Chapter 6 Exercises

6.23

Let $\bar{X}$ denote the mean of a random sample of size 25 from a gamma-type distribution with $\alpha =4$ and $\beta >0$. Use the central limit theorem to find an approximate 0.954 confidence Interval for $\mu$, the mean of the gamma distribution.

Hint: Base the confidence interval on the random variable $(\bar{X}-4\beta )/(4{\beta }^2/25{)}^{1/2}=5\bar{X}/2\beta -10$

Answer Untuk distribusi gamma dengan parameter $\alpha =4$ dan $\beta$, diketahui:

$$\begin{array}{rl}E[X]& =\alpha \beta =4\beta =\mu \\ \text{Var}(X)& =\alpha {\beta }^2=4{\beta }^2\end{array}$$

Berdasarkan teorema limit pusat, untuk sampel berukuran $n=25$:

$$\begin{array}{rl}\frac{\bar{X}-\mu }{\sqrt{\text{Var}(\bar{X})}}& =\frac{\bar{X}-4\beta }{\sqrt{4{\beta }^2/25}}\\ & =\frac{\bar{X}-4\beta }{2\beta /5}\\ & =\frac{5\bar{X}}{2\beta }-10\stackrel{D}{\to }N(0,1)\end{array}$$

Karena diketahui $\mu =4\beta$, untuk confidence interval 95.4%, dengan $z_{0.023}=2$, dapat diperoleh:

$$\begin{array}{rl}0.954& =P\left(-2<\frac{5\bar{X}}{2\beta }-10<2\right)\\ & =P\left(-2<\frac{5\bar{X}}{2\beta }-10<2\right)\\ & =P\left(8<\frac{5\bar{X}}{2\beta }<12\right)\\ & =P\left(\frac{16\beta }{5}<\bar{X}<\frac{24\beta }{5}\right)\\ & =P\left(\frac{5\bar{X}}{24}<\beta <\frac{5\bar{X}}{16}\right)\\ & =P\left(\frac{5\bar{X}}{6}<4\beta <\frac{5\bar{X}}{4}\right)\\ & =P\left(\frac{5\bar{X}}{6}<\mu <\frac{5\bar{X}}{4}\right)\end{array}$$

$\therefore$ Confidence interval 95.4% untuk $\mu$ adalah $\boxed{\left(\frac{5\bar{X}}{6},\frac{5\bar{X}}{4}\right)}$

6.26

It is known that a random variable $X$ has a Poisson distribution with parameter $\mu$. A sample of 200 observations from this population has a mean equal to 3.4. Construct an approximate 90 percent confidence Interval for $\mu$.

Answer

Untuk distribusi Poisson dengan parameter $\mu$, diketahui:

$$\begin{array}{rl}E[X]& =\mu \\ \text{Var}(X)& =\mu \end{array}$$

Berdasarkan teorema limit pusat, untuk sampel berukuran besar $n=200$: $\frac{\bar{X}-\mu }{\sqrt{\mu /n}}\stackrel{D}{\to }N(0,1)$

Untuk confidence interval 90%, dengan $\alpha =0.10$ sehingga $z_{0.05}=1.645$, dapat diperoleh: $P\left(-1.645<\frac{\bar{X}-\mu }{\sqrt{\mu /n}}<1.645\right)=0.90$

Dengan $\bar{x}=3.4$ dan $n=200$, kita perlu menyelesaikan:

$$\begin{array}{rl}-1.645& <\frac{3.4-\mu }{\sqrt{\mu /200}}<1.645\\ -1.645\sqrt{\frac{\mu }{200}}& <3.4-\mu <1.645\sqrt{\frac{\mu }{200}}\end{array}$$

Karena $n$ besar, dapat diaproksimasikan dengan $\sqrt{\mu /n}\approx \sqrt{\bar{x}/n}=\sqrt{3.4/200}=\sqrt{0.017}=0.1304$

Sehingga:

$$\begin{array}{rl}-1.645(0.1304)& <3.4-\mu <1.645(0.1304)\\ -0.2145& <3.4-\mu <0.2145\\ 3.4-0.2145& <\mu <3.4+0.2145\\ 3.1855& <\mu <3.6145\end{array}$$

$\therefore$ Confidence interval 90% untuk $\mu$ adalah $\boxed{(3.185,3.615)}$

6.27

Let $Y_1<Y_2<\cdots <Y_n$ denote the order statistics of a random sample of size $n$ from a distribution that has p.d.f. $f(x)=3x^2/{\theta }^3,0<x<\theta$, zero elsewhere.

(a) Show that $\mathrm{Pr}(c<Y_n/\theta <1)=1-c^{3n}$, where $0<c<1$ (b) If $n$ is 4 and if the observed value of $Y$, is 2.3, what is a 95 percent confidence interval for 8?

Answer

6.27.a

Diketahui cdf dari $X$:

$F(x)={\int }_0^x\frac{3t^2}{{\theta }^3}dt=\frac{x^3}{{\theta }^3},0<x<\theta$

Berdasarkan CDF of order statistics, maksimum $Y_n$ distribusinya adalah: $F_{Y_n}(y)=[F(y){]}^n={\left(\frac{y^3}{{\theta }^3}\right)}^n=\frac{y^{3n}}{{\theta }^{3n}}$

Maka:

$$\begin{array}{rl}P(c<Y_n/\theta <1)& =P(c\theta <Y_n<\theta )\\ & =F_{Y_n}(\theta )-F_{Y_n}(c\theta )\\ & =\frac{{\theta }^{3n}}{{\theta }^{3n}}-\frac{(c\theta {)}^{3n}}{{\theta }^{3n}}\\ & =\boxed{1-c^{3n}}\end{array}$$

$\therefore$ Terbukti bahwa $\mathrm{Pr}(c<Y_n/\theta <1)=1-c^{3n}$

6.27.b

Untuk $n=4$ dan confidence itnerval 95%, berarti:

$$\begin{array}{rl}P(c<Y_4/\theta <1)& =1-c^{12}=0.95\\ ⟺c^{12}& =0.05\\ ⟺c& =(0.05{)}^{1/12}\\ & =0.7943\end{array}$$

Karena $P(c<Y_4/\theta <1)=0.95$ dan $Y_4=2.3$, dapat peroleh:

$$\begin{array}{rl}& \\ c& <\frac{Y_4}{\theta }<1\\ \frac{Y_4}{1}& <\theta <\frac{Y_4}{c}\\ Y_4& <\theta <\frac{Y_4}{0.7943}\\ 2.3& <\theta <\frac{2.3}{0.7943}\\ 2.3& <\theta <2.896\end{array}$$

$\therefore$ Confidence interval 95% untuk $\theta$ adalah $\boxed{(2.3,2.896)}$

6.29

Let $X_1,X_2,\dots ,X_n$ be a random sample from a gamma distribution with known parameter $\alpha =3$ and unknown $\beta >0$. Discuss the construction of a confidence interval for $\beta$.

Hint: What is the distribution of $2{\sum }_{i=1}^n{X}_i/\beta$? Follow the procedure outlined in Exercise 6.28.

Answer

Dari Gamma Distribution Relationships, diketahui bahwa jika $X_i\sim \text{Gamma}(\alpha ,\beta )$, maka: $\frac{2X_i}{\beta }\sim {\chi }^2(2\alpha )$

Sehingga untuk $\alpha =3$: $\frac{2X_i}{\beta }\sim {\chi }^2(6)$

Karena $X_1,X_2,\dots ,X_n$ saling bebas, maka:

$$\begin{array}{rl}\frac{2{\sum }_{i=1}^n{X}_i}{\beta }& =\sum _{i=1}^n\frac{2X_i}{\beta }\\ & \sim {\chi }^2(6n)\end{array}$$

Untuk konstruksi confidence interval $(1-\alpha )100\%$ :

$$\begin{array}{rl}1-\alpha & =P\left({\chi }_{\alpha /2,6n}^2<\frac{2{\sum }_{i=1}^n{X}_i}{\beta }<{\chi }_{1-\alpha /2,6n}^2\right)\\ & =P\left({\chi }_{\alpha /2,6n}^2<\frac{2{\sum }_{i=1}^n{X}_i}{\beta }<{\chi }_{1-\alpha /2,6n}^2\right)\\ & =\boxed{P\left(\frac{2{\sum }_{i=1}^n{X}_i}{{\chi }_{1-\alpha /2,6n}^2}<\beta <\frac{2{\sum }_{i=1}^n{X}_i}{{\chi }_{\alpha /2,6n}^2}\right)}\end{array}$$

$\therefore$ Confidence interval $(1-\alpha )100\%$ untuk $\beta$ adalah $\boxed{\left(\frac{2{\sum }_{i=1}^n{X}_i}{{\chi }_{1-\alpha /2,6n}^2},\frac{2{\sum }_{i=1}^n{X}_i}{{\chi }_{\alpha /2,6n}^2}\right)}$

6.35

Let $X$ and $Y$ be the means of two independent random sample, each of size $n$, from the respective distributions $N({\mu }_1,{\sigma }^2)$ and $N({\mu }_2,{\sigma }^2)$, where the common variance is known. Find $n$ such that

$\mathrm{Pr}(\bar{X}-\bar{Y}-\sigma /5<{\mu }_1-{\mu }_2<\bar{X}-\bar{Y}+\sigma /5)=0.90$

Answer

Karena $\bar{X}\sim N({\mu }_1,{\sigma }^2/n)$ dan $\bar{Y}\sim N({\mu }_2,{\sigma }^2/n)$ saling bebas, maka:

$$\begin{array}{rl}E[\bar{X}-\bar{Y}]& ={\mu }_1-{\mu }_2\\ & \\ \text{Var}(\bar{X}-\bar{Y})& =\text{Var}(\bar{X})+\text{Var}(\bar{Y})\\ & =\frac{{\sigma }^2}{n}+\frac{{\sigma }^2}{n}\\ & =\frac{2{\sigma }^2}{n}\end{array}$$

Sehingga $\bar{X}-\bar{Y}\sim N\left({\mu }_1-{\mu }_2,\frac{2{\sigma }^2}{n}\right)$

Perhatikan bahwa $\frac{(\bar{X}-\bar{Y})-({\mu }_1-{\mu }_2)}{\sqrt{2{\sigma }^2/n}}=\frac{(\bar{X}-\bar{Y})-({\mu }_1-{\mu }_2)}{\sigma \sqrt{2/n}}\sim N(0,1)$

Sehingga

$$\begin{array}{rl}0.90& =P(\bar{X}-\bar{Y}-\sigma /5<{\mu }_1-{\mu }_2<\bar{X}-\bar{Y}+\sigma /5)\\ & =P(-\sigma /5<(\bar{X}-\bar{Y})-({\mu }_1-{\mu }_2)<\sigma /5)\\ & =P\left(\frac{-\sigma /5}{\sigma \sqrt{2/n}}<\frac{(\bar{X}-\bar{Y})-({\mu }_1-{\mu }_2)}{\sigma \sqrt{2/n}}<\frac{\sigma /5}{\sigma \sqrt{2/n}}\right)\\ & =P\left(\frac{-1/5}{\sqrt{2/n}}<Z<\frac{1/5}{\sqrt{2/n}}\right)\\ & =P\left(-\frac{\sqrt{n}}{5\sqrt{2}}<Z<\frac{\sqrt{n}}{5\sqrt{2}}\right)\end{array}$$

Untuk confidence interval 90%, $z_{0.05}=1.645$. Sehingga:

$$\begin{array}{rl}\frac{\sqrt{n}}{5\sqrt{2}}& =1.645\\ \sqrt{n}& =1.645\times 5\sqrt{2}\\ \sqrt{n}& =8.225\sqrt{2}\\ \sqrt{n}& =11.632\\ n& =135.3\end{array}$$

$\therefore$ Nilai $n$ yang diperlukan adalah $\boxed{n=136}$ (dibulatkan ke atas)