数理解析研究所講究録 1273 巻 2002 年 138-147

138

Note on Sample Size Determination for
Detecting the Worst Component of aMultivariate Exponential Distribution
筑波大学・数学研究科

青木充 (Mitsuru Aoki)

Institute of Mathematics, University of Tsukuba
フィデリティ投信株式会社

甲斐正樹 (Masaki Kai)

Insurance Distribution, Fidelity Investments
筑波大学・数学系

青嶋

誠 (Makoto Aoshima)

Institute of Mathematics, University of Tsukuba
1. INTRODUCTION

It is well known that an exponential distribution plays an important role in the fieldof survival analysis. Amultivariate exponential distribution was given by Marshall andOlkin (1967), which is much of interest in both theoretical developments and applications. Especially, Proschan and Sullo (1976) considered a $k+1$ -parameter version of the
Marshall-Olkin multivariate exponential distribution and studied likelihood estimation of
its parameters. We consider aselection problem on components of the $k+1$ -parameter
exponential distribution.
Let, $r=1$ , 2, be random samples from the multivariate exponential (MVE) distribution whose survival function is given by$(X_{1\mathrm{r}}, X_{2\mathrm{r}}, \ldots, X_{k\mathrm{r}})$$\ldots$$P(X_{1t}>x_{1}, X_{2\mathrm{r}}>x_{2}, \ldots, X_{kt}>x_{k})$$= \exp\{-\lambda_{1}x_{1}-\lambda_{2}x_{2}-\ldots-\lambda_{k}x_{k}-\lambda_{0}\max(x_{1},x_{2}, \ldots,x_{k})\}$$(i=1, 2, \ldots, k)$ andwhere $x:>0$ ,. Marshall and Olkin (1967) derived
this distribution under the assumption that failure is caused by $k+1$ types of Poissonshocks on asystem containing components. We consider the problem of detecting the$(i=1, 2, \ldots, k)$ which is viewed as the hazard of theworst component with respect tocomponent in lifetime analysis. We define that the -th component is the worst component
if. Note that$\lambda_{:}>0$$\lambda_{0}\geq 0$$k$$\lambda_{:}$$j$$\lambda_{j}=\max(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k})$$E(X_{\dot{l}T})=1/(\lambda:+\lambda_{0})$,$Var(X_{\dot{l}\Gamma})=1/(\lambda_{i}+\lambda_{0})^{2}$,$Cov(X_{ir}, X_{j\mathrm{r}})=\lambda_{0}/(\lambda:+\lambda_{0})(\lambda_{j}+\lambda_{0})(\lambda:+\lambda_{j}+\lambda_{0})$, $i\neq j$ .

139
The worst component has the smallest mean. Throughout this paper, we assume that the-th component is the worst component, that is,
$k$

(1.1)

$\lambda_{k}>\lambda_{i}(i=1,2, \ldots, k-1)$

are mutually independent if and only,
, ,without loss of generality. Since, the problem is reduced to the problem of detecting the worst one out of
if.
univariate exponential populations when
We seek aprocedure which detects one of the components as the worst component.
Acorrect decision (CD) occurs when the detected component is the worst component.
by $P(CD|R)$ , we require
Denoting the probability of acorrect decision (PCD) using$X_{1r}$$\mathrm{X}2\mathrm{r}$$\ldots$$X_{kr}$$k$$\lambda_{0}=0$$\lambda_{0}=0$

$R$

$R$

that
$P(CD|R)\geq P^{\star}$whenever$\lambda_{k}/\lambda_{i}\geq\delta^{\star}(i=1, \ldots, k-1)$

(1.2)

,

and $P^{\star}\in(k^{-1},1)$ are specified by the experimenter in advance. The special
wherecase of this problem when $k=2$ was considered by Hyakutake (1992) and Aoshima andChen (1999). So, in the present paper, we assume that $k\geq 3$ . When $k\geq 3$ , the former
paper by the authors in 2002 has tackled the problem of selecting the best component of
the MVE distribution. It would be interesting to observe how much difference the sample
size makes between both the situations of selecting the best component and detecting
the worst component. Also, from the decision theoretical point of view, this kind of
fixed-size estimation on scale parameters is desirable to research in sequential analysis
especially based on atw0-stage sampling scheme since it must deal with abit outside
Stein’s original framework. This is amotivation of this short note. Note that detecting the.
worst component with respect to is equivalent to that with respect toThen, we have from Arnold (1968) that$\delta^{\star}(>1)$$p_{i}=\lambda_{i}/\Sigma_{\ell=0}^{k}\lambda_{\ell}$$\lambda_{i}$$P(X_{1r}=X_{2r}=\ldots=X_{kr})=p_{0}$ ,$P(X_{ir}<X_{i’r}, i’=1, \ldots, i-1, i+1, \ldots, k)=p_{i}$The preference zone
, , $k-1)$ .

$\lambda_{k}/\lambda_{i}\geq\delta^{\star}(i=1, \ldots, k-1)$, $i=1,2$ ,

, .
$k$

$\ldots$in (1.2) is equivalent to(1.3)$p_{k}/p_{i}\geq\delta^{\star}(i=$$1$$\ldots$

2. SINGLE-STAGE PROCEDURES WHEN Po IS KNOWN
is known and consider two single-stage procedures
In this section, we assume thatfor detecting the component associated with . In order to meet requirement (1.2), we need to establish the least favorable configuration (LFC) of
is minimized subject to the condition thatfor which the PCD for$(i=1, \ldots, k-1)$ . The LFC for aprocedure would be given by$p_{0}$$p_{k}$$(\mathrm{R}\mathrm{e}, \ell=1,2)$$\lambda=$$(\lambda_{0}., \lambda_{1}, \ldots, \lambda_{k})$$R_{\ell}$$\lambda_{k}/\lambda_{i}\geq\delta^{\star}$$\lambda_{0}$,$\lambda_{1}=\ldots=\lambda_{k-1}=\lambda_{k}/\delta^{\star}(\delta^{\star}>1)$

.

(2.1)

140
We shall evaluate the PCD forunder the LFC given by (2.1), which we will denote by. Under the LFC, we note that
$R_{\ell}$$P^{\star}(CD|R_{\ell})$$p_{1}=\ldots=p_{k-1}=p_{k}/\delta^{\star}$

,

(2.2)

$p_{k}=\delta^{\star}(1-p_{0})/(\delta^{\star}+k-1)$

2.1 Procedure.

$R_{1}$

We consider the following rule:Take asample of observations. Let :and
denote the number of observations in the regions {x $\{<x:’\}$ and $\{x_{1}=\ldots=x_{k}\}$ ,respectively. Detect the component that attained the largest count amongasthe worst component; use randomization to break ties for the first place.
We note thathas amultinomial distribution withand $k+1$
cell probabilities
in view of (1.3). Then, the ruleis exactly reduced to therule given by Aoshima, Chen and Panchapakesan (2001) for selecting the most probable
cell whereis the count in the nuisance cell. When is large, they gave an approximate
minimum sample size needed to satisfy requirement (1.2) as, where $[x]$denotes the greatest integer less than . Here, $u_{1}>0$ issuch that$(R_{1})$

$n$

$n$

$n_{0}$$(n_{1}, \ldots, n_{k})$$(n_{1}, \ldots, n_{k}, n_{0})$$n= \sum_{\ell=0}^{k}n_{\ell}$$(p_{1}, \ldots,p_{k},p_{0})$$R_{1}$$R$$n$$n_{0}$$n_{0}^{(1)}=[u_{1}^{2}/c_{1}^{2}]+1$$x$$u_{1}=u_{1}(p_{0}, k, \delta^{\star}, P^{\star})$$\int_{-\infty}^{\infty}\Phi^{k-1}(\frac{x\sqrt{\rho_{1}}+u_{1}}{\sqrt{1-\rho_{1}}})d\Phi(x)=P^{\star}$with$\rho_{1}=\frac{p_{0}(\delta^{\star}-1)^{2}+(1+k)\delta^{\star}-1}{p_{0}(\delta^{\star}-1)^{2}+\delta^{\star}(2+k)-2+k}$,and, (say).

(2.3)$c_{1}= \{\frac{(1-p_{0})(\delta^{\star}-1)^{2}}{p_{0}(\delta^{\star}-1)^{2}+\delta^{\star}(2+k)-2+k}\}^{1/2}=c_{1}(p_{0}, k, \delta^{\star})$Since$u_{1}$$[u_{1}^{\star 2}/c_{1}^{2}]+1$

depends on
throughwhereis$p_{0}$$u_{1}^{\star}>0$$\rho_{1}$, they gave an upper bound forsuch thatas$n_{0}^{(1)}$$u_{1}^{\star}=u_{1}^{\star}(k, \delta^{\star}, P^{\star})$$\int_{-\infty}^{\infty}\Phi^{k-1}(\frac{x\sqrt{\rho_{1}^{\star}}+u_{1}^{\star}}{\sqrt{1-\rho_{1}^{\star}}})d\Phi(x)=P^{\star}$with

The following table was given by Aoshima, Chen and Panchapakesan (2001).
Table 2.1. Values of$u_{1}^{\star}$.

$\rho_{1}^{\star}=\frac{-1+\delta^{\star}+k\delta^{\star}}{-2+k+2\delta^{\star}+k\delta^{\star}}$when $P^{\star}=0.90$$n_{1}^{(1)}=$

141
is unknown, we need to estimate
Whensample.

$n_{1}^{(1)}$$p_{0}$

2.2 Procedurethrough an estimate po based on apilot$R_{2}$

If acomplete data set (without cesoring) is available, we can give another rule (R2)is the exponential
which is based on the marginal distribution. The distribution of. The selection problem with respect to is equivalent
distribution with parameter.
is an unbiased estimate of, then. Let
to that tois also the INT estimator. Then, the rule
Proschan and Sullo (1976) showed thatis described: Take asample of observations. Detect the component associated withas the worst component.
It is easy to see that under the LFC, the PCD is written for large that
$X_{i\mathrm{r}}$$\lambda_{i}$$\lambda_{i}+\lambda_{0}$$1/(\lambda_{i}+\lambda_{0})$$\overline{X}_{in}$$\overline{X}_{in}=\sum_{r=1}^{n}X_{i\mathrm{r}}/n$$\lambda_{i}+\lambda_{0}$$\overline{X}_{in}$$n$$R_{2}$$\overline{X}_{jn}=\min(\overline{X}_{1n}, \ldots, \overline{X}_{kn})$$n$$P^{\star}(CD|R_{2})=P(\overline{X}_{kn}<\overline{X}_{in}, i=1, \ldots, k-1)$$\simeq\int_{-\infty}^{\infty}\Phi^{k-1}(\frac{x\sqrt{\rho_{2}}+c_{2}\sqrt{n}}{\sqrt{1-\rho_{2}}})d\Phi(x)$where$\rho_{2}=\rho_{2}(p_{0}, k, \delta^{\star})>0$

(2.4)

,

is defined as$\rho_{2}=\{2(1+\delta^{\star})+p\mathrm{o}(-13+\delta^{\star}4+k(7+\delta^{\star}+\delta^{\star}2+\delta^{\star}3))$$+p_{0}^{2}(25-16\delta^{\star}+8\delta^{\star}2-6\delta^{\star}3+\delta^{\star}4+4k(-6+2\delta^{\star}-2\delta^{\star}2+\delta^{\star 3})+k^{2}(5+3\delta^{\star}2))$$+p_{0}^{3}(-2(7-6\delta^{\star}+\delta^{\star^{2}})+k(17-3\delta^{\star}-5\delta^{\star^{2}}+\delta^{\star}3)+k^{2}(-5-6\delta^{\star}+3\delta^{\star^{2}})+2k^{3}\delta^{\star})\}$$(2-3p_{0}+kp0+\delta^{\star}p_{0})^{-1}(1+\delta^{\star}-3p_{0}+2kp\mathrm{o}+\delta^{\star}p_{0})^{-1}(1+\delta^{\star^{2}}-2p_{0}+kp_{0}+k\delta^{\star}p\mathrm{o})^{-1}$and$c_{2}= \{\frac{(1-p\mathrm{o})(\delta^{\star}-1)^{2}(1+\delta^{\star}-2p_{0}+kp_{0})}{(1+\delta^{\star}-3p0+2kp0+\delta^{\star}p\mathrm{o})(1+\delta^{\star^{2}}-2p_{0}+kp_{0}+k\delta^{\star}p\mathrm{o})}\}^{1/2}$$=c_{2}(p_{0}, k, \delta^{\star})$So, if we solve for$li_{2}$(2.5), (say).

$=u_{2}(p_{0}, k, \delta^{\star}, P^{\star})>0$

from the equation,$\int_{-\infty}^{\infty}\Phi^{k-1}(\frac{x\sqrt{\rho_{2}}+u_{2}}{\sqrt{1-\rho_{2}}})d\Phi(x)=P^{\star}$an approximate minimum sample size needed to satisfy requirement (1.2) is given by.
. Noting
through , we consider some upper bounds for
depends on
Since, we haveis increasing in for any fixedthat$n_{0}^{(2)}=[u_{2}^{2}/c_{2}^{2}]+1$$n_{0}^{(2)}$$u_{2}$$P\mathrm{o}$$722=\rho_{2}(p_{0}, k, \delta^{\star})$$\rho_{2}$$\delta^{\star}>1$$p_{0}$$\rho_{2}=\rho_{2}(p_{0}, k, \delta^{\star})>\rho_{2}(0, k, \delta^{\star})=\frac{1}{1+\delta^{\star^{2}}}=\rho_{2}^{\star}$(say).

142
By using Slepian’s inequality, the integral in (2.4) is bounded below by$\int_{-\infty}^{\infty}\Phi^{k-1}(\frac{x\sqrt{\rho_{2}^{\star}}+c_{2}\sqrt{n}}{\sqrt{1-\rho_{2}^{\star}}})d\Phi(x)$Then, we have an upper bound forffom by solving the equation$n_{0}^{(2)}$as$n_{1}^{(2)}=[u_{2}^{\star 2}/e]+1$

.
where

$u_{2}^{\star}>0$

is given freely$P\mathrm{o}$$\int_{-\infty}^{\infty}\Phi^{k-1}(\frac{x\sqrt{\rho_{2}^{\star}}+u_{2}^{\star}}{\sqrt{1-\rho_{2}^{\star}}})d\Phi(x)=P^{\star}$

The following table gives values of$u_{2}^{\star}$for $k=3(1)10$ and $\delta^{\star}=1.5(0.5)3.0$ when $P^{\star}=0.90$ .

Table 2.2. Values ofWhen$p_{0}$.

is unknown, we need to estimate

when $7”=0.90$$u_{2}^{\star}$through an estimate$n_{1}^{(2)}$$\mathrm{p}\mathrm{O}$based on apilot

sample.

Remark 1. When we note thatforis obtained assolving the equation$\delta^{\star}\geq\rho 2(1-\hslash)^{-1}\geq\delta^{\star-2}$$n_{2}^{(2)}=[(1-\rho_{2})\tilde{u}_{2}^{2}/d]+1$$n_{0}^{(2)}$wherein (2.4), another upper boundis given freely fromby$\tilde{u}_{2}>0$$p_{0}$.

$\int_{-\infty}^{0}\Phi^{k-1}(x\sqrt{\delta^{\star}}+\tilde{u}_{2})d\Phi(x)+\int_{0}^{\infty}\Phi^{k-1}(\frac{x}{\delta^{\star}}+\tilde{u}_{2})d\Phi(x)=P^{\star}$

3. TWO-STAGE PROCEDURES
Before considering tw0-stage procedure for dealing with the case when is unknown,
let us compare the efficiencies of the procedures
andin terms of the required
sample size for given . We calculated the values of the ratios
andfor$p_{0}=0.1(0.1)0.9$ when $k=3(1)10$ , $\delta^{\star}=1.5(0.5)3.0$ and $P^{\star}=0.90$ . The findings of such
survey were as follows: The ratio
is increasing in for $p_{0}<0.4$ , decreasing in for$p_{0}\geq 0.5$ and decreasing inand . The ratio
is increasing in and decreasing
in and . When
,
;then
for
exceptfor the cases that $k=9$ and 10 for $p_{0}=0.4$ . Whenand 2.5,$p_{0}\leq
0.3$
for
except for the cases that $k=3$ and 4for $p_{0}=0.3$ when;then
$p_{0}$$R_{1}$$R_{2}$$n_{1}^{(1)}/n_{1}^{(2)}$$p_{0}$$n_{1}^{(1)}/n_{1}^{(2)}$$\delta^{\star}$$p_{0}$$\delta^{\star}$$p_{0}$$\delta^{\star}=1.5$$n_{1}^{(1)}/n_{1}^{(2)}>1$$n_{1}^{(1)}/n_{2}^{(2)}$

$k$

$k$

$n_{1}^{(1)}/n_{2}^{(2)}$$P\mathrm{o}\leq 0.4$$k$$n_{1}^{(1)}/n_{1}^{(2)}>n_{1}^{(1)}/n_{2}^{(2)}$$\delta^{\star}=2.0$$n_{1}^{(1)}/n_{1}^{(2)}>1$$\delta^{\star}=2.5$

143.
for $p_{0}\leq 0.2$ ; then,and 2.5,, $p_{0}\leq 0.3$ foris small (say, $p_{0}\leq 0.4$ forConsequently, whenseems to be the most preferable forwith), the procedure
and $p_{0}\leq 0.2$ for
consideration to start thinking atw0-stage procedure. Then, for other cases, that is ismight be considered. We will comparewithmoderate or large, the procedurein the tw0-stage procedure described below.
later by estimatingwith$n_{1}^{(1)}/n_{1}^{(2)}>n_{1}^{(1)}/n_{2}^{(2)}$. When$\delta^{\star}=3.0$$n_{1}^{(1)}/n_{1}^{(2)}>n_{1}^{(1)}/n_{2}^{(2)}$$n_{1}^{(1)}/n_{1}^{(2)}>1$$\delta^{\star}=2.0$$\delta^{\star}=1.5$$p_{0}$$\delta^{\star}=3.0$$R_{2}$$n_{1}^{(2)}$$p_{0}$$R_{1}$$R_{2}$$n_{1}^{(1)}$$R_{1}$$p_{0}$For Procedure$R_{1}$

We apply one of the procedures given by Aoshima, Chen and Panchapakesan (2001),
which is to select the most probable cell of amultinomial distribution in the presence of:is based on
anuisance cell, for the present problem. The following procedure
denote the count ofFirst, take asample of size $m$ , which is moderately large. Letbywith $\hat{p}_{0}=m_{0}/m$ as in (2.3). Define$\{x_{1}=\ldots=x_{k}\}$ . Compute$n_{1}^{(1)}$$(S_{1})$$m\circ$$N^{(1)}$$\hat{c}_{1}=c_{1}(\hat{p}_{0}, k, \delta^{\star})$$N^{(1)}= \max\{m$ ,$[u_{1}^{\star 2}/\hat{c}_{1}^{2}]+1\}$

(3.1)

.

Next, take an additional sample of size $N^{(1)}-m$ . On the basis of the total sample ofbe the count of $\{x_{i}<x_{i’}\}$ for $i=1$ , , . Then, detect the component
, let
size
as the worst component; usethat attained the largest count among$N^{(1)}$$k$$N_{i}^{(1)}$$\ldots$$(N_{1}^{(1)}, N_{2}^{(1)}, \ldots, N_{k}^{(1)})$

randomization to break ties for the first place.
When $m$ is moderately large, it is shown that Procedure

For Procedure$S_{1}$

satisfies requirement (1.2).

$R_{2}$: First, take asample
We propose the following tw0-stage procedure (S2) based ondenote the count of $\{x_{1}=\ldots=x_{k}\}$ . Computeof size $m$ , which is moderately large. Letbywith $\hat{p}_{0}=m_{0}/m$ as in (2.5). Define$n_{1}^{(2)}$$m_{0}$$N^{(2)}$$\hat{c}_{2}=c_{2}(\hat{p}_{0}, k, \delta^{\star})$$N^{(2)}= \max\{m$ ,$[u_{2}^{\star 2}/\hat{c}_{2}^{2}]+1\}$

(3.2)

.

Next, take an additional sample of size $N^{(2)}-m$ . Then, on the basis of the total sample, detect the component associated with the smallest sample mean amongof sizeas the worst component.
also satisfies requirement (1.2) when $m$ is moderIt can be shown that Procedure$N^{(2)}$$(\overline{X}_{1N^{(2)}},\overline{X}_{2N^{(2)}}, \ldots, \overline{X}_{kN^{(2)}})$$S_{2}$

ately large.
through several simand
Now, let us investigate into efficincies of Procedures
ulation studies. We estimated the PCS and the expected sample size by conducting thesimulation with 10,000 ( $=R$ , say) trials for each procedure. The following result given by$S_{1}$$S_{2}$

144
Proschan and Sullo (1976, Thorem 2.1) was used for generation of -variate exponential
random number: Let , , ,be independent exponential random numbers with parameters , , , , respectively. Then, $(T_{1}, \ldots, T_{k})=(\min(U_{1}, U_{0})$ , , $\min(U_{k}, U_{0}))$ has
the -variate exponential distribution with parameters. Now, we set $k=4$
and. Then, under the, the parameters are written as,,,, $12$ ). Set(
and $P^{\star}=0.90$ Then, for every
case, we used the values of and respectively given in Tables 2.1 and 2.2. We started
andwith apilot sample of size $m=[0.8 \min\{n_{0}^{(1)}, n_{0}^{(2)}\}]+1$ for $p_{0}=0.1(0.1)0.9$ ,
where
and
are considered as optimal fixed-sample sizes.
For each procedure , let us write
as . Letbe the observed value of $N$ and$p_{f}=1$ (or 0) according as acorrect decision occurs (or does not occur). We
denote,and.
The quantities and respectively estimate $E(N)$ and $P(CD)$ , whileand sffi) standfor their corresponding estimated standard errors. In Table 3.1, we report the values of
,
,
,
, ,
and
for each . For each , the upper line gives
those values forand the lower line gives for .
$k$$U_{0}$$U_{1}$$\ldots$$\lambda_{0}$$\lambda_{1}$$U_{k}$$\lambda_{k}$$\ldots$$\ldots$$k$$(\lambda_{0}, \lambda_{1}, \ldots, \lambda_{k})$$\lambda_{4}=12$$\frac{12p\mathrm{o}(3+\delta^{\star})}{\delta^{*}(1-p\mathrm{o})}$$\mathrm{L}\mathrm{F}\mathrm{C}$$(\lambda_{0}, \lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4})=$$12/\delta^{\star}$$12/\delta^{\star}$$u_{1}^{\star}$$S_{1}$$\delta^{\star}=1.5,2.0$$12/\delta^{\star}$$u_{2}^{\star}$$S_{2}$$n_{0}^{(1)}$$n_{0}^{(2)}$$S$$\overline{n}=\sum_{\mathrm{r}=1}^{R}n_{f}/R$$n\circ$$n^{\star}$$n_{f}$$s^{2}( \overline{n})=\sum_{f=1}^{R}(n_{f}-\overline{n})^{2}/(R^{2}-R),\overline{p}=\sum_{r=1}^{R}p_{f}/R$$s^{2}(\overline{p})=\overline{p}(1-\overline{p})/R$$\overline{n}$$\overline{p}$$n^{\star}$$P\mathrm{o}$$m,\overline{p}$$s(\overline{p})$$\overline{n}$$s(\overline{n})$$s(\overline{n})$$(\overline{n}-n^{\star})/n^{\star}$$\delta^{\star}$$P\mathrm{o}$$S_{1}$$S_{2}$

Table 3.1. Estimated $P(CD)$ and $E(N)$ for(the upper) andwith 10,000 trials when $7”=0.90$ and $k=4$$S_{1}$$\underline{\delta^{\star}=1.5}$$S_{2}$(the lower)

145$\underline{\delta^{\star}=2.0}$$n^{\star}$$p_{0}$

44.38
24.23
0.2 50.22
34.81
0.3 57.73
48.84

0.1

$m$

$\overline{p}$0.9106
0.9311
0.9121
0.92360.91290.9251202840$s(\overline{p})$0.00290.00250.00280.00270.00280.0026

45.25
26.15
51.16
38.17
59.14
54.51

0.9154 0.0028 69.58

0.4 67.74 55
67.99
0.5 81.76 66
95.22
0.6 102.8 83
136.6
0.7 137.8 111

0.92850.91720.92700.92220.92630.91800.9250206.0
0.8 207.9 167 0.9196
0.9193345.5
0.9 418.1 335 0.9271
0.9250765.4

0.00260.00280.00260.00270.00260.00270.00260.00270.00270.00260.0026$s(\overline{n})$$(\overline{n}-n^{\star})/n^{\star}$

0.03950.07300.05500.11020.06850.14270.08210.18180.11050.24880.15410.34520.22260.50140.36070.81350.77381.742

0.01940.07920.0187
0.09650.02430.1161$\overline{n}$

76.74
84.22
108.4
106.3
156.6
142.7
239.3
215.9
404.0
434.7
904.7

0.02720.12880.03010.13790.03400.14700.03520.16190.03880.16940.03960.1821

andProm these tables, we can observe that the proposed tw0-stage procedures
andwork well as expected. The arguments about the comparison of efficiencies ofseems to hold for and as well. We consequently recommend the experimenter that,once after taking apilot sample, examine the value of :If looks small, saywould be sufficient for such situation.
proceed ;otherwise, even
$S_{1}$$S_{2}$$R_{1}$$R_{2}$$S_{2}$$\hat{p}0$$S_{2}$$S_{1}$$\hat{P}0$$\hat{P}0\leq 0.3$

$S_{1}$

4. EXAMPLE
In the field of manufacturing of cellular phone, it is important to examine the durabilityof products. Acellular phone might be accidentally exposed to various situations, such as
vibration and ashock, in many cases into alife. The consumers are easy to request thatthe weight of acellular phone should be light, so no portions can be reinforced in order topursue durability. Now, the experimenters would conduct an oscillating experiment to test
durability of their products and to detect the lowest part of durability of the products.
Here, it is considered that acellular phone has four divided parts, that is “receivingpart” , “display part, “operation part” and “mouthpiece part”, which are controlled by
the “power supply part”. If the power supply part breaks, the function of the wholecellular phone stops accordingly. We suppose that the lifetime model of acellular phone
is considered as the hazard related tofollows the MVE distribution, where $k=4$ andfailure of the “power supply part”.
$\lambda_{0}$146
Let us apply atw0-stage procedure proposed in Section 3for data analysis in thissituation to detect the lowest part of durability. Survival time dataarerecorded in unit $\min$ . for (receiving part, display part, operation part, mouthpiece part).
From the experimental side, the difference was set as
and the confidence was set
as $7”=0.9$ . We start with apilot sample of size $m=30$ .
$(x_{1}, x_{2}, x_{3}, x_{4})$$\delta^{\star}=2$Table 4.1. Survival Data of Four Components (1st Stage)
(24.3
(25.3
(29.0
(5.4
(4.8
(7.1
(26.1
(14.4
(15.5
(18.2
(5.0
(16.9
(35.3
(20.5
(22.0

8.9
4.6
12.3
5.4
18.0
7.1
11.6
10.1
15.5
11.8
15.5
16.9

29.7
20.5
13.9

24.3,
15.3,
6.5,
5.4,

23.1
7.1
26.1
14.4
15.5
15.5
27.7

16.9
13.4,
20.5,
22.0,

7.3 )
20.5 )
21.4 )
5.4 )
26.1 )

14.7

27.3
21.5
22.9
30.1

14.7
33.7
21.5
26.2
8.6

16.5, 14.3,
7.1 )
10.1 )20.9 15.3
14.4 )16.6 11.5
15.5 )
14.4 14.4,
12.2 )
23.7 19.3,
27.7 )
21.2 17.5,
8.9,
16.9 )
23.9
7.1,
35.3 )
7.1
20.5 )
28.3 11.8,
22.0 )
18.4 25.1,

14.7,
12.9,
21.5,
47.6,
36.1,
14.4,
13.0,
16.6,
14.4,
10.1,
21.0,
11.8,
6.5,
19.0,
29.7,

14.7)
16.4)
21.5)
16.8)
36.2)
32.5)
20.9 )
12.2 )
14.4)
23.7)
10.0)
29.9)
7.1)
28.3)
29.7)

We observe that the power supply part breaks $m_{0}=8$ times when $x_{1}=x_{2}=x_{3}=x_{4}$ .
Then, we have $\mathrm{p}0=8/30=0.2666$ . Let us proceed with Procedure . From Table 2.2,
we find thatfor the present case when $(k, \delta^{\star})=(4,2)$ . Then, the observed
value ofis given by$S_{2}$$u_{2}^{\star}=1.796$$N^{(2)}$$n= \max\{30$ , $[ \frac{4.866\cross 7.666}{(1-0.2666)\cross 3.533}u_{2}^{\star}2]+1\}=47$

according as (3.2). We need to do 17 additional tests and then additional data are taken
as follows.

Table 4.2. Suvival Data of Four Components (2nd Stage)
(13.9
(14.2
(9.5
9.3,
(
(21.7
(15.5
(7.5
(18.0
(14.4

13.9,
13.2,
23.7
9.3,

22.6
9.3
18.2
12.2
9.7

13.9,
4.6,
23.7

9.3
28.8
20.5
18.2
18.0
12.2

13.9)
14.2)
18.7)
6.5)
28.8)
21.8)
18.2)
10.1)
25.1)

(13.9
(
4.3,
(14.6
(17.4
(17.1
(
7.8
(9.8
(14.6

12.3
9.9
14.6
10.7
10.7
7.8

15.1,
9.9,
14.6,
17.4,
25.5,
7.8,

10.1
14.6

13.4,
14.6,

15.1)
9.9)
14.6)
17.4)
25.5)
7.8)
13.4)
14.6)

147
By combing the date of both the stages, we have the overall sample means (16.87, 14.33,
17.29, 18.13) for suvival time of each component. Then, the second component “display
part” is detected as the worst component for durability with confidence $P^{\star}=0.9$ .
incidentally for this data set, it comes to the same conclusion
Remark 2. If we apply
is moderate or large, the experiment should beto detect the display part. When
based on the sample means.
is small, we use
reexamined in those cases. When
However, when we detect the component which has the minimum lifetime, in some cases
for such purpose, the experimenter might take some censoring once after recording theminimum time along with the component number. Then, only incomplete data is available
for statistician. So, it would be necessary to consider such situation as well in the next
$S_{1}$$p_{0}$$p_{0}$$S_{2}$

stage of this research.

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