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

False. CLT sirf $\bar{X}$ ki shape ke Normal hone se concerned hai, lekin z-test ko denominator mein ek known $\sigma$ bhi chahiye. Unknown $\sigma$ aur chhote $n$ ke saath estimated $s$ extra noise inject karta hai, isliye Student's t-distribution zaroori hai.

False. Ek bada p-value matlab hai ki data $H_0$ ke saath consistent hai — tum uske khilaf evidence dhundhne mein fail ho gaye. Evidence ka na hona, absence ka evidence nahi hai; kaafi saari doosri hypotheses bhi consistent hoti hain.

True. Jab $n$ badhta hai, $s\to \sigma$, toh denominator noisy rehna band ho jaata hai aur Student's t-distribution ki moti tails sikodte sikodte standard Normal Distribution ke saath mil jaati hain.

False. Har term $(O_i-E_i{)}^2/E_i$ hai: numerator squared hai aur $E_i>0$, isliye har term $\ge 0$ hai aur sum kabhi negative nahi ho sakta. Over- aur under-prediction dono positively add hote hain.

True (convention ke mutabiq). Bade $s^2$ ko upar rakhne se $F\ge 1$ enforce hota hai, isliye hum sirf F-distribution ki upper tail consult karte hain, jo standard critical-value tables mein listed hoti hai.

False. $\mathrm{SE}=\sigma /\sqrt{n}$, isliye yeh $n$ ke saath nahi balki $\sqrt{n}$ ke saath shrink karta hai. $n$ ko double karne se SE $1/\sqrt{2}\approx 0.71$ se multiply ho jaata hai; SE ko half karne ke liye $n$ ko chaar guna karna padta hai.

Partly false. Bessel's Correction variance $s^2$ ko unbiased banata hai ($\mathbb{E}[s^2]={\sigma }^2$). Kyunki square root ek nonlinear function hai, $s$ khud thoda biased rehta hai — lekin $s^2$ woh quantity hai jiske liye correction design ki gayi hai.

False. Har sample variance (scale tak) ek ${\chi }^2$ hai jo apne degrees of freedom se divided hai; ek $F$ definition se do independent aisi quantities ka ratio hai. F-test literally "do chi-squareds ki race" hai.

Raw difference tab tak meaningless hai jab tak spread se compare na kiya jaaye. Tumhe "kitne SEs door hai" jaanne ke liye standard error se divide karna padega; $3$ enormous bhi ho sakta hai ya trivial bhi, depending on $\sigma /\sqrt{n}$.

${\chi }^2$ approximation rely karta hai har cell ke roughly Normal hone par, jiske liye $E_i\gtrsim 5$ chahiye. Chhote expected counts statistic ko inflate karte hain aur false rejections dete hain; pehle chhoti categories ko pool karo.

Usi data se estimate kiya gaya har parameter ek degree of freedom cost karta hai: $\nu =k-1-m=6-1-2=3$. $m$ bhool jaane se tumhara critical value bahut bada ho jaata hai aur reject karne mein fail hone ki bias aa jaati hai. Dekho Degrees of Freedom.

Critical value $1.96$ standard Normal Distribution se hai, $t$ se nahi. $\nu =4$ ke liye two-sided $5\%$ critical value $2.78$ hai; $1.96$ use karna fat tails ko ignore karta hai aur $H_0$ ko zyada aasani se reject karta hai.

Reject karne mein fail hona phir bhi ek Type II error ho sakta hai — ek real effect miss karna kyunki sample bahut chhota ya noisy tha. Dekho Type I and Type II Errors. Non-rejection correctness ki guarantee nahi hai.

$\alpha$ Type I error rate hai — ek true $H_0$ ko reject karne ka chance. $\alpha =0.5$ set karne ka matlab hai tum aadhe time galat reject karoge. Miss rate (Type II) ko kam karna $n$ ya effect size badhane se hota hai, $\alpha$ inflate karne se nahi.

Mean ka variance ${\sigma }^2/n$ hai, ${\sigma }^2$ nahi. $n$ independent values ka average karna variance ko $n$ factor se shrink karta hai, giving $\mathrm{SE}=\sigma /\sqrt{n}$. ${\sigma }^2$ use karna noise ko bahut zyada overstate karta hai.

Divide karne se difference standardize ho jaata hai "number of standard errors" mein — ek unitless quantity jo problems mein comparable hai aur ek known reference distribution se seedha padhni jaati hai. Dekho Hypothesis Testing.

$\sigma$ ko estimate $s$ se replace karne par randomness ka ek doosra source add ho jaata hai (denominator ab bhi wobble karta hai). Centre se door jaane ke zyada tarike matlab tails mein zyada probability — exactly Student's t-distribution.

Counts ke liye, ek cell roughly Poisson hoti hai mean $E_i$ ke saath, isliye uska variance bhi $\approx E_i$ hota hai. $E_i$ se divide karna $(O_i-E_i)$ ko ek standardized $z$-jaisi quantity mein badal deta hai; squaring aur summing se ek Chi-squared Distribution milta hai.

Jab $n-1$ deviations pata hon, toh sum-to-zero rule se aakhri wala forced ho jaata hai. Sirf $n-1$ deviations freely vary kar sakte hain, isliye ek-sample $t$ mein $\nu =n-1$ hota hai.

Ek $F$ do independent variance estimates ka ratio hai, jisme se har ek ki apni noisiness hai. ${\nu }_1=n_1-1$ aur ${\nu }_2=n_2-1$ dono shape karte hain ki ratio luck se 1 se kitna door ja sakta hai. Dekho F-distribution.

Humhe "luck akela kya produce karta hai" ke liye ek fixed yardstick chahiye. $H_0$ assume karna statistic ko ek known distribution deta hai; phir hum dekhte hain ki humara observed value us yardstick ke against kitna extreme hai, taaki p-value compute ho sake.

Counts $\sum O_i=\sum E_i=n$ se bande hain: pehle $k-1$ cells fix karo aur aakhri determine ho jaata hai. Woh single constraint $k$ degrees of freedom mein se ek hata deta hai, giving $\nu =k-1$.

$t$-test assume karta hai ki $\bar{X}$ roughly Normal hai. Chhote $n$ ke saath Central Limit Theorem "kick in" nahi kiya hota, isliye heavy skew reference distribution ko tod deta hai aur p-value unreliable ho jaata hai; bade $n$ ya nonparametric test consider karo.

$F=s_1^2/s_2^2=1$, F-distribution ke centre mein baitha hai — sabse kam surprising value. Tum wahan $H_0$ kabhi reject nahi karoge; equal spreads exactly wahi hain jo $H_0$ predict karta hai.

$Z=(\bar{x}-{\mu }_0)/\mathrm{SE}=0$. Zero standard errors door matlab $H_0$ ke saath perfect agreement aur sabse bada possible p-value; claim ke khilaf koi evidence nahi hai.

Yeh standard Normal Distribution $\mathcal{N}(0,1)$ mein converge karta hai. Moti tails, jo noisy $s$ ki wajah se hoti hain, gayab ho jaati hain jab sample size badhne ke saath $s\to \sigma$ ho jaata hai.

Woh cell $(O_i-E_i{)}^2/E_i=0$ contribute karta hai — zero surprise. Perfectly matched cells statistic mein kuch nahi add karti, isliye sirf mismatches total build up karte hain.

Nahi — do-variance F-test spreads ($s_1^2/s_2^2$) compare karta hai, centres nahi. Kaafi saare means compare karna F-statistic ko alag tarike se use karta hai, ANOVA ke andar; do-variance test mean mein shift ke liye blind hai.

Tum boundary par ho: observed statistic exactly critical value par baitha hai. Usual "reject if p-value $<\alpha$" rule se tum reject nahi karoge; conventions alag-alag hain, lekin yeh borderline signal karta hai ki evidence exactly chosen threshold par hai.

Sirf tab agar underlying population khud Normal ho. Known $\sigma$ ke saath denominator fixed hai, lekin chhote $n$ ke liye CLT non-Normal $\bar{X}$ ko rescue nahi kar sakta; raw data ki Normality tab essential hai.