 3}^{2} = 7.81$ . Compare: $0.62 < 7.81$ ⇒ fail to reject. Model comfortably fit karta hai. Size rule check karo: har $E_{i} \geq 5$ (sabse chhota $8$ hai), toh χ² approximation valid hai — koi pooling needed nahi.

Recall One-line self-check

Yahan har test (signal) ÷ (noise) hai, "luck kya karta hai" se compare kiya gaya ::: haan — z/t ek mean standardize karte hain, χ² standardized squared count-errors sum karta hai, F do variance estimates ratio karta hai; har ek ka apna reference distribution hai aur apna degrees-of-freedom bookkeeping hai.