Tolerance Limits and Sample-Size Determination Using Weibull Trimmed Data

Tolerance limits are extensively employed in some statistical fields, including statistical quality control, economics, medical and pharmaceutical statistics, environmental monitoring, and reliability analysis. In essence, a tolerance interval describes the behavior of a fraction of individuals. Roughly speaking, the tolerance limits are bounds within which one expects a stated proportion of the population to lie. Two basic types of such limits have received considerable attention, $\beta$-content and $\beta$-expectation tolerance limits; see Wilks [1], Guttman [2] and Fernández [3] and references therein. Succinctly, a $\beta$-content tolerance interval contains at least $100\beta \%$ of the population with certain confidence, whereas a $\beta$-expectation tolerance interval covers, on the average, a fraction $\beta$ of the population.

In life-testing and reliability analysis, the tolerance limits are frequently computed from a complete or right-censored sample. In this paper, the available empirical information is provided by a trimmed sample, i.e., it is assumed that determined proportions $q_1$ and $q_2$ of the smallest and largest observations have been eliminated or censored. These kinds of data are frequently used in several areas of statistical practice for deriving robust inferential procedures and detecting influential observations, e.g., Prescott [4], Huber [5], Healy [6], Welsh [7], Wilcox [8], and Fernández [9,10]. In various situations, some extreme sample values may not be recorded due to restrictions on data collection (generally for reasons of economy of money, time, and effort), experimental difficulties or some other extraordinary reasons, or be discarded (especially when some observations are poorly known or the presence of outliers is suspected). In particular, a known number of observations in an ordered sample might be missing at either end (single censoring) or at both ends (double censoring) in failure censored experiments. Specifically, double censoring has been treated by many authors in the statistical literature (among others, Healy [11], Prescott [12], Schneider [13], Bhattacharyya [14], LaRiccia [15], Schneider and Weissfeld [16], Fernández [17,18], Escobar and Meeker [19], Upadhyay and Shastri [20], and Ali Mousa [21]).

The Weibull distribution provides a versatile statistical model for analyzing time-to-event data, which is useful in many fields, including biometry, economics, engineering, management, and the environmental, actuarial, and social sciences. In survival and reliability analysis, this distribution plays a prominent role and has successfully been used to describe animal and human disease mortality, as well as the reliability of both components and equipments in industrial applications. This probability model has many practical applications; e.g., Chen et al. [22], Tsai et al. [23], Aslam et al. [24], Fernández [25], Roy [26], Almongy et al. [27], and Algarni [28]. If the Weibull shape parameter $\alpha =1,$ the distribution is exponential, which plays an notable role in engineering; see Fernández et al. [29], Lee et al. [30], Chen and Yang [31], and Yousef et al. [32]. The random variable is Rayleigh distributed when $\alpha =2.$ This case is also important in various areas; see Aminzadeh [33,34], Raqab and Madi [35], Fernández [36], and Lee et al. [37].

This paper is devoted to deriving tolerance limits using a trimmed sample drawn from a Weibull $W\left(\theta ,\alpha \right)$ population. It is assumed that $\alpha$ is appropriately chosen while the Weibull scale parameter, $\theta ,$ is unknown. The conditionality principle, proposed primarily by Fisher, is adopted when some of the smallest observations have been disregarded. The related problem of determining minimum sample sizes is also tackled. In the exponential case, Fernández [38,39] presented optimal two-sided tolerance intervals and tolerance limits for k-out-of-n systems, respectively. On the basis of a complete Rayleigh sample (i.e., $\alpha =2),$ Aminzadeh [33] found $\beta$-expectation tolerance limits and discussed the determination of sample size to control stability of coverage, whereas Aminzadeh [34] derived approximate tolerance limits when $\theta$ depends on a set of explanatory variables. Weibull tolerance limits based on complete samples were obtained in Thoman et al. [40].

The structure of the remainder of this work is as follows. The sampling distribution of a Weibull trimmed sample is provided in the next section. Section 3 presents $\beta$-content tolerance limits based on $W\left(\theta ,\alpha \right)$ trimmed data. In addition, the problem of determining optimal sample sizes is discussed. Mean-coverage tolerance limits are derived in Section 4. Optimal sampling plans for setting $\beta$-expectation tolerance limits are also deduced. The corresponding unconditional and conditional bounds are compared in Section 3 and Section 4 when the lower trimming proportion, $q_1,$ is positive, whereas Section 5 includes several numerical examples, reported by Sarhan and Greenberg [41], Meeker and Escobar [42], and Lee and Wang [43], for illustrative purposes. Finally, Section 6 offers some concluding remarks.

The probability density function (pdf) of a random variable X which has a Weibull distribution with positive parameters $\theta$ and $\alpha ,$ i.e., $X\sim W\left(\theta ,\alpha \right),$ is defined by Its k-th moment is obtained to be $E[X^k\mid \theta ,\alpha ]={\theta }^k\mathsf{\Gamma }(1+k/\alpha ),$ $k=1,2,\dots ,$ where $\mathsf{\Gamma }(·)$ is the well-known gamma function. The parameter $\alpha$ controls the shape of the density whereas $\theta$ determines its scaling. Since the hazard rate is $h(x\mid \theta ,\alpha )=\left(\alpha /{\theta }^{\alpha }\right)x^{\alpha -1}$ for $x>0,$ the Weibull law may be used to model the survival distribution of a population with increasing $(\alpha >1),$ decreasing $(\alpha <1),$ or constant $(\alpha =1)$ risk. Examples of increasing and decreasing hazard rates are, respectively, patients with lung cancer and patients who undergo successful major surgery. Davis [44] reports several cases in which a constant risk is reasonable, including payroll check errors and radar set component failures.

In many practical applications, Weibull distributions with $\alpha$ in the range 1 to 3 seem appropriate. If $3\le \alpha \le 4,$ the $W\left(\theta ,\alpha \right)$ pdf has a near normal shape; for large $\alpha$ (e.g., $\alpha \ge 10),$ the shape of the density is close to that of the (smallest) extreme value density. The Weibull density becomes more symmetric as $\alpha$ grows. In Weibull data analysis it is quite habitual to assume that the shape parameter, $\alpha ,$ is a known constant. Among other authors, Soland [45], Tsokos and Rao [46], Lawless [47], and Nordman and Meeker [48] provide justifications. The $\alpha$ value may come from previous or related data or may be a widely accepted value for the problem, or even an expert guess. In reliability analysis, $\alpha$ is often tied to the failure mechanism of the product and so engineers might have some knowledge of it. Klinger et al. [49] provide tables of Weibull parameters for various devices. Abernethy [50] supplies useful information about past experiments with Weibull data. Several situations in which it is appropriate to consider that $\alpha$ is constant are described in Nordman and Meeker [48]. Among many others, Danziger [51], Tsokos and Canavos [52], Moore and Bilikam [53], Kwon [54], and Zhang and Meeker [55] also utilize a given Weibull shape parameter.

Consider a random sample of size n from a Weibull distribution (1) with unknown scale parameter $\theta \in \mathsf{\Theta }=\left(0,\infty \right)$, and let $X_{r:n},\dots ,X_{s:n}$ be the ordered observations remaining when the $\left(r-1\right)$ smallest observations and the $\left(n-s\right)$ largest observations have been discarded or censored, where $1\le r\le s\le n.$ The trimming proportions $q_1=\left(r-1\right)/n$ and $q_2=1-s/n,$ as well as the shape parameter $\alpha ,$ are assumed to be predetermined constants.

The pdf of the $\left(q_1,q_2\right)$-trimmed sample $\mathbf{X}=\left(X_{r:n},\dots ,X_{s:n}\right)$ at $\mathbf{x}=\left(x_{r:n},\dots ,x_{s:n}\right)$ is then defined by for $0<x_{r:n}<\cdots <x_{s:n},$ where $T\left(\mathbf{x}\right)$ is the observed value of Clearly, T is sufficient when $r=1,$ whereas the sample evidence is contained in the sufficient statistic $\left(X_{r:n},T\right)$ if $r>1.$

The maximum likelihood estimator (MLE) of $\theta ,$ denoted by $\widehat{\theta }\equiv \widehat{\theta }\left(\mathbf{X}\right),$ can be derived from the equation $\partial ln{f}_{\mathbf{X}}(\mathbf{X}\mid \theta ,\alpha )/\partial \theta =0.$ It is well-known that $\widehat{\theta }$ is the unique solution to the equation See, e.g., Theorem 1 in Fernández et al. [29]. Therefore, the MLE of $\theta$ is given explicitly by $\widehat{\theta }={\left(T/s\right)}^{1/\alpha }$ when $r=1.$ Otherwise, $\widehat{\theta }$ must be found upon using an iterative procedure.

Given the Weibull $\left(q_1,q_2\right)$-trimmed sample $\mathbf{X}=\left(X_{r:n},\dots ,X_{s:n}\right)$ and $\beta ,$$\gamma \in \left(0,1\right),$ a statistic $L_{\beta ,\gamma }\equiv L_{\beta ,\gamma }\left(\mathbf{X}\right)$ is called a lower $\beta$-content tolerance limit at level of confidence $\gamma$ (or simply a lower $\left(\beta ,\gamma \right)$-TL for short) of the $W\left(\theta ,\alpha \right)$ distribution if for all $\theta >0,$ where $Pr{}^{\mathbf{X}\mid \theta ,\alpha }\left\{·\right\}$ and $Pr{}^{X\mid \theta ,\alpha }\left(·\right)$ refer to the respective sampling distribution of $\mathbf{X}$ and X under the nominal values of $\theta$ and $\alpha ,$ which are defined in (1) and (2), respectively.

According to (4), one may guarantee with confidence $\gamma$ that at least $100\beta \%$ of population measurements will exceed $L_{\beta ,\gamma }.$ In other words, with confidence $\gamma ,$ the probability that a future observation of $X\sim W\left(\theta ,\alpha \right)$ will surpass $L_{\beta ,\gamma }$ is at least $\beta .$ Clearly, an upper $\left(\beta ,\gamma \right)$-TL, $U_{\beta ,\gamma }\equiv U_{\beta ,\gamma }\left(\mathbf{X}\right),$ is provided by $L_{1-\beta ,1-\gamma }.$ In this manner, one can be $100\gamma \%$ confident that at least $100\beta \%$ of Weibull $W\left(\theta ,\alpha \right)$ observations will be less than $U_{\beta ,\gamma }.$

Assuming that $1<r=s,$ as $X_{r:n}$ is minimal sufficient for $\theta ,$ it is logical to consider a lower $\left(\beta ,\gamma \right)$-TL for the form $L_{\beta ,\gamma }=C_{\beta ,\gamma }{X}_{r:n},$ where, from (4), $C_{\beta ,\gamma }$ must satisfy Since $exp\left(-X_{r:n}^{\alpha }/{\theta }^{\alpha }\right)\sim Beta\left(n-r+1,r\right),$ it follows that ${\beta }^{1/{\left(C_{\beta ,\gamma }\right)}^{\alpha }}$ is merely the $\left(1-\gamma \right)$-quantile of the $Beta(n-r+1,r)$ distribution. Thus, $C_{\beta ,\gamma }$ satisfies the equation Alternatively, $C_{\beta ,\gamma }$ may be expressed explicitly as where $F_{2r,2\left(n-r+1\right);\gamma }$ denotes the $\gamma$-quantile of the F-distribution with $2r$ and $2(n-r+1)$ degrees of freedom (df). In particular, $C_{\beta ,\gamma }={\{nln\beta /ln(1-\gamma )\}}^{1/\alpha }$ when $r=s=1,$ whereas $C_{\beta ,\gamma }={\{ln\beta /ln(1-{\gamma }^{1/n})\}}^{1/\alpha }$ if $r=s=n.$

If $1=r\le s,$ it is clear that T is minimal sufficient for $\theta ,$ which implies that it is sensible to assume that $L_{\beta ,\gamma }$ is proportional to $T^{1/\alpha },$ i.e., $L_{\beta ,\gamma }=C_{\beta ,\gamma }{T}^{1/\alpha }.$ In this situation, it can be shown that $2T/{\theta }^{\alpha }\sim {\chi }_{2s}^2,$ where ${\chi }_{2s}^2$ represents the chi-square distribution with $2s$ df. Observe that, letting $X_{0:n}\equiv 0,$ the pivotal $2T/{\theta }^{\alpha }$ coincides with ${\sum }_{i=1}^s{Z}_i,$ where $Z_i=2\left(n-i+1\right)\left(X_{i:n}^{\alpha }-X_{i-1:n}^{\alpha }\right)/{\theta }^{\alpha },$ $i=1,\dots ,s,$ are mutually independent ${\chi }_2^2$ variables. Since, in view of (4), it turns out that $C_{\beta ,\gamma }={(-2ln\beta /{\chi }_{2s;\gamma }^2)}^{1/\alpha }.$ Consequently, $L_{\beta ,\gamma }={(-2Tln\beta /{\chi }_{2s;\gamma }^2)}^{1/\alpha }.$

Given the Weibull $\left(q_1,q_2\right)$-trimmed sample $\mathbf{X}=(X_{r:n},\dots ,X_{s:n})$ and $\beta \in \left(0,1\right),$ a statistic ${\mathcal{L}}_{\beta }\equiv {\mathcal{L}}_{\beta }\left(\mathbf{X}\right)$ is called a lower $\beta$-expectation tolerance limit (or lower $\beta$-ETL for simplicity) of the $W\left(\theta ,\alpha \right)$ distribution if it satisfies for all $\theta >0.$ In this way, the probability that a future observation of X will surpass ${\mathcal{L}}_{\beta }$ is expected to be $\beta .$ Obviously, an upper $\beta$-ETL, ${\mathcal{U}}_{\beta }\equiv {\mathcal{U}}_{\beta }\left(\mathbf{X}\right),$ is given by ${\mathcal{L}}_{1-\beta }.$

Provided that $\beta \in \left(0,1\right),$ our purpose is determining $\beta$-ETLs based on the trimmed sample $\mathbf{X}=(X_{r:n},\dots ,X_{s:n})$ drawn from $X\sim W\left(\theta ,\alpha \right).$

Suppose that $1<r=s,$ in which case $X_{r:n}$ is minimal sufficient for $\theta .$ It is therefore rational to consider a lower $\beta$-ETL of the form ${\mathcal{L}}_{\beta }=D_{\beta }{X}_{r:n},$ where, from (11), the constant $D_{\beta }$ must satisfy Since $exp\left(-X_{r:n}^{\alpha }/{\theta }^{\alpha }\right)\sim Beta\left(n-r+1,r\right),$ $D_{\beta }$ is the unique positive solution to the following equation in D where $B\left(·,·\right)$ is the beta function, i.e., $D_{\beta }={\zeta }^{-1}\left(\beta \right)$. Note that $\zeta \left(·\right)$ is continuous and decreasing with $\zeta \left(0\right)=1$ and $\zeta \left(+\infty \right)=0.$ Therefore, $D_{\beta }$ is the positive constant which satisfies the equation Observe that $D_{\beta }={\{n(1-\beta )/\beta \}}^{1/\alpha }$ when $r=s=1.$ In general, as it is clear that The above lower and upper bounds on $D_{\beta }$ might serve as a starting point for iterative interpolation methods, such as regula falsi. In addition, when $r/n$ is small.

If $1=r\le s,$ as T is minimal sufficient for $\theta ,$ it is evident that ${\mathcal{L}}_{\beta }=D_{\beta }{T}^{1/\alpha }$ is an appropriate lower $\beta$-ETL. Since $2T/{\theta }^{\alpha }\sim {\chi }_{2s}^2,$ the tolerance factor is given by $D_{\beta }={({\beta }^{-1/s}-1)}^{1/\alpha }$.