Chapter9_exercise.lyx

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\begin_body

\begin_layout Part*
Chapter 9.
 Testing Hypotheses
\end_layout

\begin_layout Chapter*
9.1 Problems of Testing Hypotheses
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $X\sim f(x|\beta)=\beta e^{-\beta x}$
\end_inset

, 
\begin_inset Formula $H_{0}:\beta\ge1$
\end_inset

, 
\begin_inset Formula $H_{1}:\beta<1$
\end_inset

.
 
\begin_inset Formula $S_{1}=[1,+\infty)$
\end_inset

.
\end_layout

\begin_layout Exercise*
Power function: 
\begin_inset Formula $\pi(\theta|\delta)=Pr(X\in S_{1}|\beta)=\int_{1}^{+\infty}\beta e^{-\beta x}dx=e^{-\beta}$
\end_inset

.
\end_layout

\begin_layout Exercise*
Size: 
\begin_inset Formula $\sup_{\beta\in\Omega_{0}}\pi(\beta|\delta)$
\end_inset

, 
\begin_inset Formula $\Omega_{0}=[1,+\infty)$
\end_inset

 and 
\begin_inset Formula $\pi(\beta|\delta)=e^{-\beta}$
\end_inset

, we can see 
\begin_inset Formula $\pi(\beta|\delta)$
\end_inset

 is decreasing function so achieve maximum at 
\begin_inset Formula $\beta=1$
\end_inset

 so Size = 
\begin_inset Formula $e^{-1}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
2.
 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from uniform function on the interval 
\begin_inset Formula $[0,\theta]$
\end_inset

, the following hypotheses are to be tested:
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\theta\geq2
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\theta<2
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $Y_{n}=\max\{X_{1},...,X_{n}\}$
\end_inset

, test procedure such that the critical region contains all the outcomes
 for which 
\begin_inset Formula $Y_{n}\leq1.5$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $Y_{n}=\max\{X_{1},...,X_{n}\}$
\end_inset

.
 
\begin_inset Formula $F_{Y_{n}}(y)=(\frac{y}{\theta})^{n}$
\end_inset

, so 
\begin_inset Formula $F_{Y_{n}}^{'}(y)=f(y)=\frac{n}{\theta}(\frac{y}{\theta})^{n-1}$
\end_inset


\end_layout

\begin_layout Exercise*
a.
 Power function: 
\begin_inset Formula $\pi(\theta|\delta)=Pr(T\in R|\beta)$
\end_inset

.
 
\begin_inset Formula $T=Y_{n}$
\end_inset

, 
\begin_inset Formula $R=(-\infty,1.5]$
\end_inset

.
\end_layout

\begin_layout Exercise*
So, 
\begin_inset Formula $\pi(\theta|\delta)=\int_{0}^{1.5}f(y)dy=\int_{0}^{1.5}\frac{n}{\theta}(\frac{y}{\theta})^{n-1}dy=(\frac{y}{\theta})^{n}|_{0}^{1.5}=(\frac{1.5}{\theta})^{n}$
\end_inset


\end_layout

\begin_layout Exercise*
b.
 Size = 
\begin_inset Formula $\sup_{\theta\in\Omega_{0}}\pi(\theta|\delta)=\sup_{\theta\in\Omega_{0}}(\frac{y}{\theta})^{1.5}$
\end_inset

, 
\begin_inset Formula $\Omega_{0}=[2,+\infty)$
\end_inset

.
 So size=
\begin_inset Formula $(\frac{2}{\theta})^{1.5}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
4.
 Following 9.1.8: 
\begin_inset Formula $c=\Phi^{-1}(1-\alpha_{0}/2)\sigma n^{-1/2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example*
6.
 We can have example: 
\begin_inset Formula $P(X\ge3.5|\theta)=1$
\end_inset

 when 
\begin_inset Formula $\theta\geq4$
\end_inset

, because 
\begin_inset Formula $X\in[3.5,4.5]$
\end_inset

 when 
\begin_inset Formula $\theta=4$
\end_inset

, so it equals 1 when 
\begin_inset Formula $\theta>4$
\end_inset

.
 And 
\begin_inset Formula $P(X\ge3.5|\theta)=0$
\end_inset

 when 
\begin_inset Formula $\theta\leq3$
\end_inset

..
 So 
\begin_inset Formula $\delta$
\end_inset

: rejects 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $X\geq3.5$
\end_inset

.
 
\begin_inset Formula $S_{1}:X\geq3.5$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 For 
\begin_inset Formula $\theta\leq4$
\end_inset

.
 We have 2 situation: 
\begin_inset Formula $\theta\leq2.9$
\end_inset

: 
\begin_inset Formula $\pi(\theta|\delta)=1=\pi(\theta|\delta^{*})$
\end_inset

.
 If 
\begin_inset Formula $2.9<\theta\leq4$
\end_inset

, 
\begin_inset Formula $\pi(\theta|\delta)=(\frac{2.9}{\theta})^{n}=\pi(\theta|\delta^{*})$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
b.
 For 
\begin_inset Formula $\theta>4$
\end_inset

, with 
\begin_inset Formula $\theta>4.5$
\end_inset

, 
\begin_inset Formula $\pi(\theta|\delta^{*})=(\frac{2.9}{\theta})^{n}+1-(\frac{4.5}{\theta})^{n}<(\frac{2.9}{\theta})^{n}+1-(\frac{4}{\theta})^{n}=\pi(\theta|\delta)$
\end_inset

.
 If 
\begin_inset Formula $4<\theta\leq4.5$
\end_inset

, 
\begin_inset Formula $\pi(\theta|\delta^{*})=(\frac{2.9}{\theta})^{n}<(\frac{2.9}{\theta})^{n}+1-(\frac{4}{\theta})^{n}=\pi(\theta|\delta)$
\end_inset

.
\end_layout

\begin_layout Exercise*
c.
 We remind the point after Definition 9.1.7: If we have our choice between
 several tests, we would like to choose a test 
\begin_inset Formula $\delta$
\end_inset

 that has small probability error.
 That is, we would like the power function 
\begin_inset Formula $\pi(\theta|\delta)$
\end_inset

 to be low for values of 
\begin_inset Formula $\theta\in\Omega_{0}$
\end_inset

, and we would like 
\begin_inset Formula $\pi(\theta|\delta)$
\end_inset

 to be high for 
\begin_inset Formula $\theta\in\Omega_{1}$
\end_inset

.
 So from this point, we choose 
\begin_inset Formula $\delta$
\end_inset

.
 because 
\begin_inset Formula $\pi(\theta|\delta)=\pi(\theta|\delta^{*})$
\end_inset

 for all sitatuon of 
\begin_inset Formula $S_{0}$
\end_inset

 and 
\begin_inset Formula $\pi(\theta|\delta)>\pi(\theta|\delta^{*})$
\end_inset

 for all situation of 
\begin_inset Formula $S_{1}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 In Exercise 8, assume that 
\begin_inset Formula $Z=z$
\end_inset

 is observed.
 Find a formular for p-value.
\end_layout

\begin_layout Exercise*
From b.
 of Exercise 8, 
\begin_inset Formula $z\geq\Phi^{-1}(1-\alpha_{0})$
\end_inset

, then find 
\begin_inset Formula $\alpha_{0}$
\end_inset

 for this formula with known 
\begin_inset Formula $z$
\end_inset

.
 We see table at the end of book to see quantile function of gaussian distributi
on.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Remind: p-value is smallest level 
\begin_inset Formula $\alpha_{0}$
\end_inset

 such that we would reject the null-hypotheses at level 
\begin_inset Formula $\alpha_{0}$
\end_inset

 with observed data.
\end_layout

\begin_layout Exercise*
13.
 
\begin_inset Formula $X$
\end_inset

 have Poisson distribution with mean 
\begin_inset Formula $\theta$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\theta\leq1.0
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\theta>1.0
\]

\end_inset


\end_layout

\begin_layout Exercise*
Let 
\begin_inset Formula $\delta_{c}$
\end_inset

 be the test that rejects 
\begin_inset Formula $H_{0}$
\end_inset

 if 
\begin_inset Formula $X\geq c$
\end_inset

.
 Find c to make size 
\begin_inset Formula $\delta_{c}$
\end_inset

 as close as possible to 
\begin_inset Formula $0.1$
\end_inset

 without being larger than 0.1
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
14.
 
\begin_inset Formula $X_{1},..,X_{n}$
\end_inset

 be i.i.d.
 with exponential distribution with parameter 
\begin_inset Formula $\theta$
\end_inset

.
 Suppose that we wish to test the hypotheses
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\theta\geq\theta_{0}
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\theta<\theta_{0}
\]

\end_inset


\end_layout

\begin_layout Exercise*
Let 
\begin_inset Formula $X=\sum_{i=1}^{n}X_{i}$
\end_inset

.
 Let 
\begin_inset Formula $\delta_{c}$
\end_inset

 be the test that rejects 
\begin_inset Formula $H_{0}$
\end_inset

 if 
\begin_inset Formula $X\ge c$
\end_inset

.
\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $X$
\end_inset

 is gamma distribution with 
\begin_inset Formula $\alpha=n$
\end_inset

, 
\begin_inset Formula $\beta=\theta$
\end_inset

: 
\begin_inset Formula $\frac{\theta^{n}}{\Gamma(n)}x^{n-1}e^{-\theta x}$
\end_inset


\end_layout

\begin_layout Exercise*
a.
 Show that 
\begin_inset Formula $\pi(\theta|\delta_{c})$
\end_inset

 is a decreasing function of 
\begin_inset Formula $\theta$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $\pi(\theta|\delta_{c})=Pr(X\geq c|\theta)=\int_{c}^{\infty}\frac{\theta^{n}}{\Gamma(n)}x^{n-1}e^{-\theta x}dx$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
15.
 Remind : If all of our tests are of the form 
\begin_inset Quotes eld
\end_inset

reject the null hypothesis when 
\begin_inset Formula $T\geq c$
\end_inset


\begin_inset Quotes erd
\end_inset

 for a single test statistic 
\begin_inset Formula $T$
\end_inset

, There is a straightforward way to compute p-values.
 For each 
\begin_inset Formula $t$
\end_inset

, let 
\begin_inset Formula $\delta_{t}$
\end_inset

 be the test that rejects 
\begin_inset Formula $H_{0}$
\end_inset

 if 
\begin_inset Formula $T\geq t$
\end_inset

, Then p-value when 
\begin_inset Formula $T=t$
\end_inset

 is observed is the size of test 
\begin_inset Formula $\delta_{t}$
\end_inset

 .
 That is, the p-value equals 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
\sup_{\theta\in\Omega_{0}}\pi(\theta|\delta_{t})=\sup_{\theta\in\Omega_{0}}Pr(T\geq t|\theta)\text{ (9.1.12)}
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $Pr(X\ge c|\theta)=\frac{\theta-c}{\theta}=1-\frac{c}{\theta}$
\end_inset

, this is increasing function of 
\begin_inset Formula $\theta$
\end_inset

.
 So it achieve 
\begin_inset Formula $\sup$
\end_inset

 at 
\begin_inset Formula $\theta=1$
\end_inset

.
 and 
\begin_inset Formula $Pr(X\ge c|\theta)=1-c$
\end_inset

 if 
\begin_inset Formula $0<=c<=1$
\end_inset

 and 
\begin_inset Formula $0$
\end_inset

 if 
\begin_inset Formula $c>1$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
16.
 
\begin_inset Formula $Pr[\frac{A}{c_{2}}\leq\sigma^{2}\le\frac{A}{c_{1}}]$
\end_inset

 with 
\begin_inset Formula $A=\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}$
\end_inset

, we can choose 
\begin_inset Formula $c_{1}=\chi_{n-1}^{-1}(\frac{1+\gamma}{2})$
\end_inset

 and 
\begin_inset Formula $c_{2}=-\chi_{n-1}^{-1}(\frac{1+\gamma}{2})$
\end_inset

.
 So that is interval.
 So size=
\begin_inset Formula $\alpha_{0}=1-\gamma$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
Chapter 9.2 Testing Simple Hypotheses
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $f_{0}(x)$
\end_inset

 is p.f.
 of the Bernoulli distribution with parameter 0.3, 
\begin_inset Formula $f_{1}(x)$
\end_inset

 is p.f.
 of the Bernoulli distribution with parameter 0.6.
 Suppose single observation 
\begin_inset Formula $X$
\end_inset

 is taken from a distribution for which the p.d.f is either 
\begin_inset Formula $f_{0}(x)$
\end_inset

 of 
\begin_inset Formula $f_{1}(x)$
\end_inset

, and the following simple hypotheses are to be tested: 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:f(x)=f_{0}(x)
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:f(x)=f_{1}(x)
\]

\end_inset


\end_layout

\begin_layout Standard
Find the test procedure 
\begin_inset Formula $\delta$
\end_inset

 for which the value of 
\begin_inset Formula $\alpha(\delta)+\beta(\delta)$
\end_inset

 is a minimum.
\end_layout

\begin_layout Standard

\series bold
Solution: 
\series default
We kown that observation 
\begin_inset Formula $X$
\end_inset

 have 2 possible value : 
\begin_inset Formula $\{0,1\}$
\end_inset

.
 So we have 2 test procedure: 
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\delta_{1}:\text{ reject \ensuremath{H_{0}} when \ensuremath{X=1}}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\delta_{2}:\text{ reject \ensuremath{H_{0}} when \ensuremath{X=0}}
\]

\end_inset


\end_layout

\begin_layout Standard
With 
\begin_inset Formula $\delta_{1}$
\end_inset

: 
\begin_inset Formula $\alpha(\delta_{1})=Pr(X=1|\theta=0.3)=0.3;\beta(\delta_{1})=Pr(X=0|\theta=0.6)=0.4$
\end_inset

.
 So 
\begin_inset Formula $\alpha(\delta_{1})+\beta(\delta_{1})=0.7$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\delta_{2}:\alpha(\delta_{2})+\beta(\delta_{2})=Pr(X=0|\theta=0.3)+Pr(X=1|\theta=0.6)=0.7+0.6=1.3$
\end_inset


\end_layout

\begin_layout Standard
So we can see 
\begin_inset Formula $\delta_{1}$
\end_inset

 have lower error.
 So 
\begin_inset Formula $\delta$
\end_inset

 is reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $X=1$
\end_inset

and dont reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $X=0$
\end_inset


\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $f_{0}(x)$
\end_inset

 and 
\begin_inset Formula $f_{1}(x)$
\end_inset


\end_layout

\begin_layout Exercise*
Simple hypotheses are to be tested: 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:f(x)=f_{0}(x)
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:f(x)=f_{1}(x)
\]

\end_inset


\end_layout

\begin_layout Exercise*
a.
 Describe a test procedure for which the value of 
\begin_inset Formula $\alpha(\delta)+2\beta(\delta)$
\end_inset

 is a minimum.
 
\end_layout

\begin_layout Exercise*
From 
\series bold
Theorem 9.2.1.
 
\series default
The test procedure achive minimum when reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $\frac{f_{1}(x)}{f_{0}(x)}>\frac{a}{b}=\frac{1}{2}$
\end_inset

, so 
\end_layout

\begin_layout Exercise*
With 
\begin_inset Formula $0\leq x\leq1$
\end_inset

, 
\begin_inset Formula $\frac{f_{1}(x)}{f_{0}(x)}>\frac{1}{2}$
\end_inset

 
\begin_inset Formula $\rightarrow2x>\frac{1}{2}$
\end_inset


\begin_inset Formula $\rightarrow$
\end_inset


\begin_inset Formula $x>\frac{1}{4}$
\end_inset

.
\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $\delta$
\end_inset

 : reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $x>1/4$
\end_inset

 and accepted 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $x\le1/4$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
b.
 With above 
\begin_inset Formula $\delta$
\end_inset

, 
\begin_inset Formula $\alpha(\delta)=Pr(x>1/4|f=f_{0})=3/4$
\end_inset

; 
\begin_inset Formula $\beta(\delta)=Pr(x\le1/4|f=f_{1})=\int_{0}^{1/4}2xdx=x^{2}|_{0}^{1/4}=1/16.$
\end_inset


\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $\alpha(\delta)+\beta(\delta)=3/4+1/16=0.8125$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
3.
 a.
 
\begin_inset Formula $\delta:$
\end_inset

 reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $X\ge1.5$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
b.
 
\begin_inset Formula $\alpha(\delta)+\beta(\delta)=Pr(X\ge1.5|f_{0})+Pr(X<1.5|f_{1})=0+1=1$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
4.
 
\begin_inset Formula $\alpha(\delta)\le0.1$
\end_inset

.
 Testing hypothesis reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $\frac{f_{1}(x)}{f_{0}(x)}>k$
\end_inset

, 
\begin_inset Formula $x>\frac{k}{2}$
\end_inset

 .
 For 
\begin_inset Formula $\alpha(\delta)\le0.1$
\end_inset

, mimimum of 
\begin_inset Formula $\beta(\delta)$
\end_inset

 happen when 
\begin_inset Formula $\alpha(\delta)=0.1$
\end_inset

, So 
\begin_inset Formula $k=1.8$
\end_inset

.
 So 
\begin_inset Formula $\delta$
\end_inset

 : reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $x>0.9$
\end_inset

.
 
\begin_inset Formula $\beta(\delta)=Pr(x<0.9|f_{1})$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
5.
 
\begin_inset Formula $Pr(\bar{X}_{n}\le k^{'}|\theta=5.0)=0.05$
\end_inset

.
 So 
\begin_inset Formula $\frac{k^{'}-5.0}{1/\sqrt{n}}=-\Phi^{-1}(0.95)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
6.
 From Example 9.2.6..
 
\begin_inset Formula $\bar{Y}_{n}\le\frac{k/n-\log[(1-p_{1})/(1-p_{0})]}{\log(p_{1}/p_{0})-\log[(1-p_{1})/(1-p_{0})]}$
\end_inset

 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 From 
\begin_inset Formula $\frac{f_{1}}{f_{0}}\ge k$
\end_inset

, we have 
\begin_inset Formula $\sum_{i=1}^{n}(X_{i}-\mu)^{2}>c$
\end_inset

.
 
\begin_inset Formula $c=\frac{-1/2log(2/3)+k}{1/4-1/6}$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $\alpha(\delta)=Pr(\sum_{i=1}^{n}(X_{i}-\mu)^{2}>k^{'}|\sigma^{2}=2)=0.05$
\end_inset

.
 So 
\begin_inset Formula $\frac{k^{'}}{2}=\chi_{n}^{-1}(0.95)$
\end_inset

.
\end_layout

\begin_layout Exercise*
b 
\begin_inset Formula $n=8$
\end_inset

, 
\begin_inset Formula $k=2*\chi_{8}^{-1}(0.95)=2*15.51=31.02$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 from the Poisson distribution with unknown mean 
\begin_inset Formula $\lambda$
\end_inset

.
 
\begin_inset Formula $\lambda_{1}>\lambda_{0}>0$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\lambda=\lambda_{0}
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\lambda=\lambda_{1}
\]

\end_inset


\end_layout

\begin_layout Exercise*
a.
 
\begin_inset Formula $\alpha(\delta)+\beta(\delta)$
\end_inset

 is minimized when 
\begin_inset Formula $\delta:$
\end_inset

reject 
\begin_inset Formula $H_{0}$
\end_inset

when 
\begin_inset Formula $\frac{f_{1}}{f_{0}}>k=1$
\end_inset

, then 
\begin_inset Formula $\bar{X}_{n}\geq\frac{1/n-(\lambda_{0}-\lambda_{1})}{\log(\lambda_{1}/\lambda_{0})}$
\end_inset


\end_layout

\begin_layout Exercise*
c.
 
\begin_inset Formula $\lambda_{0}=1/4$
\end_inset

, 
\begin_inset Formula $\lambda_{1}=1/2$
\end_inset

, 
\begin_inset Formula $n=20$
\end_inset

.
 from a.
\begin_inset Formula $\bar{X}_{n}>0.43$
\end_inset


\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $\alpha(\delta)=Pr(\bar{X}_{n}>0.43|\lambda=\lambda_{0}=\frac{1}{4})$
\end_inset

 and 
\begin_inset Formula $\beta(\delta)=Pr(\bar{X}_{n}<0.43|\lambda=\lambda_{1}=\frac{1}{2})$
\end_inset

 .
 Following Central limit theorem, 
\begin_inset Formula $\bar{X}_{n}$
\end_inset

 is normal distribiton have 
\begin_inset Formula $\mu=\lambda$
\end_inset

, 
\begin_inset Formula $\sigma^{2}=\frac{\sigma_{x}^{2}}{n}$
\end_inset

.
 Because of Poisson distribution so 
\begin_inset Formula $\sigma_{x}=\lambda$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
11.
 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 fomr a random sample from the normal distribution with unknown mean 
\begin_inset Formula $\mu$
\end_inset

 and known 
\begin_inset Formula $\sigma=2$
\end_inset

, the following simple hypotheses are to be tested: 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\mu=-1
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\mu=1
\]

\end_inset


\end_layout

\begin_layout Exercise*
Determine minimum value of 
\begin_inset Formula $\alpha(\delta)+\beta(\delta)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $X_{i}=\frac{1}{\sqrt{2\pi}2}exp(\frac{-1}{2}\frac{(x-\mu)^{2}}{4})$
\end_inset

.
 so 
\begin_inset Formula $f_{1}(X_{1},...,X_{n})=(\frac{1}{\sqrt{2\pi}2})^{n}exp(\frac{-1}{8}[\sum_{i=1}^{n}(x_{i}-1)^{2}])$
\end_inset

.
\end_layout

\begin_layout Standard
and 
\begin_inset Formula $f_{0}(X_{1},..,X_{n})=(\frac{1}{\sqrt{2\pi}2})^{n}exp(\frac{-1}{8}[\sum_{i=1}^{n}(x_{i}+1)^{2}])$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Formula $\alpha(\delta)+\beta(\delta)$
\end_inset

 minimum when procedure test 
\begin_inset Formula $\delta$
\end_inset

: reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $\frac{f_{1}}{f_{0}}>1$
\end_inset

, 
\begin_inset Formula $exp(\frac{1}{8}[\sum_{i=1}^{n}(x_{i}+1)^{2}-\sum_{i=1}^{n}(x_{i}-1)^{2}])>1$
\end_inset

, 
\begin_inset Formula $\sum_{i=1}^{n}(x_{i}+1)^{2}-\sum_{i=1}^{n}(x_{i}-1)^{2}>0$
\end_inset

, then 
\begin_inset Formula $\bar{X}_{n}>0$
\end_inset

.
\end_layout

\begin_layout Standard
a, 
\begin_inset Formula $n=1$
\end_inset

.
 
\begin_inset Formula $\bar{X}_{n}$
\end_inset

 have 
\begin_inset Formula $\mu^{'}=\mu$
\end_inset

, 
\begin_inset Formula $\sigma^{'2}=\frac{4}{n}$
\end_inset

.
 
\begin_inset Formula $Pr(\bar{X}_{n}>0|\mu=-1)$
\end_inset

.
 
\begin_inset Formula $\bar{X_{n}}\sim N(-1,2)$
\end_inset

.
 so 
\begin_inset Formula $\alpha(\delta)=1-\Phi^{-1}(0.5)=1-0.6915$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\beta(\delta)=Pr(\bar{X}_{n}<0|\mu=1),$
\end_inset

 in this situation, 
\begin_inset Formula $\bar{X}_{n}\sim N(1,2)$
\end_inset

, 
\begin_inset Formula $\beta(\delta)=\Phi^{-1}(-0.5)=1-0.6915$
\end_inset


\end_layout

\begin_layout Standard
so 
\begin_inset Formula $\alpha(\delta)+\beta(\delta)=0.6170$
\end_inset


\end_layout

\begin_layout Chapter*
9.3.
 Uniformly Most Powerful Tests.
\end_layout

\begin_layout Exercise*
1.
 With 
\begin_inset Formula $\lambda_{1}$
\end_inset

 , 
\begin_inset Formula $\lambda_{2}$
\end_inset

 .
 Poisson distribution: 
\begin_inset Formula $f(x)=\frac{e^{-\lambda}\lambda^{x}}{x!}$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $f(X)=\frac{e^{-n\lambda}\lambda^{\sum x_{i}}}{\prod_{i}^{n}x_{i}!}$
\end_inset

.
 
\begin_inset Formula $\frac{f_{2}(x)}{f_{1}(x)}=\frac{e^{n\lambda_{1}}\lambda_{1}^{\sum x_{i}}}{e^{n\lambda_{2}}\lambda_{2}^{\sum x_{i}}}$
\end_inset

.
 So this ratio only depend on vector x through 
\begin_inset Formula $T(x)=\sum_{i}x_{i}$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
So this joint p.f.
 has a monotone likelihood ratio in statistic 
\begin_inset Formula $\sum_{i=1}^{n}x_{i}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
2-5 same exercise 1.9
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
8.
 From Exercise 2.
 the distribution has MLR in 
\begin_inset Formula $T(X)=\sum_{i=1}^{n}(X_{i}-\mu)^{2}$
\end_inset

, we can convert increasing to decreasing or otherwise easily.
 Therefore, by Theorem 9.3.1, a test procedure that rejects 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $Y\geq c$
\end_inset

 will be a UMP test of hypotheses.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9.
 Because of 
\begin_inset Formula $\mu=0$
\end_inset

, so 
\begin_inset Formula $T(X)=\sum_{i=1}^{n}X_{i}^{2}$
\end_inset

.
 The size of this test is 
\begin_inset Formula $\alpha_{0}=Pr(T\ge2|\sigma^{2}=2)$
\end_inset

.
 With 
\begin_inset Formula $n=10,\alpha_{0}=0.05$
\end_inset

.
 
\begin_inset Formula $Pr(\sum_{i=1}^{10}X_{i}^{2}\ge c|\sigma^{2}=2)=0.05\rightarrow Pr(\sum_{i=1}^{10}X_{i}^{2}/2\ge c/2|\sigma^{2}=2)=0.05$
\end_inset

.
 We have 
\begin_inset Formula $\sum_{i=1}^{10}X_{i}^{2}/2$
\end_inset

 is 
\begin_inset Formula $\chi_{10}$
\end_inset

 distribution.
 so 
\begin_inset Formula $c=2\chi_{10}^{-1}(0.95)=2*18.31=36.62$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 We have 
\begin_inset Formula $Y=\sum_{i=1}^{20}X_{i}$
\end_inset

 is statistic make this distribution monotone likelihood ratio.
 It has Binomial distribution, we prove have 
\begin_inset Formula $c$
\end_inset

 integer : 
\begin_inset Formula $Pr(Y\ge c|p=1/2)=0.0577$
\end_inset

, from table we can see 
\begin_inset Formula $c=14$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
11.
 First we prove 
\begin_inset Formula $T=\sum X_{i}$
\end_inset

 is statistic make this distribution monotone likelihood ratio.
 then prove 
\begin_inset Formula $Pr(T\ge c|\lambda=1)=0.0143$
\end_inset

 with 
\begin_inset Formula $n=10$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
12.
 Prove of theorem 9.3.1.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
13.We already prove 
\begin_inset Formula $T(X)=\sum_{i}X_{i}$
\end_inset

 is statistic from 
\begin_inset Formula $f_{1}/f_{0}$
\end_inset

 make this distribution monotone 
\end_layout

\begin_layout Exercise*
a.
 
\begin_inset Formula $Pr(\bar{X}_{4}\leq c|\mu=10)=0.1$
\end_inset

.
 
\begin_inset Formula $\bar{X}_{4}$
\end_inset

 has normal distribution 
\begin_inset Formula $N(10,0.5).$
\end_inset

 So 
\begin_inset Formula $\frac{c-\mu_{0}}{0.5}=-\Phi^{-1}(0.9)$
\end_inset

 or from (9.3.20) 
\begin_inset Formula $c=\mu_{0}-\Phi^{-1}(0.9)*0.5=9.36$
\end_inset

.
 Reject when 
\begin_inset Formula $\bar{X}_{4}\le c$
\end_inset


\end_layout

\begin_layout Exercise*
c.
 
\begin_inset Formula $Pr(\bar{X}_{n}\geq9.36|\mu=11)=1-\Phi(-3.28)=\Phi(3.28)\sim0.9995$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
14.
 we find 
\begin_inset Formula $\alpha_{0}$
\end_inset

 for 
\begin_inset Formula $c$
\end_inset

 in integer.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
15.
 We already have in example 9.3.4: exponential random vairbale has decraase
 MLR in 
\begin_inset Formula $T=\sum_{i}X_{i}$
\end_inset

.
 so it is increasingly with 
\begin_inset Formula $T=-\sum x_{i}$
\end_inset

.
 but 
\begin_inset Formula $H_{0}:\beta\ge1/2$
\end_inset

.
 So 
\begin_inset Formula $T\le c$
\end_inset

 so 
\begin_inset Formula $\bar{X_{n}}\ge c$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
17.
 Ratio likelihood depends both paramter 
\begin_inset Formula $\theta$
\end_inset

 and 
\begin_inset Formula $X$
\end_inset

.
 So it is not monotone 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
19.
 From Theorem 9.3.1 
\begin_inset Formula $\pi(\theta|\delta^{*})$
\end_inset

 is monotone increasing function of 
\begin_inset Formula $\theta$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
a.
 
\begin_inset Formula $\mu\geq0.5$
\end_inset

.
 For 
\begin_inset Formula $\pi(\mu|\delta^{*})\ge0.9$
\end_inset

, when 
\begin_inset Formula $min(\pi(\mu|\delta^{*}))\geq0.9$
\end_inset

, it happen when 
\begin_inset Formula $\mu=0.5$
\end_inset

 becasue of increasing function of 
\begin_inset Formula $\theta$
\end_inset

.
 So repacle 
\begin_inset Formula $\mu=0.5$
\end_inset

 and( 
\begin_inset Formula $\alpha_{0}=0.025$
\end_inset

, we can find c following 
\begin_inset Formula $n$
\end_inset

).
 then 
\begin_inset Formula $Pr(T\ge c|\mu=0.5)\ge0.9$
\end_inset

.
 Find smallest 
\begin_inset Formula $n$
\end_inset

 from here.
\end_layout

\begin_layout Exercise*
b.
 same a.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
9.4 Two-Sided Alternatives
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from the normal distribution with unknown 
\begin_inset Formula $\mu$
\end_inset

 and known variance 
\begin_inset Formula $1$
\end_inset

.
 and it is desired to test the following hypotheses for a given number 
\begin_inset Formula $\mu_{0}$
\end_inset

:
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\mu=\mu_{0}
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\mu\neq\mu_{0}
\]

\end_inset


\end_layout

\begin_layout Exercise*
Consider a test procedure 
\begin_inset Formula $\delta$
\end_inset

 such that the hypothesis 
\begin_inset Formula $H_{0}$
\end_inset

 is rejected if either 
\begin_inset Formula $\bar{X}_{n}\le c_{1}$
\end_inset

 or 
\begin_inset Formula $\bar{X}_{n}\ge c_{2}$
\end_inset

 , and let 
\begin_inset Formula $\pi(\mu|\delta)$
\end_inset

 denote the power function of 
\begin_inset Formula $\delta$
\end_inset

.
 Determine the values of the constants 
\begin_inset Formula $c_{1}$
\end_inset

 and 
\begin_inset Formula $c_{2}$
\end_inset

 such that 
\begin_inset Formula $\pi(\mu_{0}|\delta)=0.1$
\end_inset

 and the function 
\begin_inset Formula $\pi(\mu|\delta)$
\end_inset

 is symmetric with respect to the point 
\begin_inset Formula $\mu=\mu_{0}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
From exercise 9.4.1, we have 
\begin_inset Formula $\alpha_{1}=\alpha_{2}=0.05$
\end_inset

 is make 
\begin_inset Formula $\pi(\mu|\delta)$
\end_inset

 symmetric with respect to the point 
\begin_inset Formula $\mu=\mu_{0}$
\end_inset

.
 We have 
\begin_inset Formula $\bar{X}_{n}$
\end_inset

 have mean is 
\begin_inset Formula $\mu$
\end_inset

 and 
\begin_inset Formula $\sigma=\frac{\sigma}{\sqrt{n}}$
\end_inset

.
 
\end_layout

\begin_layout Solution*
\begin_inset Formula $Pr(\bar{X}_{n}\le c_{1}|\mu=\mu_{0})=0.05$
\end_inset

 and 
\begin_inset Formula $Pr(\bar{X}_{n}\ge c_{2}|\mu=\mu_{0})$
\end_inset

.
 So 
\begin_inset Formula $c_{1}=\mu_{0}+\Phi^{-1}(0.05)*\frac{\sigma}{\sqrt{n}}=\mu_{0}-\Phi^{-1}(0.95)*\frac{\sigma}{\sqrt{n}}=\mu_{0}-1.64*\frac{\sigma}{\sqrt{n}}$
\end_inset

.
 and 
\begin_inset Formula $c_{2}=\mu_{0}+\Phi^{-1}(0.95)*\frac{\sigma}{\sqrt{n}}$
\end_inset

 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
2.
 Conditions of Exercise 
\begin_inset Formula $1$
\end_inset

.
 
\begin_inset Formula $c_{1}=\mu_{0}-1.96n^{-1/2}$
\end_inset

.
 So 
\begin_inset Formula $Pr(\bar{X}_{n}>c_{1}|\mu=\mu_{0})=\Phi(-1.96)=1-\Phi(1.96)=1-0.975=0.025$
\end_inset

.
 So 
\begin_inset Formula $\alpha_{2}=0.1-0.025=0.075$
\end_inset

.
 So 
\begin_inset Formula $c_{2}=\mu_{0}+\Phi^{-1}(0.925)n^{-1/2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
4.
 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from the normal distribution with unknown mean 
\begin_inset Formula $\mu$
\end_inset

 and known variance 
\begin_inset Formula $1$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:0.1\le\mu\le0.2
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\mu<0.1\text{ or }\mu>0.2
\]

\end_inset


\begin_inset Formula $n=23$
\end_inset

.
 Find 
\begin_inset Formula $c_{1}$
\end_inset

 and 
\begin_inset Formula $c_{2}$
\end_inset

 such that 
\begin_inset Formula $\pi(0.1|\delta)=\pi(0.2|\delta)=0.07$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
It is like Example 9.4.5.
 Let 
\begin_inset Formula $c_{1}=0.15-c$
\end_inset

 and 
\begin_inset Formula $c_{2}=0.15+c$
\end_inset

.
 Then 
\begin_inset Formula $\pi(0.1|\delta)=\pi(0.2|\delta)=\Phi[n^{1/2}(0.05-c)]+1-\Phi[n^{1/2}(0.05+c)]=0.07$
\end_inset


\end_layout

\begin_layout Standard
with 
\begin_inset Formula $n=25$
\end_inset

, 
\begin_inset Formula $\Phi[5(0.05-c)]+1-\Phi[5(0.05+c)]=0.07$
\end_inset

, 
\begin_inset Formula $\Phi(0.25+5c)-\Phi(0.25-5c)=0.93$
\end_inset

.
 From this, we can find 
\begin_inset Formula $c$
\end_inset

.
\end_layout

\begin_layout Exercise*
6.
 Suppose that 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from the uniform distribution on the interval 
\begin_inset Formula $[0,\theta]$
\end_inset

, where the value of 
\begin_inset Formula $\theta$
\end_inset

 is unknown.
 Test the following hypotheses:
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\theta\leq3
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\theta>3
\]

\end_inset


\end_layout

\begin_layout Exercise*
a.
 Show that for each level of significance 
\begin_inset Formula $\alpha_{0}$
\end_inset

 (
\begin_inset Formula $0\le\alpha_{0}<1$
\end_inset

), there exists a UMP test that specifies that 
\begin_inset Formula $H_{0}$
\end_inset

 should be rejected if 
\begin_inset Formula $\max\{X_{1},...,X_{n}\}\ge c$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
For 
\begin_inset Formula $\theta=\theta_{1}$
\end_inset

, 
\begin_inset Formula $f_{1}(X)=\frac{X}{\theta_{1}}$
\end_inset

 if 
\begin_inset Formula $X\le\theta_{1}$
\end_inset

 and 
\begin_inset Formula $0$
\end_inset

 if 
\begin_inset Formula $X>\theta_{1}$
\end_inset

.
\end_layout

\begin_layout Exercise*
for 
\begin_inset Formula $X=X_{i}$
\end_inset

, we have 
\begin_inset Formula $f(X)=\frac{X}{\theta}$
\end_inset

 if 
\begin_inset Formula $X\le\theta$
\end_inset

 or 
\begin_inset Formula $0$
\end_inset

 if 
\begin_inset Formula $X>\theta$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $f(X_{1},...,X_{n})=0$
\end_inset

 if 
\begin_inset Formula $max(X_{1},...,X_{n})>\theta$
\end_inset

 of 
\begin_inset Formula $\frac{\prod_{i}X_{i}}{\theta^{n}}$
\end_inset

 if 
\begin_inset Formula $max(X_{1},...,X_{n})\le\theta$
\end_inset

.
\end_layout

\begin_layout Exercise*
Following definition 9.3.2, suppose 
\begin_inset Formula $\theta_{1},\theta_{2}\in\Omega$
\end_inset

, and 
\begin_inset Formula $\theta_{1}<\theta_{2}$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
\begin_inset Formula $\frac{f_{2}(X_{1},...,X_{n})}{f_{1}(X_{1},....,X_{n})}=\frac{\theta_{1}^{n}}{\theta_{2}^{n}}$
\end_inset

 if 
\begin_inset Formula $max(X_{1},...,X_{n})\le\theta_{1}$
\end_inset

 ; or 
\begin_inset Formula $\infty$
\end_inset

 if 
\begin_inset Formula $\theta_{1}<max(X_{1},...,X_{n})$
\end_inset

 
\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $\frac{f_{2}}{f_{1}}$
\end_inset

 is increasing for 
\begin_inset Formula $max(X_{1},..,X_{n})$
\end_inset

 and ratio is depends 
\begin_inset Formula $x$
\end_inset

 through 
\begin_inset Formula $max(X_{1},..,X_{n})$
\end_inset

.
 So 
\begin_inset Formula $max(X_{1},...,X_{n})$
\end_inset

 is increasing MLR of hypothesis tesing.
 So UMP tests specifies that 
\begin_inset Formula $H_{0}$
\end_inset

 should be rejected if 
\begin_inset Formula $max(X_{1},..,X_{n})\ge c$
\end_inset

.
\end_layout

\begin_layout Exercise*
b.
 Determine the value of 
\begin_inset Formula $c$
\end_inset

 for each possible value of 
\begin_inset Formula $\alpha_{0}$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $Pr(max\{X_{1},...,X_{n}\}\geq c)=\alpha_{0}$
\end_inset

, 
\begin_inset Formula $Pr(max\{X_{1},...,X_{n}\}\le c)=1-\alpha_{0}$
\end_inset

.
 
\begin_inset Formula $[Pr(X\le c)]^{n}=1-\alpha_{0}$
\end_inset

, 
\begin_inset Formula $\alpha_{0}=1-(\frac{c}{\theta_{0}})^{n}$
\end_inset

.
 If 
\begin_inset Formula $max(X_{1}...X_{n})>\theta$
\end_inset

 then 
\begin_inset Formula $\alpha_{0}=1$
\end_inset

.
 With 
\begin_inset Formula $\theta_{0}=3$
\end_inset

, because we have increasing MLR, so level achieve at 
\begin_inset Formula $\theta_{0}=3$
\end_inset

 (boundary of 
\begin_inset Formula $\Omega_{0}$
\end_inset

)
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 If we have 
\begin_inset Formula $\alpha_{0},n$
\end_inset

, From exercise 6.b we can found 
\begin_inset Formula $c$
\end_inset

.
 so 
\begin_inset Formula $\pi(\theta|\delta)=Pr(max\{X_{1},...,X_{n}\}\geq c|\theta)=1-(\frac{c}{\theta})^{n}$
\end_inset

.
 From this function we sketch with 
\begin_inset Formula $\theta$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
8.
 Similar to exercise 6.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9 Similar Exercise 7.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 Condition on Exercise 6.
 but suppose now that desired to test the following hypotheses:
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\theta=3
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\theta\neq3
\]

\end_inset


\end_layout

\begin_layout Standard
Consider the test procedure 
\begin_inset Formula $\delta$
\end_inset

 such that the hypothesis 
\begin_inset Formula $H_{0}$
\end_inset

 is rejected if either 
\begin_inset Formula $max(X_{1},..,X_{n})\le c_{1}$
\end_inset

 or 
\begin_inset Formula $max\{X_{1},...,X_{n}\}\ge c_{2}$
\end_inset

 and let 
\begin_inset Formula $\pi(\theta|\delta)$
\end_inset

 denote the power function of 
\begin_inset Formula $\delta$
\end_inset

.
\end_layout

\begin_layout Standard
a.
 Determine the values of the constants 
\begin_inset Formula $c_{1}$
\end_inset

 and 
\begin_inset Formula $c_{2}$
\end_inset

 such that 
\begin_inset Formula $\pi(3|\delta)=0.05$
\end_inset

 and 
\begin_inset Formula $\delta$
\end_inset

 is unbiased.
\end_layout

\begin_layout Standard
Remind that Unbiased Test is a test procedure 
\begin_inset Formula $\delta$
\end_inset

, if for every 
\begin_inset Formula $\theta\in\Omega_{0}$
\end_inset

 and 
\begin_inset Formula $\theta^{'}\in\Omega_{1}$
\end_inset

, we have 
\begin_inset Formula $\pi(\theta|\delta)\le\pi(\theta^{'}|\delta)$
\end_inset

.
\end_layout

\begin_layout Standard
We have 
\begin_inset Formula $\Omega_{0}:\theta=3$
\end_inset

, so 
\begin_inset Formula $\pi_{\theta=3}(\theta|\delta)=Pr(max(X_{1},..,X_{n})\le c_{1})+Pr(max\{X_{1},...,X_{n}\}\ge c_{2})$
\end_inset

 
\begin_inset Formula $=Pr(max(X_{1},..,X_{n})\le c_{1})+1-Pr(max\{X_{1},...,X_{n}\}\le c_{2})$
\end_inset


\end_layout

\begin_layout Standard
From Eq.(9.4.12), we have equations:
\end_layout

\begin_layout Standard
\begin_inset Formula $A_{\theta=3}=(\frac{c_{1}}{\theta})^{n}+1-(\frac{c_{2}}{\theta})^{n}=0.05$
\end_inset

 and 
\begin_inset Formula $\frac{dA}{d\theta}|_{\theta=3}=\frac{-nc_{1}^{n}}{3^{n+1}}+\frac{nc_{2}^{n}}{3^{n+1}}=0$
\end_inset

, From these equations, we can find 
\begin_inset Formula $c_{1},c_{2}$
\end_inset

 depends on 
\begin_inset Formula $n$
\end_inset

.
\end_layout

\begin_layout Standard
We have 
\begin_inset Formula $\frac{dA}{d\theta}|_{\theta=3}$
\end_inset

 because we want 
\begin_inset Formula $\pi_{\theta\neq3}(\theta|\delta)>\pi_{\theta=3}(\theta|\delta)$
\end_inset

.
 So 
\begin_inset Formula $\theta=3$
\end_inset

 is minimum point of function 
\begin_inset Formula $\pi(\theta|\delta)$
\end_inset

.
\end_layout

\begin_layout Exercise*
11.
 Like Exercise 10, we have 2 equations
\end_layout

\begin_layout Exercise*
\begin_inset Formula $\pi(\mu_{0}|\delta)=0.1$
\end_inset

 and 
\begin_inset Formula $\frac{d\pi(\mu|\delta)}{d\mu}|_{\mu=\mu_{0}}=0$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $Pr(\bar{X}_{n}\le c_{1}|\text{\ensuremath{\mu_{0}}})+Pr(\bar{X}_{n}\ge c_{2}|\text{\ensuremath{\mu_{0}}})=0.1$
\end_inset


\end_layout

\begin_layout Exercise*
We have 
\begin_inset Formula $A=Pr(\bar{X}_{n}\le c_{1}|\text{\ensuremath{\mu}})+Pr(\bar{X}_{n}\ge c_{2}|\text{\ensuremath{\mu}})=\Phi(\frac{c_{1}-\mu}{n^{-1/2}})+1-\Phi(\frac{c_{2}-\mu}{n^{-1/2}})$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $\frac{dA}{d\mu}=-n^{1/2}f(\frac{c_{1}-\mu}{n^{-1/2}})+n^{1/2}f(\frac{c_{2}-\mu}{n^{-1/2}})=0(1)$
\end_inset

 with 
\begin_inset Formula $f$
\end_inset

 is p.d.f having c.d.f is 
\begin_inset Formula $\Phi$
\end_inset

.
 From 
\begin_inset Formula $(1)$
\end_inset

, 
\begin_inset Formula $f(\frac{c_{1}-\mu}{n^{-1/2}})=f(\frac{c_{2}-\mu}{n^{-1/2}})$
\end_inset

, Because 
\begin_inset Formula $f$
\end_inset

 is gaussion distribution, so 
\begin_inset Formula $c_{1},c_{2}$
\end_inset

 is symmetric value between 
\begin_inset Formula $\mu$
\end_inset

.
 Because 
\begin_inset Formula $\mu=\mu_{0}$
\end_inset

, 
\begin_inset Formula $c_{1},c_{2}$
\end_inset

 is symmetric between 
\begin_inset Formula $\mu_{0},$
\end_inset

 so we turn problem 11 to problem exercise 1.
 then 
\begin_inset Formula $c_{1}=\mu_{0}-1.645n^{-1.2}$
\end_inset

, 
\begin_inset Formula $c_{2}=\mu_{0}+1.645n^{-1.2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
12.
 a.
 
\begin_inset Formula $F^{-1}(c_{1})+1-F^{-1}(c_{2})=\alpha_{0}$
\end_inset

 .
 
\begin_inset Formula $F$
\end_inset

 is cdf of expnetial with 
\begin_inset Formula $\beta=1$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
13.
 From 
\begin_inset Formula $G(x;n,\theta)=\int_{0}^{x}\frac{\theta^{n}}{(n-1)!}t^{n-1}exp(-t\theta)dt$
\end_inset

 we can derive to this 2 terms.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
9.5.
 The 
\begin_inset Formula $t$
\end_inset

 test
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $n=10$
\end_inset

 samples in Example 8.5.4.
 The hypotheses:
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\mu\le1.2
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\mu>1.2
\]

\end_inset


\end_layout

\begin_layout Exercise*
a, Perform the level 
\begin_inset Formula $\alpha_{0}=0.05$
\end_inset

 test of these hypotheses
\end_layout

\begin_layout Exercise*
\begin_inset Formula $\bar{x}=1.379$
\end_inset

, 
\begin_inset Formula $\sigma^{'}=0.3277$
\end_inset

.
 We reject 
\begin_inset Formula $H_{0}$
\end_inset

 if 
\begin_inset Formula $n^{1/2}(\bar{X}_{n}-\mu_{0})/\sigma^{'}>T_{n-1}^{-1}(1-\alpha_{0})$
\end_inset

.
 
\begin_inset Formula $T_{9}^{-1}(0.95)=1.833$
\end_inset

.
 
\begin_inset Formula $n^{1/2}(\bar{X}_{n}-\mu_{0})/\sigma^{'}=3(1.379-1.2)/0.32277=1.6637<1.833$
\end_inset

.
 So we don't reject null hypotheses.
\end_layout

\begin_layout Exercise*
b.
 Compute the 
\begin_inset Formula $p-$
\end_inset

value.
 
\begin_inset Formula $=1-T_{9}(1.6637)=1-0.925=0.075$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
2.
 Nine observations are selected at random from the normal distribution with
 unknown mean 
\begin_inset Formula $\mu$
\end_inset

 and unknown variance 
\begin_inset Formula $\sigma^{2}$
\end_inset

.
 
\begin_inset Formula $\bar{X}_{n}=22$
\end_inset

 and 
\begin_inset Formula $\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}=72$
\end_inset


\end_layout

\begin_layout Standard
a.
 
\begin_inset Formula $H_{0}:\text{\ensuremath{\mu_{0}\le20} }$
\end_inset

 and 
\begin_inset Formula $H_{1}:\mu>20$
\end_inset

.
 We reject 
\begin_inset Formula $H_{0}$
\end_inset

 if 
\begin_inset Formula $n^{1/2}(\bar{X}_{n}-\mu_{0})/\sigma^{'}>T_{n-1}^{-1}(1-\alpha_{0})$
\end_inset

.
 
\begin_inset Formula $T_{n-1}^{-1}(1-\alpha_{0})=T_{8}^{-1}(0.95)=1.86$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $n^{1/2}(\bar{X}_{n}-\mu_{0})/\sigma^{'}=0.7<1.86$
\end_inset

.
 So we don't reject null hypotheses.
\end_layout

\begin_layout Standard
b.
 Remind : For the two-sided hypothesis, The test has the form 
\begin_inset Quotes eld
\end_inset

reject 
\begin_inset Formula $H_{0}$
\end_inset

 if 
\begin_inset Formula $|U|\ge T_{n-1}^{-1}(1-\alpha_{0}/2)$
\end_inset

.
 
\begin_inset Formula $U=\frac{n^{1/2}(\bar{X}_{n}-\mu_{0})}{\sigma^{'}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $U=0.7$
\end_inset

.
 
\begin_inset Formula $T_{8}^{-1}(1-\frac{0.05}{2})=T_{8}^{-1}(0.975)=2.304>0.7$
\end_inset

.
 So we don't reject null hypotheses.
\end_layout

\begin_layout Standard
c.
 From the data, construct the observed confidence interval for 
\begin_inset Formula $\mu$
\end_inset

 with confidence coefficient 0.95.
 
\end_layout

\begin_layout Standard
\begin_inset Formula $U\in[-T_{8}^{-1}(-0.975),T_{8}^{-1}(0.975)]$
\end_inset


\begin_inset Formula $=[-2.306,2.306]$
\end_inset

.
 So 
\begin_inset Formula $\frac{n^{1/2}(\bar{X}_{n}-\mu_{0})}{\sigma^{'}}\in[-2.306,2.306]$
\end_inset

, 
\begin_inset Formula $\mu_{0}\in[\bar{X}_{n}-\frac{\sigma^{'}}{n^{1/2}}2.306,\bar{X}_{n}+\frac{\sigma^{'}}{n^{1/2}}2.306]$
\end_inset

.
\end_layout

\begin_layout Exercise*
3.
 
\begin_inset Formula $X_{i}=15.6;18.6;18.3;20.1;21.5;18.4;19.1;20.4;19.0$
\end_inset

 .
\end_layout

\begin_layout Exercise*
\begin_inset Formula $n=9$
\end_inset

, 
\begin_inset Formula $\bar{X}_{n}=19.0$
\end_inset

, 
\begin_inset Formula $\sigma^{'}=[\frac{\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}}{n-1}]^{1/2}=1.653$
\end_inset

.
 
\begin_inset Formula $\frac{n^{1/2}(\bar{X}_{n}-\mu_{0})}{\sigma^{'}}=-1.81$
\end_inset

.
 
\begin_inset Formula $\alpha_{0}=0.05$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
We test in two-side test.
 
\begin_inset Formula $H_{0}:\mu=20$
\end_inset

, 
\begin_inset Formula $H_{1}:\mu\neq20$
\end_inset

.
 
\begin_inset Formula $T_{8}^{-1}(1-0.05/2)=2.304>|1.81|$
\end_inset

.
 So we dont reject null hypotheses.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
4.
 
\begin_inset Formula $\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}=\sum_{i=1}^{n}X_{i}^{2}-2\bar{X}_{n}\sum_{i=1}^{n}X_{i}+n\bar{X}_{n}=43.7-2\frac{-11.2}{8}11.2-11.2=63.86$
\end_inset

.
 So we can find 
\begin_inset Formula $U$
\end_inset

, then compare 
\begin_inset Formula $|U|$
\end_inset

 with 
\begin_inset Formula $T_{7}^{-1}(1.-0.1/2)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
5.
 Conditions of Exercise 4.
 And this 
\begin_inset Formula $t$
\end_inset

 test is not to be symmetric and hypothesis 
\begin_inset Formula $H_{0}$
\end_inset

 is to be rejected if either 
\begin_inset Formula $U\le c_{1}$
\end_inset

 or 
\begin_inset Formula $U\ge c_{2}$
\end_inset

.
 
\begin_inset Formula $Pr(U\le c_{1})=0.01$
\end_inset

 and 
\begin_inset Formula $Pr(U\ge c_{2})=0.09$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
From condition, we reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $-T_{8}^{-1}(0.99)\le U\le T_{8}^{-1}(0.91)$
\end_inset

.
 
\end_layout

\begin_layout Standard
From Exercise 
\begin_inset Formula $4$
\end_inset

, 
\begin_inset Formula $U=\frac{n^{1/2}(\bar{X}_{n}-\mu_{0})}{\sigma^{'}}=\frac{\sqrt{8}(-11.2/8-0)}{\sqrt{63.86}}=-3.964$
\end_inset

.
 
\begin_inset Formula $-T_{8}^{-1}(0.99)=-2.896$
\end_inset

, 
\begin_inset Formula $T_{8}^{-1}(0.91)=1.397$
\end_inset

.
 So we dont reject 
\begin_inset Formula $H_{0}$
\end_inset

.
\end_layout

\begin_layout Exercise*
6.
 
\begin_inset Formula $\pi(\mu,\sigma^{2}|\delta)=Pr(U\ge c|\delta)=1-T_{n-1}(c|\psi)$
\end_inset

.
 
\begin_inset Formula $\psi=\frac{n^{1/2}(\mu-\mu_{0})}{\sigma}$
\end_inset

.
 Becase 
\begin_inset Formula $\frac{\mu_{1}-\mu_{0}}{\sigma_{1}}=\frac{\mu_{2}-\mu_{0}}{\sigma_{2}}$
\end_inset

, so 
\begin_inset Formula $\psi_{1}=\psi_{2}$
\end_inset

.
 So 
\begin_inset Formula $T_{n-1}(c|\psi_{1})=T_{n-1}(c|\psi_{2})$
\end_inset

.
 So 
\begin_inset Formula $\pi(\mu_{1},\sigma_{1}^{2}|\delta)=\pi(\mu_{2},\sigma_{2}^{2}|\delta)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 We can convert 
\begin_inset Formula $Y_{i}$
\end_inset

 to 
\begin_inset Formula $X_{i}=Y_{i}+\mu_{0}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
8.
 
\begin_inset Formula $c=\chi_{n}^{-1}(1-\alpha_{0})$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9.
 We compare 
\begin_inset Formula $\chi_{9}^{-1}(0.95)$
\end_inset

 vs 
\begin_inset Formula $\frac{S_{n}^{2}}{\sigma_{0}^{2}}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 From 
\begin_inset Formula $Pr(S_{n}^{2}\le c_{1})=Pr(S_{n}^{2}\ge c_{2})=0.025$
\end_inset

, 
\begin_inset Formula $Pr(S_{n}^{2}/\sigma^{2}\le c_{1}/\sigma^{2})=Pr(S_{n}^{2}/\sigma^{2}\ge c_{2}/\sigma^{2})=0.025$
\end_inset

, but 
\begin_inset Formula $\chi^{2}$
\end_inset

 is not symmetric , so 
\begin_inset Formula $c_{1}=\chi_{9}^{-1}(-0.025)*4=2.18*4$
\end_inset

 , 
\begin_inset Formula $c_{2}=\chi_{9}^{-1}(0.975)*4=19.02*4$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
12.
 
\begin_inset Formula $p-value=1-T_{n-1}(u)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
13.
\begin_inset Formula $U=n^{1/2}(\bar{X}_{n}-\mu_{0})/\sigma^{'}$
\end_inset

, 
\begin_inset Formula $n=170$
\end_inset

 .
 
\begin_inset Formula $U=169^{1/2}(3.2-3)/0.3=8.67$
\end_inset

, 
\begin_inset Formula $p-value=1-T_{169}(8.67)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
14-16 same for finding 
\begin_inset Formula $p-value$
\end_inset

 of two-sided.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
17-18 same for Example 9.5.12.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
9.6 Comparing the Means of Two Normal Distribution
\end_layout

\begin_layout Exercise*
1.
 In Example 9.6.3, On other location, Island Thorns, had five samples 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 with an average aluminum oxide percentage of 
\begin_inset Formula $\bar{X}=18.18$
\end_inset

 with 
\begin_inset Formula $\sum_{i=1}^{5}(X_{i}-\bar{X})^{2}=12.61$
\end_inset

.
 
\begin_inset Formula $Y_{1},...,Y_{5}$
\end_inset

 be the five sample from Ashley Rails in Example 9.6.3.
 Test the null hypothesis that mean aluminum oxide percentages at Ashley
 Rails and Island Thorns are the same versus the alternative that they are
 different at level 
\begin_inset Formula $\alpha_{0}=0.05$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
From 
\series bold
Theorem 9.6.1.
 
\series default

\begin_inset Formula $U=\frac{(m+n-2)^{1/2}(\bar{X}_{m}-\bar{Y}_{n})}{(\frac{1}{m}+\frac{1}{n})^{1/2}(S_{X}^{2}+S_{Y}^{2})^{1/2}}$
\end_inset

, the distribution of 
\begin_inset Formula $U$
\end_inset

 is the 
\begin_inset Formula $t$
\end_inset

 distribution with 
\begin_inset Formula $m+n-2$
\end_inset

 degrees of freedom.
 From Eq.
 (9.6.9), 
\begin_inset Formula $\alpha_{0}$
\end_inset

 two-sided 
\begin_inset Formula $t$
\end_inset

 test rejects 
\begin_inset Formula $H_{0}$
\end_inset

 if 
\begin_inset Formula $|U|\ge c$
\end_inset

 where 
\begin_inset Formula $c=T_{m+n-2}^{-1}(1-\alpha_{0}/2)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $m=n=5$
\end_inset

, 
\begin_inset Formula $\bar{X}_{m}=18.18$
\end_inset

, 
\begin_inset Formula $S_{X}^{2}=12.61$
\end_inset

, 
\begin_inset Formula $\bar{Y}_{n}=17.32$
\end_inset

, 
\begin_inset Formula $\bar{S}_{Y}^{2}=11.01$
\end_inset

.
 
\begin_inset Formula $U=0.791$
\end_inset

.
 
\begin_inset Formula $c=T_{8}^{-1}(0.975)=2.306$
\end_inset

.
 So 
\begin_inset Formula $|U|\le c$
\end_inset

 so we don't reject 
\begin_inset Formula $H_{0}.$
\end_inset


\end_layout

\begin_layout Exercise*
2.
 
\begin_inset Formula $H_{0}:\bar{A}\ge\bar{B}$
\end_inset

, 
\begin_inset Formula $H_{1}:\bar{A}\le\bar{B}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
4.
 
\begin_inset Formula $U=\frac{(m+n-2)^{1/2}(\bar{X}_{m}-\bar{Y}_{n})}{(1/m+k/n)^{1/2}(S_{X}^{2}+\frac{S_{Y}^{2}}{k})^{1/2}}$
\end_inset

.
 
\begin_inset Formula $X_{i}\sim N(\mu_{1},\sigma_{1}^{2})$
\end_inset

, 
\begin_inset Formula $Y_{i}\sim N(\mu_{1},k\sigma_{1}^{2})$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
\begin_inset Formula $Z_{1}=\bar{X}_{m}-\bar{Y}_{n}\sim N(0,(1/m+k/n)\sigma_{1}^{2})$
\end_inset

.
 So, 
\begin_inset Formula $Z=\frac{Z_{1}}{(1/m+k/n)^{1/2}\sigma_{1}}\sim N(0,1)$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $W=(S_{X}^{2}+\frac{S_{Y}^{2}}{k})/\sigma_{1}^{2}\sim\chi_{m+n-2}^{2}$
\end_inset

.
 So 
\begin_inset Formula $U=\frac{Z}{\sqrt{\frac{W}{(m+n-2)}}}$
\end_inset

 is 
\begin_inset Formula $T$
\end_inset

 distribution with 
\begin_inset Formula $m+n-2$
\end_inset

 degree of freedom.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
5.
 Apply fomular 
\begin_inset Formula $U$
\end_inset

 of exercise 
\begin_inset Formula $4$
\end_inset

 to test.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
6.
 Change 
\begin_inset Formula $U=\frac{(m+n-2)^{1/2}(\bar{X}_{m}-\bar{Y}_{n}-\lambda)}{(\frac{1}{m}+\frac{1}{n})^{1/2}(S_{X}^{2}+S_{Y}^{2})^{1/2}.}$
\end_inset

 is 
\begin_inset Formula $T$
\end_inset

 distribution with 
\begin_inset Formula $m+n-2$
\end_inset

 degree of freedom.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 Using 
\begin_inset Formula $U$
\end_inset

 in exercise 6, 
\begin_inset Formula $T_{8}^{-1}(-0.95)\le U\le T_{8}^{-1}(0.95)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9.
 Imagine Symetry distribution of T distribution.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 Test X is Calcium and 
\begin_inset Formula $Y$
\end_inset

 is Placebo for testing.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
11.
 
\begin_inset Formula $n=43$
\end_inset

, 
\begin_inset Formula $\bar{X}_{43}=8.56$
\end_inset

, 
\begin_inset Formula $S_{X}^{2}=2745.7$
\end_inset

.
 
\begin_inset Formula $m=35$
\end_inset

, 
\begin_inset Formula $\bar{Y}_{35}=5.551$
\end_inset

, 
\begin_inset Formula $\bar{S}_{y}=783.9$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $H_{0}:\mu_{X}\le\mu_{Y}$
\end_inset

, 
\begin_inset Formula $H_{1}:\mu_{X}\ge\mu_{Y}$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $U=\frac{(m+n-2)^{1/2}(\bar{X}_{m}-\bar{Y}_{n})}{(1/m+1/n)^{1/2}(S_{X}^{2}+S_{Y}^{2})^{1/2}}=1.94$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $\alpha_{0}=0.01$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
Following Theorem 9.6.1, 
\begin_inset Formula $\delta$
\end_inset

 rejects 
\begin_inset Formula $H_{0}$
\end_inset

 if 
\begin_inset Formula $U\ge T_{m+n-2}^{-1}(1-\alpha_{0})$
\end_inset

.
 
\begin_inset Formula $T_{76}^{-1}(0.99)=$
\end_inset


\end_layout

\begin_layout Exercise*
b.
 Test hypotheses at level of significance 
\begin_inset Formula $\alpha_{0}=0.01$
\end_inset

, using Welch's approximate test.
\end_layout

\begin_layout Exercise*
Following Welch, he using statistic: 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
V=\frac{\bar{X}_{m}-\bar{Y}_{n}}{(\frac{S_{X}^{2}}{m(m-1)}+\frac{S_{Y}^{2}}{n(n-1)})^{1/2}}
\]

\end_inset


\end_layout

\begin_layout Exercise*
having degree of freedoms:
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
v
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
12.
 
\begin_inset Formula $W=\frac{S_{X}^{2}}{m(m-1)}+\frac{S_{Y}^{2}}{n(n-1)}$
\end_inset

.
 
\begin_inset Formula $\mu(W)=\frac{1}{m-1}\mu(\frac{S_{X}^{2}}{m})+\frac{1}{n-1}\mu(\frac{S_{Y}^{2}}{n})$
\end_inset

.
 For 
\begin_inset Formula $\frac{S_{X}^{2}}{m}$
\end_inset

 and 
\begin_inset Formula $\frac{S_{Y}^{2}}{n}$
\end_inset

 is 
\begin_inset Formula $\chi^{2}$
\end_inset

 distribution with 
\begin_inset Formula $m-1$
\end_inset

 and 
\begin_inset Formula $n-1$
\end_inset

 degree of freedom.
\end_layout

\begin_layout Exercise*
From Equation (9.6.11), We already found M.L.E for maximizing denominator, and
 find 
\begin_inset Formula $\mu_{1},\mu_{2},\sigma^{2}$
\end_inset

 in sitatuon 
\begin_inset Formula $\mu_{1}=\mu_{2}$
\end_inset

.
 So it's same Section The Two-Sample 
\begin_inset Formula $t$
\end_inset

 Test as a Likelihood Ratio Test.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
The F Distribution
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $\bar{X}_{43}=8.56$
\end_inset

, 
\begin_inset Formula $S_{X}^{2}=2745.7$
\end_inset

; 
\begin_inset Formula $\bar{Y}_{35}=5.551;S_{Y}^{2}=783.9$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $H_{0}:\sigma_{X}^{2}\ge\sigma_{Y}^{2}$
\end_inset

, 
\begin_inset Formula $H_{1}:\sigma_{X}^{2}<\sigma_{Y}^{2}$
\end_inset

.
 We can write: 
\begin_inset Formula $H_{0}:\sigma_{Y}^{2}\le\sigma_{X}^{2}$
\end_inset

, 
\begin_inset Formula $H_{1}:\sigma_{Y}^{2}>\sigma_{X}^{2}$
\end_inset


\end_layout

\begin_layout Exercise*
Level significant 
\begin_inset Formula $\alpha_{0}=0.05$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
From Theorem 9.7.4.
 Let 
\begin_inset Formula $c$
\end_inset

 be a 
\begin_inset Formula $1-\alpha_{0}$
\end_inset

 quantile of the 
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $m-1$
\end_inset

 and 
\begin_inset Formula $n-1$
\end_inset

 degrees of freedom.
 
\begin_inset Formula $\delta$
\end_inset

 be test that rejects 
\begin_inset Formula $H_{0}$
\end_inset

 in (9.7.3) when 
\begin_inset Formula $V\ge c$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $34$
\end_inset

 and 
\begin_inset Formula $42$
\end_inset

 degrees of freedom.
 
\begin_inset Formula $c=F_{34,42}^{-1}(0.95).$
\end_inset

 Then we check whether 
\begin_inset Formula $V\ge c$
\end_inset

.
\end_layout

\begin_layout Exercise*
2.
 Suppose that a random variable 
\begin_inset Formula $X$
\end_inset

 has the 
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $3,8$
\end_inset

 degrees of freedom.
 Determine value 
\begin_inset Formula $c:Pr(X>c)=0.975$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
\begin_inset Formula $X\sim F_{3,8}$
\end_inset

.
 
\begin_inset Formula $Y=\frac{1}{X}\sim F_{8,3}$
\end_inset

.
 
\begin_inset Formula $Pr(X>c)=P(\frac{1}{X}<\frac{1}{c})=Pr(Y<\frac{1}{c})=0.975$
\end_inset

.
 So, 
\begin_inset Formula $\frac{1}{c}=F_{8,3}^{-1}(0.975)=14.54$
\end_inset

.
 So 
\begin_inset Formula $c=\frac{1}{14.54}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
3.
 Suppose that a random variable 
\begin_inset Formula $X$
\end_inset

 has 
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $1,8$
\end_inset

 degrees of freedom.
 Use table of 
\begin_inset Formula $t$
\end_inset

 distribution to determine the value of 
\begin_inset Formula $c$
\end_inset

 such that 
\begin_inset Formula $Pr(X>c)=0.3$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
From Theorem 9.7.2.
 
\begin_inset Formula $Y$
\end_inset

 has the 
\begin_inset Formula $t$
\end_inset

 distribution with 
\begin_inset Formula $n$
\end_inset

 degrees of freedom, then 
\begin_inset Formula $Y^{2}$
\end_inset

 has the 
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $1$
\end_inset

 and 
\begin_inset Formula $n$
\end_inset

 degrees of freedom.
 
\begin_inset Formula $Y:Y^{2}=X$
\end_inset

, so 
\begin_inset Formula $Y$
\end_inset

 is 
\begin_inset Formula $t$
\end_inset

 distribution with 
\begin_inset Formula $8$
\end_inset

 degree of freedom.
 
\begin_inset Formula $Pr(X>c)=Pr(Y>c^{0.5})+Pr(Y<-c^{0.5})=2T_{8}(-c^{0.5})=0.3$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $T_{8}(X>c^{0.5})=0.15$
\end_inset

, 
\begin_inset Formula $T_{8}(X\le c^{0.5})=0.85$
\end_inset

.
 So 
\begin_inset Formula $c^{0.5}=1.108$
\end_inset

.
 
\begin_inset Formula $c=1.108^{2}=1.228$
\end_inset

.
\end_layout

\begin_layout Exercise*
4.
 Suppose that a random variable 
\begin_inset Formula $X$
\end_inset

 has the 
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $m$
\end_inset

 and 
\begin_inset Formula $n$
\end_inset

 degrees of freedom (
\begin_inset Formula $n>2)$
\end_inset

.
 Show that 
\begin_inset Formula $E(X)=n/(n-2)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $X$
\end_inset

 is 
\begin_inset Formula $F$
\end_inset

 distribution, so 
\begin_inset Formula $X=\frac{U_{1}/m}{U_{2}/n}$
\end_inset

 where 
\begin_inset Formula $U_{1},U_{2}$
\end_inset

 is 
\begin_inset Formula $\chi^{2}$
\end_inset

 distribution with 
\begin_inset Formula $m,n$
\end_inset

 degree of freedom.
 
\end_layout

\begin_layout Standard
https://www.statlect.com/probability-distributions/F-distribution#hid5
\end_layout

\begin_layout Exercise*
5.
 What is the value of the median of the 
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $m$
\end_inset

 and 
\begin_inset Formula $n$
\end_inset

 degrees of freedom when 
\begin_inset Formula $m=n$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Median = 
\begin_inset Formula $F_{m,m}^{-1}(0.5)$
\end_inset


\end_layout

\begin_layout Exercise*
6.
 
\begin_inset Formula $X$
\end_inset

 has 
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $m$
\end_inset

 and 
\begin_inset Formula $n$
\end_inset

 degrees of freedom.
 Show that the variable 
\begin_inset Formula $mX/(mX+n)$
\end_inset

 has the beta distribution with parameters 
\begin_inset Formula $\alpha=m/2$
\end_inset

 and 
\begin_inset Formula $\beta=n/2$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $Y=\frac{mX}{mX+n}$
\end_inset

, 
\begin_inset Formula $X=\frac{nY}{m(Y-1)}$
\end_inset

, we use function of a random variable.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 Consider two different normal distributions for which both means 
\begin_inset Formula $\mu_{1}$
\end_inset

 and 
\begin_inset Formula $\mu_{2}$
\end_inset

 and the variances 
\begin_inset Formula $\sigma_{1}^{2}$
\end_inset

 and 
\begin_inset Formula $\sigma_{2}^{2}$
\end_inset


\end_layout

\begin_layout Exercise*
are unknown, suppose the following hypotheses:
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\sigma_{1}^{2}\le\sigma_{2}^{2}
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\sigma_{1}^{2}>\sigma_{2}^{2}
\]

\end_inset


\end_layout

\begin_layout Exercise*
Suppose random sample consisting of 
\begin_inset Formula $16$
\end_inset

 observations for the first normal distribution yeilds the values 
\begin_inset Formula $\sum_{i=1}^{16}X_{i}=84$
\end_inset

, 
\begin_inset Formula $\sum_{i=1}^{16}X_{i}^{2}=563$
\end_inset

, and an independent random sample consisting of 
\begin_inset Formula $10$
\end_inset

 observations from second normal distribution yeilds the values 
\begin_inset Formula $\sum_{i=1}^{10}Y_{i}=18$
\end_inset

 and 
\begin_inset Formula $\sum_{i=1}^{10}Y_{i}^{2}=72$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
a.
 What is MLE's of 
\begin_inset Formula $\sigma_{1}^{2}$
\end_inset

 and 
\begin_inset Formula $\sigma_{2}^{2}$
\end_inset

.
\end_layout

\begin_layout Exercise*
From Example 7.5.6, M.L.E.
 of 
\begin_inset Formula $\theta=(\mu,\sigma^{2})$
\end_inset

 is 
\begin_inset Formula $\hat{\theta}=(\hat{\mu},\hat{\sigma^{2}})=(\bar{X}_{n},\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2})$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $\sigma^{2}=\sum_{i=1}^{n}X_{i}^{2}-2\bar{X}_{n}\sum_{i=1}^{n}X_{i}+\bar{X}_{n}^{2}$
\end_inset


\end_layout

\begin_layout Exercise*
b.
 We find 
\begin_inset Formula $F_{15,9}^{-1}(0.95)$
\end_inset

 and compare with 
\begin_inset Formula $F=\frac{S_{X}^{2}/15}{S_{Y}^{2}/9}$
\end_inset

 with this number, if 
\begin_inset Formula $V\ge c$
\end_inset

, we can reject 
\begin_inset Formula $H_{0}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
8.
 
\begin_inset Formula $H_{0}:\sigma_{1}^{2}\le3\sigma_{2}^{2}$
\end_inset

, 
\begin_inset Formula $H_{1}:\sigma_{1}^{2}>3\sigma_{2}^{2}$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
We change formulate of 
\begin_inset Formula $V=\frac{S_{X}^{2}/(m-1)}{S_{Y}^{2}/(n-1)}$
\end_inset

 from (9.7.4) to 
\begin_inset Formula $V=\frac{S_{X}^{2}/(m-1)}{3S_{Y}^{2}/(n-1)}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9.
 Following (9.7.7), The most convenient choice of 
\begin_inset Formula $c_{1}$
\end_inset

 and 
\begin_inset Formula $c_{2}$
\end_inset

 is the the makes 
\begin_inset Formula $Pr(V\le c_{1})=Pr(V\ge c_{2})=\alpha_{0}/2$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
\begin_inset Formula $c_{1}$
\end_inset

 and 
\begin_inset Formula $c_{2}$
\end_inset

 to be 
\begin_inset Formula $\alpha_{0}/2$
\end_inset

 and 
\begin_inset Formula $1-\alpha_{0}/2$
\end_inset

 quantiles of 
\begin_inset Formula $F_{15,9}^{-1}$
\end_inset

.
 
\begin_inset Formula $c_{1}=F_{15,9}^{-1}(0.025)=$
\end_inset

, 
\begin_inset Formula $c_{1}=F_{15,9}^{-1}(0.975)=3.77$
\end_inset


\end_layout

\begin_layout Exercise*
From Example 9.7.2, Determining the 0.05 Quantile of an 
\begin_inset Formula $F$
\end_inset

 Distribution, 
\begin_inset Formula $Pr(X<x)=0.05$
\end_inset

, 
\begin_inset Formula $Pr(\frac{1}{X}>\frac{1}{x})=0.05$
\end_inset

, then 
\begin_inset Formula $Pr(1/X)=0.95$
\end_inset

, 
\begin_inset Formula $X$
\end_inset

 has 
\begin_inset Formula $F_{15,9}$
\end_inset

, so 
\begin_inset Formula $\frac{1}{X}$
\end_inset

 has 
\begin_inset Formula $F_{9,15}$
\end_inset

, so 
\begin_inset Formula $\frac{1}{x}=F_{9,15}^{-1}(0.95)=2.59$
\end_inset

.
 so 
\begin_inset Formula $c_{1}=1/2.59=0.386$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 
\begin_inset Formula $H_{0}:\sigma_{1}^{2}=r\sigma_{2}^{2}$
\end_inset

, 
\begin_inset Formula $H_{1}:\sigma_{1}^{2}\ne r\sigma_{2}^{2}$
\end_inset

.
 
\begin_inset Formula $V_{1}=\frac{S_{X}^{2}/(m-1)}{rS_{Y}^{2}/(n-1)}=\frac{V}{r}$
\end_inset

 compare with 
\begin_inset Formula $F_{15,9}^{-1}(0.95)$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
11.
 
\begin_inset Formula $V_{1}=\frac{S_{X}^{2}/(m-1)}{rS_{Y}^{2}/(n-1)}$
\end_inset

 is 
\begin_inset Formula $F$
\end_inset

 distribution with 
\begin_inset Formula $15$
\end_inset

, 
\begin_inset Formula $9$
\end_inset

 degree of freedoms.
 
\begin_inset Formula $0.95$
\end_inset

 confidence coefficient is 
\begin_inset Formula $F_{15,9}^{-1}(0.025)\le V_{1}\le F_{15,9}^{-1}(0.975)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
14.
 
\begin_inset Formula $\sigma_{1}^{2}=2\sigma_{2}^{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
9.8.
 Bayes Test Procedures.
\end_layout

\begin_layout Exercise*
1.
 In certain industrial process can be either in control or out of control,
 and that at any specified time the prior probability that it will be in
 control is 
\begin_inset Formula $0.9$
\end_inset

, and the prior probability that it will be out control is 
\begin_inset Formula $0.1$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{0}:\text{ In control}
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
H_{1}:\text{ Out of control}
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $X$
\end_inset

 is output of the process is to be taken, 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
P(X|H_{0})\sim N(50,1)
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
P(X|H_{1})\sim N(52,1)
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Tabular
<lyxtabular version="3" rows="3" columns="3">
<features tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
choose 
\begin_inset Formula $H_{0}$
\end_inset

 
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
choose 
\begin_inset Formula $H_{1}$
\end_inset

 
\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $H_{0}$
\end_inset

 true
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $1000$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $H_{1}$
\end_inset

true
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $18000$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0$
\end_inset


\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Exercise*
Find a test procedure for which the expected loss will be a minimum.
 For what values of 
\begin_inset Formula $X$
\end_inset

 should it be decided that the process is out of control.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
We want to minimize loss in equation (9.8.5) 
\begin_inset Formula $r(\delta)=\xi_{0}w_{0}\alpha(\delta)+\xi_{1}w_{1}\beta(\delta)$
\end_inset

.
 Following Theorem 9.2.1, optimal 
\begin_inset Formula $\delta^{*}$
\end_inset

 is not reject 
\begin_inset Formula $H_{0}$
\end_inset

 whenever 
\begin_inset Formula $\xi_{0}w_{0}f_{0}(x)>\xi_{1}w_{1}f_{x}(x)$
\end_inset

, then 
\end_layout

\begin_layout Standard
\begin_inset Formula $f_{0}=\frac{1}{\sqrt{2\pi}}e^{-(x-50)^{2}/2}$
\end_inset

, 
\begin_inset Formula $f_{1}=\frac{1}{\sqrt{2\pi}}e^{-(x-52)^{2}/2}$
\end_inset

.
 
\begin_inset Formula $\frac{f_{1}(x)}{f_{0}(x)}>\frac{0.9*1000}{0.1*18000}=162$
\end_inset

, 
\begin_inset Formula $\xi_{0}=0.9.\xi_{1}=0.1,w_{0}=1000,w_{1}=18000$
\end_inset

,
\end_layout

\begin_layout Standard
\begin_inset Formula $\frac{f_{1}(x)}{f_{0}(x)}=\frac{e^{-(x-52)^{2}/2}}{e^{-(x-50)^{2}/2}}>162$
\end_inset

, 
\begin_inset Formula $-(x-52)^{2}/2+(x-50)^{2}/2>\log(162)$
\end_inset

, 
\begin_inset Formula $4x>2*\log(162)-50^{2}+52^{2}$
\end_inset

, 
\begin_inset Formula $x>50.6$
\end_inset


\end_layout

\begin_layout Exercise*
3.
 
\begin_inset Formula $H_{0}:\text{ Failures are caused by mirror defects}$
\end_inset

, 
\begin_inset Formula $H_{1}:\text{ Failures are caused by maor defects}$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $\xi_{0}=0.8$
\end_inset

, 
\begin_inset Formula $\xi_{1}=0.2$
\end_inset

.
\end_layout

\begin_layout Exercise*
When a failure occurs, 
\begin_inset Formula $n$
\end_inset

 independent souding 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 are made on the system.
 
\begin_inset Formula $f(X_{i}|H_{0})\sim Poisson(3)$
\end_inset

, 
\begin_inset Formula $f(X_{i}|H_{1})\sim Poisson(7)$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Tabular
<lyxtabular version="3" rows="3" columns="3">
<features tabularvalignment="middle">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<column alignment="center" valignment="top">
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout

\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
choose 
\begin_inset Formula $H_{0}$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
choose 
\begin_inset Formula $H_{1}$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $H_{0}$
\end_inset

 true
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $400$
\end_inset


\end_layout

\end_inset
</cell>
</row>
<row>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $H_{1}$
\end_inset

true
\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $2500$
\end_inset


\end_layout

\end_inset
</cell>
<cell alignment="center" valignment="top" topline="true" bottomline="true" leftline="true" rightline="true" usebox="none">
\begin_inset Text

\begin_layout Plain Layout
\begin_inset Formula $0$
\end_inset


\end_layout

\end_inset
</cell>
</row>
</lyxtabular>

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $w_{0}=400$
\end_inset

, 
\begin_inset Formula $w_{1}=2500$
\end_inset

, 
\begin_inset Formula $f(x_{1},...,x_{n}|\lambda)=\frac{e^{-n\lambda}\lambda^{\sum_{i}X_{i}}}{\prod X_{i}!}$
\end_inset

.
 
\begin_inset Formula $\frac{f(x_{1},..,x_{n}|7)}{f(x_{1},..,x_{n}|3)}=e^{-n4}(\frac{7}{3})^{\sum_{i}X_{i}}>\frac{w_{0}\xi_{o}}{w_{1}\xi_{1}}=\frac{400*0.8}{2500*0.2}=0.64$
\end_inset

.
 
\begin_inset Formula $\sum_{i}X_{i}>(4n+log(0.64))/log(7/3)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
5.
 Posterior probability of 
\begin_inset Formula $\theta=\theta_{0}$
\end_inset


\end_layout

\begin_layout Exercise*
a.
 
\begin_inset Formula $P(\theta=\theta_{0}|X)=\frac{P(X|\theta_{0})P(\theta_{0})}{P(X|\theta_{1})P(\theta_{1})+P(X|\theta_{0})P(\theta_{0})}=\frac{f_{0}(X)\xi_{0}}{f_{0}(X)\xi_{0}+f_{1}(X)\xi_{1}}$
\end_inset

 (because we just have two 
\begin_inset Formula $w_{0},w_{1}$
\end_inset

)
\end_layout

\begin_layout Exercise*
b.
 When we want minimize (9.8.5) 
\begin_inset Formula $r(\delta)=\xi_{0}w_{0}\alpha(\delta)+\xi_{1}w_{1}\beta(\delta)$
\end_inset

, Following Theorem 9.2.1, we reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $\xi_{0}w_{0}f_{0}(x)<\xi_{1}w_{1}f_{1}(x)$
\end_inset

, 
\begin_inset Formula $\frac{\xi_{1}f(x|\theta_{1})w_{1}}{\xi_{0}f(x|\theta_{0})w_{0}}>1$
\end_inset

, Following section 
\series bold
Tests Based on the Posterior Distribution 
\series default
, we choose 
\begin_inset Formula $d_{0}$
\end_inset

 when 
\begin_inset Formula $r(d_{0}|x)<r(d_{1}|x)$
\end_inset

, or 
\begin_inset Formula $\frac{r(d_{1}|x)}{r(d_{0}|x)}>1$
\end_inset

, 
\begin_inset Formula $\frac{w_{1}Pr(H_{1}|x)}{w_{0}Pr(H_{0}|x)}=\frac{w_{1}Pr(x|H_{1})P(H_{1})}{w_{0}P(x|H_{0})P(H_{0})}=\frac{\xi_{1}f(x|\theta_{1})w_{1}}{\xi_{0}f(x|\theta_{0})\xi_{0}}$
\end_inset

 , it is same as result of minimizing 
\begin_inset Formula $r(\delta)$
\end_inset

, so we can conclude.
\end_layout

\begin_layout Exercise*
c.As we known, if we tests on Posterior distribution we use: 
\begin_inset Formula $r(d_{1}|x)>r(d_{0}|x)$
\end_inset

, 
\begin_inset Formula $w_{1}(1-P(H_{0}|x))>w_{0}P(H_{0}|x))$
\end_inset

, then 
\begin_inset Formula $P(H_{0}|x)<\frac{w_{1}}{w_{0}+w_{1}}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
6.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 Suppose that we have a situation in which the Bayes test that reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $Pr(H_{0}\text{ true}|x)\le\alpha_{0}$
\end_inset

 is the same as level 
\begin_inset Formula $\alpha_{0}$
\end_inset

 test of 
\begin_inset Formula $H_{0}$
\end_inset

 for all 
\begin_inset Formula $\alpha_{0}$
\end_inset

.
 Prove that the 
\begin_inset Formula $p-value$
\end_inset

 equals the posterior probability that 
\begin_inset Formula $H_{0}$
\end_inset

 is true.
\end_layout

\begin_layout Exercise*

\series bold
Remind: 
\series default
Level of test is defined in (9.1.6), 
\begin_inset Formula $p-value=Pr(H_{0}\text{ true}|x)$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $P-value$
\end_inset

 is the smallest level 
\begin_inset Formula $\alpha_{0}$
\end_inset

 such that we would reject the null-hypothesis at level 
\begin_inset Formula $\alpha_{0}$
\end_inset

 with the observed data.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $p-values$
\end_inset

 equals 
\begin_inset Formula $\sup_{\theta\in\Omega_{0}}\pi(\theta|\delta_{t})=\sup_{\theta\in\Omega_{0}}Pr(T\ge t|\theta)=\sup_{\theta\in\Omega_{0}}Pr(H_{0}\text{ true}|x)$
\end_inset

 in equation (9.1.12), but 
\begin_inset Formula $\Omega_{0}=\{\theta_{0}\}$
\end_inset

 so 
\begin_inset Formula $\sup_{\theta\in\Omega_{0}}Pr(H_{0}\text{ true}|x)=Pr(H_{0}\text{ true}|x)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 Apply Theorem 9.8.2
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
9.9 Foundational Issues
\end_layout

\begin_layout Exercise*
1.
 Suppose that a single observation 
\begin_inset Formula $X$
\end_inset

 is taken from the normal distribution with unknown mean 
\begin_inset Formula $\mu$
\end_inset

 and known variance is 
\begin_inset Formula $1$
\end_inset

.
 Suppose that it is known that the value of 
\begin_inset Formula $\mu$
\end_inset

 must be 
\begin_inset Formula $-5,0,5$
\end_inset

 and it is desired to test the following hypotheses at the level of significance
 
\begin_inset Formula $0.05$
\end_inset

:
\end_layout

\begin_layout Exercise*
\begin_inset Formula $H_{0}:\mu=0$
\end_inset

, 
\begin_inset Formula $H_{1}:\mu=-5\text{ or }\mu=5$
\end_inset


\end_layout

\begin_layout Exercise*
Suppose also that the test procedure to be used specifies rejecting 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $|X|>c$
\end_inset

, where the constant 
\begin_inset Formula $c$
\end_inset

 is chosen so that 
\begin_inset Formula $Pr(|X|>c|\mu=0)=0.05$
\end_inset

.
\end_layout

\begin_layout Exercise*
a.
 Find the value of 
\begin_inset Formula $c$
\end_inset

, and show that if 
\begin_inset Formula $X=2$
\end_inset

, then 
\begin_inset Formula $H_{0}$
\end_inset

 will be rejected.
\end_layout

\begin_layout Exercise*
b.
 Show that if 
\begin_inset Formula $X=2$
\end_inset

, then the value of the likelihood function at 
\begin_inset Formula $\mu=0$
\end_inset

 is 12.2 times as large as its value at 
\begin_inset Formula $\mu=5$
\end_inset

 and is 
\begin_inset Formula $5.9x10^{9}$
\end_inset

 times as large as its value at 
\begin_inset Formula $\mu=-5$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
a.
 
\begin_inset Formula $\delta:$
\end_inset

 Reject 
\begin_inset Formula $H_{0}$
\end_inset

 when 
\begin_inset Formula $|X|>c$
\end_inset

.
 We have 
\begin_inset Formula $Pr(X>c|\mu=0)+Pr(X<-c|\mu=0)=0.05$
\end_inset

, so 
\begin_inset Formula $Pr(X>c|\mu=0)=0.05/2=0.025$
\end_inset

, so 
\begin_inset Formula $c=F^{-1}(0.975)$
\end_inset

, 
\begin_inset Formula $F$
\end_inset

 is c.d.f of 
\begin_inset Formula $Normal(0,1)$
\end_inset

, 
\begin_inset Formula $c=1.96$
\end_inset

, 
\begin_inset Formula $X=2>c=1.96$
\end_inset

, so 
\begin_inset Formula $H_{0}$
\end_inset

 will be rejected.
\end_layout

\begin_layout Standard
b.
 pdf of normal distribution: 
\begin_inset Formula $f(x|\mu)=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}$
\end_inset

, so 
\begin_inset Formula $\frac{f(2|0)}{f(2|5)}\sim12.2$
\end_inset


\end_layout

\begin_layout Exercise*
3.
 Conditions of Exercise 2, Suppose now that :
\end_layout

\begin_layout Exercise*
\begin_inset Formula $H_{0}:\mu\le0$
\end_inset

, 
\begin_inset Formula $H_{1}:\mu>0$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
We compute 
\begin_inset Formula $p-value$
\end_inset

 of 
\begin_inset Formula $\bar{X}_{n}=0.03$
\end_inset

.
 
\begin_inset Formula $\bar{X}_{n}\sim N(0,\sigma=1/100)$
\end_inset

, so 
\begin_inset Formula $p-value=1-F(0.03)$
\end_inset

 where 
\begin_inset Formula $F$
\end_inset

 is cdf of 
\begin_inset Formula $N(0,1/100)$
\end_inset

.
 so 
\begin_inset Formula $p-value=0.013$
\end_inset


\end_layout

\begin_layout Exercise*
4.
 
\begin_inset Formula $G=19\pi(0|\delta)+1-\pi(0.5|\delta)=19F(\frac{c_{n}}{1/100})+2-F(\frac{c_{n}-5}{1/100})$
\end_inset

, We want to minimize 
\begin_inset Formula $G$
\end_inset

 we find 
\begin_inset Formula $\frac{dG}{dc_{n}}=19f(..)*1/100-f(..)*1/100=0$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
5.
 
\begin_inset Formula $19\pi(0|\delta)=1-\pi(0.5|\delta)$
\end_inset

.
 
\begin_inset Formula $19\pi(0|\delta)$
\end_inset

 is increasing function, 
\begin_inset Formula $1-\pi(0.5|\delta)$
\end_inset

 is decreasing function so we can solve this equation by derivation left
 and right part.
 
\end_layout

\end_body
\end_document