qnorm(0.9,0,1)*sqrt(5)
2.865636417229
랜덤표본
수업 과제_카이제곱 분포
1.
\(X_1 \sim \chi^2_m\), \(X_2 \sim \chi^2_n\) 이고 서로 독립이면 \(X_1 + X_2 \sim \chi^2_{m+n}\)
answer
\(Y = X_1 + X_2\)
\(M_Y(t) = M_{X_1 + X_2}(t) = M_{X_1}(t) \times M_{X_2}(t) = (1-2t)^{-\frac{m+n}{2}}\)
적률생성함수의 유일성에 의하여 \(X_1 + X_2 \sim \chi^2_{m+n}\)
2.
서로 독립인 \(X_1\)과 \(X_2\)에 대하여 \(Y = X_1 + X_2\) 라고 할 떄 \(Y \sim \chi^2_m\), \(X_1 \sim \chi^2_n\)이면 \(X_2 \sim \chi^2_{m-n}\)
answer
\(X_2 = Y - X_1\)
\(M_Y (t) = (1-2t)^{\frac{m}{2}}\)
\(M_{X_1}(t) = (1-2t)^{-\frac{n}{2}}\)
\(M_{X_2}(t) = M_{Y-X_1}(t) = E(e^{(y-X_1)t}) = E(e^{Yt} \times e^{-X_1t}) = \frac{M_Y(t)}{M_{X_1}(t)} = (1-2t)^{-\frac{m-n}{2}}\)
\(\therefore X_2 \sim \chi^2_{m-n}\)
6.
모분포 \(N(0,25)\)로부터 랜덤표본 \(X_1,X_2, \dots , X_5\)를 얻었다고 할 때,
(1)
\(P(\bar{X}_5 < c) = 0.90\)을 만족하는 상수 \(c\)값을 구하라.
answer
\(\bar{X}_5 \sim N(0,5)\)
\(P(\bar{X}_t <c) = 0.90\)
\(c = 2.8656\)
(2)
\(P(\frac{1}{5}\sum^5_{i=1} X^2_i <c) = 0.90\)을 만족하는 상수 \(c\)값을 구하라.
answer
\(P(\frac{1}{5} \sum^5_{i=1} X^2_i <c) = 0.90\)
\(\star\)
\(\sum^5_{i=1} X^2_i \sim \chi^2_5\)
\(\frac{1}{5} \sum^5_{i=1} X^2_i \sim \chi^2_1\)
\(\star\)
\(E(\frac{1}{5}\sum^5_{i=1} X^2_i ) =\frac{1}{5} \times 5 = 1\)
\(var(\frac{1}{5}\sum^5_{i=1} X^2_i) = 2\)
\(\therefore c = 2.7055\)
qchisq(0.9,df=1)
2.70554345409542
7.
\(X_1,X-2,\dots,X_n\) 이 \(N(\mu,\sigma^2)\)로부터 구한 랜덤표본이라고 하자.
이때 \(\frac{\sum^n_{i=1}(X_i - \bar{X}_n)^2}{n}\)의 분산을 구하라.
answer
\(S_n^2 = \frac{\sum^n_{i=1} (X_i - \bar{X}_n)^2}{(n-1)}\)
\(\frac{(n-1)S_n^2}{\sigma^2} = \frac{\sum^n_{i=1}(X_i - \bar{X})^2}{\sigma^2} \sim \chi^2_{n-1}\)
\(var(\frac{\sum^n_{i=1}(X_i - \bar{X})^2}{n}) = var(\frac{\sum^n_{i=1}(X_i - \bar{X})^2}{\sigma^2} \times \frac{\sigma^2}{n}) = \frac{\sigma^4}{n^2} var(\frac{\sum^n_{i=1}(X_i - \bar{X})^2}{\sigma^2}) = \frac{\sigma^4}{n^2} \times 2(n-1)\)
12.
\(X_1,X_2, \dots, X_{10}\)이 \(N(\mu,25)\)으로부터 구한 랜덤표본이라고 하자,
이떄 \(P(4< \frac{\sum^{10}_{i=1}(X_i - \bar{X}_{10})^2}{10} <14)\)를 계산하라.
answer
\(S_{10}^2 = \frac{\sum^{10}_{i=1} (X_i - \bar{X}_{10})^2}{9}\), \(\frac{9S_{10}^2}{25} \sim \chi^2_9\)
\(\frac{\sum^{10}_{i=1}(X_i - \bar{X})^2}{10} = \frac{9S^2_{10}}{10} = \frac{9S^2_{10}}{25} \times \frac{25}{10}\)
\(P(4< \frac{\sum^{10}_{i=1}(X_i - \bar{X}_{10})^2}{10} <14)\)
\(= P(4 \times \frac{10}{25} < \frac{9 S^2_{10}}{25} < 14 \times \frac{10}{25})\)
\(= P(\frac{9S^2_{10}}{25} < 14 \times \frac{10}{25}) - P(\frac{9S^2_{10}}{25} < 4 \times \frac{10}{25}) = 0.2171\)
pchisq(14*10/25,df=9) - pchisq(4*10/25,df=9)
0.217146972939312
13.
\(X_1,X_2,\dots,X_{16}\)이 \(N(5,25)\)으로부터 구한 랜덤표본이라고 하자.
이때 \(P(0<\bar{X}_{16}<6,20<\sum^{16}_{i=1}\frac{(X_i - \bar{X}_{16})^2}{16} <30)\)을 계산하라.
answer
\(\bar{X}_{16} \sim N(5,\frac{25}{16})\)
\(P(0<\bar{X}_{16} < 6) = P(-4<\frac{\bar{X}_{16} - 5}{25/16} < 2) = 0.9772\)
pnorm(2,0,1) - pnorm(-4,0,1)
0.977218196809988
\(S^2_{16} = \frac{\sum^{16}_{i=1}(X_i - \bar{X}_{16})^2}{15}\)
\(\frac{15S^2_{16}}{25} = \frac{3S^2_{16}}{5} \sim \chi^2_{15}\)
\(P(20<\sum^{16}_{i=1}\frac{(X_i - \bar{X}_{16})^2}{16} <30)\)
\(= P(20 \times \frac{16}{25} < \frac{(X_i - \bar{X}_{16})^2}{25} < 30 \times \frac{16}{25})\)
\(P(12.8)<\chi^2_{15} <19.2) = 0.4129\)
pchisq(19.2,df=15) - pchisq(12.8,df=15)
0.412929051104458
\(\bar{X}_{16}\)과 \(\sum^{16}_{i=1}(X_i - \bar{X}_{16})^2\)은 독립
\(P(0<\bar{X}_{16}<6,20<\frac{(X_i - \bar{X}_{16})^2}{16}<30) = 0.9772 \times 0.4129 = 0.4035\)
(pnorm(2,0,1) - pnorm(-4,0,1)) * (pchisq(19.2,df=15) - pchisq(12.8,df=15))
0.403521782730758
14.
\(X_1,X_2,\dots, X_{16}\)과 \(Y_1,Y_2,\dots, Y_{25}\)이 각각 \(N(0,9)\)와 \(N(2,16)\)으로부터 구한 서로 독립인 랜덤표본이라고 하자, 이때
\(\star \bar{X}_{16} \sim N(0,\frac{9}{16})\), \(\bar{Y}_{25} \sim N(2,\frac{16}{25})\), 독립
(1)
\(\bar{X}_{16} - \bar{Y}_{25}\)의 분포를 구하라.
answer
\(E(\bar{X}_{16} - \bar{Y}_{25}) = E(0-2) = -2\)
\(Var(\bar{X}_{16} - \bar{Y}_{25}) = \frac{9}{16} + \frac{26}{25} = \frac{481}{400}\)
\(M_{\bar{X}_{16} - \bar{Y}_{25}}(t) = exp(t(0-2) + \frac{t^2}{2}(\frac{9}{16} + \frac{16}{25}))\)
\(\bar{X}_{16} - \bar{Y}_{25} \sim N(-2,\frac{481}{400})\)
(2)
\(P(\bar{X}_{16} - \bar{Y}_{25} >0)\)를 계산하라.
answer
\(P(\bar{X}_{16} - \bar{Y}_{25} > 0) = 1-P(\bar{X}_{16} - \bar{Y}_{25} <0) = 0.0341\)
1-pnorm(0,-2,sqrt(481/400))
0.0340879047226217
17.
모분포 \(N(0,1)\)로부터 랜덤표본 \(X_1,X_2,\dots, X_{50}\)을 얻었다고 할 떄,
(1)
\(P(X_1 - X_2 <2)\)을 계산하라.
answer
\(P(X_1 - X_2 < 2)\), \(X_1 - X_2 \sim N(0,2)\)
\(= P(\frac{(X_1 - X_2) - 0}{\sqrt{2}} <\sqrt{2}) = P(z<\sqrt{2}) = 0.9214\)
pnorm(sqrt(2),0,1)
0.921350396474857
(2)
\(P(X_1 + X_2 <2)\)을 계산하라.
answer
\(P(X_1 + X_2 < 2)\), \(X_1 + X_2 \sim N(0,2)\)
\(= P(\frac{(X_1 + X_2) - 0}{\sqrt{2}} <\sqrt{2}) = P(z<\sqrt{2}) = 0.9214\)
pnorm(sqrt(2),0,1)
0.921350396474857
(3)
\(P(X^2_1 + \dots + X^2_{50} <60)\)을 계산하라.
answer
\(P(X^2_1 + \dots + X^2_{50}<60 ) = 0.8428\)
\(\star X^2_1 + \dots X^2_{50} \sim \chi^2_{50}\)
pchisq(60,df=50)
0.842757972761609
(4)
\(P(40<X^2_1 + \dots + X^2_{50} <60)\)을 계산하라.
answer
\(P(40<X^2_1 + X^2_{50} <60) = P(X^2_1 + \dots + X^2_{50}<60) - P(X^2_1 + \dots X^2_{50} <40) = 0.6860\)
pchisq(60,df=50) - pchisq(40,df=50)
0.685985350935371
(5)
\(P(X^2_1 + \dots + X^2_{50} + c) = 0.9\) 를 만족하는 상수 \(c\)를 구하라.
answer
\(P(X^2_1 + \dots + X^2_{50} < 50+ c) = 0.9\), \(50+c = 63.1671\), \(c=13.1671\)
qchisq(0.9,df=50)
63.1671210057263