Confidence intervals — derivation for mean, proportion: Exercises

4.9.19 · D4 · Maths › Probability Theory & Statistics › Confidence intervals — derivation for mean, proportion

Do recipes ka ek quick reminder jo tum har jagah reuse karoge:

Game ki shape yeh hai — woh picture jo har problem draw karti hai:

Figure — Confidence intervals — derivation for mean, proportion

Tall pastel curve sampling distribution hai. Critical value woh jagah mark karta hai jahan hum har side se tail probability $α /2$ cut off karte hain; ise standard error se multiply karo aur tumhe half-width $E$ (yaani margin of error) milta hai jo hum estimate ke dono sides par lagate hain.

Level 1 — Recognition

L1.1 Har scenario ke liye, bolo ki teen mein se kaun sa formula apply hota hai (z-mean / t-mean / z-proportion): (i) $σ = 4$ known, $n = 50$ , mean estimate karna hai. (ii) $σ$ unknown, $n = 12$ , sample SD diya hua hai, mean estimate karna hai. (iii) 500 mein se 300 voters ne "yes" kaha, $p$ estimate karna hai.

L1.2 Ek 90% confidence interval. Iska $α$ kya hoga, aur kaun sa critical value $z_{α /2}$ lookup karoge?

Recall Solution L1.1

(i) $σ$ known hai → z-mean formula use karo $\overset{ˉ}{X} \pm z_{α /2} σ / n$ . (ii) $σ$ unknown hai aur $n$ small hai → t-mean formula use karo $n - 1 = 11$ degrees of freedom ke saath. (iii) Trials mein se successes ki count → yeh ek proportion hai, $\overset{p}{^} \pm z_{α /2} \overset{p}{^} (1 - \overset{p}{^}) / n$ use karo.

Recall Solution L1.2

Confidence level hai $1 - α = 0.90$ , toh $α = 0.10$ . Hum $α$ ko do tails mein split karte hain, toh har tail mein $α /2 = 0.05$ hai. Critical value hai $z_{0.05} = 1.645$ . (Note: $1.645$ two-tailed 90% value hai — yeh same hi one-sided 95% value bhi hai, jo ek famous trap ka source hai; neeche dekho.)

Level 2 — Application

L2.1 Ek thermostat ki readings ka known $σ = 0. 8^{\circ}$ C hai. $n = 64$ readings ka sample average $\overset{ˉ}{X} = 21. 5^{\circ}$ C hai. True mean temperature ke liye 99% confidence interval banao.

L2.2 $n = 25$ batteries ke ek sample ka mean life $\overset{ˉ}{X} = 40$ hours aur sample SD $s = 5$ hours hai ( $σ$ unknown). 95% CI banao. $t_{0.025, 24} = 2.064$ use karo.

L2.3 Ek survey mein, $200$ mein se $48$ logon ke paas electric car hai. True proportion $p$ ke liye 95% CI banao.

Recall Solution L2.1

Step — standard error. $SE = \frac{σ}{n} = \frac{0.8}{64} = \frac{0.8}{8} = 0.1$ . $n$ kyun? Hum mean bound kar rahe hain, jiska spread $σ / n$ hai, na ki ek reading ka spread. Step — critical value. 99% → $α = 0.01$ → $α /2 = 0.005$ → $z_{0.005} = 2.576$ . Step — margin. $E = 2.576 \times 0.1 = 0.2576$ . Answer. $21.5 \pm 0.2576 = [21.2424, 21.7576]^{\circ}$ C.

Recall Solution L2.2

Step — t kyun. $σ$ unknown hai aur $n = 25$ thoda small hai, toh $n - 1 = 24$ df ke saath t use karo. Step — standard error. $SE = s / n = 5/ 25 = 5/5 = 1$ . Step — margin. $E = t_{0.025, 24} \times SE = 2.064 \times 1 = 2.064$ . Answer. $40 \pm 2.064 = [37.936, 42.064]$ hours.

Recall Solution L2.3

Step — point estimate. $\overset{p}{^} = 48/200 = 0.24$ . Normality check: $n \overset{p}{^} = 48 \geq 5$ ✓ aur $n (1 - \overset{p}{^}) = 152 \geq 5$ ✓. Step — standard error. $SE = \frac{p ^ ( 1 - p ^ )}{n} = \frac{0.24 \times 0.76}{200} = \frac{0.1824}{200} = 0.000912 = 0.030199 \dots$ Step — margin. $E = 1.96 \times 0.030199 = 0.059191$ . Answer. $0.24 \pm 0.0592 = [0.1808, 0.2992]$ .

Level 3 — Analysis

L3.1 Do labs same mean estimate kar rahi hain. Lab A: $n = 100$ , 95% CI half-width $E_{A}$ . Lab B same $σ$ use karti hai lekin $n = 400$ . Bina numbers ke, Lab B ka margin Lab A se kitne factor se chhota hai? Formula ke zariye samjhao.

L3.2 Ek student ne 95% CI $[65.74, 74.26]$ compute kiya (yeh parent note ka example (b) hai, $\overset{ˉ}{X} = 70$ , $s = 8$ , $n = 16$ ). Phir woh claim karta hai: "True mean ke 65.74 aur 74.26 ke beech hone ki 95% probability hai." Kya yeh sahi hai? Bilkul sahi interpretation batao.

L3.3 L3.2 ka interval galat tarike se recompute karo jaise tumne $t_{0.025, 15} = 2.131$ ki jagah $z_{0.025} = 1.96$ use kiya ho. Half-width kitne score points se understate hoti hai? Honest interval kaun sa hai, aur kyun?

Recall Solution L3.1

Margin hai $E = z_{α /2} σ / n$ . Sirf $n$ change hota hai. $\frac{E _{B}}{E _{A}} = \frac{1/ 400}{1/ 100} = \frac{100}{400} = \frac{10}{20} = \frac{1}{2} .$ Lab B ka margin Lab A ka aadha hai. Lesson: Margin ko aadha karne ke liye sample size chaar guna karni padti hai — precision ki cost $n$ mein quadratically hai. (Yahi lever Sample Size Determination mein use hota hai.)

Recall Solution L3.2

Sahi nahi hai. Jab numbers $[65.74, 74.26]$ fix ho jaate hain aur true $μ$ ek fixed (chahe unknown) constant hai, toh $μ$ ya toh andar hai ya bahar — is interval ke liye probability $0$ ya $1$ hai, "95%" bacha hi nahi. Sahi interpretation: "Agar hum puri sampling procedure kai baar repeat karen, toh is tarah banaye gaye lagbhag 95% intervals true mean $μ$ ko contain karengeوت" 95% method ki long-run reliability describe karta hai, ek realized interval ki nahi.

Recall Solution L3.3

Standard error dono ke liye same hai: $s / n = 8/ 16 = 2$ (jo multiplier choose karo usse yeh change nahi hota).

Honest ( $t$ ): $E_{t} = 2.131 \times 2 = 4.262$ .
Wrong ( $z$ ): $E_{z} = 1.96 \times 2 = 3.92$ . Understatement $= 4.262 - 3.92 = 0.342$ score points on half-width. $t$ -interval honest hai. Kyunki $s$ sirf $σ$ ka ek estimate hai aur small $n$ ke liye kam ho sakta hai, true uncertainty badi hoti hai; t ke fatter tails interval ko compensate karne ke liye wide karte hain. $z$ use karna pretend karta hai ki tum $σ$ exactly jaante ho.

Level 4 — Synthesis

L4.1 (invert the margin — mean). Tumhe mean ke liye 95% CI chahiye jiska margin $E = 0.5$ se zyada na ho, aur tumhe pata hai $σ = 3$ . Smallest sample size $n$ kya hai? (Margin formula ko $n$ ke liye solve karo, phir round up karo.)

L4.2 (invert the margin — proportion, worst case). Tumhe proportion ke liye 95% CI chahiye margin $\leq 0.03$ ke saath, lekin tumhare paas $p$ ka koi prior estimate nahi hai. $\overset{p}{^} (1 - \overset{p}{^})$ ki worst-case value use karo. Smallest $n$ dhundho.

L4.3 (CI ↔ test). Mean ke liye ek 99% CI $[48.2, 51.8]$ aata hai. Ek colleague $μ_{0} = 50$ hypothesize karta hai; doosra $μ_{0} = 47$ hypothesize karta hai. Dono ke liye, kya level $α = 0.01$ par do-sided test $μ_{0}$ reject karega? Sirf interval use karke explain karo.

Recall Solution L4.1

$E = z_{α /2} \frac{σ}{n}$ se shuru karo. $n$ ke liye solve karo: $n = \frac{z _{α /2} σ}{E} ⟹ n = (\frac{z _{α /2} σ}{E})^{2} .$ Plug in $z_{0.025} = 1.96$ , $σ = 3$ , $E = 0.5$ : $n = (\frac{1.96 \times 3}{0.5})^{2} = (\frac{5.88}{0.5})^{2} = (11.76)^{2} = 138.2976.$ Sample size ek whole number honi chahiye aur margin at most $0.5$ hona chahiye, toh round up karo: $n = 139$ . (Round up kyun, down kyun nahi? $n = 138$ se margin thodi $0.5$ se badi ho jaati hai, jo requirement violate karta hai.)

Recall Solution L4.2

Margin: $E = z_{α /2} \overset{p}{^} (1 - \overset{p}{^}) / n$ . Product $\overset{p}{^} (1 - \overset{p}{^})$ sabse bada $\overset{p}{^} = 0.5$ par hota hai, jahan yeh $0.25$ hota hai. Yeh worst case use karne se guarantee milti hai ki margin $p$ chahe jo bhi nikle. $n = \frac{z _{α /2}^{2} p ^ ( 1 - p ^ )}{E ^{2}} = \frac{( 1.96 ) ^{2} ( 0.25 )}{( 0.03 ) ^{2}} = \frac{3.8416 \times 0.25}{0.0009} = \frac{0.9604}{0.0009} = 1067.11.$ Round up karo: $n = 1068$ . (Isliye "±3% margin" wale national polls roughly hazaar logon ko sample karte hain.)

Recall Solution L4.3

Rule: ek two-sided level- $α$ test $μ_{0}$ ko reject karta hai exactly tab jab $μ_{0}$ corresponding $(1 - α)$ CI ke bahar hota hai.

$μ_{0} = 50$ : kya $50 \in [48.2, 51.8]$ hai? Haan. Toh hum fail to reject karte hain — 50 ek plausible mean hai.
$μ_{0} = 47$ : kya $47 \in [48.2, 51.8]$ hai? Nahi ( $47 < 48.2$ ). Toh hum $α = 0.01$ par $μ_{0} = 47$ ko reject karte hain. Interval wahi set hai jiske null values tum reject nahi karte — yahi CI↔test duality hai.

Level 5 — Mastery

L5.1 (full proportion build with all checks). Ek quality auditor $n = 400$ items inspect karta hai aur $25$ defective paata hai. (a) Normal-approximation conditions verify karo. (b) Defect rate $p$ ke liye 90% CI banao. (c) Factory guarantee karti hai "defect rate 8% se kam hai." Kya tumhara CI us guarantee ko rule out karta hai, ya 8% abhi bhi plausible hai?

L5.2 (comparing methods on one dataset). $n = 9$ , $\overset{ˉ}{X} = 100$ aur $s = 12$ wale sample ke liye, 95% CI do tarike se banao: (a) sahi tarike se $t_{0.025, 8} = 2.306$ ke saath; (b) galat tarike se $z_{0.025} = 1.96$ ke saath. Dono half-widths report karo aur percentage bolo jitna galat method true half-width ko understate karta hai.

Recall Solution L5.1

(a) Conditions. $\overset{p}{^} = 25/400 = 0.0625$ . Check karo $n \overset{p}{^} = 400 \times 0.0625 = 25 \geq 5$ ✓ aur $n (1 - \overset{p}{^}) = 400 \times 0.9375 = 375 \geq 5$ ✓. Normal approximation valid hai. (b) 90% CI. $α = 0.10$ , $z_{0.05} = 1.645$ . $SE = \frac{0.0625 \times 0.9375}{400} = \frac{0.05859375}{400} = 0.0001464844 = 0.0121030.$ $E = 1.645 \times 0.0121030 = 0.019909$ . CI $= 0.0625 \pm 0.019909 = [0.042591, 0.082409]$ . (c) Kya 8% rule out hai? Interval $0.0824 > 0.08$ tak jaata hai. Kyunki $0.08$ interval ke andar hai, hum is confidence par true defect rate ke 8% hone ko rule out nahi kar sakte — yeh plausible rehta hai. Guarantee statistically contradict nahi hui hai (na hi confirm) is sample se.

Recall Solution L5.2

Standard error dono ke liye common hai: $SE = s / n = 12/ 9 = 12/3 = 4$ . (a) Correct (t): $E_{t} = 2.306 \times 4 = 9.224$ . CI $= 100 \pm 9.224 = [90.776, 109.224]$ . (b) Wrong (z): $E_{z} = 1.96 \times 4 = 7.84$ . CI $= 100 \pm 7.84 = [92.16, 107.84]$ . Understatement percentage: $\frac{E _{t} - E _{z}}{E _{t}} = \frac{9.224 - 7.84}{9.224} = \frac{1.384}{9.224} = 0.15004 \dots \approx 15.0%$ . $z$ -interval lagbhag 15% zyada narrow hai — $n = 9$ ke liye serious overconfidence. Small samples $z$ -shortcut ko sabse zyada punish karte hain, kyunki $t_{n - 1}$ ke tails bahut fatter hote hain jab $n - 1$ tiny ho.

Recap

Recall D4 ke one-line takeaways

$z_{α /2}$ lookup karne se pehle $α$ halve karo (L1).
Width standard error $σ / n$ mein hoti hai, $σ$ mein nahi (L2).
Confidence procedure ke baare mein hai, aur $t$ honest choice hai jab $σ$ estimate kiya gaya ho (L3).
Margin formula invert karo aur target precision hit karne ke liye $n$ round up karo; proportions ke liye worst-case $p = 0.5$ use karo (L4).
Hamesha normality conditions check karo aur claims ko poore interval ke against judge karo (L5).

Connections

Confidence intervals — derivation for mean, proportion — parent derivations jo yeh exercises drill karti hain.
Central Limit Theorem — isliye sampling distribution pehli jagah normal hai.
Standard Error — $σ / n$ jis par har problem hinges karti hai.
Student's t-distribution — L2, L3, L5 mein fatter-tailed critical values.
Hypothesis Testing — L4.3 mein use hone wali CI↔test duality.
Bernoulli & Binomial Distributions — har proportion problem ke peeche $p (1 - p)$ variance.
Sample Size Determination — L4.1 aur L4.2 ka margin-inversion.