Tugas Kelompok 4

Chapter 6

6.40

Let $X_1,X_2$ be a random sample of size $n=2$ from the distribution having p.d.f. $f(x;\theta )=(1/\theta )e^{-x/\theta }$, $0<x<\infty$, zero elsewhere. We reject $H_0:\theta =2$ and accept $H_1:\theta =1$ if the observed values of $X_1,X_2$, say $x_1,x_2$ are such that

$$\frac{f(x_1;2)f(x_2;2)}{f(x_1;1)f(x_2;1)}\le \frac{1}{2}$$

Here $\Omega =\{\theta :\theta =1,2\}$. Find the significance level of the test and the power of the test when $H_0$ is false.

Answer

Diketahui $f(x;\theta )$ berasal dari distribusi eksponensial dengan mean $\theta$.

fungsi likelihood untuk sampel acak $X_1,X_2$ adalah:

$$\begin{array}{rl}L(\theta ;x_1,x_2)& =f(x_1;\theta )\cdot f(x_2;\theta )\\ & =\left(\frac{1}{\theta }{e}^{-x_1/\theta }\right)\cdot \left(\frac{1}{\theta }{e}^{-x_2/\theta }\right)\\ & =\frac{1}{{\theta }^2}{e}^{-x_1/\theta }{e}^{-x_2/\theta }\\ & =\frac{1}{{\theta }^2}\exp \left(-\frac{x_1+x_2}{\theta }\right)\end{array}$$

Daerah kritis (critical region) C didefinisikan oleh:

$$\begin{array}{rl}\frac{L(2;x_1,x_2)}{L(1;x_1,x_2)}& \le \frac{1}{2}\\ \frac{(1/4)e^{-(x_1+x_2)/2}}{(1/1)e^{-(x_1+x_2)/1}}& \le \frac{1}{2}\\ \frac{1}{4}{e}^{(x_1+x_2)/2}& \le \frac{1}{2}\\ e^{(x_1+x_2)/2}& \le 2\\ \frac{x_1+x_2}{2}& \le \ln (2)\\ x_1+x_2& \le 2\ln (2)\end{array}$$

Jadi, daerah kritisnya adalah $C=\{(x_1,x_2):x_1+x_2\le 2\ln (2)\}$.

Tingkat signifikansi ($\alpha$) adalah probabilitas menolak $H_0$ ketika $H_0$ benar, yaitu ketika $\theta =2$, sehingga $\alpha =P_{H_0}(X_1+X_2\le 2\ln (2))$.

Ketika $\theta =2$, $X_1,X_2\sim \text{iid Gamma}(1,2)$. Maka, $Y=X_1+X_2\sim \text{Gamma}(2,2)$ dengan pdf $g(y)=\frac{y}{4}{e}^{-y/2}$ untuk $y>0$.

$$\begin{array}{rl}\alpha & ={\int }_0^{2\ln (2)}\frac{y}{4}{e}^{-y/2}dy\\ & ={\left[-\frac{y}{2}{e}^{-y/2}-e^{-y/2}\right]}_0^{2\ln (2)}\\ & =\left(-\frac{2\ln (2)}{2}{e}^{-\ln (2)}-e^{-\ln (2)}\right)-(-0-e^0)\\ & =\left(-\ln (2)\cdot \frac{1}{2}-\frac{1}{2}\right)+1\\ & =\frac{1-\ln (2)}{2}\approx 0.1534\end{array}$$

Power dari tes adalah probabilitas menolak $H_0$ ketika $H_1$ benar, yaitu ketika $\theta =1$, sehingga Power = $P_{H_1}(X_1+X_2\le 2\ln (2))$.

Ketika $\theta =1$, $X_1,X_2\sim \text{iid Gamma}(1,1)$. Maka, $Y=X_1+X_2\sim \text{Gamma}(2,1)$ dengan pdf $g(y)=y{e}^{-y}$ untuk $y>0$.

$$\begin{array}{rl}\text{Power}& ={\int }_0^{2\ln (2)}y{e}^{-y}dy\\ & ={\left[-y{e}^{-y}-e^{-y}\right]}_0^{2\ln (2)}\\ & =\left(-2\ln (2)e^{-2\ln (2)}-e^{-2\ln (2)}\right)-(-0-e^0)\\ & =\left(-2\ln (2)\cdot \frac{1}{4}-\frac{1}{4}\right)+1\\ & =-\frac{\ln (2)}{2}-\frac{1}{4}+1\\ & =\frac{3-2\ln (2)}{4}\approx 0.4035\end{array}$$

$\therefore$ Tingkat signifikansi tes adalah $\boxed{\alpha =\frac{1-\ln (2)}{2}\approx 0.1534}$ dan power tes adalah $\boxed{\frac{3-2\ln (2)}{4}\approx 0.4035}$.

Chapter 9

9.12

Let $X$ have a p.d.f. of the form $f(x;\theta )=1/\theta$, $0<x<\theta$, zero elsewhere. Let $Y_1<Y_2<Y_3<Y_4$ denote the order statistics of a random sample of size $4$ from this distribution. Let the observed value of $Y_4$ be $y_4$. We reject $H_0:\theta =1$ and accept $H_1:\theta \ne 1$ if either $y_4\le \frac{1}{2}$ or $y_4\ge 1$. Find the power function $K(\theta )$, $0<\theta$, of the test.

Answer

Diketahui $f(x;\theta )$ berasal dari distribusi uniform $U(0,\theta )$. CDF-nya adalah $F(x)=x/\theta$ untuk $0<x<\theta$.

CDF dari statistik terurut terbesar, $Y_4$, dari sampel ukuran $n=4$ adalah: $G(y_4)=[F(y_4){]}^4=(y_4/\theta {)}^4,0<y_4<\theta$

Daerah penolakan adalah $C=\{y_4:y_4\le 1/2\text{ atau }{y}_4\ge 1\}$. $K(\theta )=P(Y_4\le 1/2|\theta )+P(Y_4\ge 1|\theta )$

Kasus 1: $0<\theta \le 1/2$

• $P(Y_4\le 1/2)=1$
• $P(Y_4\ge 1)=0$, karena $Y_4<\theta \le 1/2$. Jadi, $K(\theta )=1$.

Kasus 2: $1/2<\theta <1$

• $P(Y_4\le 1/2)=G(1/2)=((1/2)/\theta {)}^4=1/(16{\theta }^4)$.
• $P(Y_4\ge 1)=0$, karena $Y_4<\theta <1$. Jadi, $K(\theta )=1/(16{\theta }^4)$.

Kasus 3: $\theta \ge 1$

• $P(Y_4\le 1/2)=G(1/2)=((1/2)/\theta {)}^4=1/(16{\theta }^4)$.
• $P(Y_4\ge 1)=1-P(Y_4<1)=1-G(1)=1-(1/\theta {)}^4$. Jadi, $K(\theta )=\frac{1}{16{\theta }^4}+1-\frac{1}{{\theta }^4}=1-\frac{15}{16{\theta }^4}$.

Menggabungkan semua kasus, fungsi power adalah: $K(\theta )=\{\begin{array}{ll}1& 0<\theta \le 1/2\\ 1/(16{\theta }^4)& 1/2<\theta <1\\ 1-15/(16{\theta }^4)& \theta \ge 1\end{array}$

$\therefore$ Fungsi power dari tes tersebut adalah $\boxed{K(\theta )=\{\begin{array}{ll}1& 0<\theta \le 1/2\\ 1/(16{\theta }^4)& 1/2<\theta <1\\ 1-15/(16{\theta }^4)& \theta \ge 1\end{array}}$.

9.14

Consider the two normal distributions $N({\mu }_1,400)$ and $N({\mu }_2,225)$. Let $\theta ={\mu }_1-{\mu }_2$. Let $\bar{x}$ and $\bar{y}$ denote the observed means of two independent random samples, each of size $n$, from these two distributions. We reject $H_0:\theta =0$ and accept $H_1:\theta >0$ if and only if $\bar{x}-\bar{y}\ge c$. If $K(\theta )$ is the power function of this test, find $n$ and $c$ so that $K(0)=0.05$ and $K(10)=0.90$, approximately.

Answer

Diketahui $\bar{X}\sim N({\mu }_1,400/n)$ dan $\bar{Y}\sim N({\mu }_2,225/n)$.

Misalkan $W=\bar{X}-\bar{Y}$. Karena $\bar{X}$ dan $\bar{Y}$ independen, $W$ berdistribusi normal dengan:

• $E[W]=E[\bar{X}]-E[\bar{Y}]={\mu }_1-{\mu }_2=\theta$
• $\text{Var}(W)=\text{Var}(\bar{X})+\text{Var}(\bar{Y})=\frac{400}{n}+\frac{225}{n}=\frac{625}{n}$ Sehingga, $W\sim N(\theta ,625/n)$.

Fungsi power $K(\theta )$ dari tes yang menolak $H_0$ jika $W\ge c$ adalah: $K(\theta )=P(W\ge c)=P\left(\frac{W-\theta }{\sqrt{625/n}}\ge \frac{c-\theta }{\sqrt{625/n}}\right)=P\left(Z\ge \frac{c-\theta }{25/\sqrt{n}}\right)$ dimana $Z\sim N(0,1)$.

Diberikan dua kondisi:

1. $K(0)=0.05$

   $P\left(Z\ge \frac{c}{25/\sqrt{n}}\right)=0.05$.

   Dari tabel normal standar, $z_{0.05}\approx 1.645$.

   Maka, $\frac{c}{25/\sqrt{n}}=1.645⟹c=1.645\frac{25}{\sqrt{n}}$ (1)

2. $K(10)=0.90$.

   $P\left(Z\ge \frac{c-10}{25/\sqrt{n}}\right)=0.90$.

   Dari tabel, nilai z yang sesuai adalah $z_{0.90}=-z_{0.10}\approx -1.282$.

   Maka, $\frac{c-10}{25/\sqrt{n}}=-1.282⟹c-10=-1.282\frac{25}{\sqrt{n}}$ (2)

Substitusikan (1) ke dalam (2):

$$\begin{array}{rl}1.645\frac{25}{\sqrt{n}}-10& =-1.282\frac{25}{\sqrt{n}}\\ (1.645+1.282)\frac{25}{\sqrt{n}}& =10\\ 2.927\frac{25}{\sqrt{n}}& =10\\ \sqrt{n}& =\frac{2.927\times 25}{10}=7.3175\\ n& =(7.3175{)}^2\approx 53.54\end{array}$$

Karena $n$ harus bilangan bulat, $n$ dapat dibulatkan ke atas menjadi $n=54$.

Lalu, $c$ dapat dicari menggunakan $n=54$ dalam persamaan (1): $c=1.645\frac{25}{\sqrt{54}}\approx 1.645\times \frac{25}{7.348}\approx 5.596$

$\therefore$ Nilai $n$ dan $c$ yang memenuhi adalah $\boxed{n=54}$ dan $\boxed{c\approx 5.596}$.

9.18

Let $X_1,X_2,\dots ,X_n$ denote a random sample from a normal distribution $N(\theta ,16)$. Find the sample size $n$ and a uniformly most powerful test of $H_0:\theta =25$ against $H_1:\theta <25$ with power function $K(\theta )$ so that approximately $K(25)=0.10$ and $K(23)=0.90$.

Answer

Diketahui $X_i\sim N(\theta ,16)$, sehingga mean sampel $\bar{X}\sim N(\theta ,16/n)$.

Berdasarkan Teorema Karlin-Rubin (karena keluarga distribusi normal memiliki Monotone Likelihood Ration (mlr)), uniformly most powerful test adalah dengan menolak $H_0$ jika statistik cukup $\bar{X}$ kurang dari suatu nilai kritis $c$.

Bentuk tes UMP: Tolak $H_0$ jika $\bar{X}\le c$.

Fungsi power $K(\theta )$ adalah: $K(\theta )=P(\bar{X}\le c|\theta )=P\left(\frac{\bar{X}-\theta }{4/\sqrt{n}}\le \frac{c-\theta }{4/\sqrt{n}}\right)=\Phi \left(\frac{c-\theta }{4/\sqrt{n}}\right)$ dimana $\Phi$ adalah CDF dari distribusi normal standar.

Diberikan dua kondisi:

1. $K(25)=0.10$ (tingkat signifikansi $\alpha$).

   $\Phi \left(\frac{c-25}{4/\sqrt{n}}\right)=0.10$.

   Dari tabel normal, $z_{0.10}\approx -1.282$.

   Maka, $\frac{c-25}{4/\sqrt{n}}=-1.282⟹c=25-1.282\frac{4}{\sqrt{n}}$ (1)

2. $K(23)=0.90$.

   $\Phi \left(\frac{c-23}{4/\sqrt{n}}\right)=0.90$.

   Dari tabel normal, $z_{0.90}\approx 1.282$.

   Maka, $\frac{c-23}{4/\sqrt{n}}=1.282⟹c=23+1.282\frac{4}{\sqrt{n}}$ (2)

Sehingga dapat diperoleh

$$\begin{array}{rl}25-1.282\frac{4}{\sqrt{n}}& =23+1.282\frac{4}{\sqrt{n}}\\ 2& =2\left(1.282\frac{4}{\sqrt{n}}\right)\\ 1& =\frac{5.128}{\sqrt{n}}\\ \sqrt{n}& =5.128\\ n& =(5.128{)}^2\approx 26.29\end{array}$$

Karena $n$ harus bilangan bulat, kita bulatkan ke atas menjadi $n=27$.

Cari nilai $c$ menggunakan $n=27$ dalam persamaan (2): $c=23+1.282\frac{4}{\sqrt{27}}\approx 23+1.282\frac{4}{5.196}\approx 23+0.987\approx 23.987$

$\therefore$ Ukuran sampel yang dibutuhkan adalah $\boxed{n=27}$, dan tes UMP adalah $\boxed{\text{Tolak }{H}_0\text{ jika }\bar{X}\le 24}$.