Assessing the lifetime performance index with Lomax distribution based on progressive type I interval censored sample

Abstract

In manufacturing industry, the lifetime performance index $C_L$ is applied to evaluate the larger-the-better quality features of products. It can quickly show whether the lifetime performance of products meets the desired level. In this article, first we obtain the maximum likelihood estimator of $C_L$ with two unknown parameters in the Lomax distribution on the basis of progressive type I interval censored sample. With the MLE we proposed, some asymptotic confidence intervals of $C_L$ are discussed by using the delta method. Furthermore, the MLE of $C_L$ is used to establish the hypothesis test procedure under a given lower specification limit L. In addition, we also conduct a hypothesis test procedure when the scale parameter in the Lomax distribution is given. Finally, we illustrate the proposed inspection procedures through a real example. The testing procedure algorithms presented in this paper are efficient and easy to implement.

1. Introduction

Process capability analysis is an efficient method and is widely applied to measure the potential performance of products. People have developed various methods to evaluate the performance of products. Process capability analysis has some advantages: continuously observing the process performance of products through process capability indices to be sure so that the desired level of products manufactured is met; providing information to engineers and designers in order to improve the product quality and design better products; giving a basis to reduce the cost when the products in the experiments fail. In manufacturing, people usually use the process capability indices to evaluate whether the performance of products reach the required level. For example, process capability index $C_L$ is one of the process capability indices and is applied to assess lifetime performance of products, where L is the given lower specification limit. And the larger $C_L$ of products, the better the quality of them. Lee et al. [13] obtained Bayes estimator of $C_L$ under Rayleigh distribution based on the progressive type II right censoring. Ahmadi et al. [2] obtained the maximum likelihood estimator of $C_L$ which is used to propose a confidence interval under Weibull distribution on the basis of progressive first-failure censoring schemes. Wu et al. [16] derived the maximum likelihood estimator of $C_L$ with two-parameter Burr XII distribution based on the progressive type II right censoring scheme. Wu et al. [17] established a hypothesis testing procedure for $C_L$ based on the progressive type I interval censored sample when the products follow Burr XII distribution. Wu et al. [18] built a new hypothesis testing procedure for $C_L$ based on the progressive type I interval censoring when the lifetimes of products follow Rayleigh distribution.

In lifetime testing experiment, we cannot always be there to monitor the lifetimes of the products in the experiment. Because of time constraints (or lack of funds, experimental materials or other difficulties), we can use censored samples in practice. So, in this article, we consider the case based on the progressive type I interval censoring scheme. The following is a brief introduction. Assume that we have n same products being tested for lifetime at time 0. Let $(t_1,\dots ,t_m)$ be the scheduled inspection time in which $t_m$ is scheduled to end the experiment. Let $(p_1,\dots ,p_m)$ be predetermined removal percentage of the rest of survival units at time $t_i$, $i=1,\dots ,m$, where $p_m=1$. In the ith time interval $(t_{i-1},t_i]$, we will observe the number of failure units $X_i$, and then randomly remove the $R_i=[(n-\sum _{j=1}^i{X}_j-\sum _{j=1}^{i-1}{R}_j)p_i]$ units from the remaining $n-\sum _{j=1}^i{X}_j-\sum _{j=1}^{i-1}{R}_j$ survival units, where $[k]$ represents the largest integer less than or equal to k. During the last time interval $(t_{m-1},t_m]$, the number of failure units $X_m$ is observed, and the remaining $n-\sum _{i=1}^m{X}_i-\sum _{i=1}^{m-1}{R}_i$ are all removed. Then at the time $t_m$, the life test is ended. For progressive type I interval censoring scheme, Balakrishnan and Cramer [6] conducted a comprehensive and up-to-date discussion. Chen and Lio [8] derived the maximum likelihood estimator of the parameters in the generalized exponential distribution under progressive type I interval censoring scheme. Kaushik et al. [11] developed classical and the Bayesian estimation of unknown parameters in the Weibull distribution based on the progressive type I interval censored samples with beta-binomial removals. Aggarwala [1] considered likelihood point and interval estimation based on progressive type I interval censored sample.

The Lomax distribution is often applied in the fields of business, economics, actuarial science, bioscience and engineering. Estimation of parameters or confidence intervals for Lomax distribution gradually gained increasingly more attention from scholars, but estimation of lifetime performance index with Lomax distribution has rarely been studied. Cramer and Schmiedt [9] obtained the maximum likelihood estimator (MLE) and the expected Fisher information matrix under Lomax distribution under progressive type II censoring schemes. Okasha [15] computed the estimators of unknown parameters in the Lomax distribution through the E-Bayesian method based on the type II censoring schemes. Alzahrani and Alsobhi [3] derived the maximum likelihood estimator (MLE) and Bayes estimators under the Lomax distribution on basis of general progressive censored data.

In this paper, we suppose the lifetime $(Z)$ of products following Lomax distribution. Then the probability density function and cumulative distribution function are

$$f(z)=\frac{\beta }{\lambda }{(1+\frac{z}{\lambda })}^{-(\beta +1)},z>0,\beta >0,\lambda >0$$ (1)

and

$$F(z)=1-{(1+\frac{z}{\lambda })}^{-\beta },z>0,\beta >0,\lambda >0$$ (2)

respectively, where λ and β are the scale parameter and the shape parameter, respectively. The failure rate function is given by

$$h(z)=\frac{\beta }{\lambda +z}.$$ (3)

The novelty of this paper is that we propose a testing procedure algorithm for lifetime performance index based on the progressive type I interval censoring in the case where the two parameters are unknown in the Lomax distribution. The rest of this article is organized as follows. In Section 2, we describe some features of lifetime performance index and find out the conforming rate of products with two unknown parameters in the Lomax distribution. We also discuss the maximum likelihood estimator of lifetime performance index using different iterative calculation methods on the basis of progressive type I interval censoring scheme. With the MLE we proposed, we obtain some asymptotic confidence intervals of $C_L$ by using the delta method. In Section 3, we conduct a hypothesis testing procedure to determine whether the lifetime performance index is up to the required level based on two unknown parameters in the Lomax distribution. In addition, a hypothesis testing procedure for lifetime performance index based on the known scale parameter is established in this section. In Section 4, we analyzed a real numerical example in which the data set being analyzed represents lifetime or property loss. Usually, it is difficult to directly judge whether the data set has reached the desired level, but through the testing procedure algorithms proposed in this paper, we can make a quick judgment. Section 5 makes the conclusion.

2. Some inferences about lifetime performance index based on two unknown parameters

In this section, first we introduce the lifetime performance index and the conforming rate. We also discuss the maximum likelihood estimator of lifetime performance index. Finally, we obtain some asymptotic confidence intervals of $C_L$ by using the delta method.

2.1. The lifetime performance index and the conforming rate

Generally speaking, the lifetime of different products is not identical. Suppose the lifetime of products can be modeled by the Lomax distribution (2). The longer the lifetime of a product, the better the quality of it. So the lifetime is the larger-the-better quality feature of products. In general, the lifetime needs to surpass L unit times to ensure commercial profitability and meets consumers' demands, where L is called the lower specification limit. As we know, Montgomery [14] proposed a process capability index $C_L$ to assess the larger-the-better quality feature of products. And $C_L$ is given by

$$C_L=\frac{\mu -L}{\sigma }$$ (4)

where μ and σ are the process mean and the process standard deviation respectively, and L is a predetermined low specification limit. $C_L$ is able to be applied to estimate the lifetime performance of products. According to Equation (1), the process mean and variance of the Lomax distribution can be obtained by $E(z)=\lambda /\beta -1$ and $Var(z)=\beta {\lambda }^2/(\beta -2)(\beta -1{)}^2$ respectively. Therefore, we get $\mu =\lambda /\beta -1$ and $\sigma =\sqrt{Var(z)}=\lambda /\beta -1\sqrt{\beta /\beta -2}$. Assuming that $L_Z$ is the lower specification limit set in advance for the lifetime variable Z, the lifetime performance index $C_L$ is derived by

$$C_L=\frac{\mu -L}{\sigma }=\frac{\frac{\lambda }{\beta -1}-L_Z}{\frac{\lambda }{\beta -1}\sqrt{\frac{\beta }{\beta -2}}}=\sqrt{\frac{\beta -2}{\beta }}(1-\frac{\beta -1}{\lambda }{L}_Z).$$ (5)

We can observe that $C_L>0$ as $\beta -1/\lambda <1/L_Z$ and $C_L<0$ as $\beta -1/\lambda >1/L_Z$. The plots of $C_L$ with respect to λ and β are presented in Figures 1 and 2. By observing Figures 1 and 2 and Equation (3), it is easy to know that the lifetime performance index $C_L$ becomes larger as the failure rate decreases. Therefore, it is suitable for $C_L$ to represent the lifetime performance of products. When the lifetime Z of a product surpasses the specification lower limit $L_Z$, we can know that the product is a conforming one. So, the conforming rate is given by:

$$C_r=P(Z\ge L_Z)={\left(\frac{\lambda }{\lambda +L_Z}\right)}^{\beta }={\left(\frac{(\beta -1)\sqrt{\beta -2}}{\beta \sqrt{\beta -2}-C_L\sqrt{\beta }}\right)}^{\beta },-\mathrm{\infty }<C_L<\sqrt{\beta (\beta -2)}.$$ (6)

The relations between $C_r$ and $C_L$ are plotted in Figure 3. According to Figure 3, it is clear that with the increase of the lifetime performance index $C_L$, the conforming rate $C_r$ is also increasing gradually. In Table 1, we give the values of the lifetime performance index $C_L$ and their corresponding conforming rate $C_r$ for $\beta =2.5$ and $\beta =4.5$. From Table 1, when $\beta =4.5$, if we want the conforming rate $C_r$ over 0.887139, the lifetime performance index must be more than 0.675.

Figure 1.

Lifetime performance index vs. λ

Figure 2.

Lifetime performance index vs. β

Figure 3.

The conforming rate vs. lifetime performance index

Table 1. The lifetime performance index $C_L$ and its corresponding conforming rate $C_r$ for $\beta =2.5$ and $\beta =4.5$.

$\beta =2.5$
$C_L$	$C_r$	$C_L$	$C_r$	$C_L$	$C_r$
$-\mathrm{\infty }$	0.000000	0.050	0.312646	0.275	0.564810
−3.000	0.010710	0.075	0.331718	0.300	0.608958
−2.500	0.014802	0.100	0.352460	0.325	0.658091
−2.000	0.021469	0.125	0.375064	0.375	0.774451
−1.500	0.033234	0.150	0.399750	0.425	0.921784
−1.000	0.056453	0.175	0.426773	0.440	0.973614
−0.500	0.110674	0.200	0.456424	0.443	0.984468
0.000	0.278855	0.225	0.489042	0.446	0.995492
0.025	0.295074	0.250	0.525017	0.447	0.999204
$\beta =4.5$
$-\mathrm{\infty }$	0.000000	0.075	0.357311	0.525	0.694264
−3.000	0.018205	0.125	0.382891	0.575	0.752269
−2.500	0.026325	0.175	0.410746	0.625	0.816308
−2.000	0.039342	0.225	0.489042	0.650	0.850823
−1.500	0.061158	0.275	0.564810	0.675	0.887139
−1.000	0.099747	0.325	0.510541	0.700	0.925366
−0.500	0.172699	0.375	0.550249	0.725	0.965628
0.000	0.322738	0.425	0.593797	0.735	0.982329
0.025	0.333788	0.475	0.641633	0.745	0.999386

2.2. Maximum likelihood estimation of the lifetime performance index

In this section, we consider the maximum likelihood estimate of $C_L$. Let $(X_1,X_2,\cdot \cdot \cdot ,X_m)$ be a progressive type I interval censored sample with the censoring scheme $(R_1,R_2,\dots ,R_m)$ which is randomly removed from the rest of survival units at pre-set times $(t_1,t_2,\dots ,t_m)$ as the removal percentages is $(p_1,p_2,\dots ,p_m)$. Then the likelihood function on basis of the progressive type I interval censoring scheme is derived by

$$\begin{array}{rl}L(\beta ,\lambda )& \propto \prod _{i=1}^m{(F(t_i)-F(t_{i-1}))}^{X_i}{(1-F(t_i))}^{R_i}\\ & =\prod _{i=1}^m{[{(1+\frac{t_{i-1}}{\lambda })}^{-\beta }-{(1+\frac{t_i}{\lambda })}^{-\beta }]}^{X_i}{(1+\frac{t_i}{\lambda })}^{-\beta R_i}\end{array}$$ (7)

Then we get the log-likelihood function by

$$\ln L(\beta ,\lambda )=\sum _{i=1}^m[X_i\ln ({(1+\frac{t_{i-1}}{\lambda })}^{-\beta }-{(1+\frac{t_i}{\lambda })}^{-\beta })-\beta R_i\ln (1+\frac{t_i}{\lambda })].$$ (8)

By making the derivatives of the log-likelihood function for β or λ equal to 0, the likelihood equations can be obtained by

$$\sum _{i=1}^m\frac{X_i[-{(1+\frac{t_{i-1}}{\lambda })}^{-\beta }\ln (1+\frac{t_{i-1}}{\lambda })+{(1+\frac{t_i}{\lambda })}^{-\beta }\ln (1+\frac{t_i}{\lambda })]}{{(1+\frac{t_{i-1}}{\lambda })}^{-\beta }-{(1+\frac{t_i}{\lambda })}^{-\beta }}=\sum _{i=1}^m{R}_i\ln (1+\frac{t_i}{\lambda })$$ (9)

and

$$\sum _{i=1}^m\frac{X_i[-{(1+\frac{t_{i-1}}{\lambda })}^{-\beta -1}\frac{\beta t_{i-1}}{{\lambda }^2}+{(1+\frac{t_i}{\lambda })}^{-\beta -1}\frac{\beta t_i}{{\lambda }^2}]}{{(1+\frac{t_{i-1}}{\lambda })}^{-\beta }-{(1+\frac{t_i}{\lambda })}^{-\beta }}=\sum _{i=1}^m\frac{\beta R_i{t}_i}{{\lambda }^2+\lambda t_i}.$$ (10)

The MLEs of β and λ are the solutions of the above likelihood equations, denoted by ${\hat{\beta }}_{ML}$ and ${\hat{\lambda }}_{ML}$, respectively. Due to the invariance principle of MLE, we have the MLE of $C_L$ as follows:

$${\hat{C}}_{L_{ML}}=\sqrt{\frac{{\hat{\beta }}_{ML}-2}{{\hat{\beta }}_{ML}}}(1-\frac{{\hat{\beta }}_{ML}-1}{{\hat{\lambda }}_{ML}}{L}_Z).$$ (11)

However, the above likelihood equations have no closed form of solutions, we can use an iterative numerical search to get the MLEs. To simplify the calculation, the mid-point approximation method and the EM algorithm can been used to obtain the estimates of β and λ.

2.3. The performance of the MLE of $C_L$

In order to observe the performance of the maximum likelihood estimator of $C_L$, we conduct a simulation study and list the results in Table 2. We assume that not only are the time intervals equal, but the removal percentages are equal. That is, $t_i-t_{i-1}=t$ and $p_1=p_2=\cdots =p_{m-1}=p$, $i=1,2,\dots ,m$. In addition, we set the size of samples to $n=50(50)150$ and the number of inspection times to $m=5(5)20$. The removal percentages are set to $p=0.05(0.05)0.15$ and the length of time interval is t = 4. The lower specification limit is given by $L_Z=0.09$. We set two unknown parameters β and λ to 2.5 and 5.5, respectively. By observing Table 2, we know that the Biases and MSEs are small, so the estimator of the lifetime performance index $C_L$ performs well.

Table 2. Biases and MSEs for the estimates of the lifetime performance index $C_L$ for $\beta =2.5$, $\lambda =5.5$, t = 10 and $L_Z=0.09$.

m	n	p	Biases	MSEs	m	n	p	Biases	MSEs
5	50	0.05	0.273959	0.298005	15	50	0.05	0.187407	0.390846
		0.10	0.278333	0.278767			0.10	0.296329	0.224318
		0.15	0.262607	0.312861			0.15	0.320434	0.240585
	100	0.05	0.224238	0.107587		100	0.05	0.237168	0.107907
		0.10	0.280048	0.129274			0.10	0.232252	0.104605
		0.15	0.252339	0.134609			0.15	0.284517	0.129033
	150	0.05	0.196995	0.089290		150	0.05	0.214361	0.093250
		0.10	0.196893	0.101803			0.10	0.229503	0.092358
		0.15	0.196667	0.100805			0.15	0.166511	0.088992
10	50	0.05	0.194059	0.361560	20	50	0.05	0.272127	0.346234
		0.10	0.302449	0.287759			0.10	0.316847	0.157899
		0.15	0.367305	0.189041			0.15	0.349127	0.184381
	100	0.05	0.256107	0.133049		100	0.05	0.221664	0.118737
		0.10	0.211911	0.100891			0.10	0.217807	0.117896
		0.15	0.183279	0.105792			0.15	0.283242	0.138400
	150	0.05	0.185112	0.085821		150	0.05	0.178789	0.102298
		0.10	0.159119	0.070340			0.10	0.194052	0.086236
		0.15	0.188183	0.092236			0.15	0.156739	0.083693

2.4. Asymptotic CI of the lifetime performance index

The $100(1-\zeta )\mathrm{\%}$ confidence interval of the lifetime performance index $C_L$ can be obtained by applying the delta method. In order to obtain an approximate estimator of the variance of $C_L$, we need to calculate the inverse of the Fisher information matrix. The Fisher information matrix $I=I(\beta ,\lambda )$ is given by the expectation of the negative second derivative of the log-likelihood function (8). Then we have

$$I(\beta ,\lambda )=E\left[\begin{array}{cc}-{\displaystyle \frac{d^2\ln L}{d{\beta }^2}}& -{\displaystyle \frac{d^2\ln L}{d\beta d\lambda }}\\ -{\displaystyle \frac{d^2\ln L}{d\lambda d\beta }}& -{\displaystyle \frac{d^2\ln L}{d{\lambda }^2}}\end{array}\right].$$

Therefore, we consider the observed Fisher information matrix, which is taken at point $(\beta ,\lambda )=({\hat{\beta }}_{ML},{\hat{\lambda }}_{ML})$. So, the inverse of the observed Fisher information matrix is given by

$$I^{-1}({\hat{\beta }}_{ML},{\hat{\lambda }}_{ML})=E{\left[\begin{array}{cc}-{\displaystyle \frac{d^2\ln L}{d{\beta }^2}}& -{\displaystyle \frac{d^2\ln L}{d\beta d\lambda }}\\ -{\displaystyle \frac{d^2\ln L}{d\lambda d\beta }}& -{\displaystyle \frac{d^2\ln L}{d{\lambda }^2}}\end{array}\right]}_{(\beta ,\lambda )=({\hat{\beta }}_{ML},{\hat{\lambda }}_{ML})}^{-1}$$

Furthermore, let

$$Q^T=(\frac{d{C}_L}{d\beta },\frac{d{C}_L}{d\lambda }),$$

where

$$\begin{array}{rl}\frac{d{C}_L}{d\beta }& =\frac{\lambda -(\beta -1)L_Z}{\lambda \beta \sqrt{\beta (\beta -2)}}-L_Z\sqrt{\frac{\beta -2}{\beta }},\\ \frac{d{C}_L}{d\lambda }& =\frac{\sqrt{\beta (\beta -2)}(\beta -1)L_Z}{\beta {\lambda }^2}.\end{array}$$

Then the asymptotic estimator of $Var({\widehat{C_L}}_{ML})$ is defined as

$$\widehat{Var}({\widehat{C_L}}_{ML})\simeq {\left[Q^T{I}^{-1}Q\right]}_{(\beta ,\lambda )=({\hat{\beta }}_{ML},{\hat{\lambda }}_{ML})}.$$ (12)

According to Budhiraja et al. [7], we know that the consistency and asymptotic normality of the MLEs on the basis of progressive type I interval censoring are proved. Therefore, by using the delta method, we have the following relationship:

$$\frac{{\widehat{C_L}}_{ML}-C_L}{\sqrt{\widehat{Var}({\widehat{C_L}}_{ML})}}\sim N(0,1).$$ (13)

Then we can derive the $100(1-\zeta )\mathrm{\%}$ asymptotic confidence interval of $C_L$ by

$${\widehat{C_L}}_{ML}\pm W_{\zeta /2}\sqrt{\widehat{Var}({\widehat{C_L}}_{ML})},$$ (14)

where $W_{\zeta /2}$ is the $\zeta /2$ percentile of the standard normal distribution.

In addition, Krishnamoorthy and Lin [12] discussed two transformations as follows:

(Logit transformation) Let $g({\widehat{C_L}}_{ML})=\ln ({\widehat{C_L}}_{ML}/1-{\widehat{C_L}}_{ML})$. Then we can get $g^{\prime }({\widehat{C_L}}_{ML})=1/{\widehat{C_L}}_{ML}(1-{\widehat{C_L}}_{ML})$. Using the delta theorem, we can obtain the $100(1-\zeta )\mathrm{\%}$ confidence interval of $g(C_L)$ by

$$\ln \left(\frac{{\widehat{C_L}}_{ML}}{1-{\widehat{C_L}}_{ML}}\right)\pm W_{\zeta /2}\frac{\sqrt{\widehat{Var}({\widehat{C_L}}_{ML})}}{{\widehat{C_L}}_{ML}(1-{\widehat{C_L}}_{ML})}.$$ (15)

If the upper and lower bounds of the above confidence interval are represented by LB and UB, respectively, the confidence interval of $C_L$ can be expressed as

$$[{\mathrm{e}}^{LB}{(1+{\mathrm{e}}^{LB})}^{-1},{\mathrm{e}}^{UB}{(1+{\mathrm{e}}^{UB})}^{-1}].$$ (16)

(Arc sine transformation) Let $g({\widehat{C_L}}_{ML})=\arcsin (\sqrt{{\widehat{C_L}}_{ML}})$. Then we have $g^{\prime }({\widehat{C_L}}_{ML})=1/2\sqrt{{\widehat{C_L}}_{ML}(1-{\widehat{C_L}}_{ML})}$. Using the delta theorem, we can obtain the $100(1-\zeta )\mathrm{\%}$ confidence interval of $g(C_L)$ by

$$\arcsin \left(\sqrt{{\widehat{C_L}}_{ML}}\right)\pm W_{\zeta /2}\sqrt{\frac{\widehat{Var}({\widehat{C_L}}_{ML})}{4{\widehat{C_L}}_{ML}(1-{\widehat{C_L}}_{ML})}}.$$ (17)

Then the confidence interval of $C_L$ can be obtained by

$$[{\sin }^2(LB),{\sin }^2(UB)],$$ (18)

where LB and UB are given in (15).

3. Testing procedure algorithm for the lifetime performance index

In this part, first we propose a hypothesis testing procedure to determine whether the lifetime performance index is up to the required level when the two parameters in the Lomax distribution are unknown. Then we consider the hypothesis testing procedure based on a known scale parameter.

First, we set $c^{\ast }$ as required level. If the lifetime performance index exceeds $c^{\ast }$, the process is said to be conforming. The hypothesis test is constructed as follows: We conduct the null hypothesis and the alternative hypothesis $H_0:C_L\le c^{\ast }$ and $H_1:C_L>c^{\ast }$, respectively. Assume that the specified significance level is α, and ${\hat{C}}_{L_{ML}}$ is the test statistic, then the critical value $C_0^{\ast }$ is given by the following computational process:

$$\begin{array}{rl}P(\widehat{C_L}>C_0^{\ast }|C_L\le c^{\ast })& =P(\frac{{\hat{C}}_{L_{ML}}-C_L}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}}>\frac{C_0^{\ast }-C_L}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}}|C_L\le c^{\ast })\\ & =P(W>\frac{C_0^{\ast }-C_L}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}}|C_L\le c^{\ast })\le \alpha ,\end{array}$$

where $W={\hat{C}}_{L_{ML}}-C_L/\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}\sim N(0,1)$.

It is observed that

$$\underset{C_L\le c^{\ast }}{sup}P(W>\frac{C_0^{\ast }-C_L}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}})=\alpha .$$

Because $C_0^{\ast }-C_L/\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}$ is an increasing function when $C_L\le c^{\ast }$, the conditional probability $P(W>C_0^{\ast }-C_L/\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})})$ reaches the minimum at the point $C_L=c^{\ast }$. Then we have

$$\begin{array}{rl}& P(W>\frac{C_0^{\ast }-C_L}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}}|C_L=c^{\ast })=\alpha \\ & \Rightarrow W_{1-\alpha }=\frac{C_0^{\ast }-C_L}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}}\end{array}$$

where $W_{1-\alpha }$ is defined as the αth percentile of the standard normal distribution $N(0,1)$.

Then we know that the critical value can be given by

$$C_0^{\ast }=W_{1-\alpha }\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}+c^{\ast }.$$ (19)

And the rejection region is represented as $\{{\hat{C}}_{L_{ML}}|{\hat{C}}_{L_{ML}}>C_0^{\ast }\}$.

In summary, the testing procedure algorithm for $C_L$ we propose is conducted as follows:

1. Set a given lower specification limit $L_Z$. With the progressive type I interval censoring scheme $(R_1,R_2,\dots ,R_m)$, get the censored sample $(X_1,X_2,\dots ,X_m)$ at the time $(t_1,t_2,\dots ,t_m)$ set in advance from the Lomax distribution.

2. Specify the desired level $c^{\ast }$ and conduct the null hypothesis and the alternative hypothesis as $H_0:C_L\le c^{\ast }$ and $H_1:C_L>c^{\ast }$, respectively.

3. Obtain the MLEs of two unknown parameters, denoted as ${\hat{\beta }}_{ML}$ and ${\hat{\lambda }}_{ML}$, then compute the value of test statistic ${\hat{C}}_{L_{ML}}=\sqrt{{\hat{\beta }}_{ML}-2/{\hat{\beta }}_{ML}}(1-\frac{{\hat{\beta }}_{ML}-1}{{\hat{\lambda }}_{ML}}{L}_Z)$.

4. Set the significance level α to get the critical value $C_0^{\ast }$ by using Equation (19).

5. The decision rule is drawn: If ${\hat{C}}_{L_{ML}}>C_0^{\ast }$, then it can be declared that the lifetime performance index of the products reaches the desired level.

The lifetime performance of products can be easily accessed on the basis of the above test process. Importantly, the power of a hypothesis test is the probability of rejecting the null hypothesis under the condition that the alternative hypothesis is true. Then the power $h(c_1^{\ast })$ of this statistic test at the point $C_L=c_1^{\ast }>c^{\ast }$ can be obtained by

$$\begin{array}{rl}h(c_1^{\ast })& =P({\hat{C}}_{L_{ML}}>C_0^{\ast }|C_L=c_1^{\ast })\\ & =P(\frac{{\hat{C}}_{L_{ML}}-C_L}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}}>\frac{W_{1-\alpha }\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}+c^{\ast }-c_1^{\ast }}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}}|C_L=c_1^{\ast })\\ & =1-\mathrm{\Phi }\left(\frac{W_{1-\alpha }\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}+c^{\ast }-c_1^{\ast }}{\sqrt{\widehat{Var}({\hat{C}}_{L_{ML}})}}\right),\end{array}$$ (20)

where $\mathrm{\Phi }(\cdot )$ is the cumulative distribution function of the standard normal distribution $N(0,1)$. According to Aggarwala [1], we set up the algorithm to calculate the power $h(c_1^{\ast })$ as follows.

In order to observe the trend of the power $h(c_1^{\ast })$ in Equation (20) under different impact factors, we list the values of $h(c_1^{\ast })$ under different influence factors in tables. We assume that not only are the time intervals equal, but the removal percentages are equal. That is, $t_i-t_{i-1}=t$ and $p_1=p_2=\cdots =p_{m-1}=p$, $i=1,2,\dots ,m$. The impact factors are set as follows: $c_1^{\ast }=0.74(0.02)0.92$, $\alpha =0.02(0.03)0.08$, $n=50(50)150$, $m=5(5)20$, $p=0.05(0.05)0.15$, t = 4 and $L_Z=0.09$. Due to space limit, we only present the values of $h(c_1^{\ast })$ for $\alpha =0.02$ in Table 3 and we plot Figure 4 to make the data more intuitive. Then we have the following findings:

1. As shown in Figure 4, power function $h(c_1^{\ast })$ is an increasing function of n when $m=5,\alpha =0.02,p=0.1$ are fixed.

2. We can observe that power function $h(c_1^{\ast })$ is an increasing function of m when $n=100,\alpha =0.02,p=0.1$ are fixed.

3. Power function $h(c_1^{\ast })$ is gradually increasing for the removal percentage p when $n=100,m=5,\alpha =0.02$ are fixed.

4. As $c_1^{\ast }$ gradually increases, power function $h(c_1^{\ast })$ is incremented for α when $n=100,m= 5,p=0.1$is fixed.

5. Therefore, it is easy to know that the power function is increasing regardless of n, m, p, or α.

Table 3. The values of $h(c_1^{\ast })$ for $\alpha =0.02$, $c^{\ast }=0.8$, t = 4 and $L_Z=0.09$.

			$c_1^{\ast }$
m	n	p	0.74	0.76	0.78	0.80	0.82	0.84	0.86	0.88	0.90	0.92
5	50	0.05	0.961722	0.968928	0.974973	0.980000	0.984143	0.987527	0.990267	0.992465	0.994214	0.995593
		0.10	0.975830	0.977294	0.978683	0.980000	0.981247	0.982427	0.983543	0.984598	0.985595	0.986535
		0.15	0.978371	0.978926	0.979469	0.980000	0.980519	0.981027	0.981524	0.982010	0.982484	0.982948
	100	0.05	0.955186	0.965327	0.973502	0.980000	0.985093	0.989028	0.992027	0.994279	0.995948	0.997167
		0.10	0.961786	0.968964	0.974988	0.980000	0.984133	0.987510	0.990247	0.992444	0.994193	0.995573
		0.15	0.967222	0.972076	0.976315	0.980000	0.983187	0.985928	0.988276	0.990276	0.991971	0.993400
	150	0.05	0.927232	0.951028	0.968154	0.980000	0.987876	0.992909	0.995999	0.997824	0.998859	0.999424
		0.10	0.926131	0.950491	0.967965	0.980000	0.987959	0.993012	0.996092	0.997895	0.998908	0.999455
		0.15	0.957101	0.966369	0.973921	0.980000	0.984832	0.988624	0.991564	0.993815	0.995517	0.996787
10	50	0.05	0.979968	0.979979	0.979989	0.980000	0.980011	0.980021	0.980032	0.980043	0.980054	0.980064
		0.10	0.979998	0.979998	0.979999	0.980000	0.980001	0.980002	0.980002	0.980003	0.980004	0.980005
		0.15	0.979988	0.979992	0.979996	0.980000	0.980004	0.980008	0.980012	0.980015	0.980019	0.980023
	100	0.05	0.932789	0.953765	0.969128	0.980000	0.987434	0.992346	0.995481	0.997415	0.998567	0.999231
		0.10	0.973096	0.975591	0.977888	0.980000	0.981938	0.983714	0.985338	0.986821	0.988173	0.989402
		0.15	0.973918	0.976098	0.978122	0.980000	0.981739	0.983348	0.984834	0.986204	0.987466	0.988627
	150	0.05	0.957591	0.966637	0.974030	0.980000	0.984763	0.988517	0.991440	0.993688	0.995398	0.996681
		0.10	0.965241	0.970928	0.975819	0.980000	0.983551	0.986547	0.989061	0.991155	0.992889	0.994316
		0.15	0.975085	0.976825	0.978462	0.980000	0.981444	0.982798	0.984067	0.985254	0.986365	0.987403
15	50	0.05	0.973231	0.975673	0.977926	0.980000	0.981906	0.983655	0.985258	0.986723	0.988061	0.989280
		0.10	0.973631	0.975920	0.978040	0.980000	0.981810	0.983478	0.985013	0.986425	0.987720	0.988906
		0.15	0.979992	0.979995	0.979997	0.980000	0.980003	0.980005	0.980008	0.980010	0.980013	0.980015
	100	0.05	0.826903	0.905338	0.953829	0.980000	0.992332	0.997404	0.999226	0.999797	0.999953	0.999991
		0.10	0.971512	0.974624	0.977447	0.980000	0.982304	0.984377	0.986238	0.987905	0.989394	0.990720
		0.15	0.974846	0.976675	0.978392	0.980000	0.981506	0.982914	0.984229	0.985456	0.986601	0.987667
	150	0.05	0.954275	0.964835	0.973306	0.980000	0.985212	0.989211	0.992233	0.994484	0.996135	0.997329
		0.10	0.965001	0.97079	0.975760	0.980000	0.983593	0.986619	0.989150	0.991254	0.992991	0.994416
		0.15	0.966174	0.971466	0.976051	0.980000	0.983382	0.986263	0.988702	0.990755	0.992474	0.993905
20	50	0.05	0.936621	0.955678	0.969821	0.980000	0.987104	0.991912	0.995067	0.997075	0.998314	0.999056
		0.10	0.970196	0.973833	0.977091	0.980000	0.982590	0.984887	0.986920	0.988711	0.990285	0.991664
		0.15	0.971254	0.974469	0.977376	0.980000	0.982361	0.984480	0.986376	0.988069	0.989577	0.990915
	100	0.05	0.926869	0.950851	0.968091	0.980000	0.987903	0.992943	0.996030	0.997848	0.998876	0.999434
		0.10	0.936560	0.955647	0.969810	0.980000	0.987109	0.991919	0.995074	0.997081	0.998319	0.999059
		0.15	0.972635	0.975308	0.977758	0.980000	0.982047	0.983913	0.985610	0.987150	0.988546	0.989808
	150	0.05	0.894150	0.935390	0.962907	0.980000	0.989884	0.995206	0.997872	0.999117	0.999657	0.999876
		0.10	0.955554	0.965526	0.973582	0.980000	0.985044	0.988953	0.991941	0.994194	0.995870	0.997099
		0.15	0.938721	0.956736	0.970209	0.980000	0.986913	0.991656	0.994817	0.996864	0.998153	0.998940

Figure 4.

Power function for $\alpha =0.02$, m = 5, p = 0.1.

Furthermore, we discuss a testing procedure algorithm for $C_L$ in the case where the scale parameter λ is given in the Lomax distribution. According to Aggarwala [1] and Wu et al. [17], we have the likelihood equation

$$\sum _{i=1}^m[\frac{X_i(y_i-y_{i-1}){\mathrm{e}}^{-\beta (y_i-y_{i-1})}}{1-{\mathrm{e}}^{-\beta (y_i-y_{i-1})}}-(y_i{R}_i+y_{i-1}{X}_i)]=0,$$ (21)

where $y_i=\log [1+t_i/\lambda ]$. By solving Equation (21), we have the MLE of β, denoted as $\hat{\beta }$. It is known that $\hat{\beta }\underset{m\to \mathrm{\infty }}{\overset{d}{\to }}N(\beta ,I^{-1}(\beta ))$, where $I^{-1}(\beta )$ is the inverse of the Fisher information matrix $I(\beta )$. The Fisher information matrix $I(\beta )$ is given by

$$I(\beta )=\frac{n}{{\beta }^2}\sum _{i=1}^m\frac{{\ln }^2(1-k_i)(1-k_i)}{k_i}\prod _{j=1}^{i-1}(1-p_j)(1-k_j),$$ (22)

where $k_i=1-{\mathrm{e}}^{-\beta (y_i-y_{i-1})}$, $i=1,2,\dots ,m$. Therefore, it is known that

$$\widehat{C_L}=1-\hat{\beta }L\underset{m\to \mathrm{\infty }}{\overset{d}{\to }}N(C_L,Var(\widehat{C_L})),$$ (23)

through the Delta method from Balakrishnan [5], where $Var(\widehat{C_L})=L^2{I}^{-1}(\hat{\beta })$.

Set c as required level. If the lifetime performance index exceeds c, the process is said to be conforming. The hypothesis test is constructed as follows:

The null hypothesis and the alternative hypothesis are $H_0:C_L\le c$ and $H_1:C_L>c$, respectively. Assume that the specified significance level is α, and the MLE of $C_L$ is the test statistic. According to Wu et al. [17], the critical value $C_0$ is given by

$$C_0=1-L(W_{\alpha }\sqrt{I^{-1}({\beta }_0)}+{\beta }_0),$$ (24)

where ${\beta }_0=1-c/L$ and $W_{\alpha }$ is defined as the αth percentile of the standard normal distribution $N(0,1)$. And the rejection region is represented as $\{\widehat{C_L}|\widehat{C_L}>C_0\}$.

The testing procedure algorithm for $C_L$ is conducted as follows:

1. Set a given lower specification limit $L_Z$, and obtain the lower specification limit L by using equation $L=L_Y=\ln (1+L_Z/\lambda )$. With the progressive type I interval censoring scheme $(R_1,R_2,\dots ,R_m)$, get the censored sample $(X_1,X_2,\dots ,X_m)$ at the time $(t_1,t_2,\dots ,t_m)$ set in advance from the Lomax distribution.

2. Specify the desired level c and conduct the null hypothesis and the alternative hypothesis as $H_0:C_L\le c$ and $H_1:C_L>c$, respectively.

3. Derive the MLE of β which is denoted as $\hat{\beta }$ and compute the value of test statistic $\widehat{C_L}=1-\hat{\beta }L$.

4. Set the significance level α to get the critical value $C_0$ by using Equation (24).

5. The decisive rule is drawn: If $\widehat{C_L}>C_0$, then it can be declared that the lifetime performance index of the products reaches the desired level.

In addition, the power $h(c_1)$ at the point $C_L=c_1>c$ can be obtained as (see Wu et al. [17])

$$h(c_1)=P(\widehat{C_L}>C_0|c_1=1-{\beta }_{1}L)=\mathrm{\Phi }\left(\frac{W_{\alpha }\sqrt{I^{-1}({\beta }_0)}+{\beta }_0-{\beta }_1}{\sqrt{I^{-1}({\beta }_1)}}\right),$$ (25)

where ${\beta }_0=1-c/L$, ${\beta }_1=1-c_1/L$ and $\mathrm{\Phi }(\cdot )$ is the cumulative distribution function of the standard normal distribution $N(0,1)$.

In order to observe the trend of the power $h(c_1)$ in Equation (25) under different impact factors, we list the values of $h(c_1)$ under different influence factors in Table 4. The impact factors are set as follows: $c_1=0.87(0.01)0.96$, $\alpha =0.02(0.03)0.08$, $n=50(50)150$, $m=5(5)20$, $p=0.05(0.05)0.15$, t = 0.1 and L = 0.09. Due to space limit, we only present the values of $h(c_1)$ for $\alpha =0.02$ in Table 4 and we plot Figure 5 to make the data more intuitive. Then we have the following findings:

1. Power function $h(c_1)$ is an increasing function of n when $m=5,\alpha =0.02,p=0.1$ are fixed.

2. We can observe that power function $h(c_1)$ is an increasing function of m when $n=100,\alpha =0.02,p=0.1$ are fixed.

3. We know that power function $h(c_1)$ is gradually increasing for the removal percentage p when $n=100,m=5,\alpha =0.02$ are fixed.

4. Figure 5 shows that as $c_1$ gradually increases, power function $h(c_1)$ is incremented for α when n = 100, m = 5, p = 0.1 is fixed.

5. Therefore, it is easy to know that the power function is increasing regardless of n, m, p, or α.

Table 4. The values of $h(c_1)$ for $\alpha =0.02$, c = 0.9, t = 0.1 and L = 0.09.

			$c_1$
m	n	p	0.87	0.88	0.89	0.90	0.91	0.92	0.93	0.94	0.95	0.96
5	50	0.05	0.930634	0.950755	0.967255	0.980000	0.989088	0.994899	0.998087	0.999483	0.999917	0.999994
		0.10	0.934899	0.953029	0.968115	0.980000	0.988704	0.994474	0.997797	0.999352	0.999883	0.999991
		0.15	0.938743	0.955102	0.968911	0.980000	0.988331	0.994043	0.997486	0.999201	0.999838	0.999985
	100	0.05	0.911184	0.940655	0.963585	0.980000	0.990525	0.996336	0.998940	0.999801	0.999981	0.999999
		0.10	0.918406	0.944360	0.964906	0.980000	0.990042	0.995881	0.998692	0.999720	0.999968	0.999998
		0.15	0.924822	0.947694	0.966119	0.980000	0.989565	0.995405	0.998410	0.999616	0.999948	0.999997
	150	0.05	0.893701	0.931857	0.960546	0.980000	0.991513	0.997181	0.999340	0.999909	0.999995	1.000000
		0.10	0.903750	0.936888	0.962268	0.980000	0.990973	0.996734	0.999138	0.999858	0.999989	1.000000
		0.15	0.912596	0.941376	0.963840	0.980000	0.990435	0.996254	0.998897	0.999788	0.999979	0.999999
10	50	0.05	0.920091	0.945231	0.965221	0.980000	0.989922	0.995764	0.998625	0.999697	0.999964	0.999999
		0.10	0.932008	0.951485	0.967530	0.980000	0.988968	0.994768	0.997999	0.999445	0.999980	0.999994
		0.15	0.940749	0.956194	0.969335	0.980000	0.988124	0.993798	0.997301	0.999106	0.999808	0.999981
	100	0.05	0.892920	0.931469	0.960415	0.980000	0.991552	0.997212	0.999353	0.999912	0.999995	1.000000
		0.10	0.913519	0.941848	0.964008	0.980000	0.990375	0.996198	0.998867	0.999779	0.999978	0.999999
		0.15	0.928128	0.949430	0.966761	0.980000	0.989300	0.995127	0.998236	0.999547	0.999933	0.999996
	150	0.05	0.867954	0.919209	0.956372	0.980000	0.992641	0.998001	0.999649	0.999968	0.999999	1.000000
		0.10	0.896956	0.933480	0.961097	0.980000	0.991345	0.997046	0.999281	0.999895	0.999993	1.000000
		0.15	0.917119	0.943696	0.964668	0.980000	0.990132	0.995968	0.998741	0.999737	0.999971	0.999999
15	50	0.05	0.915969	0.943105	0.964456	0.980000	0.990211	0.996044	0.998783	0.999751	0.999974	0.999999
		0.10	0.934657	0.952900	0.968066	0.980000	0.988727	0.994500	0.997815	0.999361	0.999885	0.999991
		0.15	0.945940	0.959051	0.970463	0.980000	0.987546	0.993085	0.996737	0.998792	0.999696	0.999961
	100	0.05	0.885645	0.927868	0.959207	0.980000	0.991900	0.997480	0.999462	0.999935	0.999997	1.000000
		0.10	0.917994	0.944148	0.964830	0.980000	0.990071	0.995909	0.998708	0.999726	0.999969	0.999999
		0.15	0.936553	0.953919	0.968456	0.980000	0.988547	0.994295	0.997669	0.999292	0.999866	0.999989
	150	0.05	0.857597	0.914192	0.954768	0.980000	0.993017	0.998241	0.999723	0.999978	1.000000	1.000000
		0.10	0.903175	0.936599	0.962168	0.980000	0.991005	0.996762	0.999151	0.999862	0.999990	1.000000
		0.15	0.928527	0.949641	0.966840	0.980000	0.989267	0.995092	0.998213	0.999537	0.999930	0.999996
20	50	0.05	0.915444	0.942836	0.964360	0.980000	0.990246	0.996077	0.998802	0.999757	0.999975	0.999999
		0.10	0.938939	0.955209	0.968953	0.980000	0.988311	0.994020	0.997468	0.999192	0.999836	0.999985
		0.15	0.950790	0.961771	0.971563	0.980000	0.986941	0.992296	0.996064	0.998378	0.999524	0.999923
	100	0.05	0.884712	0.927408	0.959055	0.980000	0.991943	0.997511	0.999474	0.999937	0.999997	1.000000
		0.10	0.925138	0.947860	0.966181	0.980000	0.989541	0.995379	0.998394	0.999611	0.999947	0.999997
		0.15	0.944231	0.958105	0.970087	0.980000	0.987744	0.993333	0.996938	0.998908	0.999739	0.999969
	150	0.05	0.856263	0.913548	0.954564	0.980000	0.993063	0.998269	0.999731	0.999979	1.000000	1.000000
		0.10	0.913025	0.941596	0.963918	0.980000	0.990407	0.996227	0.998883	0.999784	0.999979	0.999999
		0.15	0.938744	0.955104	0.968912	0.980000	0.988331	0.994043	0.997486	0.999209	0.999838	0.999985

Figure 5.

Power function for n = 100, m = 5, p = 0.1.

4. Real data analysis

In this section, we analyze a real data set to illustrate the methods we proposed above and to verify how our testing procedure algorithms work in practice.

Asl et al. [4] considered a data set in Table 5 showing that the losses exceeding $5,000,000 due to major hurricanes. The data set is given by Helu et al. [10].

Table 5. The losses exceeding $5,000,000 due to major hurricanes.

i	1	2	3	4	5	6	7	8	9	10
$t_i$	6766	7123	10562	14474	15351	16983	18383	19030	25304	29112
i	11	12	13	14	15	16	17	18	19	20
$t_i$	30146	33727	40596	41409	47905	49397	52600	59917	63123	77809
i	21	22	23	24	25	26	27	28	29	30
$t_i$	102942	103217	123680	140136	192013	198446	227338	329511	361200	421680
i	31	32	33	34	35
$t_i$	513586	545778	750389	863881	1638000

First we consider the case where the scale parameter λ is given. To simplify the calculation, we reduced the elements in the data set by a factor of 10,000, for example, 6766 is converted to 0.6766. We obtain the value of λ based on Gini statistic. As shown in the following Figure 6, when λ is 9.8000, the corresponding p-value is.9971. We can know that the Lomax distribution fits the data set well in Table 5. Therefore, we assume the losses (shrinking 10,000 times) exceeding $5,000,000 due to major hurricanes follow the Lomax distribution with the scale parameter $\lambda =9.8000$.

Figure 6.

p-Value vs. λ

1. Let the lower specification limit $L_Z=0.9229$, so the corresponding lower specification limit L = 0.09. Then we can get a progressive type I interval censored sample $X=(21,3,3,1,0\ast 5,1,0\ast 16)$ and corresponding the censoring scheme $R=(4,2,0\ast 24)$ at the time $t=(3.5,7,10.5,\dots ,91)$.

2. Specify the desired level c = 0.8, then we can conduct the null hypothesis and the alternative hypothesis as $H_0:C_L\le 0.8$ and $H_1:C_L>0.8$ respectively.

3. By using Equation (21), we have the MLE of β, that is $\hat{\beta }=2.7770$. Then we get $\widehat{C_L}=1-\hat{\beta }L=0.7501$.

4. Set the significance level $\alpha =0.02$. We are able to get the critical value $C_0=0.8977$ by using Equation (24).

5. Due to $\widehat{C_L}=0.7501<C_0=0.8977$, it can be declared that the lifetime performance index of the products doesn't reach the desired level.

Next we consider the case where both parameters in the Lomax distribution are unknown. We assume that the number of observations is m = 26, the average length of the observation interval t = 3.5 (minutes) and the removal percentages of the remaining survival units $p=(0.3\ast 25,1)$. Then we can establish the testing procedure for $C_L$ as follows:

1. Set the lower specification limit $L_Z=0.9229$. Then we obtain a progressive type I interval censored sample $X=(21,3,3,1,0\ast 5,1,0\ast 16)$ and corresponding the censoring scheme $R=(4,2,0\ast 24)$ at the time $t=(3.5,7,10.5,\dots ,91)$.

2. Specify the desired level $c^{\ast }=0.8$ and conduct the null hypothesis and the alternative hypothesis as $H_0:C_L\le 0.8$ and $H_1:C_L>0.8$ respectively.

3. Obtain the MLEs of two unknown parameters, that is, ${\hat{\beta }}_{ML}=2.1809$ and ${\hat{\lambda }}_{ML}=7.0675$. So we have ${\hat{C}}_{L_{ML}}=0.2436$.

4. Set the significance level $\alpha =0.02$. Then we can get the critical value $C_0^{\ast }=2.2923$ by using Equation (19).

5. Due to ${\hat{C}}_{L_{ML}}=0.2436<C_0^{\ast }=2.2923$, it can be declared that the lifetime performance index of the products doesn't reach the desired level.

5. Conclusion

In this paper, we use the lifetime performance index $C_L$ to evaluate products following Lomax distribution based on progressive type I interval censoring schemes. Compared with complete samples, progressive type I interval censored samples have more advantages in many practical cases. We obtain the MLE and some CIs of $C_L$ with two unknown parameters in the Lomax distribution on basis of progressive type I interval censored samples. Then we use the MLE to establish a hypothesis test procedure under a given lower specification limit L. Furthermore, we analyze the effect of m, n, p and α on the power function and we know that the power function is increasing regardless of n, m, p, or α. In addition, we discuss a hypothesis test procedure for $C_L$ in the case where a scale parameter in the Lomax distribution is known. Through a real data set, we show that the proposed testing procedures are able to determine whether a product meets the required level and work well.

Disclosure statement

No potential conflict of interest was reported by the authors.