Chi-squared, t, F distributions — definition, degrees of freedom: Worked examples

4.9.9 · D3 · Maths › Probability Theory & Statistics › Chi-squared, t, F distributions — definition, degrees of fre

Yeh page parent topic ka case-crunching drill room hai. Definitions wahan hain. Yahan hum har scenario ko dhundte hain jo ek problem throw kar sakti hai — small aur large degrees of freedom, degenerate inputs, limiting behaviour, ek real-world word problem, aur ek exam-style trap — aur har ek ko zero se work karte hain.

Shuru karne se pehle, kuch plain-word groundwork taaki koi bhi symbol unexplained na rahe:

Recall Teeno distributions ka matlab kya hai (ek line mein)

$χ_{k}^{2}$ ::: $k$ independent squared standard normals $Z_{i}^{2}$ ko add karo; total squared wobble measure karta hai. $t_{k}$ ::: ek standard normal $Z$ ko $k$ df se bane estimated standard deviation se divide karo; ek bell with fatter tails. $F_{d_{1}, d_{2}}$ ::: do chi-squareds ka ratio, pehle har ek ko apne df se divide karo; poochta hai "ek wobble doosre se kitne times bada hai?"

Scenario matrix

Is topic ke har sawaal ka ek case class hota hai. Pehle poori table padho — neeche ke examples un cells se labelled hain jo woh hit karte hain, aur milke woh har cell fill karte hain.

Cell	Case class	Kya tricky hai	Example
A	$χ^{2}$ mean/variance, small $k$	bas $E = k$ , $Var = 2 k$ apply karo	Ex 1
B	ek constraint se df ( $n$ vs $n - 1$ )	yaad rakhna ki $\overset{x}{ˉ}$ estimate karna ek df kharach karta hai	Ex 2
C	ek $t$ statistic banana (unknown $σ$ )	$t$ kyun, $Z$ kyun nahi	Ex 3
D	$t^{2} = F_{1, k}$ identity	$Z^{2}$ ko $χ_{1}^{2}$ pehchaanna	Ex 4
E	$F$ do variances ke liye + reciprocal swap	kaunsa df numerator hai; $1/ F$ symmetry	Ex 5
F	Degenerate / boundary inputs ( $k = 1$ , $k = 2$ , $s^{2} = 0$ )	formulas jo $\infty$ ya undefined dete hain	Ex 6
G	Limiting behaviour ( $k \to \infty$ )	$t_{k} \to Z$ , $F \to$ narrow, kyun	Ex 7
H	Real-world word problem ( $n$ , $s$ , df extract karna hai)	English ko $\overset{x}{ˉ}, s, n$ mein translate karna	Ex 8
I	Independent $χ^{2}$ ki Additivity	$χ_{a}^{2} + χ_{b}^{2} = χ_{a + b}^{2}$	Ex 9
J	Exam twist: $t$ ka variance, $F$ ka mean, jab woh exist nahi karte	$(k - 2)$ aur $(d_{2} - 2)$ ke traps	Ex 10

Dhyan do ki yahan angles ki tarah koi "quadrant/sign" cells nahi hain — yeh distributions $[0, \infty)$ par rehti hain ( $χ^{2}$ , $F$ ) ya pure $R$ par ( $t$ ). Humare liye "sign" ka analogue df ki boundary values hain ( $k = 1, 2$ ; $d_{2} \leq 2$ ), jahan means aur variances exist karna band kar dete hain. Cell F aur J unki raksha karte hain.

Neeche ki figure woh geography draw karti hai: black curves $χ_{5}^{2}$ aur $F_{5, 8}$ hain, dono left edge $0$ par pinned (woh kabhi negative nahi ja sakte — woh squares ke sums se bane hain), jabki red curve $t_{5}$ hai, $0$ ke baare mein symmetric aur dono taraf spread. Yeh picture dhyan mein rakho: $χ^{2}$ aur $F$ one-sided hain; $t$ two-sided hai. Neeche ka har example is map par kahin na kahin baith ta hai.

Figure — Chi-squared, t, F distributions — definition, degrees of freedom

Worked examples

Worked example Example 2 — Cell B: sample variance se

$χ^{2}$ , aur df $= n - 1$ kyun $N (μ, σ^{2})$ se $n = 12$ ka sample deta hai $s^{2} = 5$ ; true $σ^{2} = 4$ hai. Test statistic ki value aur distribution do.

Forecast: kya df $12$ hoga ya $11$ ?

Step 1. Statistic $\frac{( n - 1 ) s ^{2}}{σ ^{2}} = \frac{11 \cdot 5}{4} = \frac{55}{4} = 13.75$ . Yeh step kyun? Yeh standard result hai $\frac{( n - 1 ) s ^{2}}{σ ^{2}} \sim χ_{n - 1}^{2}$ Sample Variance and Bessel's Correction se — $s^{2}$ aur $σ^{2}$ ka woh combination jo ek clean chi-squared deta hai.

Step 2. Distribution hai $χ_{n - 1}^{2} = χ_{11}^{2}$ . Yeh step kyun? $\overset{x}{ˉ}$ compute karna $\sum (x_{i} - \overset{x}{ˉ}) = 0$ force karta hai: ek deviation baaki gyarah se determined hai, toh sirf $11$ free hain. Df $=$ data $-$ estimated parameters $= 12 - 1$ .

Verify: observed $13.75$ $χ_{11}^{2}$ ke mean (jo $11$ hai) ke paas baith ta hai — plausible, extreme nahi, toh $σ^{2} = 4$ ke against $s^{2} = 5$ unremarkable hai. ✓

Worked example Example 3 — Cell C: unknown

$σ$ ke saath ek $t$ statistic banao $n = 25$ , sample mean $\overset{x}{ˉ} = 103$ , sample sd $s = 10$ . Test karo ki $μ = 100$ hai ya nahi.

Forecast: $Z$ ya $t$ ? Aur kaunsa df?

Step 1. Standard error $= s / n = 10/ 25 = 10/5 = 2$ . Yeh step kyun? $\overset{x}{ˉ}$ ka spread $σ / n$ hai; kyunki $σ$ unknown hai hum uska estimate $s / n$ use karte hain.

Step 2. $t = \frac{x ˉ - μ}{s / n} = \frac{103 - 100}{2} = 1.5$ . Yeh step kyun? Hum mean ko standardise karte hain, lekin denominator ab random hai ( $s$ se bana hai), aur yahi cheez ise $Z$ nahi balki $t$ banati hai.

Step 3. Reference distribution: $t_{24}$ , kyunki $k = n - 1 = 24$ . Yeh step kyun? Chi-squared jo denominator mein chupi hai, $\frac{( n - 1 ) s ^{2}}{σ ^{2}}$ , $n - 1$ df carry karta hai — dekho Hypothesis Testing ki kaise yeh $t$ ek $p$ -value banta hai.

Verify: $24$ df ke saath, $t_{24}$ $Z$ se almost indistinguishable hai; $1.5$ ki value usual $\pm 2$ band ke andar hai, toh hum $μ = 100$ reject nahi karenge. ✓

Worked example Example 5 — Cell E: do variances ke liye

$F$ + reciprocal swap Sample 1: $s_{1}^{2} = 12$ , $n_{1} = 8$ . Sample 2: $s_{2}^{2} = 3$ , $n_{2} = 6$ . Equal true variances assume karo. $F$ aur uske df do, phir value do agar sample 2 upar hoti.

Forecast: $F$ guess karo aur guess karo ki swapping $1/ F$ deta hai ya nahi.

Step 1. $F = \frac{s _{1}^{2}}{s _{2}^{2}} = \frac{12}{3} = 4$ . Yeh step kyun? Equal true variances ke under, $\frac{s _{1}^{2} / σ ^{2}}{s _{2}^{2} / σ ^{2}} = \frac{s _{1}^{2}}{s _{2}^{2}}$ , aur har $s^{2} / σ^{2}$ ek $χ^{2}$ /df hai — bilkul $F$ recipe.

Step 2. Df: numerator $d_{1} = n_{1} - 1 = 7$ , denominator $d_{2} = n_{2} - 1 = 5$ . Toh $F \sim F_{7, 5}$ . Yeh step kyun? Har sample variance apna $n - 1$ carry karta hai; top variance ka df numerator df hota hai.

Step 3. Swap: $\frac{s _{2}^{2}}{s _{1}^{2}} = \frac{3}{12} = 0.25 = \frac{1}{4}$ , distributed as $F_{5, 7}$ . Yeh step kyun? Reciprocal symmetry: $1/ F_{d_{1}, d_{2}} \sim F_{d_{2}, d_{1}}$ — ratio ko invert karna df swap kar deta hai.

Verify: $0.25 \times 4 = 1$ , confirm karta hai ki dono statistics exact reciprocals hain, aur df waakai $(7, 5)$ se $(5, 7)$ swap ho gaye hain. Yahi $F$ machinery hai jo ANOVA internally run karta hai. ✓

Worked example Example 6 — Cell F: degenerate & boundary inputs

(a) Ek sample $s^{2} = 0$ deta hai. $χ^{2}$ statistic $\frac{( n - 1 ) s ^{2}}{σ ^{2}}$ kya hai? (b) Kya $Var (t_{2})$ exist karta hai? (c) Kya $E [F_{4, 2}]$ exist karta hai?

Forecast: inmen se kaunse blow up ya vanish ho jaate hain?

Step 1 (a). $s^{2} = 0$ matlab har data point $\overset{x}{ˉ}$ ke barabar hai; statistic hai $\frac{( n - 1 ) \cdot 0}{σ ^{2}} = 0$ . Yeh step kyun? Zero spread matlab zero total squared deviation — $χ^{2}$ ka sabse chhota possible value, bilkul apne support ke left edge $0$ par.

Step 2 (b). $Var (t_{k}) = \frac{k}{k - 2}$ ko $k > 2$ chahiye. $k = 2$ par denominator $0$ hai: variance infinite / undefined hai. Yeh step kyun? Itne kam df ke saath fat tails itne heavy hain ki squared values ka finite average nahi hota — mean $E [t_{2}] = 0$ abhi bhi exist karta hai (sirf $k > 1$ chahiye), lekin spread nahi karta.

Step 3 (c). $E [F_{d_{1}, d_{2}}] = \frac{d _{2}}{d _{2} - 2}$ ko $d_{2} > 2$ chahiye. Yahan $d_{2} = 2$ se $\frac{2}{0}$ milta hai: mean undefined hai. Yeh step kyun? Sirf denominator df control karta hai ki mean exist karta hai ya nahi (bottom par ek chhota $χ^{2}$ zero ke paas ho sakta hai, ratio ko $\infty$ par bhej deta hai).

Verify: teeno edges parent ke formula boxes mein parenthetical conditions se align hote hain: $E [t]$ ko $k > 1$ chahiye, $Var (t)$ ko $k > 2$ chahiye, $E [F]$ ko $d_{2} > 2$ chahiye. ✓

Worked example Example 7 — Cell G: df

$\to \infty$ hone par limiting behaviour Jab df bina bound ke badhta hai, toh (a) $t_{k}$ , (b) $Var (t_{k})$ , (c) $\frac{χ _{k}^{2}}{k}$ , aur (d) $F_{d_{1}, d_{2}}$ ka kya hota hai?

Forecast: inmen se kaunse ek bell mein converge hote hain, kaunse number $1$ mein, aur kaunse ek spike mein collapse ho jaate hain?

Step 1 (a). $t_{k} = \frac{Z}{V / k}$ jahan $V \sim χ_{k}^{2}$ . Jab $k \to \infty$ , $V / k \to 1$ , toh $t_{k} \to Z \sim N (0, 1)$ . Yeh step kyun? $V / k$ $k$ independent $Z_{i}^{2}$ 's ka average hai, har ek mean $1$ ; law of large numbers ke zariye yeh average $1$ par lock ho jaata hai, toh random denominator random rehna band ho jaata hai. (Yeh "average settling to its mean" law hai, Central Limit Theorem se alag jo us average ke fluctuations ki shape describe karta hai.)

Step 2 (b). $Var (t_{k}) = \frac{k}{k - 2} \to 1$ jab $k \to \infty$ . Yeh step kyun? $\frac{k}{k - 2} = \frac{1}{1 - 2/ k} \to \frac{1}{1 - 0} = 1$ , jo $Var (Z) = 1$ se match karta hai. Fat tails normal tails mein shrink ho jaati hain.

Step 3 (c). $\frac{χ _{k}^{2}}{k} \to 1$ (in probability). Yeh step kyun? Same law of large numbers — $k$ terms ka mean jo har ek average $1$ hai.

Step 4 (d). $F_{d_{1}, d_{2}} = \frac{U / d _{1}}{V / d _{2}} \to 1$ jab dono $d_{1}, d_{2} \to \infty$ : numerator $U / d_{1} \to 1$ aur denominator $V / d_{2} \to 1$ , toh ratio single value $1$ par collapse ho jaata hai (iska spread $\to 0$ ). Yeh step kyun? Har $χ^{2}$ ko apne df se divide karna bilkul wahi object hai Step 3 se, jo $1$ par pin ho jaata hai; do cheezein jo dono $1$ par pinned hain unka ratio $1$ par pinned hota hai. Iska mean $E [F_{d_{1}, d_{2}}] = \frac{d _{2}}{d _{2} - 2} \to 1$ centre confirm karta hai, aur density $1$ par ek spike mein narrow ho jaati hai — yahi wajah hai ki $1$ ke paas ka $F$ matlab "equal variances."

Verify: numerically trend check karo. $Var (t_{10}) = \frac{10}{8} = 1.25$ , $Var (t_{100}) = \frac{100}{98} \approx 1.0204$ — $1$ ki taraf march kar raha hai. Aur $E [F_{10, 10}] = \frac{10}{8} = 1.25$ jabki $E [F_{100, 100}] = \frac{100}{98} \approx 1.0204$ — yeh bhi $1$ ki taraf ja raha hai. ✓

Figure teeno limits ek saath dikhati hai: black $t$ curves apni heavy tails khote hain aur red normal curve par sink ho jaate hain jab $k$ badhta hai ( $t_{k} \to N (0, 1)$ ), jabki inset dashed curves $χ_{k}^{2} / k$ aur $F_{k, k}$ ke liye value $1$ par narrow ho jaate hain — "sab kuch $1$ par settle ho jaata hai" ka visual meaning yahi hai.

Worked example Example 8 — Cell H: real-world word problem

Ek coffee machine ko $250$ mL daalna chahiye. Ek inspector $n = 16$ cups nikalta hai, average $\overset{x}{ˉ} = 246$ mL aur sample standard deviation $s = 8$ mL paata hai. Appropriate test statistic set up karo aur evaluate karo aur uski reference distribution ka naam do.

Forecast: kaunsi quantities $n$ , $\overset{x}{ˉ}$ , $s$ , $μ$ se map hoti hain? Kya $σ$ known hai?

Step 1. Extract karo: $μ = 250$ (claimed mean), $\overset{x}{ˉ} = 246$ , $s = 8$ , $n = 16$ . True $σ$ nahi diya gaya. Yeh step kyun? "Supposed to pour $250$ " hypothesised mean hai; jo sd hamare paas hai woh sample sd hai, toh $σ$ unknown hai → yeh ek $t$ situation hai, $Z$ nahi.

Step 2. Standard error $= s / n = 8/ 16 = 8/4 = 2$ mL. Yeh step kyun? $16$ cups ka mean ek single cup se $16 = 4$ times kam wobble karta hai.

Step 3. $t = \frac{x ˉ - μ}{s / n} = \frac{246 - 250}{2} = \frac{- 4}{2} = - 2$ . Yeh step kyun? Negative hai kyunki machine underfill kar rahi hai; magnitude $2$ measure karta hai ki target se kitne standard errors neeche hai.

Step 4. Reference distribution $t_{15}$ ( $k = n - 1 = 15$ ). Yeh step kyun? Estimated standard deviation $s$ ek $χ_{n - 1}^{2}$ chupaata hai (humne usi data se $\overset{x}{ˉ}$ compute ki, constraint $\sum (x_{i} - \overset{x}{ˉ}) = 0$ par ek df kharach ki), toh $n = 16$ ke saath $t$ $16 - 1 = 15$ df carry karta hai — kabhi $n$ nahi.

Verify: units sahi cancel hote hain — $mL / mL$ se $t$ dimensionless rahta hai, jaise koi bhi test statistic hona chahiye. $t_{15}$ ke against $- 2$ ki value $5%$ two-sided boundary ke paas hai, toh underfill borderline-significant hai. ✓

Worked example Example 10 — Cell J: exam twist (

$t$ variance, $F$ mean, existence) (a) $Var (t_{5})$ do. (b) $E [F_{6, 10}]$ do. (c) $t_{30}$ ke liye, iska variance normal ke variance se kitne percent zyada hai?

Forecast: teeno $(k - 2)$ / $(d_{2} - 2)$ formulas mein plug hote hain — denominators dekho.

Step 1 (a). $Var (t_{5}) = \frac{k}{k - 2} = \frac{5}{3} \approx 1.6667$ . Yeh step kyun? Yahan $k = 5 > 2$ toh variance exist karta hai; answer $1$ se zyada hai, confirming normal se fatter tails.

Step 2 (b). $E [F_{6, 10}] = \frac{d _{2}}{d _{2} - 2} = \frac{10}{8} = 1.25$ . Yeh step kyun? Sirf denominator df $d_{2} = 10$ mean mein aata hai; $10 > 2$ toh exist karta hai. Numerator df $d_{1} = 6$ mean mein kabhi nahi aata — ek classic trap.

Step 3 (c). $Var (t_{30}) = \frac{30}{28} \approx 1.0714$ , toh yeh normal ke $1$ ke variance se lagbhag $7.14%$ zyada hai. Yeh step kyun? $\frac{30}{28} - 1 = \frac{2}{28} = \frac{1}{14} \approx 0.0714$ — $7%$ excess, chhota hai kyunki $30$ df already normal ke paas hai.

Verify: saare denominators apni existence conditions satisfy karte hain ( $k > 2$ , $d_{2} > 2$ ), toh inmen se koi bhi undefined nahi hai — Ex 6 se contrast karo jahan woh the. Aur $Var (t_{5}) = 1.6667 > Var (t_{30}) = 1.0714 > 1$ , correctly dikhata hai ki zyada df ⇒ normal ke paas. ✓

Recall Rapid self-test

Kaunsa cell: " $n = 20$ , unknown $σ$ , mean test karo"? ::: Cell C — ek $t_{19}$ banao. $t \sim t_{9}$ ; $t^{2}$ ki distribution? ::: $F_{1, 9}$ (Cell D). Kya $Var (t_{2})$ exist karta hai? ::: Nahi — $k > 2$ chahiye (Cell F). $χ_{3}^{2} + χ_{5}^{2}$ independent barabar hai? ::: $χ_{8}^{2}$ (Cell I). $E [F_{5, 10}]$ ? ::: $10/8 = 1.25$ (Cell J).