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\begin{center}
{\Large \textbf{Moments Inequalities for NBRUL Distributions with Hypotheses
Testing Applications}}

\bigskip

Mohamed A. W. Mahmoud, Rashad M. EL-Sagheer, Walid B. H. Etman

Mathematics Department, Faculty of Science, AL-Azhar University, Nasr City
11884, Cairo, Egypt

\bigskip
\end{center}

\begin{quotation}
\textbf{Abstract: }In this paper, moment inequalities for the new better
than renwal used in Laplace transform order ( NBRUL ) class of ageing
distributions are derived. This inequalities demonstrate that if the mean
life is finite, then all higher order moments exist. A new test for
exponentiality versus NBRUL can be constructed using thes inequalities.
Pitman's asymptotic efficiencies and critical values of the proposed test
are calculated and tabulated. The powers of this test are estimated for some
famously alternatives distributions in reliability such as Linear failure
rate,Weibull and gamma distributions. Finally, examples in different areas
are used as a practical applications of the proposed test.

\textbf{Keywords:} Classes of life distributions, NBRUL, Moments
inequalities, Life testing .
\end{quotation}

\section{Introduction}

Classes of life distributions are defined to classify the life distributions
according to their aging properties. The definitions of these classes helped
statisticians to define the test statistics. The test statistics are defined
based on definition of the classes. the main aim of constructing new tests
is to gain higher efficiencies. Many authors proposed tests for
exponentiality versus some classes of life distributions based on the moment
inequalities. Testing exponentiality against IFR, NBU and NBUE based on
moment inequalities have been studied by Ahmad \cite{8}. Ahmad and Mugdadi 
\cite{10} constructed the tests of NBUC, IFRA and DMRL depends on the moment
inequalities, while testing NRBU and RNBU based on moment inequalities have
been studied by Mahmoud et al. \cite{23}. Using the moment inequalities of
the class NBUL, Mahmoud et al. \cite{24} constructed a test statistic for
testing exponentiality versus this class.

In this paper our theme is formulates a new test statistic for testing
exponentiality against NBRUL class based on the moment inequalities and
discuss this test. The main classes of life distributions which have been
introduced in the literature are based on new better than used (NBU), new
better than used failure rate (NBUFR), new better than average failure rate
(NBAFR), new better than used renewal failure rate (NBURFR), new better than
used average renewal failure rate (NBARFR), new better than renewal used
(NBRU)\ and exponential better than used in Laplace transform order (EBUL).
Testing exponentiality against some classes of life distributions has been
introduced by many researchers from many points of views. For more details
one can refer to Bryson and Siddiqui \cite{12}, Deshpande et al. \cite{13},
Abouammoh and Ahmed \cite{3,4}, Abouammoh et al. \cite{2}, Mahmoud et al. 
\cite{25}, Kumazawa \cite{17}, Ahmed \cite{7,8}. Abouammoh and Newby \cite{5}%
, Mahmoud and Abdul Alim \cite{19,20,21}, Ahmed et al. \cite{9}, Abu-
Youssef \cite{6}, Ismail and Abu- Youssef \cite{15} and Mahmoud and Rady 
\cite{26}. Recently Atallah et al. \cite{011} developed a new method for
testing exponentiality which is more general and flexible than goodness
approach.

The rest of this paper can be organized as follows, Section 2 gives a brief
knowledge about renewal classes. In Section 3 moment inequalities for the
NBRUL class are developed. In Section 4, Testing exponentiality against
NBRUL is proposed based on moment inequalities. In Section 5, Pitman's
asymptotic efficiency (PAE) of the test for several common alternatives will
be considered. In Section 6, Monte Carlo null distribution critical points
from the null distribution for sample size n = 5(5)35; 39; 40(5)50. In
section 7, The power estimate for the test are calculated. Finally, the
application of the proposed test for real data sets are discussed in Section
8.

\section{Renewal Classes}

Let $T$ be a random variable represents life time of a device (system or
component) with a continuous life distribution $F\left( t\right) .$ Upon
arising the failure of the device, it can be substituted by a sequence of
mutually independent devices which are identically distributed with the same
life distribution $F\left( t\right) .$ The following stationary renewal
distribution constitutes the remaining life distribution of the device under
operation at time $t$.%
\begin{equation*}
W_{F}\left( t\right) =\mu _{F}^{-1}\int_{0}^{t}\overline{F}\left( u\right)
du,\text{ \ \ \ }t\geq 0,
\end{equation*}%
where $\mu _{F}=\mu =\int_{0}^{\infty }\overline{F}\left( u\right) du.$

It is easy to show that%
\begin{equation*}
\overline{W_{F}}\left( t\right) =\mu _{F}^{-1}\int_{t}^{\infty }\overline{F}%
\left( u\right) du,\text{ \ \ \ \ }t\geq 0.
\end{equation*}%
For extra details, see Barlow and Proschan \cite{11} and Abouammoh and Ahmed 
\cite{3,4}. Now we need to present the definitions of the NBRU (NWRU) and
NBRUL (NWRUL) classes of life distributions.\newline
\textbf{Definition (2.1): }Abouammoh et al. \cite{2}.

If\ $X$ is a\ random variable with survival function $\bar{F}\left( x\right)
,$ then $X$ \ is said to have new better (worse) than renewal used property,
denoted by NBRU (NWRU), if%
\begin{equation*}
\overline{W_{F}}\left( x|t\right) \leq (\geq )\overline{F}\left( x|0\right) ,%
\text{ \ \ \ \ }x\geq 0,t\geq 0,
\end{equation*}%
or%
\begin{equation*}
\overline{W_{F}}\left( x+t\right) \leq (\geq )\overline{W_{F}}\left(
t\right) \overline{F}\left( x\right) ,\text{ \ \ \ \ }x\geq 0,t\geq 0.
\end{equation*}%
Depending on the definition (2.1), Mahmoud et al. \cite{027} defined a new
class which is called new better (worse) than renewal used in Laplace
transform order NBRUL (NWBRUL) as follows\newline
\textbf{Definition (2.2):}

$X$ is said to be NBRUL (NWRUL) if%
\begin{equation*}
\int_{0}^{\infty }e^{-sx}\overline{W_{F}}\left( x+t\right) dx\leq (\geq )%
\overline{W_{F}}\left( t\right) \int_{0}^{\infty }e^{-sx}\overline{F}\left(
x\right) dx,\text{ \ \ }x,t,s\geq 0.
\end{equation*}%
It is obvious that NBRU$\Rightarrow $NBRUL$\Rightarrow $NBRUE.

\section{Moments inequalities}

In this section,\ the moment inequalities for NBRUL class are established.%
\newline
\textbf{Theorem (2.1):}\newline
Let \ $F$ be NBRUL \ life distribution \ such that all moments exist and
finite then for integer \ $r\geq 0$ and $s\geq 0$. Then%
\begin{eqnarray}
\frac{\mu _{(r+2)}}{s(r+1)(r+2)}[1-\zeta (s)] &\geq &\frac{-(-1)^{r}r!}{%
s^{r+2}}[\mu _{F}-\frac{1}{s}(1-\zeta (s))]  \notag \\
&&+\frac{r!}{s^{r+1}}\sum_{i=0}^{r}(-1)^{i}\frac{s^{r-i}}{(r-i+2)!}\mu
_{(r-i+2)},  \TCItag{1}
\end{eqnarray}%
where $\mu _{(r)}=E(X^{r}),\zeta (s)=Ee^{-sX}.$\newline
\textbf{Proof.}\newline
Since $F$ is NBRUL, then%
\begin{equation}
\int_{0}^{\infty }e^{-sx}\overline{\;W_{F}}\left( x+t\right) dx\leq 
\overline{W_{F}}\left( t\right) \int_{0}^{\infty }e^{-sx}\overline{F}\left(
x\right) dx,,\text{ \ \ }x,t\geq 0.  \tag{2}
\end{equation}%
Making use of (2), yields%
\begin{equation}
\int_{0}^{\infty }t^{r}\int_{0}^{\infty }e^{-sx}\overline{\;W_{F}}\left(
x+t\right) dxdt\leq \int_{0}^{\infty }t^{r}\overline{W_{F}}\left( t\right)
\int_{0}^{\infty }e^{-sx}\overline{F}(x)dxdt.  \tag{3}
\end{equation}%
The left hand side of (3) can be written as%
\begin{equation*}
\int_{0}^{\infty }t^{r}\overline{W_{F}}\left( t\right) \int_{0}^{\infty
}e^{-sx}\overline{F}(x)dxdt=E\int_{0}^{\infty }t^{r}\overline{W_{F}}\left(
t\right) \int_{0}^{\infty }e^{-sx}I(X>x)dxdt,
\end{equation*}%
where%
\begin{equation*}
I(X>x)=\left\{ 
\begin{array}{c}
0\;\;\;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ,x\geq X, \\ 
1\;\;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\;x<X.%
\end{array}%
\right.
\end{equation*}%
After some calculations, the left hand side of (3) is given by%
\begin{equation}
\frac{\mu _{F}^{-1}\mu _{(r+2)}}{s(r+1)(r+2)}(1-\zeta (s)).  \tag{4}
\end{equation}%
Also, the right hand side of (3) can be put in the following form%
\begin{equation}
\int_{0}^{\infty }t^{r}\overline{W_{F}}\left( t\right) \int_{0}^{\infty
}e^{-sx}\overline{F}(x)dxdt=\int_{0}^{\infty }e^{-sv}\overline{W_{F}}\left(
v\right) \int_{0}^{v}u^{r}e^{su}dudv.  \tag{5}
\end{equation}%
After some calculations (5) can be rewritten as%
\begin{eqnarray}
\int_{0}^{\infty }t^{r}\overline{W_{F}}\left( t\right) \int_{0}^{\infty
}e^{-sx}\overline{F}(x)dxdt &=&\frac{r!}{s^{r+1}}\mu
_{F}^{-1}\sum_{i=0}^{r}(-1)^{i}\frac{s^{r-i}}{(r-i+2)!}\mu _{(r-i+2)}  \notag
\\
&&-\frac{(-1)^{r}r!}{s^{r+2}}\mu _{F}^{-1}\left[ \mu _{F}-\frac{1}{s}\left(
1-\zeta (s)\right) \right] .  \TCItag{6}
\end{eqnarray}%
From (4) and (6), Eq. (1) can be proved.\newline
\textbf{Remark:} For $r=1,$ Eq.(1)\ will be reduced to%
\begin{equation}
\frac{\mu _{3}}{6s}[1-\zeta (s)]\leq \frac{1}{s^{3}}\left[ \mu -\frac{1}{s}%
\left( 1-\zeta (s)\right) \right] +\frac{1}{s^{2}}\left[ \frac{s}{6}\mu
_{\left( 3\right) }-\frac{1}{2}\mu _{\left( 2\right) }\right] ,  \tag{7}
\end{equation}

where $\mu _{(r)}=\int_{0}^{\infty }x^{r}dF(x).$

\section{Testing Against NBRUL Alternatives}

Using Inequality (7) we can test the null hypothesis \ $H_{0}:F$ \ is
exponential aganist \ $H_{1}:F$ is NBRUL and not exponential . $\delta
_{1}(s)$ has been used as follows \ 
\begin{equation}
\delta ^{(1)}(s)=\frac{1}{2s^{2}}\mu _{\left( 2\right) }-\frac{1}{6s}\mu
_{\left( 3\right) }\zeta (s)-\frac{1}{s^{4}}\zeta (s)-\frac{1}{s^{3}}\mu +%
\frac{1}{s^{4}}.  \tag{8}
\end{equation}%
Note that\ under $H_{0},\delta ^{(1)}(s)=0,$ while under $H_{1},\delta
^{(1)}(s)>0.$\newline
Let $X_{1},X_{2},X_{3},.........,X_{n}$ be a random sample from a
distribution F , the empirical estimate $\delta _{n}^{(1)}(s)$ of $\delta
^{(1)}(s)$ can be obtained as 
\begin{equation*}
\delta _{n}^{(1)}(s)=\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}[\frac{1}{%
2s^{2}}X_{i}^{2}-\frac{1}{6s}X_{i}^{3}e^{-sX_{j}}-\frac{1}{s^{4}}e^{-sX_{i}}-%
\frac{1}{s^{3}}X_{i}+\frac{1}{s^{4}}].
\end{equation*}%
To make the test invarient, let \ $\Delta _{n}^{(1)}(s)=\frac{\delta
_{n}^{(1)}(s)}{\overline{X}^{4}},$ where \ $\overline{X}$\ is the sample
mean. Then%
\begin{equation}
\Delta _{n}^{(1)}(s)=\frac{1}{n^{2}\overline{X}^{4}}\sum_{i=1}^{n}%
\sum_{j=1}^{n}[\frac{1}{2s^{2}}X_{i}^{2}-\frac{1}{6s}X_{i}^{3}e^{-sX_{j}}-%
\frac{1}{s^{4}}e^{-sX_{i}}-\frac{1}{s^{3}}X_{i}+\frac{1}{s^{4}}].  \tag{9}
\end{equation}%
One can note that $\delta ^{(1)}(s)$ is an unbiased estimator of \ $\delta
_{n}^{(1)}(s).$

Now, set%
\begin{equation}
\phi _{s}(X_{i},X_{j})=\frac{1}{2s^{2}}X_{i}^{2}-\frac{1}{6s}%
X_{i}^{3}e^{-sX_{j}}-\frac{1}{s^{4}}e^{-sX_{i}}-\frac{1}{s^{3}}X_{i}+\frac{1%
}{s^{4}},  \tag{10}
\end{equation}%
and define the symmetric Kernel%
\begin{equation*}
\psi _{s}(X_{i},X_{j})=\frac{1}{2!}\sum_{R}\phi _{s}(X_{i},X_{j}),
\end{equation*}%
where the summation is over all arrangements of \ $X_{i},X_{j}$. Then $%
\Delta _{n}^{(1)}(s)$ \ in (9) is equivalent to the $U_{n}$ -statistic given
by%
\begin{equation}
U_{n}=\frac{1}{\left( _{2}^{n}\right) }\sum_{i<j}\psi _{s}(X_{i},X_{j}) 
\tag{11}
\end{equation}%
The asymptotic normality of $\Delta _{n}^{(1)}(s)$ can be summarized in the
following theorem.\newline
\textbf{Theorem 3.1} \ \ \ (i)\ $As$ \ $n\rightarrow \infty ,\sqrt{n}(\Delta
_{n}^{(1)}(s)-\Delta ^{(1)}(s))$\ is asymptotically normal with mean\ 0 \
and variance $\sigma ^{2}(s),$ where%
\begin{eqnarray}
\sigma ^{2}(s) &=&Var\{\frac{1}{2s^{2}}X^{2}-\frac{1}{6s}X^{3}\zeta (s)-%
\frac{1}{s^{4}}e^{-sx}-\frac{1}{s^{3}}X+\frac{1}{2s^{2}}\mu _{\left(
2\right) }  \notag \\
&&-\frac{1}{6s}e^{-sx}\mu _{\left( 3\right) }-\frac{1}{s^{4}}\zeta (s)-\frac{%
1}{s^{3}}\mu +\frac{2}{s^{4}}\}.  \TCItag{12}
\end{eqnarray}%
(ii) \ Under $H_{0},$ the variance \ $\sigma _{0}^{2}(s)$ is%
\begin{equation}
\sigma _{0}^{2}(s)=\frac{19+14s+s^{2}}{(1+s)^{4}(1+2s)}  \tag{13}
\end{equation}%
\newline
\textbf{Proof.}\newline
Using standard U-statistic theory, cf. Lee \cite{18},%
\begin{equation*}
\sigma ^{2}(s)=Var\{E[\phi _{s}(X_{1},X_{2})\mid X_{1}]+E[\phi
_{s}(X_{1},X_{2})\mid X_{2}]\}.
\end{equation*}%
Recall the definition of $\phi _{s}(X_{i},X_{j})$ in (10), thus it is easy
to show that%
\begin{equation*}
E(\phi _{s}(X_{1},X_{2})\mid X_{1})=\frac{1}{2s^{2}}X^{2}-\frac{1}{6s}%
X^{3}\int_{0}^{\infty }e^{-sx}dF(x)-\frac{1}{s^{4}}e^{-sx}-\frac{1}{s^{3}}X+%
\frac{1}{s^{4}},
\end{equation*}%
and 
\begin{eqnarray*}
E(\phi _{s}(X_{1},X_{2}) &\mid &X_{2})=\frac{1}{2s^{2}}\int_{0}^{\infty
}x^{2}dF(x)-\frac{1}{6s}e^{-sx}\int_{0}^{\infty }x^{3}dF(x)-\frac{1}{s^{4}}%
\int_{0}^{\infty }e^{-sx}dF(x) \\
&&-\frac{1}{s^{3}}\int_{0}^{\infty }xdF(x)+\frac{1}{s^{4}},
\end{eqnarray*}%
therefore%
\begin{eqnarray*}
\sigma ^{2}(s) &=&Var\{\frac{1}{2s^{2}}X^{2}-\frac{1}{6s}X^{3}\int_{0}^{%
\infty }e^{-sx}dF(x)-\frac{1}{s^{4}}e^{-sx}-\frac{1}{s^{3}}X+\frac{1}{2s^{2}}%
\int_{0}^{\infty }x^{2}dF(x) \\
&&-\frac{1}{6s}e^{-sx}\int_{0}^{\infty }x^{3}dF(x)-\frac{1}{s^{4}}%
\int_{0}^{\infty }e^{-sx}dF(x)-\frac{1}{s^{3}}\int_{0}^{\infty }xdF(x)+\frac{%
2}{s^{4}}\}.
\end{eqnarray*}%
Under \ $H_{0}$%
\begin{equation*}
\sigma _{0}^{2}(s)=\frac{19+14s+s^{2}}{(1+s)^{4}(1+2s)}.
\end{equation*}

\section{The Pitman Asymptotic Efficiency}

To judge on the quality of this procedure, Pitman asymptotic efficiencies
(PAEs) are computed and compared with some other tests for the following
alternative distributions:

\begin{enumerate}
\item[(i)] The Weibull distribution: $\bar{F}_{1}(x)=e^{-x^{\theta }},x\geq
0,\theta \geq 1.$

\item[(ii)] The linear failure rate distribution (LFR): $\bar{F}%
_{2}(x)=e^{-x-\frac{\theta }{2}x^{2}},x\geq 0,\theta \geq 0.$

\item[(iii)] The Makeham distribution: $\bar{F}_{3}(x)=e^{-x-\theta \left(
x+e^{-x}-1\right) },x\geq 0,\theta \geq 0.$
\end{enumerate}

Note that For $\theta $ $=1,\bar{F}_{1}(x)$ reduces to exponential
distribution while for $\theta =0,\bar{F}_{2}(x)$and $\bar{F}_{3}(x)$
reduces to exponential distribution. The PAE is defined by:%
\begin{equation}
PAE(\Delta _{n}^{(1)}\left( s\right) )=\frac{1}{\sigma _{\circ }\left(
s\right) }\left\vert \frac{d}{d\theta }\delta _{\theta }^{(1)}(s)\right\vert
_{_{\theta \rightarrow \theta _{\circ }}}.  \tag{14}
\end{equation}%
At \ $s=5$,%
\begin{equation*}
\delta _{\theta }^{(1)}(s)=\frac{1}{2s^{2}}\mu _{\theta (2)}-\frac{1}{6s}\mu
_{\theta (3)}\zeta _{\theta }(s)-\frac{1}{s^{4}}\zeta _{\theta }(s)-\frac{1}{%
s^{3}}\mu _{\theta }+\frac{1}{s^{4}},
\end{equation*}%
\newline
where%
\begin{eqnarray*}
\mu _{\theta } &=&\int_{0}^{\infty }\overline{F}_{\theta }(u)du,\mu _{\theta
(2)}=2\int_{0}^{\infty }u\overline{F}_{\theta }(u)du,\mu _{\theta
(3)}=3\int_{0}^{\infty }u^{2}\overline{F}_{\theta }(u)du \\
\zeta _{\theta }(s) &=&E_{\theta }(e^{-su})=\int_{0}^{\infty
}e^{-su}dF_{\theta }(u)=-\int_{0}^{\infty }e^{-su}d\overline{F}_{\theta }(u).
\end{eqnarray*}%
Hence,%
\begin{equation*}
\frac{d}{d\theta }\delta _{\theta }^{(1)}(s)=\frac{1}{2s^{2}}\mu _{\theta
(2)}^{\backprime }-\frac{1}{6s}(\mu _{\theta (3)}\zeta _{\theta
}^{\backprime }(s)+\mu _{\theta (3)}^{\backprime }\zeta _{\theta }(s))-\frac{%
1}{s^{4}}\zeta _{\theta }^{\backprime }(s)-\frac{1}{s^{3}}\mu _{\theta
}^{\backprime },
\end{equation*}%
\newline
where%
\begin{eqnarray*}
\backprime &=&\frac{d}{d\theta },\mu _{\theta }^{\backprime
}=\int_{0}^{\infty }\overline{F}_{\theta }^{\backprime }(u)du,\mu _{\theta
(2)}^{\backprime }=2\int_{0}^{\infty }u\overline{F}_{\theta }^{\backprime
}(u)du, \\
\mu _{\theta (3)}^{\backprime } &=&3\int_{0}^{\infty }u^{2}\overline{F}%
_{\theta }^{\backprime }(u)du,\zeta _{\theta }^{\backprime
}(s)=-\int_{0}^{\infty }e^{-su}d\overline{F}_{\theta }^{\backprime }(u).
\end{eqnarray*}%
Upon using the definition of the PAE in (14),we obtain%
\begin{equation*}
PAE(\delta ^{(1)})=\frac{1}{\sigma _{0}}\mid \frac{1}{2s^{2}}\mu _{\theta
(2)}^{\backprime }-\frac{1}{6s}(\mu _{\theta (3)}\zeta _{\theta
}^{\backprime }(s)+\mu _{\theta (3)}^{\backprime }\zeta _{\theta }(s))-\frac{%
1}{s^{4}}\zeta _{\theta }^{\backprime }(s)-\frac{1}{s^{3}}\mu _{\theta
}^{\backprime }\mid _{\theta \rightarrow \theta _{0}}.
\end{equation*}%
When $s=5,$ this leads to:

\begin{center}
\begin{gather*}
PAE\;[\Delta _{n}^{(1)}\left( 5\right) ,Weibull]=1.04561,\ PAE[\Delta
_{n}^{(1)}\left( 5\right) ,LFR]=0.931891\ \text{and} \\
PAE\;[\Delta _{n}^{(1)}\left( 5\right) ,Makeham]=0.232973,\text{ where}%
\;\sigma _{0}(5)=0.0894239. \\
\end{gather*}

\begin{tabular}{cccllcllc}
\multicolumn{9}{l}{\textbf{Table 1. }Comparison between the PAEs of our test
and some other tests} \\ \hline
Test &  & Weibull &  &  & LFR &  &  & Makeham \\ \hline
\multicolumn{1}{l}{Kango \cite{16}} &  & 0.132 &  &  & 0.433 &  &  & 0.144
\\ 
\multicolumn{1}{l}{Mugdadi and Ahmad \cite{27}} &  & 0.170 &  &  & 0.408 & 
&  & 0.039 \\ 
\multicolumn{1}{l}{Abdel-Aziz \cite{1}} &  & 0.223 &  &  & 0.535 &  &  & 
0.184 \\ 
\multicolumn{1}{l}{Mahmoud and Abdul Alim \cite{19,20,21}} &  & 0.050 &  & 
& 0.217 &  &  & 0.144 \\ 
\multicolumn{1}{l}{Our test $\Delta _{n}^{(1)}\left( 5\right) $} &  & 1.046
&  &  & 0.932 &  &  & 0.233 \\ \hline
\end{tabular}
\end{center}

From Table 1, it is obvious that $\Delta _{n}^{(1)}(5)$ is better than the
other tests based on the PAEs .

\section{Monte Carlo Null Distribution Critical Points}

In this section the Monte Carlo null distribution critical points of $\Delta
_{n}^{(1)}\left( 5\right) $ are simulated based on $10000$ generated samples
of size $n=5(5)35,39,40(5)50.$ From the standard exponential distribution by
using Mathematica 8 program. Table 2 gives the upper percentile points of
statistic $\Delta _{n}^{(1)}\left( 5\right) $ for different confidence
levels $90\%,95\%$ and $99\%$.

\begin{center}
\begin{tabular}{ccccccccccccc}
\multicolumn{13}{l}{\textbf{Table 2. }Critical Values of the statistic\ $%
\Delta _{n}^{(1)}\left( 5\right) $} \\ \hline
n &  &  &  & 90\% &  &  &  & 95\% &  &  &  & 99\% \\ \hline
5 &  &  &  & 0.039527 &  &  &  & 0.051659 &  &  &  & 0.081158 \\ 
10 &  &  &  & 0.024073 &  &  &  & 0.029370 &  &  &  & 0.041747 \\ 
15 &  &  &  & 0.019395 &  &  &  & 0.023351 &  &  &  & 0.032516 \\ 
20 &  &  &  & 0.016674 &  &  &  & 0.019865 &  &  &  & 0.026928 \\ 
25 &  &  &  & 0.015041 &  &  &  & 0.017706 &  &  &  & 0.023554 \\ 
30 &  &  &  & 0.013927 &  &  &  & 0.016372 &  &  &  & 0.021441 \\ 
35 &  &  &  & 0.013179 &  &  &  & 0.015419 &  &  &  & 0.019853 \\ 
39 &  &  &  & 0.012648 &  &  &  & 0.014794 &  &  &  & 0.018750 \\ 
40 &  &  &  & 0.012483 &  &  &  & 0.014519 &  &  &  & 0.018562 \\ 
45 &  &  &  & 0.011991 &  &  &  & 0.013944 &  &  &  & 0.018084 \\ 
50 &  &  &  & 0.011490 &  &  &  & 0.013433 &  &  &  & 0.017040 \\ \hline
\end{tabular}
\end{center}

From Table 2, it is obvious that the critical values are decreasing as the
samples size increasing and they are increasing as the confidence levels
increasing.

\section{Power Estimates of the test $\Delta _{n}^{(1)}\left( 5\right) $}

In this section the power of our test $\Delta _{n}^{(1)}\left( 5\right) $ \
will be estimated at $(1-\alpha )\%$ \ confidence level, $\alpha =0.05$ \
with suitable parameters values of $\theta $ at \ $n=10,20$ and $30$ \ with
respect to three alternatives Linear failure rate (LFR) , Weibull and Gamma
distributions based on $10000$ samples.

\begin{center}
\begin{tabular}{ccccccccccccccccc}
\multicolumn{17}{c}{\textbf{Table 3. }The Power Estimates of $\Delta
_{n}^{(1)}\left( 5\right) $} \\ \hline
$n$ &  &  &  & $\theta $ &  &  &  & LFR &  &  &  & Weibull &  &  &  & Gamma
\\ \hline
$10$ &  &  &  & $2$ &  &  &  & $0.6674$ &  &  &  & $0.9978$ &  &  &  & $%
0.9922$ \\ 
&  &  &  & $3$ &  &  &  & $0.8501$ &  &  &  & $1.0000$ &  &  &  & $0.9991$
\\ 
&  &  &  & $4$ &  &  &  & $0.9324$ &  &  &  & $1.0000$ &  &  &  & $1.0000$
\\ 
$20$ &  &  &  & $2$ &  &  &  & $0.9360$ &  &  &  & $1.0000$ &  &  &  & $%
0.9888$ \\ 
&  &  &  & $3$ &  &  &  & $0.9816$ &  &  &  & $1.0000$ &  &  &  & $0.9988$
\\ 
&  &  &  & $4$ &  &  &  & $0.9911$ &  &  &  & $1.0000$ &  &  &  & $0.9998$
\\ 
$30$ &  &  &  & $2$ &  &  &  & $0.9828$ &  &  &  & $1.0000$ &  &  &  & $%
0.9861$ \\ 
&  &  &  & $3$ &  &  &  & $0.9944$ &  &  &  & $1.0000$ &  &  &  & $0.9992$
\\ 
&  &  &  & $4$ &  &  &  & $0.9983$ &  &  &  & $1.0000$ &  &  &  & $1.0000$
\\ \hline
\end{tabular}
\end{center}

Table 3 shows that the power estimates of our test $\Delta _{n}^{(1)}\left(
5\right) $ are good power for all alternatives and increases when the value
of the parameter $\theta $ \ and the sample sizes increasing.

\section{Applications to Real Data}

In this section, we apply our test to some real data-sets at 95\% confidence
level.

\begin{description}
\item[1-] Consider the data in Al-Gashgari et al. \cite{07} which represent $%
39$ liver cancers patients taken from Elminia cancer center Ministry of
Health -- Egypt, which entered in (1999). The ordered life times (in days)
\end{description}

\begin{center}
\ 
\begin{tabular}{cccccccccc}
10 & 14 & 14 & 14 & 14 & 14 & 15 & 17 & 18 & 20 \\ 
20 & 20 & 20 & 20 & 23 & 23 & 24 & 26 & 30 & 30 \\ 
31 & 40 & 49 & 51 & 52 & 60 & 61 & 67 & 71 & 74 \\ 
75 & 87 & 96 & 105 & 107 & 107 & 107 & 116 & 150 & 
\end{tabular}
\end{center}

\ In this case, $\Delta _{n}^{(1)}\left( 5\right) =$ $0.0000132958$ which is
less than the corresponding critical value in Table 2, then we reject H$_{1}$
which states that the data set have NBRUL property.

\begin{description}
\item[2-] Consider the real data-set given in Grubbs \cite{14} and have been
used in Shapiro \cite{28}. This data set gives the times between arrivals of 
$25$ customers at a facility.
\end{description}

\begin{center}
\begin{tabular}{cccccccccc}
1.80 & 2.89 & 2.93 & 3.03 & 3.15 & 3.43 & 3.48 & 3.57 & 3.85 & 3.92 \\ 
3.98 & 4.06 & 4.11 & 4.13 & 4.16 & 4.23 & 4.34 & 4.37 & 4.53 & 4.62 \\ 
4.65 & 4.84 & 4.91 & 4.99 & 5.17 &  &  &  &  & 
\end{tabular}
\end{center}

Since $\Delta _{n}^{(1)}\left( 5\right) =$ $0.00119843$ and this value less
than the corresponding critical value in Table 2. Then we conclude that this
data set have the exponential property.

\begin{description}
\item[3-] Consider the data in Abouammoh et \ al. \cite{2}. These data
represent $40$ patients suffering from blood cancer from one of the Ministry
of Health \ Hospital in Saudi Arabia and the ordered life times (in days):
\end{description}

\begin{center}
\begin{tabular}{cccccccccc}
115 & 181 & 255 & 418 & 441 & 461 & 516 & 739 & 743 & 789 \\ 
807 & 865 & 924 & 983 & 1024 & 1062 & 1063 & 1169 & 1191 & 1222 \\ 
1222 & 1251 & 1277 & 1290 & 1357 & 1369 & 1408 & 1455 & 1478 & 1549 \\ 
1578 & 1578 & 1599 & 1603 & 1604 & 1696 & 1735 & 1799 & 1815 & 1852%
\end{tabular}
\end{center}

Since $\Delta _{n}^{(1)}\left( 5\right) =$ $1.81685\times 10^{-8}$ and this
value less than the corresponding critical value in Table 2. Then we
conclude that this data set have the exponential property.

\section{Conclusion{}}

The NBRUL class of life distributions is considered. The moments
inequalities are derived. A new test statistics for exponentiality versus
NBRUL class is proposed based on the moment inequalities. Quality criteria
of the test is shown by the famous criterion which is Pitman asymptotic
efficiency. The upper percentiles and the power of the proposed test are
calculated and tabulated. Our test is applied to some real data to show the
usefulness of the test .

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\end{document}