Exercises — Confidence intervals

Kya decide karta hai: do sawaal — kya $\sigma$ known hai, aur kya $n$ bada hai ($\ge 30$)? Yahan $\sigma$ unknown hai aur $n=12<30$, isliye hume ==$t$-distribution== use karni hogi. Iske heavier tails $\sigma$ ki jagah $s$ use karne ki extra uncertainty ka bhugtan karte hain. Degrees of freedom $=n-1=12-1=11$.

(a) Point estimate interval ka center hai (Figure s01 mein coral line): $\bar{x}=\frac{0.804+0.936}{2}=\mathrm{0.870.}$ (b) Margin of error $E$ half-width hai — ek mint arrow ki length, center se kisi bhi end tak ki doori: $E=\frac{0.936-0.804}{2}=\mathrm{0.066.}$

99% confidence ka matlab $\alpha =1-0.99=0.01$. Us bacha hua probability ko dono tails mein equally split karte hain, toh har tail mein $\alpha /2=0.005$ aata hai. Upar wale 0.5% tail ko cut off karne wala value hai $z_{0.005}=2.576.$ Agar $\alpha /2$ ki jagah $\alpha$ use karte to sirf ek side guard hoti aur net doosri taraf se leak ho jaata.

Step 1 — standard error $SE$ (sample mean ka spread): $SE=\frac{\sigma }{\sqrt{n}}=\frac{4.0}{\sqrt{64}}=\frac{4.0}{8}=\mathrm{0.5.}$ Step 2 — margin of error $E$ ($z_{0.025}=1.96$): $E=1.96\times 0.5=\mathrm{0.98.}$ Step 3 — interval ($\bar{x}\pm E$, exactly s01 picture): $50.0\pm 0.98=[49.02,50.98].$

Step 0 — check karo ki approximation valid hai. Proportion ke liye normal approximation ko kaafi "successes" aur "failures" chahiye. Common rule of thumb hai $n\hat{p}\ge 10$ aur $n(1-\hat{p})\ge 10$ (ek looser version $\ge 5$ use karta hai): $n\hat{p}=200\times 0.72=144\ge 10,n(1-\hat{p})=200\times 0.28=56\ge 10.$ Dono comfortably bar clear kar lete hain, toh normal approximation safe hai. (Agar koi bhi 5–10 se neeche hota — jaise koi rare event — toh hum exact binomial ya Wilson interval use karte.) Step 1 — point estimate: $\hat{p}=\frac{144}{200}=\mathrm{0.72.}$ Step 2 — standard error $SE$ proportion ke liye (binomial variance $p(1-p)/n$): $SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.72\times 0.28}{200}}=\sqrt{0.001008}=\mathrm{0.03175.}$ Step 3 — margin $E$: $E=1.96\times 0.03175=0.06222$. ${\text{CI}}_{95\%}=0.72\pm 0.0622=[0.6578,0.7822].$

$n=7$, toh $df=6$; $\sigma$ unknown hai aur $n<30\Rightarrow$ $t$ use karo. $SE=\frac{s}{\sqrt{n}}=\frac{0.07}{\sqrt{7}}=\frac{0.07}{2.6458}=\mathrm{0.02646.}$ $E=2.447\times 0.02646=\mathrm{0.06474.}$ ${\text{CI}}_{95\%}=0.50\pm 0.0647=[0.4353,0.5647].$

Margin ki definition se shuru karo aur $n$ ko step by step isolate karo. Step 1 — jo pata hai woh likho: $E=z_{\alpha /2}\cdot \frac{\sigma }{\sqrt{n}}.$ Step 2 — $\sqrt{n}$ ko upar lao. $n$ denominator mein $\sqrt{n}$ ke andar chhupa hai, toh pehle dono sides ko $\sqrt{n}$ se multiply karo aur dono sides ko $E$ se divide karo taaki $\sqrt{n}$ ek taraf akela aa jaye: $\sqrt{n}=\frac{z_{\alpha /2}\sigma }{E}.$ Step 3 — square root hatao. Unknown abhi bhi  ​ ke neeche fansa hai, aur "square root" ka inverse "square" hai, toh dono sides ko square karo (square karna yahan legal hai kyunki har quantity — $z$, $\sigma$, $E$, $n$ — positive hai): $n={\left(\frac{z_{\alpha /2}\sigma }{E}\right)}^2.$ Step 4 — numbers daalo: $n={\left(\frac{1.96\times 4.0}{0.5}\right)}^2=(15.68{)}^2=\mathrm{245.86.}$ Hamesha round up karo (fractional person exist nahi karta aur neeche round karne se target miss ho jaata hai): $n=246.$

Width $\propto 1/\sqrt{n}$ hai. Width aadhi karne ke liye chahiye $\sqrt{n_{\text{new}}}=2\sqrt{n_{\text{old}}}$, yaani $n_{\text{new}}=4n_{\text{old}}=4\times 100=400.$ Sabak: precision mehengi hoti hai — error aadhi karne mein 4× zyada data lagta hai. Figure s02 ise concrete banata hai: yeh interval width ko $n$ ke against plot karta hai. Curve $1/\sqrt{n}$ ki tarah girta hai, aur do marked dots ($n=100$ coral mein, $n=400$ mint mein) ek doosre ki exactly aadhi height par hain — sirf aadha neeche aane ke liye tum chart par chaar guna daayein jaate ho.

Independent estimates ka difference → variances add hote hain: $S{E}_{\text{diff}}=\sqrt{S{E}_A^2+S{E}_B^2}=\sqrt{0.020^2+0.018^2}=\sqrt{0.000724}=\mathrm{0.02691.}$ ${\text{CI}}_{95\%}=(0.87-0.82)\pm 1.96\times 0.02691=0.05\pm 0.05274=[-0.0027,0.1027].$ Interval 0 ko include karta hai, toh hum yeh conclude nahi kar sakte ki B sach mein better hai — matched view ke liye dekho Hypothesis Testing aur A/B Testing.

Mean: $\bar{x}=(10+12+11+13+14)/5=60/5=12$. Deviations aur squares: $(-2{)}^2,(0{)}^2,(-1{)}^2,(1{)}^2,(2{)}^2=4,0,1,1,4$; sum $=10$. Sample variance (Bessel $n-1=4$): $s^2=10/4=2.5$, toh $s=1.5811$. Standard error: $SE=s/\sqrt{5}=1.5811/2.2361=0.7071$. Margin: $E=t_{0.05,4}\times SE=2.132\times 0.7071=1.5075$. Interval — center $\bar{x}=12$, dono taraf cushion $E=1.5075$: ${\text{CI}}_{90\%}=12\pm 1.5075=[12-1.5075,12+1.5075]=[10.492,13.508].$

$t$-interval approximate normality maanta hai (ya bada $n$ taaki CLT smooth kare); $n=8$ ke saath, skew aur outlier dono ko violate karte hain, toh interval mis-centred aur mis-sized hoga. Safer choice hai ek percentile bootstrap interval — 8 scores ko replacement ke saath hazaron baar resample karo aur 5th/95th percentiles padho (dekho Bootstrap Methods aur Cross-validation).

Ek baar compute ho jaane ke baad interval $[10.49,13.51]$ fixed ho jaata hai aur $\mu$ ek fixed (unknown) constant hai — true mean ya toh usme hai ya nahi, probability bacha hi nahi. 90% procedure describe karta hai: agar hum yeh sampling-and-interval recipe baar baar repeat karein, toh lagbhag 90% resulting intervals $\mu$ ko contain karenge.

Planning (expected 0.20 use karo). Proportion SE deta hai $n=\frac{z_{\alpha /2}^{2}p(1-p)}{E^2}=\frac{1.96^2\times 0.20\times 0.80}{0.01^2}=\frac{3.8416\times 0.16}{0.0001}=6146.56\Rightarrow 6147.$ Observed CI with $\hat{p}=0.22$, $n=6147$ (quick validity check: $n\hat{p}=1352\ge 10$, $n(1-\hat{p})=4795\ge 10$, toh normal approximation theek hai): $SE=\sqrt{\frac{0.22\times 0.78}{6147}}=\sqrt{2.7916\times 10^{-5}}=0.005284,$ ${\text{CI}}_{95\%}=0.22\pm 1.96\times 0.005284=0.22\pm 0.01036=[0.2096,0.2304].$ Realised margin ($\pm 1.04\%$) target se thoda zyada hai kyunki observed $\hat{p}=0.22$ ka variance planned $0.20$ se zyada hai — ek sabak: jab sure na ho toh most conservative $p$ ($=0.5$) se plan karo.

Frequentist net: interval random hai, $\mu$ ek fixed machhli hai; 95% nets jo hum phainkein unhe pakad leti hain, lekin koi ek net ya toh pakadti hai ya nahi. Bayesian pond: $\mu$ ko random treat kiya jaata hai ek probability distribution ke saath; 95% credible interval genuinely us posterior probability ka 95% hold karta hai, toh "95% probability hai ki $\mu$ andar hai" wahan legitimate hai. Coincidence: flat (uninformative) prior aur bade $n$ ke saath, posterior likelihood se dominate hota hai aur dono intervals numerically almost identical ho jaate hain.

Red flags:

1. $n=4$ bahut tiny hai. Critical value hai $t_{0.025,3}=3.182$ (jo $z=1.96$ se kaafi upar hai), jo akela net ko bahut wide bana deta hai — study badly under-powered hai.
2. $\sigma$ unknown aur data non-normal. $t$-interval approximate normality par depend karta hai (ya bade $n$ par via CLT); $n=4$ aur skew ke saath, dono nahi milte, toh interval dono mis-centred aur mis-sized hai.
3. CI mein 0 hai. Kyunki $[-2,8]$ zero ko straddle karta hai, effect statistically "no effect" se distinguishable nahi hai — tum real effect ka claim nahi kar sakte.
4. Sirf point estimate "3.0" report karna. Crisp center quote karna chhupa deta hai ki honest interval $[-2,8]$ tak failta hai, jo ek badi range hai; yeh false precision hai (L3-style "wide interval is honest" ka sabak).

Kya karein: zyada data collect karo — L3·Q2 se yaad karo ki width aadhi karne ke liye 4× sample chahiye; non-normality ki wajah se bootstrap percentile interval use karo; interval ko ek formal hypothesis test se back up karo ki effect 0 se alag hai ya nahi; aur hamesha full interval report karo, kabhi sirf point estimate nahi.