Acceptance sampling plans from truncated life test based on Tsallis q-exponential distribution

ABSTRACT

In this study, we propose a new single acceptance sampling plan from truncated life test assuming that the quality characteristics follow the Tsallis q-exponential distribution. The proposed plan is given, and then we derived the operation characteristics function and calculate the optimal sample size and producer’'s risk for some given parameter values to measure the performance of this plan. Also, a comparative study with other sampling plan is discussed to show the benefit of the proposed plan, and real data analysis is given to illustrate the applicability of the proposed plan in the industry.

1. Introduction

Acceptance sampling plan (ASP) is an important inspection tool that is used by the quality assurance managers to determine whether to accept or to reject a product based on pre-specified quality standards. ASP is widely used when the inspection process of a lot of products is costly [4]. The inspection process of a single ASP starts by selecting a simple random sample from a given lot and based on pre-assigned quality standards the decision should be made by the manager to reject the lot if some products fail to meet the pre-assigned quality standards. For any kind of products, the lifetime is expected to be very high, therefore a pre-assigned time t₀ initially set in order to stop the inspection process if the lifetime test period exceeds t₀. One objective of these tests is to set a confidence limit on the mean life and to establish a specified mean life, μ₀, with a probability of at least P* which reflects the consumer's confidence level [22,30]. The testing process on a given item is a Bernoulli experiment. Therefore, to complete the testing on a lot of products we should apply the principles of the Binomial experiment where the random variable, in this case, is the number of failures within a pre-determined time t₀. Based on the Binomial experiment, the decision is to reject a lot if and only if the number of failures within the time t₀ is greater than a specified acceptance number, ‘c’; otherwise, the lot should be accepted.

Therefore, the problem in ASP based on the truncated life test is in finding the minimum sample size n that is necessary to ensure a certain mean lifetime. Accordingly, a lot is considered bad if μ < μ₀, where μ is the true mean life of a product, and μ₀ is the specified mean life. Performing the ASP procedure has two risks; the first risk is the producer's risk (rejection of a good lot), and the second one is the consumer's risk (acceptance of a bad lot). The ASP should be designed such that both types of risks have a minimum value.

Truncated life tests of this type have been proposed by many researchers. In all previous works, usually the author develops ASP assuming that the product's lifetime is following a specific lifetime distribution such as exponential, Gamma, Weibull, and so on [23,18,27,11–14,6–9,31,34].

In a similar fashion, this article develops and discusses an ASP for truncated life test based on the Tsallis q-exponential distribution (TQED) which has not been studied yet.

The rest of this paper is structured as follows. Section 2 summarizes the q-exponential distribution and its properties. In Section 3, the proposed ASPs are developed and the operating characteristic (OC) values and the producer's risk are analyzed. The numerical results and illustrative examples are presented in Section 4. A comparative study between the proposed ASP and other sampling plans based on different distributional assumptions is discussed in Section 5. A real data analysis is given in Section 6. The work is concluded in Section 7.

2. Tsallis q-exponential distribution

The TQED is derived by maximizing the Tsallis entropy [35] as given in (1) subject to the probability density function constraint (i.e. ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)\text{d}x=1$); and a fixed mean constraint (i.e. ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf(x)\text{d}x=\text{given}$); assuming non-negative domain

$$Tsalli{s}_{Entropy}=\frac{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{[f(x)]}^q\text{d}x-1}{1-q};q\ne 1$$ (1)

where q is a real parameter [24,25].

The TQED is a generalization of some lifetime distributions such as Lomax distribution, and it is a particular case of the generalized type II Pareto distribution. Moreover, as ‘q →1’, the TQED converges to the ordinary exponential distribution with parameter $\lambda$. Also, it is equivalent to the Burr-type distributions or to the Zipf–Mandelbrot-type asymptotic power-law distribution when $q>1$ [19,32,33,10]. Therefore, TQED is a generalization of these distributions in the sense that q < 1. Moreover, choosing a suitable value of q then TQED will be a good representation of both short and long-tailed distributions which could be considered an advantage over many distributions in the context of ASP [26].

The TQED [35,20] has the probability density function (pdf):

$$f(x,\lambda ,q)=(2-q)\lambda e_q(-\lambda x)\text{where }x\in \{\begin{array}{ll}(0,\mathrm{\infty })& \text{for }1\le q\\ \left[0,\frac{1}{\sqrt{\lambda (1-q)}}\right]& \text{for }q<1\end{array}$$

given that the shape parameter q < 2 and the scale parameter $\lambda >0$ and

$$e_q(x)=\{\begin{array}{ll}{(1+(1-q)x)}^{\frac{1}{1-q}}& \text{if }q\ne 1\\ e^x& \text{if }q=1\end{array}$$

which can be simplified to

$$f(x,\lambda ,q)=(2-q)\lambda [1+(q-1)\lambda x{]}^{\frac{1}{1-q}}$$ (2)

Figure 1 shows the shape of the pdf of the TQED for some parameter values.

Figure 1.

pdf of TQED for some given parameter values.

The mathematical formula of the corresponding cumulative distribution function (cdf) is

$$F(x,\lambda ,q)=1-e_{\acute{q}}\left(-\lambda x/\acute{q}\right)\text{where }\acute{q}=\frac{1}{2-q}$$

which can be simplified to

$$F(x,\lambda ,q)=1-[1+(q-1)\lambda x{]}^{\frac{2-q}{1-q}}$$ (3)

Figure 2 provides an illustration of the TQED cdf for some parameter values

Figure 2.

cdf of TQED for some given parameter values.

The rth moment about the origin of the TQED is

$$E(X^r)=\{\begin{array}{lll}{\displaystyle \frac{\mathrm{\Gamma }(r+1)\mathrm{\Gamma }\left({\displaystyle \frac{r-q(r+1)+2}{q-1}}\right)}{\mathrm{\Gamma }\left({\displaystyle \frac{1}{q-1}}-1\right){(\lambda (q-1))}^r}};& q\ge 1{\displaystyle \frac{\mathrm{\Gamma }(r+1)\mathrm{\Gamma }\left(2+{\displaystyle \frac{1}{1-q}}\right)}{\mathrm{\Gamma }\left(r+2+{\displaystyle \frac{1}{1-q}}\right){(\lambda (1-q))}^r}};& q\le 1\end{array}$$

Accordingly, the mean and the variance are given as, respectively:

$$\mu =\frac{1}{\lambda (3-2q)}\text{for}q<\frac{3}{2}$$ (4)

and

$${\sigma }^2=\frac{q-2}{{(2q-3)}^2(3q-4){\lambda }^2}\text{for}q<\frac{4}{3}$$

the q values should be chosen less than 1.3. In this article, q = 1.2 and $\lambda$ = 1 will be used to develop the ASP.

3. Design of the ASP

Suppose that the lifetime of the products being tested follows the TQED as given in (2) and the specified mean lifetime is ${\mu }_0$. Now, the ASP problem is to find the optimal sample size that ensures an actual average life (μ) such that no more than c units fail to pass the test period (t). To perform the test according to this plan, a random sample of size n units is selected from a lot. If μ_o can be obtained with a pre-assigned probability, $P^{\ast }$, as specified by the consumer, then the lot is accepted. If not, then it is rejected. Figure 3 provides a visual illustration of the ASP.

Figure 3.

ASP algorithm (source: [2,3]).

The ASP based on truncated life tests consists of the following:

1. The sample size: number of units n to be drawn from the lot.

2. The test duration t: the maximum test duration time.

3. The benchmark of defective (d) units: c, where if d ≤ c defectives out of n occur at the end of the test period $t_0$, the lot is accepted.

4. The minimum ratio $t/{\mu }_o$: where ${\mu }_0$ is the quality parameter of the product life; and t is the maximum test duration.

5. The ASP parameters will be (n, c, $t/{\mu }_o$).

3.1. Optimal sample size of the ASP (n, c, $t/{\mu }_o$)

Suppose that 1 − $P^{\ast }$ where $P^{\ast }\in (0,1)$ is the confidence level of rejecting a lot when ${\mu }_0\le \mu$ with probability $P^{\ast }$. Also, suppose that the lot size is large enough to use the binomial distribution. Then the problem of the ASP (n, c, $t/{\mu }_o$) is to find the minimum sample size n such that the number of failures d does not exceed c, the function up to acceptance number, c, to ensure that $\mu >{\mu }_0$ satisfies the following inequality:

$$\sum _{i=0}^c\left(\begin{array}{c}n\\ i\end{array}\right)p^i(1-p{)}^{n-i}\le 1-P^{\ast }$$ (5)

where

$$p=F(t,{\mu }_0)=1-{\left[1+\frac{(q-1)}{3-2q}\frac{t}{{\mu }_0}\right]}^{\frac{2-q}{1-q}}$$

it should be noted that the probability of success of finding a defective item in a given lot being observed during the test process time t is $p=F(t;{\mu }_0)$, which is a monotonically increasing function of $t/{\mu }_o$. Then for the ASP (n, c, $t/{\mu }_o$) and inequality (4), we assure that $F(t;\mu )\le F(t;{\mu }_0)$ with probability $P^{\ast }$, which implies that ${\mu }_0\le \mu$.

The results for this plan when the lifetime distribution is TQED with q = 1.2 that satisfying (4) are given in Table 1, where the initial values of $t/{\mu }_0=0.\text{628},$ 0.942, 1.257, 1.571, 2.356, 3.141, 3.927, 4.712, when $P^{\ast }=0.75$, 0.9, 0.95, 0.99 and $c=0,1,2,\dots ,10$. The values of $t/{\mu }_0$ and $P^{\ast }$, presented in this work, are the same as the corresponding values are found in the literature for similar sampling plan but with different model [1,16,17,5,21,2,3,28]

Table 1. Minimum sample size to assert that the mean life exceeds a given value ${\mu }_0$ with probability $P^{\ast }$ and acceptance number c based on binomial probabilities when q = 1.2.

		$t/{\mu }_0$
$P^{\ast }$	c	0.628	0.942	1.257	1.571	2.356	3.141	3.927	4.712
0.75	0	2	2	1	1	1	1	1	1
	1	5	4	3	3	2	2	2	2
	2	7	5	5	4	4	3	3	3
	3	9	7	6	5	5	4	4	4
	4	11	9	7	7	6	6	5	5
	5	13	10	9	8	7	7	6	6
	6	15	12	10	9	8	8	7	7
	7	17	13	12	11	9	9	8	8
	8	19	15	13	12	11	10	10	9
	9	21	17	14	13	12	11	11	10
	10	23	18	16	15	13	12	12	11
0.9	0	4	3	2	2	1	1	1	1
	1	6	5	4	3	3	3	2	2
	2	9	7	6	5	4	4	4	3
	3	11	8	7	6	5	5	5	4
	4	13	10	9	8	7	6	6	6
	5	15	12	10	9	8	7	7	7
	6	18	14	12	10	9	8	8	8
	7	20	15	13	12	10	9	9	9
	8	22	17	15	13	11	11	10	10
	9	24	19	16	15	13	12	11	11
	10	26	20	18	16	14	13	12	12
0.95	0	4	3	3	2	2	2	1	1
	1	7	5	4	4	3	3	3	2
	2	10	7	6	5	5	4	4	4
	3	12	9	8	7	6	5	5	5
	4	15	11	9	8	7	6	6	6
	5	17	13	11	10	8	8	7	7
	6	19	15	13	11	10	9	8	8
	7	22	17	14	13	11	10	9	9
	8	24	18	16	14	12	11	10	10
	9	26	20	17	15	13	12	12	11
	10	28	22	19	17	14	13	13	12
0.99	0	7	5	4	3	2	2	2	2
	1	10	7	6	5	4	3	3	3
	2	13	9	8	7	5	5	4	4
	3	15	11	9	8	7	6	6	5
	4	18	13	11	10	8	7	7	6
	5	20	15	13	11	9	8	8	8
	6	23	17	14	13	11	10	9	9
	7	25	19	16	14	12	11	10	10
	8	28	21	18	16	13	12	11	11
	9	30	23	19	17	14	13	12	12
	10	32	25	21	19	16	14	13	13

3.2. OC function of the ASP (n, c, $t/{\mu }_o$)

The OC function of the ASP (n, c, $t/{\mu }_o$) is defined as the probability of accepting a lot:

$$OC(p)=P_r(\text{Accepting a lot})=\sum _{i=0}^c\left(\begin{array}{c}n\\ i\end{array}\right)p^i(1-p{)}^{n-i}$$

for simplification, the $OC(p)$ can be rewritten by using the incomplete beta function ${\mathrm{B}}_p(a,b)$ as

$$OC(p)=1-{\mathrm{B}}_p(c+1,n-c)$$

Noting that ${\mathrm{B}}_p(c+1,n-c)$ is an increasing function of $p=F(t;\mu )$ which is given in (5). Moreover, for a fixed t, the p, itself, is a monotonically decreasing function of $\mu \ge {\mu }_0$. Table 2 presents the OC function values as a function of $\mu \ge {\mu }_0$ for the ASP (n, c = 2, $t/{\mu }_o$).

Table 2. OC function values for the sampling plan $(n,c=2,t/{\mu }_0)$ for a given probability $P^{\ast }$.

		$\mu /{\mu }_0$
P*	n	$t/{\mu }_0$	2	4	6	8	10	12
0.75	7	0. 628	0.581885	0.877457	0.950573	0.975556	0.986212	0.991483
	5	1.257	0.607854	0.884687	0.953247	0.976795	0.986878	0.991879
	5	1.571	0.438824	0.797001	0.910081	0.953159	0.972693	0.982745
	4	2.356	0.515188	0.833608	0.927576	0.962589	0.978297	0.986329
	4	3.141	0.288702	0.669583	0.833674	0.906609	0.942825	0.962608
	3	3.927	0.459896	0.778065	0.893179	0.941328	0.964539	0.977002
	3	4.712	0.34931	0.688675	0.837338	0.906077	0.941303	0.961011
	3	0.942	0.267176	0.604286	0.778027	0.865934	0.913598	0.941314
0.90	9	0.628	0.389256	0.778097	0.902402	0.949459	0.970666	0.981526
	7	0.942	0.332262	0.73352	0.877457	0.934923	0.961617	0.975556
	6	1.257	0.284572	0.689587	0.851141	0.919012	0.951481	0.968763
	5	1.571	0.307895	0.701952	0.85687	0.921974	0.953176	0.969813
	4	2.356	0.288702	0.669583	0.833674	0.906609	0.942825	0.962608
	4	3.141	0.15913	0.515369	0.724769	0.833707	0.893107	0.927627
	4	3.927	0.088797	0.387704	0.615733	0.752389	0.833647	0.883707
	3	4.712	0.267176	0.604286	0.778027	0.865934	0.913598	0.941314
0.95	10	0.628	0.310686	0.723836	0.873377	0.932938	0.960528	0.974903
	7	0.942	0.332262	0.73352	0.877457	0.934923	0.961617	0.975556
	6	1.257	0.284572	0.689587	0.851141	0.919012	0.951481	0.968763
	5	1.571	0.307895	0.701952	0.85687	0.921974	0.953176	0.969813
	5	2.356	0.120916	0.477869	0.702053	0.820113	0.884585	0.922011
	4	3.141	0.15913	0.515369	0.724769	0.833707	0.893107	0.927627
	4	3.927	0.088797	0.387704	0.615733	0.752389	0.833647	0.883707
	4	4.712	0.050782	0.288702	0.515309	0.669583	0.768984	0.833674
0.99	13	0.628	0.146975	0.559468	0.772782	0.87157	0.921247	0.948499
	9	0.942	0.161159	0.570226	0.778097	0.87434	0.922818	0.949459
	8	1.257	0.105725	0.479754	0.711523	0.829587	0.892403	0.928167
	7	1.571	0.090984	0.4448	0.682041	0.808306	0.877293	0.91727
	5	2.356	0.120916	0.477869	0.702053	0.820113	0.884585	0.922011
	5	3.141	0.047472	0.308073	0.547764	0.702104	0.797179	0.856961
	4	3.927	0.088797	0.387704	0.615733	0.752389	0.833647	0.883707
	4	4.712	0.050782	0.288702	0.515309	0.669583	0.768984	0.833674

3.3. Producer's risk of the ASP (n, c, $t/{\mu }_o$)

Unlike the $OC(p)$, the producer's risk $P_r(p)$ is defined as the probability of the consumer rejecting the lot when $\mu >{\mu }_0$. This is given by the following:

$$P_r(p)=P_r(\text{Rejecting a lot})={\mathrm{B}}_p(c+1,n-c)$$

Therefore, for the ASP (n, c, $t/{\mu }_o$) with a known value $P_r(p)$, say $\gamma$, the minimum ratio value ‘$\mu /{\mu }_0$’ is of the researcher interest, to ensure that the producer's risk is at most equal to $\gamma$. This problem can be analyzed by solving the following equation:

$$P_r(p)={\mathrm{B}}_p(c+1,n-c)\le \gamma$$ (6)

where $p=F((t/{\mu }_0)({\mu }_0/\mu ))$.

The smallest values of the ratio $\mu /{\mu }_0$ at confidence level $P^{\ast }$ for the ASP (n, c, $t/{\mu }_o$) under the TQED when q = 1.2 are given in Table 3.

Table 3. Minimum ratio of the true mean life to specified mean life for the acceptance of a lot with producer's risk of 0.05 with q = 1.2.

		$t/{\mu }_0$
P*	C	0.628	0.942	1.257	1.571	2.356	3.141	3.927	4.712
0.75	0	32.545	48.817	32.466	40.576	60.851	81.126	101.427	121.702
	1	10.426	12.073	11.317	14.143	12.024	16.03	20.041	24.047
	2	5.971	5.831	7.781	7.07	10.603	8.601	10.753	12.903
	3	4.427	4.765	5.088	4.737	7.104	6.032	7.541	9.048
	4	3.658	4.184	3.81	4.761	5.417	7.221	5.938	7.125
	5	3.199	3.318	3.757	3.841	4.432	5.909	4.978	5.973
	6	2.894	3.163	3.148	3.245	3.789	5.051	4.337	5.204
	7	2.677	2.713	3.176	3.407	3.335	4.446	3.877	4.652
	8	2.513	2.665	2.802	3.019	3.782	3.996	4.995	4.235
	9	2.385	2.622	2.516	2.722	3.428	3.647	4.56	3.908
	10	2.282	2.37	2.586	2.864	3.147	3.369	4.212	3.644
0.90	0	65.194	73.304	65.141	81.413	60.851	81.126	101.427	121.702
	1	12.796	15.639	16.11	14.143	21.21	28.277	20.041	24.047
	2	8.037	8.957	9.874	9.725	10.603	14.136	17.673	12.903
	3	5.667	5.706	6.358	6.359	7.104	9.471	11.84	9.048
	4	4.52	4.837	5.583	5.878	7.14	7.221	9.028	10.833
	5	3.85	4.308	4.427	4.695	5.759	5.909	7.387	8.864
	6	3.671	3.951	4.22	3.935	4.866	5.051	6.315	7.577
	7	3.319	3.367	3.62	3.97	4.242	4.446	5.558	6.669
	8	3.058	3.22	3.556	3.502	3.782	5.042	4.995	5.994
	9	2.857	3.102	3.175	3.559	4.082	4.57	4.56	5.471
	10	2.698	2.794	3.162	3.231	3.73	4.195	4.212	5.053
0.95	0	65.194	73.304	97.816	81.413	122.093	162.774	101.427	121.702
	1	15.161	15.639	16.11	20.134	21.21	28.277	35.353	24.047
	2	9.067	8.957	9.874	9.725	14.584	14.136	17.673	21.205
	3	6.284	6.64	7.613	7.946	9.537	9.471	11.84	14.207
	4	5.378	5.487	5.583	5.878	7.14	7.221	9.028	10.833
	5	4.498	4.799	5.09	5.533	5.759	7.678	7.387	8.864
	6	3.929	4.341	4.748	4.609	5.901	6.488	6.315	7.577
	7	3.745	4.015	4.058	4.524	5.109	5.656	5.558	6.669
	8	3.419	3.495	3.927	3.976	4.527	5.042	4.995	5.994
	9	3.17	3.34	3.498	3.559	4.082	4.57	5.713	5.471
	10	2.974	3.214	3.446	3.594	3.73	4.195	5.244	5.053
0.99	0	114.167	122.277	130.49	122.25	122.093	162.774	203.506	244.186
	1	22.246	22.742	25.611	26.081	30.194	28.277	35.353	42.42
	2	12.15	12.055	14.021	14.937	14.584	19.443	17.673	21.205
	3	8.133	8.5	8.861	9.515	11.916	12.714	15.896	14.207
	4	6.661	6.78	7.322	8.067	8.814	9.519	11.901	10.833
	5	5.467	5.775	6.403	6.362	7.041	7.678	9.6	11.518
	6	4.958	5.119	5.271	5.933	6.912	7.866	8.111	9.732
	7	4.382	4.657	4.926	5.072	5.953	6.811	7.071	8.484
	8	4.14	4.315	4.664	4.908	5.251	6.035	6.303	7.563
	9	3.795	4.05	4.139	4.372	4.716	5.442	5.713	6.855
	10	3.523	3.839	4.009	4.307	4.846	4.973	5.244	6.293

4. Results description

In this article, the results of proposed ASP based on TQED are given in Tables 1–3 for the smallest sample size, OC values and the producer risk, respectively.

For example, assume that the researcher aims to ensure that the product's mean lifetime is at least 1000 h, with probability $P^{\ast }=0.90$ when $c=2$, such that the lifetime-distribution test is at least t = 1257 h; that is,$t/{\mu }_0=1.257$. Then from Table 1, the optimal sample size for this plan is 6, accordingly, we can use the appropriated ASP (n, c, $t/{\mu }_o$) = (6, 2, $1.257$), which means that a simple random sample of size 6 items should be selected from a lot of products, and if at most 2 items fail in meeting the quality standards before the specified time, t, within the test period (1000 h), then the lot is accepted with probability 0.90 with the assurance that $t/1.257\le \mu$.

Based on the minimum sample size obtained in Table 1, the results in Table 2 summarize the values of the $OC(p)$ for the ASP (n, c = 2, $t/{\mu }_o$). As an example, when $P^{\ast }=0.90$, the $OC(p)$ values for the ASP (6, 2, $1.257$) are re-calculated as follows:

$\mu /{\mu }_0$	2	4	6	8	10	12
OC(p)	0.284572	0.689587	0.851141	0.919012	0.951481	0.968763
Producer's risk	0.715428	0.310413	0.148859	0.080988	0.048519	0.031237

The plan indicates that the lot is accepted if out of 6 items, less than or equal to 2 items fail before the time t. Now, if the true mean is six times as the specified mean $\mu /{\mu }_0=6$ then we are assured that the lot will be accepted under this ASP with probability equal to 0.851141 and the producer's risk is about 0.148859. The probability of accepting a lot under the ASP (6, 2, $1.257$) will be more than 0.90 if and only if the true mean is eight times or more than the specified mean. According to the ASP design, and under the assumption that the producer risk $\gamma$ is equal to 0.05, the minimum ratios $(\mu /{\mu }_0)$ values for the ASP (n, c, $t/{\mu }_o$) are given in Table 3. As an example, suppose we are using the ASP (6, 2, $1.257$) with the consumer's risk equals to 10% ($P^{\ast }=0.90$), then from Table 3, we found that the minimum value of $\mu /{\mu }_0$ is $9.874$. It implies that, when c = 2, the lot with 6 items will be rejected with probability less than or equal to $\gamma$ = 0.05.

5. Comparative study

In this section, the advantages of the proposed plan (TQED) are compared with other ASP under various types of distributions. The comparison criterion will be the cost of inspection based on the sample size of the ASP. We said that a sampling plan with a smaller sample size is more efficient in reducing the cost of inspection compared to other sampling plans. The proposed ASP will be compared with ASP proposed by Aslam et al. [15], for the generalized exponential distribution (GED), Balakrishnan et al. [18] for the generalized Birnbaum–Saunders distribution (GBSD) and Sampath and Lalitha [29] for the hybrid exponential distribution (HED). The comparisons were made at confidence level $P^{\ast }=0.9$ for the ASP (n, c, $t/{\mu }_o$) where c = 2, 4, 6, 8; and under the initial values of $t/{\mu }_0=0.\text{628},$ 0.942, 1.257, 1.571, 2.356, 3.141, 3.927, 4.712. The comparison results are given in Figure 4.

Figure 4.

Comparisons between the sample sizes obtained by different ASP.

It can be observed that the proposed ASP produced much smaller sample size when the ratio $t/{\mu }_0$ is less than 3.141. However, it is equivalent or comparable with the other ASP whenever $t/{\mu }_0$ is more than 3.141. Accordingly, the proposed ASP is more efficient than the other plans and it worth to be used by the decision makers who seek a lot of product with minimum sample size.

6. Real data application

To illustrate the proposed ASP, we considered the data studied by Xia et al. [36]. The data consist of 30 breaking strengths (in megapascals, MPa) of jute fiber measured using a gauge of length 10 mm. The data are given as follows:

$$\begin{array}{l}693.73,704.66,323.83,778.17,123.06,637.66,383.43,151.48,\\ 108.94,50.16,671.49,183.16,257.44,727.23,\\ 291.27,101.15,376.42,163.40,141.38,700.74,262.90,353.24,422.11,\\ 43.93,590.48,212.13,303.90,506.60,\\ 530.55,177.25\end{array}$$

Now suppose that the manufacturing company assured that the mean strength of the jute fiber at a gauge of length 10 mm is 423 MPa, then the research question will be ‘what is the acceptable ASP (n, c, $t/{\mu }_o$) with P* = 0.99, for this manufacturing company?’. To solve such problem the data analyst should use the following steps:

• Step 1. Test if the data fit the model. For this step, goodness of fit measures and the Kolmogorov–Smirnov (K–S) test should be implemented. In this step, a p-value more than 0.05 indicates a good fit to the model.

• Step 2. Estimate the model parameters. One estimation method should be used such as the maximum likelihood estimation (MLE) method, or method of moment (MOM); to estimate the model parameters.

• Step 3. Estimate the average lifetime of the product based on the results obtained in step 2.

• Step 4. Use the existing table results (if possible) or re-compute the ASP parameters using the estimated model parameters.

Several goodness of fit criterion were used to test if the data fit the model including, minimum value of the function –log (likelihood) (−2MLL),Cramér–von Misse (CvM), Akaike information criteria (AIC), Bayesian information criteria (BIC), consistent Akaike information criteria (CAIC), Hannan–Quinn information criteria (HQIC) and two distribution tests; K–S and Anderson–Darling (A–D). The goodness of fit results was acceptable (−2MLL = 211.7295, CvM = 0.05965915, AIC = 427.459, BIC = 430.2614, CAIC = 427.9034, HQIC = 428.3555, A–D = 0. 4822257, K–S = 0. 2058419, P-value = 0. 1362742). The results indicate an excellent fit with K–S distance value between the empirical and the theoretical TQED equal to 0. 2058419 with P-value equal to 0.1362742. Moreover, the MLE is used to estimate the TQED parameters. The results showed that $\hat{q}=1.3485(0.1309)\text{ and }\hat{\lambda }=0.0049(0.0015)$, where the values between brackets are the standard deviation value of the estimator. Therefore, using (4) the mean life can be estimated as

$$\hat{\mu }=\frac{1}{\hat{\lambda }(3-2\hat{q})}=\frac{1}{0.0049(3-2(1.3485))}=673.5367$$

Now, based on our study, we have $\hat{q}=1.3485,\hat{\lambda }=0.0049\text{ and }\hat{\mu }=\mathrm{673.5367.}$ Also, it is assumed that t = 423 MPa. Therefore,

$$\frac{t}{{\mu }_0}=\frac{423}{673.5367}=0.628$$

based on the estimated values and the given minimum sample size values in Table 1 that corresponds to the with P* = 0.99 and $t/{\mu }_0=0.628$, we obtained n = 30 when c = 9, therefore, the optimal ASP will be ASP (30, 9, 0.628).

7. Concluding remarks

In this paper, ASP for TQED is developed assuming that the mean lifetime is a pre-assigned quality parameter. The proposed new ASP parameters, minimum sample size, OC function and the producer risk are obtained simultaneously. The ASP numerical results are discussed and illustrated using numerical examples.

Based on the results, the developed ASP is advised to be used by consumers as well as the producers who aim to minimize production process cost. In similar fashion future research on single ASP could be developed using other q-exponential family distributions.

Correction Statement

This article has been republished with minor changes. These changes do not impact the academic content of the article.

Disclosure statement

No potential conflict of interest was reported by the authors.

ORCID

Amjad D. Al-Nasser http://orcid.org/0000-0001-7515-2357