Quality Control Testing with Experimental Practical Illustrations under the Modified Lindley Distribution Using Single, Double, and Multiple Acceptance Sampling Plans

Abstract

Quality control testing under acceptance sampling plans involves inspecting a representative sample of products or materials from a larger lot or batch to determine whether the lot meets predetermined quality standards. In this research, the modified Lindley distribution is used as a model for lifetime study. When a life test is amputated at a pre-appropriated time to decide on the admission or refusal of the submitted batches, the problems of the single, double, and multiple (three and four stages) acceptance sampling strategies are introduced. The optimal sample sizes are computed for single, double, and multiple acceptance sampling plans to ensure that the veritable mean life is greater than the prescribed mean life at the stipulated consumer’s risk. The operating characteristic functions are investigated at diverse quality levels. For single, double, and multiple acceptance sampling plans, the minimal ratios of the veritable mean life to the prescribed mean life at the established percent of the producer’s risk are obtained. To demonstrate the uses of single, double, and multiple, some numerical experiments are presented.

1. Introduction

Acceptance sampling plans are widely used in quality control and are an important tool for managing risk in manufacturing and service industries. An acceptance sampling plan involves testing a sample of items from a larger batch or population and making a decision based on the results of the test. The decision is typically whether to accept or reject the entire batch or population based on the quality of the sample. The importance of acceptance sampling plans lies in their ability to provide a cost-effective and efficient means of ensuring product quality. Sampling allows manufacturers to test a portion of a batch rather than inspecting every single item, which can be time-consuming and expensive. Sampling also reduces the risk of accepting a batch with low quality or rejecting a batch that meets the required standards, which can result in unnecessary costs and delays.

Acceptance sampling plans and statistical quality control (SQC) are closely related concepts in the field of quality management. Acceptance sampling plans are a type of quality control technique that involves inspecting a sample of products to determine whether they meet a specified quality standard, while SQC is a broader approach to quality control that involves using statistical methods to monitor and control the quality of a process. The relationship between acceptance sampling plans and SQC is that acceptance sampling plans can be used as a tool within SQC to monitor and control the quality of a process. For example, an acceptance sampling plan may be used to inspect a sample of products at a specific point in the manufacturing process to ensure that the quality of the products meets the required standards.

Acceptance sampling is a methodology for checking the output process’s end result, its input, or its initial product. Each batch is randomly sampled, and a choice is made regarding whether to accept or reject the batch based on the information provided by the sample. Acceptance sampling occurs between pre-examination and complete examination. There are two approaches to accept sampling in the literature. The first methodology, acceptance sampling by characteristics (attributes), counts the number of nonconforming units after classifying the product as conforming or nonconforming (defective) based on a set of criteria. The alternative approach is acceptance sampling by variables, which is performed if the item inspection results in continuous measurement.

An acceptance methodological approach is typically a statistical instrument employed in quality management. Determining a certain number of items for testing enables a company to assess the quality of a specific batch of goods. The quality of one particular sample will serve as the standard for the entire collection of products. The business is unable to always check each and every one of its products. Certain products might be available for inspection at a reasonable cost or in a reasonable amount of time. However, a thorough examination of every component of the product can harm it or render it inappropriate for sale. As a result, examining a tiny sample would be sufficient without causing harm to all items. Acceptance sampling involves two types of decisions based on the sample: accepting or rejecting the batch. Additionally, two types of mistakes are associated with each decision. When the confirming batch is rejected based on sample results, this is known as a type I error, whereas a type II error, also known as a customer’s risk, occurs when the consumer rejects the good batch (for more details see Balakrishnan et al. [1], Aslam et al. [2] and Carolino and Barao [3]).

The producer desires the acceptance of “confirming” batches with high probability, and the consumer desires the acceptance of “nonconforming” batches with a small probability ($1-\varkappa *$). In the literature, many researchers have investigated acceptance sampling plans based on lifetime distributions, such as Gupta and Groll [4], Aslam et al. [2], and Balakrishnan et al. [1]. Following Wu et al. [5], when a product’s lifetime follows a two-parameter Lindley distribution, an acceptance sampling plan is created for a truncated life test. For various combinations of Lindley-distributed parameters, the required minimum sample size for inspection and the threshold number of failures for lot acceptance are determined, with the contracted quality-and-risk standard levels approved by the supplier and the buyer.

Other relevant works may be useful such as Mahmood et al. [6] (for acceptance sampling plans based on Topp–Leone Gompertz distribution) and Tripathi et al. [7] (for double and group acceptance sampling plans for a truncated life test based on inverse log-logistic distribution). Ahmed and Yousof [8] (for new group acceptance sampling plans based on percentiles for the Weibull Fréchet model) and Ahmed et al. [9] (for a novel g family for single acceptance sampling plan with application in quality and risk decisions). Al-Omari and Al-Nasser [10] studied the two-parameter quasi-Lindley distribution in acceptance sampling plans from truncated life tests. Saha et al. [11] presented the acceptance sampling inspection plan for the Lindley and power Lindley distributed quality characteristics.

Typically, in industry and other fields, the effectual implementation of statistical methods aids in the improvement of production quality control. Controlling the production process in order to obtain a sufficient product for the primary purpose for which it was created is what quality control entails.

Both the manufacturer and the consumer aim to ensure that the submitted batches are conforming to the acceptance sampling strategy. There are two methods for doing so, as follows:

*

Complete inspection: this entails inspecting every unit in the batch, removing defective items, and replacing them with conforming units.

*

The second method entails taking a random specimen or more from the batch, inspecting the units, and determining whether to accept or refuse the batch.

*

The sorts of plans for this study are single, double, and multiple examination sampling plans, according to the number of drawn specimens.

2. The Modified Lindley Distribution Acceptance Sampling Plans

According to these methods, this paper aims at studying acceptance sampling plans when the lifetime of the product is a random variable that has a new probability distribution of the modified Lindley (ML) distribution (see Chesneau et al. [12]). Its main feature is to operate a simple trade-off between the exponential and Lindley distributions, offering an interesting alternative to these two well-established distributions. According to Chesneau et al. [12], the probability density function (PDF) of ML can be written as

$$f_{\psi }\left(𝓍\right)=\frac{{\psi }^{-1}}{1+{\psi }^{-1}}\left[2\frac{𝓍}{\psi }-1+e^{\frac{𝓍}{\psi }}\left(1+{\psi }^{-1}\right)\right]e^{-2\frac{𝓍}{\psi }}|𝓍>0,\psi >0.$$

(1)

The corresponding cumulative distribution function (CDF) can be written as

$$F_{\psi }\left(𝓍\right)=1-e^{-\frac{𝓍}{\psi }}\left(1+\frac{\frac{𝓍}{\psi }}{{\psi }^{-1}+1}{e}^{-\frac{𝓍}{\psi }}\right).$$

(2)

Chesneau et al. [12] studied the main properties of the ML distribution, with a special emphasis on its moments, reliability parameter, and asymptotic distributions of the extreme order statistics. Then, inferential considerations were explored. The mean and the variance of $X$ can be deduced as

$$E\left(X\right)=\frac{5+4{\psi }^{-1}}{4{\psi }^{-1}\left(1+{\psi }^{-1}\right)},$$

and

$$Va𝓇\left(X\right)=\frac{\left(3+4{\psi }^{-1}\right)\left(5+4{\psi }^{-1}\right)}{16{\psi }^{-2}{\left(1+{\psi }^{-1}\right)}^2}.$$

In this article, we have chosen the ML distribution for two reasons:

• It is a simple model with only one parameter.
• All its moments are exits.

Although the Lindley distribution (and many of its generalizations) has received a great deal of study and analysis in the field of sampling acceptance plans (see Al-Omari and Al-Nasser [10], Wu et al. [5], and Saha et al. [11]), the current article has a feature that is not found in all the aforementioned statistical literature, which is that, in this work, we have studied and analyzed three different types of previewing in the admission plans, and we dealt with each case with applied examples and numerical explanations.

This main study objective is to offer novel single, double, and multiple acceptance sampling designs based on the ML distribution, along with its applications. When a manufacturer’s quality is evaluated using a lifespan random variable, its corresponding variance (or, if one exists, the scale parameter) may be regarded as the quality parameter. Hence, life tests can be subjected to the acceptance sampling techniques (see Balakrishnan et al. [1].

Some common applications of acceptance sampling plans are as follows:

• Acceptance sampling plans can be used to ensure that raw materials meet the required quality standards before they are used in the manufacturing process.
• Finished product inspection: acceptance sampling plans can be used to inspect finished products to ensure that they meet the required quality standards before they are shipped to customers.
• Process control: acceptance sampling plans can be used to monitor the quality of a manufacturing process and make adjustments as needed to ensure that the final products meet the required quality standards.
• Supplier quality management: acceptance sampling plans can be used to evaluate the quality of products received from suppliers and ensure that they meet the required standards before they are used in the manufacturing process.

In this paper, the probability distribution of a lifetime random variable is considered as ML distribution. We are motivated to consider the ML distribution in the field of quality control for the following reasons:

• All moments of the ML distribution exist; hence, it will be easy to apply the new model in quality control.
• It is the first time that the ML distribution will be used for acceptance sampling plan analysis.
• The most popular and easiest sampling plans to utilize are single, double, and multiple acceptance sampling plans; however, they are not the most efficient in terms of the average number of samples required. It is possible that, by considering the ML distribution, additional authors will be drawn to other innovative models.
• Under the ML distribution, the proposed single, double, and multiple acceptance sampling plans are beneficial for quality control and could be rapid cures for production.
• Other related extensions of the ML distribution, such as the ML exponential model and truncated ML versions, can be employed with the results obtained.

3. The ML Distribution for the Single Acceptance Sampling Plans

The examination sampling strategy is a statistical instrument utilized in quality control. It allows a business to evaluate the quality of a batch of items by picking out a particular number of specimens for testing. Presume that the product’s lifetime follows the ML distribution. One purpose of reliability tests is to provide a confidence limit for the mean residual life. The consumer risk is therefore desired to be determined with a defined mean life with a probability of at least ϰ*. We accept a certain mean life if and only if the number of recorded nonconforming does not override a defined examination limit, ${\mathcal{A}}_{\mathsf{𝕔}}$. As a result, the test is terminated at time x or when the ($1+{\mathcal{A}}_{\mathsf{𝕔}}$)th nonconforming occurs, whichever happens first.

We are focusing our efforts on obtaining sampling strategies for such truncated reliability tests and their accompanying decision rules. In other words, we want to identify the optimal sample size required to meet our main purpose.

Whether the units in the batch are designated as faulty or non-defective based on a life testing experiment, the acceptance sampling strategy must take the following factors into account:

i. 

“$𝓃$”: the number of units under test;

ii. 

“${\mathcal{A}}_{\mathsf{𝕔}}$”: the examination limit;

iii. 

“$\frac{𝓍}{{\psi }_0}$”: the plan ratio, where the denominator is the prescribed mean life, and the numerator is the maximum test duration.

For $X ~$ ML($\psi$), ${\mu }_0$ will represent the mean of the model, that is, ${\mu }_0={\psi }_0$, and $\psi$ refers to the mean life, and it presents the scale parameter of the ML distribution or the quality parameter.

The choice of X, ${\mathcal{A}}_{\mathsf{𝕔}}$, and n will be made at the producer’s discretion. This indicates that the probability of rejecting a good batch, i.e., the acceptance of the submitted batch, will be dependent on the hypothesis test:

$$H_0=\psi \ge {\psi }_0 \mathrm{versus} H_1=\psi <{\psi }_0,$$

(3)

where $\psi$ and ${\psi }_0$ are veritable and specified mean life, respectively. This hypothesis is tested in the inspection sampling plan based on the number of faulty units from the sample in a pre-determined time. The batch is rejected if the number of faulty units exceeds the examination limit, ${\mathcal{A}}_{\mathsf{𝕔}}$. The batch will be accepted only if there is sufficient evidence that $\psi$ ≥ ${\psi }_0$ at a specific level of consumer risk. The consumer’s risk is fixed and cannot surpass 1 − $\varkappa *$, where $\varkappa *$ is the level of confidence.

3.1. Sample Size of the Single Sampling Plans

The sample size of single sampling plans is determined based on a combination of factors, including the desired level of confidence, the acceptable quality level (AQL), and the lot size. The AQL is the maximum proportion of defective items that is considered acceptable in the population of interest. The lot size is the total number of items in the population that are available for sampling. To determine the sample size for a single sampling plan, the following steps are typically followed:

• Determine the desired level of confidence: The level of confidence is the probability of correctly accepting a lot that meets the AQL or correctly rejecting a lot that does not meet the AQL. The level of confidence is typically set at 95% or 99%.
• Determine the AQL: The AQL is the maximum proportion of defective items that is considered acceptable. This is typically expressed as a percentage.
• Determine the lot size: The lot size is the total number of items in the population that are available for sampling.
• Use a sampling plan table: Single sampling plans are typically presented in tables that provide the sample size for a given lot size, AQL, and level of confidence. These tables are based on statistical calculations that take into account the variability of the population and the desired level of confidence.
• Select the appropriate sample size: using the sampling plan table, the appropriate sample size can be selected based on the lot size, AQL, and level of confidence.

When we talk about a particular batch, we usually indicate that it is a very large batch; hence, the binomial model can be examined. Furthermore, accepting or rejecting a batch is similar to accepting or rejecting the quality parameter hypothesis.

The size of a single sampling plan can be expressed as follows: given a number, $0<\varkappa *<1$, the mean value ${\psi }_0$, the experimental time $X$, and the examination limit ${\mathcal{A}}_{\mathsf{𝕔}}$, we want to find the smallest positive integer, $𝓃$, so that if the observed number of faulty units does not exceed ${\mathcal{A}}_{\mathsf{𝕔}}$, it is ensured that $\psi \ge {\psi }_0$ with probability $\varkappa *$. In that case, for a given ${\psi }_0$, $X$, $\varkappa *$, and ${\mathcal{A}}_{\mathsf{𝕔}}$, we need to find $𝓃$, the smallest positive integer, which satisfies the inequality represented by the following formula:

$$\sum _{𝓇=0}^{{\mathcal{A}}_{\mathsf{𝕔}}}\left(\begin{array}{c}𝓃\\ 𝓇\end{array}\right){𝓅}^{𝓇}{\left(1-𝓅\right)}^{𝓃-𝓇}\le 1-\varkappa *$$

(4)

where ${\mathcal{A}}_{\mathsf{𝕔}}\le 𝓃$ refers to the examination limit, and $𝓅=F_{{\psi }_0}\left(𝓍\right)$, given in Equation (2), is a monotonically increasing function of $\frac{𝓍}{{\psi }_0}$ and a decreasing function of ${\psi }_0$. Additionally, $𝓅=F_{{\psi }_0}\left(𝓍\right)$ is the probability of a failure in time $𝓍$ if the veritable mean life is

$$\mathrm{E}\left(X\right)=\frac{5+4\frac{1}{{\psi }_0}}{4\frac{1}{{\psi }_0}\left(1+\frac{1}{{\psi }_0}\right)}.$$

Since the CDF depends only on the ratio $\frac{𝓍}{{\psi }_0}$, the experimenter needs to specify only this ratio. If the observed number of failures is less than or equal to ${\mathcal{A}}_{\mathsf{𝕔}}$, then, from inequality (4), we can make the confidence statement that $F_{{\psi }_0}\left(𝓍\right)\ge F_{\psi }\left(𝓍\right)$ with probability $\varkappa *$ (see Gupta and Groll [4] for more details). The optimal value of sample size satisfying inequality (4) for

$$\varkappa *=0.75, 0.90, 0.95, 0.99$$

and

$$\frac{𝓍}{{\psi }_0}=0.650, 0.910, 1.000, 1.585, 2.000, 3.500, 4.000 \mathrm{and} 6.000$$

were determined and are presented in

Table 1. Values of $𝓃$ to be examined for a certain time $𝓍$.

$\mathit{\varkappa }*$	${\mathcal{A}}_{\mathsf{𝕔}}$	$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$
		0.650	0.910	1.000	1.585	2.000	3.500	4.000	6.000
0.75	0	4	2	2	1	1	1	1	1
	1	14	10	4	3	2	2	1	1
	2	21	15	15	9	8	7	6	6
	3	22	21	17	15	13	10	8	11
	4	27	23	18	16	16	14	12	12
	5	30	24	21	19	17	16	13	15
	6	33	25	25	23	20	18	17	17
	7	36	26	26	24	21	19	18	18
	8	40	28	28	27	25	23	21	19
	9	43	29	29	28	27	25	22	19
	10	47	30	29	28	27	26	23	20
0.90	0	6	4	3	2	1	1	1	1
	1	15	11	6	4	3	2	2	1
	2	25	16	16	10	10	9	9	8
	3	27	23	18	15	15	11	8	11
	4	29	24	19	17	18	15	12	12
	5	34	25	22	20	20	17	15	15
	6	33	26	26	26	23	19	18	17
	7	37	28	26	26	24	21	19	19
	8	41	28	28	28	29	24	22	19
	9	48	29	29	31	31	26	23	19
	10	52	32	29	32	31	26	25	20
0.95	0	8	5	4	2	2	1	1	1
	1	18	13	7	4	3	2	2	2
	2	29	19	18	13	12	10	9	8
	3	30	25	19	16	15	13	10	11
	4	31	26	19	18	19	15	14	13
	5	35	27	23	21	21	17	15	16
	6	36	29	27	27	24	20	18	18
	7	40	29	27	27	25	22	19	20
	8	45	33	29	30	30	26	23	21
	9	50	37	29	33	32	27	24	21
	10	54	41	31	33	32	28	25	23
0.99	0	13	8	7	4	3	1	1	1
	1	19	14	10	6	4	3	2	2
	2	30	20	19	13	12	11	10	9
	3	34	26	19	17	16	14	11	11
	4	34	27	20	18	19	15	14	12
	5	39	28	24	22	22	18	16	17
	6	44	29	28	28	25	21	18	18
	7	48	31	28	29	26	23	19	21
	8	53	34	34	30	31	27	23	24
	9	57	40	38	34	33	28	24	26
	10	61	44	42	34	33	29	26	29

below. Table 1 provides the values of sample size $𝓃$ to be examined for a time $𝓍$ in order to confirm that the mean life overrides a given value, ${\psi }_0$, with probability $\varkappa *$ and the acceptance level ${\mathcal{A}}_{\mathsf{𝕔}}$.

Table 1 shows that the higher the examination limit, the larger the sample size required to provide more accurate results. Additionally, the greater the confidence level, the greater the sample size needed to attain more trusted results. There is an inverse relationship between the experiment time ratio and sample size: the greater the experiment time ratio, the smaller the required sample size.

3.2. Operating Characteristic Function

The Operating Characteristic (OC) function is a statistical tool used to evaluate the performance of an acceptance sampling plan. It is an important tool for quality control and is widely used in manufacturing and service industries. The OC function is a graph that shows the probability of accepting a batch or population of items as a function of its quality level. The quality level is typically measured by the percentage of defective items in the batch. The OC function can be used to evaluate the effectiveness of different acceptance sampling plans and to determine the optimal sample size and acceptance criteria. The importance of the OC function lies in its ability to provide a quantitative measure of the performance of an acceptance sampling plan. By analyzing the OC function, manufacturers can evaluate the risks and costs associated with different sampling plans and make informed decisions about the best course of action. The OC function can be used to design acceptance sampling plans that meet specific quality objectives and risk levels. The OC function can be used to evaluate the effectiveness of different sampling plans and to compare their performance. The OC function can be used to determine the optimal sample size and acceptance criteria that will provide the desired level of quality and risk. The OC function can be used to monitor the performance of the sampling plan over time and to make adjustments as needed. Generally, the OC function is an important tool for managing quality and reducing risk in manufacturing and service industries. Its use is widespread in many industries around the world, and it plays a critical role in ensuring that products and services meet the required quality standards. For more details, see Wu [13], and for more new applications, see Aslam et al. [2], Balakrishnan et al. [1], Ahmed and Yousof [8], and Ahmed et al. [9].

The OC of the sampling plan $\left(𝓃,{\mathcal{A}}_{\mathsf{𝕔}},\frac{𝓍}{{\psi }_0}\right)$ is the probability of accepting a batch and is given by the following equation:

$${\mathrm{P}}_1{ =\mathrm{P}}_{𝓃,𝓅}=\sum _{𝓇=0}^{{\mathcal{A}}_{\mathsf{𝕔}}}\left(\begin{array}{c}𝓃\\ 𝓇\end{array}\right){𝓅}^{𝓇}{\left(1-𝓅\right)}^{𝓃-𝓇},$$

(5)

The OC function decreases from $1$ to $0$ as $𝓅$ increases from 0 to 1. $𝓅=F_{\psi }\left(𝓍\right)$ is treated as a function of the batch quality parameter $\psi$. The relation between the binomial and the beta distribution functions will be used to extend the definition of the binomial to non-integer values of ${\mathcal{A}}_{\mathsf{𝕔}}$ and $𝓃$ (see Hald [14]). For fixed $𝓍$, $1-F_{\psi }\left(𝓍\right)$ is a monotonically increasing function of $\psi$. It follows that the OC function is a decreasing function of $\psi$. Equation (5) shows how to compute the OC function of sampling plan. The OC function is important for deciding on a sampling plan.

For a given $\varkappa *$ and $\frac{𝓍}{{\psi }_0}$, the choice of $𝓃$ and ${\mathcal{A}}_{\mathsf{𝕔}}$ will be made based on the OC function. The OC values as a function of $\frac{\psi }{{\psi }_0}$ are presented in

Table 2. Values of OC function for the single acceptance sampling plan $\left(𝓃,{\mathcal{A}}_{\mathsf{𝕔}},\frac{𝓍}{{\psi }_0}\right)$ under a given $\varkappa *$ when ${\mathcal{A}}_{\mathsf{𝕔}}=2$.

$\mathit{\varkappa }*$	$𝓃$	$\frac{𝓍}{{\mathit{\psi }}_0}$	$\mathit{\psi }/{\mathit{\psi }}_0$
			2	3	4	5	6	7	8	9
0.75	21	0.650	0.613	0.901	0.972	0.99	0.996	0.996	0.998	0.999
	15	0.910	0.453	0.822	0.942	0.979	0.991	0.992	0.996	0.998
	15	1.000	0.371	0.772	0.92	0.97	0.987	0.991	0.996	0.998
	9	1.585	0.159	0.555	0.801	0.911	0.958	0.961	0.98	0.99
	8	2.000	0.109	0.464	0.734	0.872	0.936	0.942	0.97	0.984
	7	3.500	0.013	0.153	0.4	0.618	0.766	0.818	0.892	0.934
	6	4.000	0.010	0.129	0.356	0.573	0.729	0.741	0.838	0.897
	6	6.000	4.46 × 10−4	0.018	0.101	0.249	0.415	0.446	0.592	0.705
0.90	25	0.650	0.496	0.853	0.955	0.984	0.993	0.993	0.996	0.999
	16	0.910	0.436	0.813	0.938	0.977	0.99	0.991	0.995	0.998
	16	1.000	0.339	0.75	0.911	0.966	0.985	0.99	0.994	0.998
	10	1.585	0.205	0.609	0.833	0.927	0.966	0.973	0.98	0.994
	10	2.000	0.121	0.482	0.747	0.879	0.94	0.939	0.97	0.983
	9	3.500	0.0086	0.124	0.357	0.579	0.737	0.786	0.801	0.867
	9	4.000	2.00 × 10−4	0.017	0.11	0.281	0.465	0.62	0.735	0.816
	8	6.000	1.40 × 10−6	6.85 × 10−4	0.013	0.063	0.161	0.289	0.421	0.541
0.95	29	0.650	0.374	0.788	0.93	0.975	0.99	0.995	0.998	0.999
	19	0.910	0.305	0.729	0.903	0.962	0.984	0.993	0.996	0.998
	18	1.000	0.244	0.674	0.876	0.95	0.978	0.99	0.995	0.997
	13	1.585	0.078	0.42	0.71	0.861	0.932	0.965	0.981	0.989
	12	2.000	0.021	0.229	0.527	0.737	0.856	0.92	0.954	0.973
	10	3.500	1.26 × 10−4	0.015	0.106	0.28	0.468	0.625	0.742	0.823
	9	4.000	6.75 × 10−5	0.0094	0.078	0.227	0.405	0.565	0.69	0.781
	8	6.000	1.32 × 10−6	6.63 × 10−4	0.013	0.062	0.159	0.286	0.418	0.538
0.99	30	0.650	0.348	0.771	0.924	0.972	0.988	0.995	0.997	0.999
	20	0.910	0.281	0.710	0.894	0.959	0.982	0.992	0.996	0.998
	19	1.000	0.221	0.653	0.865	0.945	0.976	0.989	0.994	0.997
	13	1.585	0.067	0.396	0.692	0.85	0.926	0.961	0.979	0.988
	12	2.000	0.017	0.211	0.505	0.721	0.846	0.914	0.95	0.97
	11	3.500	1.03 × 10−4	0.013	0.100	0.27	0.457	0.616	0.734	0.817
	10	4.000	1.93 × 10−5	0.005	0.053	0.176	0.342	0.504	0.638	0.74
	9	6.000	3.37 × 10−8	7.85 × 10−5	0.0032	0.024	0.082	0.177	0.293	0.413

. The producer’s risk (PR) is the probability of rejecting the conforming batch when $\psi \ge {\psi }_0$ and is given by the following equation:

$$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{r}’\mathrm{s} \mathrm{R}\mathrm{i}\mathrm{s}\mathrm{k}=1-{\mathrm{P}}_1=\sum _{𝓇={\mathcal{A}}_{\mathsf{𝕔}}+1}^{𝓃}\left(\begin{array}{c}𝓃\\ 𝓇\end{array}\right){𝓅}^{𝓇}{\left(1-𝓅\right)}^{𝓃-𝓇}.$$

(6)

For a particular value of the producer’s risk $\tau$ (usually $\tau =0.05$), one may be interested in knowing what value of mean ratio $\frac{\psi }{{\psi }_0}$ will ensure the producer’s risk to be less than or equal to $\tau$. The value of $\frac{\psi }{{\psi }_0}$ is the minimum positive number >1 for which $𝓅=F_{\psi }\left(𝓍\right)$ satisfies the following inequality:

$$\sum _{𝓇=0}^{{\mathcal{A}}_{\mathsf{𝕔}}}\left(\begin{array}{c}𝓃\\ 𝓇\end{array}\right){𝓅}^{𝓇}{\left(1-𝓅\right)}^{𝓃-𝓇}\ge 0.95.$$

(7)

It is worth noting that the ratio $\frac{\psi }{{\psi }_0}$ occurs in the expression for $𝓅=F_{\psi }\left(𝓍\right)$. For instance,

$$𝓅=F_{\psi }\left(𝓍\right)=1-\mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{𝓍}{{\psi }_0}\frac{{\psi }_0}{\psi }\right)\left[1+\frac{\left(\frac{𝓍}{{\psi }_0}\frac{{\psi }_0}{\psi }\right)}{\frac{1}{{\psi }_0}+1}\mathrm{e}\mathrm{x}\mathrm{p}\left(-\frac{𝓍}{{\psi }_0}\frac{{\psi }_0}{\psi }\right)\right].$$

(8)

For a given acceptance sampling plan $\left(𝓃,{\mathcal{A}}_{\mathsf{𝕔}},\frac{𝓍}{{\psi }_0}\right)$ and $\varkappa *$, the minimum ratio of $\frac{\psi }{{\psi }_0}$, satisfying inequality (7), is computed and present in

Table 3. Minimum ratio of veritable mean life to prescribed mean life for the acceptance of a batch with the producer’s risk 0.05.

$\mathit{\varkappa }*$	${\mathcal{A}}_{\mathsf{𝕔}}$	$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$
		0.650	0.910	1.000	1.585	2.000	3.500	4.000	6.000
0.75	0	6.555	6.979	7.118	7.948	8.482	14.442	15.875	29.081
	1	4.563	5.384	5.687	6.154	6.497	10.282	12.785	23.305
	2	3.529	4.144	4.47	5.762	6.374	9.266	10.099	21.262
	3	2.733	3.732	3.819	5.214	5.99	8.665	9.141	16.08
	4	2.5	3.148	2.943	4.316	5.374	8.534	9.024	13.167
	5	2.21	2.721	2.777	4.048	4.71	7.663	8.741	12.519
	6	2.038	2.398	2.635	3.896	4.452	7.199	7.814	12.028
	7	1.921	2.139	2.349	3.497	4.219	6.952	7.689	11.002
	8	1.836	2.040	2.231	3.408	4.128	6.763	7.135	9.809
	9	1.771	1.950	2.042	3.152	3.927	6.479	6.635	8.936
	10	1.719	1.760	1.872	2.917	3.599	6.024	6.246	8.221
0.90	0	8.564	9.149	9.341	10.496	11.242	16.964	19.387	29.081
	1	4.673	5.618	7.884	8.468	9.370	12.16	14.682	25.756
	2	3.906	4.202	4.594	5.48	7.095	11.582	12.733	17.866
	3	3.049	3.883	4.435	5.236	6.524	9.065	10.822	16.205
	4	2.581	3.251	2.997	4.393	5.872	8.981	9.022	13.642
	5	2.400	2.751	2.817	4.22	5.155	8.102	8.429	12.778
	6	2.063	2.452	2.665	4.176	4.962	7.641	8.184	12.178
	7	1.936	2.143	2.354	3.753	4.736	7.440	8.036	11.096
	8	1.862	2.049	2.246	3.568	4.528	7.027	7.519	9.809
	9	1.898	1.960	2.057	3.456	4.308	6.614	7.007	8.936
	10	1.838	1.932	1.874	3.194	3.945	6.112	6.61	8.221
0.95	0	9.828	10.514	10.74	12.099	12.979	16.964	19.387	29.081
	1	5.161	6.195	7.101	8.248	9.958	14.298	16.136	19.287
	2	4.314	4.689	4.997	6.468	7.842	12.536	13.41	17.901
	3	3.243	4.115	4.582	5.533	6.694	10.572	12.01	16.247
	4	2.682	3.394	3.573	4.598	6.031	9.126	11.357	14.157
	5	2.46	2.929	3.025	4.416	5.503	8.244	8.599	13.352
	6	2.171	2.608	2.752	4.34	5.123	7.784	8.304	12.228
	7	2.064	2.322	2.449	3.884	4.699	7.398	7.931	11.779
	8	1.98	2.25	2.34	3.668	4.655	7.284	7.643	10.729
	9	1.939	2.267	2.282	3.553	4.429	6.809	7.128	9.658
	10	1.877	2.226	2.251	3.282	4.055	6.345	6.727	9.204
0.99	0	12.292	13.175	13.466	15.224	16.367	20.056	21.167	29.081
	1	5.387	6.453	6.998	7.533	8.42	16.596	17.208	24.873
	2	4.407	4.791	5.106	6.601	7.989	12.653	14.15	19.977
	3	3.491	4.178	5.163	5.651	6.827	11.05	12.136	16.28
	4	2.868	3.466	4.148	4.696	6.155	9.24	10.652	15.241
	5	2.621	2.949	3.361	4.553	5.619	8.356	9.096	14.104
	6	2.444	2.638	2.809	4.434	5.249	8.095	8.399	12.268
	7	2.415	2.441	2.601	4.049	5.122	7.956	8.354	12.22
	8	2.206	2.42	2.529	3.746	4.755	7.622	7.742	11.823
	9	2.121	2.409	2.489	3.629	4.523	7.036	7.223	11.488
	10	2.050	2.354	2.455	3.351	4.142	6.574	6.82	11.410

.

Table 2 shows that there is a direct relationship between quality level $\psi /{\psi }_0$ and probabilities of acceptance: the greater the quality level, $\psi /{\psi }_0$, the greater the probabilities of acceptance while $𝓍/{\psi }_0$ and ${\mathcal{A}}_{\mathsf{𝕔}}$ are fixed. There is an inverse relationship between probabilities of acceptance and the confidence level $\varkappa *$: the higher the $\varkappa *$, the lower the probabilities of acceptance while the $\psi /{\psi }_0$ is fixed. Additionally, the probabilities of acceptance decrease as the $𝓍/{\psi }_0$ increases while the ${\mathcal{A}}_{\mathsf{𝕔}}$ and $\psi /{\psi }_0$ are fixed.

Table 3 indicates that when the experiment time ratio increases, so does the minimum mean ratio. Furthermore, when the confidence level rises, so does the minimum mean ratio. The minimum mean ratio and examination limit have an inverse connection, while the experiment time ratio is constant. As a result, lowering the examination limit from ${\mathcal{A}}_{\mathsf{𝕔}}=2$ to ${\mathcal{A}}_{\mathsf{𝕔}}=0$ has negative effects on the producer. As a result, we discover that any reduction in quality, no matter how minor, reduces the probability of acceptance.

4. The ML Distribution for the Double Acceptance Sampling Plans

A double acceptance sampling (DAS) plan is a type of acceptance sampling plan that involves two stages of sampling and decision-making. DAS plans are designed to reduce the risk of accepting a batch with low quality or rejecting a batch that meets the required standards. The first stage of a DAS plan involves taking a smaller sample from the batch and making an initial decision based on the quality of the sample. If the initial decision to accept the batch is inconclusive, it proceeds to the second stage, in which another larger sample is taken, and a final decision is made based on the combined results of both samples. If the initial decision is to reject the batch, it is rejected without further testing.

The importance of DAS plans lies in their ability to provide a more efficient and cost-effective means of ensuring product quality than single-stage acceptance sampling plans. By using two stages of sampling and decision-making, DAS plans can reduce the risk of accepting a batch with low quality or rejecting a batch that meets the required standards, which can result in unnecessary costs and delays. The DAS plans are commonly used in industries that require high levels of quality assurance, such as pharmaceuticals, food and beverage, and electronics. DAS plans can be used to ensure that raw materials meet the required quality standards before they are used in the manufacturing processes DAS plans can be used to inspect finished products to ensure that they meet the required quality standards before they are shipped to customers. DAS plans can be used to monitor the quality of a manufacturing process and adjust as needed to ensure that the final products meet the required quality standards. DAS plans can be used to evaluate the quality of products received from suppliers and ensure that they meet the required standards before they are used in the manufacturing process. DAS plans are an important tool for managing quality and reducing risk in manufacturing and service industries.

They provide a cost-effective and efficient means of ensuring product quality, and their use is widespread in many industries around the world. When comparing the number of studies in a double sampling plan to the number of studies in a single sampling plan, the number of studies in the double sampling plan is lower. (see, for more details, Aslam et al. [2], Aslam and Jun [15], Aslam et al. [16], Gui and Xu [17], Muthulakshmi and Selvi [18], and Fallahnezhad et al. [19]). The implementation of a double inspection plan requires that a first sample of size ${𝓃}_1$ is drawn at random from the batch. Let ${𝓇}_1$ and ${𝓇}_2$ denote the number of nonconforming units in the first and second samples, respectively. The number of nonconforming is determined and contrasted to the examination limit ${\mathcal{A}}_{{\mathsf{𝕔}}_1}$ and the rejection limit ${\mathcal{A}}_{{𝓇}_1}$ for the first sample. The process of the double plan is presented in Figure 1.

It is noted from Figure 1 that the first sample is taken from a large batch and put under test for ${𝓍}_0$ units of time. When ${\mathcal{A}}_{{\mathsf{𝕔}}_1}={\mathcal{A}}_{{\mathsf{𝕔}}_2}$, the double sampling becomes the single sampling. As mentioned earlier, a product is considered to be good and accepted for consumer use if the sample information supports the hypothesis $H_0=\psi \ge {\psi }_0$; if the sample does not support this hypothesis, a batch of products are rejected. As the significance level for the test, the consumer’s risk is assessed via $1-\varkappa *$, where $\varkappa *$ is the consumer’s confidence level. The double sampling strategy is developed and tested using the parameters $\left({𝓃}_1,{𝓃}_2,{\mathcal{A}}_{{\mathsf{𝕔}}_1},{\mathcal{A}}_{{\mathsf{𝕔}}_2},\frac{𝓍}{{\psi }_0}\right)$, where $\left({\mathcal{A}}_{{\mathsf{𝕔}}_1}<{\mathcal{A}}_{{\mathsf{𝕔}}_2}\right)$. Then, the probability of accepting a batch under the double sampling plan can be obtained from

$${\mathrm{P}}_2={\mathrm{P}}_{{𝓃}_1,𝓅}+{\mathrm{P}}_{{𝓃}_2,𝓅},$$

which can be written as

$${\mathrm{P}}_2=\sum _{{𝓇}_1=0}^{{\mathcal{A}}_{c_1}}\left(\begin{array}{c}{𝓃}_1\\ {𝓇}_1\end{array}\right){𝓅}^{{𝓇}_1}{\left(1-𝓅\right)}^{{𝓃}_1-{𝓇}_1}+\sum _{{𝓇}_1=a_{c_1}+1}^{{\mathcal{A}}_{c_2}}\left(\begin{array}{c}{𝓃}_1\\ {𝓇}_1\end{array}\right){𝓅}^{{𝓇}_1}{\left(1-𝓅\right)}^{{𝓃}_1-{𝓇}_1}\left[\sum _{{𝓇}_2=0}^{{\mathcal{A}}_{c_2-{𝓇}_1}}\left(\begin{array}{c}{𝓃}_2\\ {𝓇}_2\end{array}\right){𝓅}^{{𝓇}_2}{\left(1-𝓅\right)}^{{𝓃}_2-{𝓇}_2}\right],$$

(9)

where $𝓅=F_{\psi }\left(𝓍\right)$ reflects the probability that an item will be rejected before the termination time ${𝓍}_0$, which is given in (2). For a product of quality $𝓅$, the acceptance probability of the first sample is ${\mathrm{P}}_{{𝓃}_1,𝓅}$ and the second set of (9) represents the acceptance probability of the second sample. Particularly, we are considering a life testing experiment for a double plan having the measurements of ${\mathcal{A}}_{{\mathsf{𝕔}}_1}$ = 0 and ${\mathcal{A}}_{{\mathsf{𝕔}}_2}$ = 2. According to these measurements, the batch acceptance probability of (9) reduces to

$${\mathrm{P}}_2={\left(1-𝓅\right)}^{{𝓃}_1}+{𝓃}_1{𝓅}^2{\left(1-𝓅\right)}^{{𝓃}_1+{𝓃}_2-2}\left[\frac{\left(1-𝓅\right)}{𝓅}+{𝓃}_2+\frac{\left({𝓃}_1-1\right)}{2}\right].$$

(10)

Therefore, the optimal sample sizes ${𝓃}_1$ and ${𝓃}_2$, ensuring $\psi \ge {\psi }_0$ at the consumer’s risk, $1-\varkappa *$, can be found as the solution to the following inequality:

$${\mathrm{P}}_2\le 1-\varkappa *.$$

(11)

There is more than one solution that may satisfy this inequality (11), so Aslam and Jun (2010) proposed to minimize the ASN to find ${𝓃}_1$ and ${𝓃}_2$ by putting the constraint as ${𝓃}_1\ge {𝓃}_2$. The ASN is defined as the average sample number of items inspected for a series of batches with a given incoming batch quality in order to make a decision. Notice that the average sample number in the single sampling plan with parameter $𝓃, {\mathcal{A}}_{\mathsf{𝕔}}$ is $ASN=𝓃$. This is because in each incoming batch a sample with a size of $𝓃$ is taken. That is why $𝓃$ is constant from one batch to another. For a double sampling plan, the ASN is given by the following equation:

$$ASN={𝓃}_1+{𝓃}_2\left(1-\psi \right),$$

(12)

where $\psi$ is the probability of making a decision on the first sample. The probability $\psi$ can be expressed as

$$\psi =1-\sum _{{𝓇}_1={\mathcal{A}}_{\mathsf{𝕔}1}+1}^{{\mathcal{A}}_{\mathsf{𝕔}2}}\left(\begin{array}{c}{𝓃}_1\\ {𝓇}_1\end{array}\right){𝓅}^{{𝓇}_1}{\left(1-𝓅\right)}^{{𝓃}_1-{𝓇}_1},$$

(13)

For ${\mathcal{A}}_{{\mathsf{𝕔}}_1}$ = 0 and ${\mathcal{A}}_{{\mathsf{𝕔}}_2}$ = 2, the ASN under the double plan is given as

$$ASN={𝓃}_1+{𝓃}_1{𝓃}_2𝓅{\left(1-𝓅\right)}^{{𝓃}_1-1}\left[1+\frac{𝓅\left({𝓃}_1-1\right)}{2\left(1-𝓅\right)}\right].$$

Many authors have followed the criterion of minimizing the ASN (see Balamurali et al. [20] and Jun et al. [21]). Therefore, the optimum sampling sizes for ${\mathcal{A}}_{{\mathsf{𝕔}}_1}$ = 0 and ${\mathcal{A}}_{{\mathsf{𝕔}}_2}$ = 2 in the double plan can be obtained by solving the following optimization problem:

Minimize

$$ASN={𝓃}_1+{𝓃}_1{𝓃}_2𝓅{\left(1-𝓅\right)}^{{𝓃}_1-1}\left[1+\frac{𝓅\left({𝓃}_1-1\right)}{2\left(1-𝓅\right)}\right].$$

(14)

Subject it to

$${\mathrm{P}}_2\le 1-\varkappa *$$

(15)

$$1\le {𝓃}_2\le {𝓃}_1$$

(16)

and

$${𝓃}_1 \mathrm{and} {𝓃}_2 \mathrm{positive} \mathrm{integers}.$$

The sample size that must be used for the first and second samples to satisfy the inequality (15) can be determined by the search procedure by changing the initial values of ${𝓃}_1$ and ${𝓃}_2$. The optimal value of sample sizes ${𝓃}_1$, ${𝓃}_2$ and the ASN for

$$\varkappa *=0.75, 0.90, 0.95 \mathrm{and} 0.99,$$

and

$$\frac{𝓍}{{\psi }_0}=0.650, 0.910, 1.000, 1.585, 2.000, 3.500, 4.000 \mathrm{and} \mathrm{6.000}$$

were determined and are presented in

Table 4. Minimum sample sizes and ASN for ML distribution in double sampling plan.

$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$	${\mathit{\varkappa }}^{\mathit{*}}$
	0.75			0.90			0.95			0.99
	${𝓃}_1$	${𝓃}_2$	ASN	${𝓃}_1$	${𝓃}_2$	ASN	${𝓃}_1$	${𝓃}_2$	ASN	${𝓃}_1$	${𝓃}_2$	ASN
0.650	11	9	13.8	13	13	15.3	23	5	23.2	24	5	23.7
0.910	8	8	9.5	8	8	9.7	12	6	11.9	12	5	12.2
1.000	7	6	8.6	7	7	8.5	10	5	10.3	10	5	10.5
1.585	6	4	6.6	6	3	6.0	6	3	6.5	7	3	6.6
2.000	6	3	5.8	6	2	5.8	6	2	6.1	6	2	6.2
3.500	5	1	4.7	5	1	5.0	6	1	5.6	6	1	5.6
4.000	5	1	4.7	5	1	4.6	6	1	5.5	6	1	5.6
6.000	4	1	4.1	5	1	4.6	5	1	4.9	5	1	5.0

.

From Table 4, it is seen that the sample sizes required and the ASN, respectively, increase as the confidence level increases. Additionally, the sample sizes and the ASN decrease as the termination time ratio increases. After obtaining the sample sizes ${𝓃}_1$ and ${𝓃}_2$, one may be concerned to find the probability of accepting the batch when the quality of an item is conforming enough. As mentioned previously, an item is considered to be conforming if the veritable mean life to the prescribed mean life is >1. The OC values according to (10) for different values of $\psi /{\psi }_0$, consumer’s confidence level, $\varkappa *$, and for a given sampling plan $\left({𝓃}_1,{𝓃}_2,{\mathcal{A}}_{{\mathsf{𝕔}}_1},{\mathcal{A}}_{{\mathsf{𝕔}}_2},\frac{𝓍}{{\psi }_0}\right)$ are shown in

Table 5. Values of the OC under the double acceptance sampling plan $\left({𝓃}_1,{𝓃}_2,{\mathcal{A}}_{{\mathsf{𝕔}}_1},{\mathcal{A}}_{{\mathsf{𝕔}}_2},\frac{𝓍}{{\psi }_0}\right)$ and for a given $\varkappa *$ when ${\mathcal{A}}_{{\mathsf{𝕔}}_1}$ = 0 and ${\mathcal{A}}_{{\mathsf{𝕔}}_2}$ = 2.

$\mathit{\varkappa }*$	$\frac{𝓍}{{\mathit{\psi }}_0}$	$\mathit{\psi }/{\mathit{\psi }}_0$
		2	3	4	5	6	7	8	9
0.75	0.650	0.635	0.908	0.974	0.991	0.996	0.998	0.999	1
	0.910	0.489	0.839	0.948	0.981	0.992	0.996	0.998	0.999
	1.000	0.453	0.818	0.939	0.977	0.991	0.994	0.997	0.998
	1.585	0.174	0.573	0.812	0.916	0.961	0.979	0.989	0.994
	2.000	0.128	0.493	0.755	0.883	0.942	0.966	0.982	0.989
	3.500	0.028	0.222	0.488	0.691	0.818	0.857	0.911	0.944
	4.000	0.012	0.14	0.374	0.589	0.741	0.829	0.891	0.93
	6.000	7.25 × 10−4	0.024	0.12	0.277	0.446	0.563	0.681	0.769
0.90	0.650	0.509	0.856	0.956	0.984	0.994	0.994	0.997	0.999
	0.910	0.456	0.822	0.941	0.978	0.991	0.991	0.996	0.998
	1.000	0.448	0.814	0.938	0.977	0.99	0.99	0.995	0.998
	1.585	0.287	0.688	0.875	0.947	0.976	0.976	0.988	0.994
	2.000	0.197	0.585	0.813	0.915	0.959	0.959	0.979	0.989
	3.500	0.018	0.179	0.435	0.648	0.788	0.788	0.872	0.921
	4.000	0.011	0.130	0.359	0.575	0.73	0.73	0.83	0.892
	6.000	3.45 × 10−4	0.016	0.093	0.235	0.399	0.399	0.548	0.668
0.95	0.650	0.395	0.800	0.935	0.977	0.99	0.997	0.999	0.999
	0.910	0.369	0.773	0.922	0.97	0.988	0.996	0.998	0.999
	1.000	0.363	0.766	0.918	0.968	0.987	0.996	0.996	0.997
	1.585	0.267	0.670	0.866	0.943	0.974	0.974	0.987	0.993
	2.000	0.152	0.528	0.778	0.896	0.949	0.949	0.974	0.986
	3.500	0.011	0.139	0.379	0.600	0.752	0.752	0.847	0.905
	4.000	0.0034	0.073	0.259	0.474	0.649	0.649	0.77	0.850
	6.000	1.52 × 10−4	0.0099	0.07	0.195	0.352	0.352	0.502	0.628
0.99	0.650	0.38	0.792	0.932	0.975	0.99	0.997	0.998	0.999
	0.910	0.366	0.771	0.921	0.97	0.987	0.993	0.997	0.998
	1.000	0.35	0.757	0.914	0.967	0.986	0.991	0.995	0.997
	1.585	0.224	0.629	0.844	0.933	0.969	0.963	0.985	0.992
	2.000	0.137	0.508	0.765	0.889	0.945	0.925	0.972	0.984
	3.500	0.0087	0.125	0.359	0.581	0.738	0.636	0.837	0.898
	4.000	0.0028	0.066	0.245	0.459	0.636	0.524	0.76	0.843
	6.000	1.04 × 10−4	0.0081	0.061	0.179	0.331	0.339	0.482	0.609

.

From Table 5, it is observed that the OC values increase as the quantity $\psi /{\psi }_0$ increases. Additionally, the OC values decrease as the confidence level increases. For the sample plan in question and a percent value for the producer’s risk, say, $\tau$, one could be concerned that the value of $\psi /{\psi }_0$ that will assure the producer’s risk is at most $\tau$. The smallest ratio of $\psi /{\psi }_0$ can be determined by solving the following the inequality for the producer’s risk of $\tau$, where

$${\mathrm{P}}_2\ge 1-\tau ,$$

(17)

Here, ${\mathrm{P}}_2$ is defined in (10) and $𝓅=F_{\psi }\left(𝓍\right)$ is given in (2). For a given acceptance sampling plan $\left({𝓃}_1,{𝓃}_2,{\mathcal{A}}_{{\mathsf{𝕔}}_1},{\mathcal{A}}_{{\mathsf{𝕔}}_2},\frac{𝓍}{{\psi }_0}\right)$ and $\varkappa *$, the minimum ratio of $\psi /{\psi }_0$, satisfying inequality (17), are computed and present in

Table 6. The minimum ratio of the veritable mean life to prescribed mean life for the acceptance of a batch under producer’s risk of 0.05.

$\mathit{\varkappa }*$	$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$
	0.650	0.910	1.000	1.585	2.000	3.500	4.000	6.000
0.75	3.467	4.040	4.187	5.673	6.211	8.575	9.658	13.585
0.90	3.888	4.156	4.216	5.061	5.718	8.985	9.792	14.168
0.95	4.240	4.447	4.504	5.154	6.021	9.436	10.736	14.779
0.99	4.292	4.458	4.557	5.376	6.133	9.61	10.88	15.053

.

5. The ML Distribution for Multiple Acceptance Sampling Plans

Following Wilson and Burgess [22], the multiple acceptance sampling (MAS) plan is a type of acceptance sampling plan that involves taking multiple samples from a batch and making acceptance decisions based on the combined results of all samples. MAS plans are designed to reduce the risk of accepting a batch with low quality or rejecting a batch that meets the required standards. The importance of MAS plans lies in their ability to provide a more comprehensive and reliable means of ensuring product quality than single-stage acceptance sampling plans. By taking multiple samples and making acceptance decisions based on the combined results of all samples, MAS plans can provide a more accurate assessment of the quality of the batch and reduce the risk of making incorrect acceptance decisions. MAS plans are commonly used in industries that require high levels of quality assurance, such as pharmaceuticals, food and beverage, and electronics. The MAS plans can be used to ensure that raw materials meet the required quality standards before they are used in the manufacturing process. They can be used to inspect finished products to ensure that they meet the required quality standards before they are shipped to customers. Additionally, they can be used to monitor the quality of a manufacturing process and make adjustments as needed to ensure that the final products meet the required quality standards. MAS plans are an important tool for managing quality and reducing risk in manufacturing and service industries. They provide a more comprehensive and reliable means of ensuring product quality, and their use is widespread in many industries around the world.

However, despite the fact that they achieve the minimum average sample number, they are complex systems that necessitate significant administrative effort and competent inspectors. The multiple sample plan is defined by $\left(𝓃,{\mathcal{A}}_{\mathsf{𝕔}},{\mathcal{A}}_{𝓇}\right)$, where $𝓃=\left({𝓃}_1,\dots ,{𝓃}_k\right)$ is a sequence of sample size, and ${𝓃}_1,\dots ,{𝓃}_k\ge 1$, ${\mathcal{A}}_{\mathsf{𝕔}}=\left({\mathcal{A}}_{{\mathsf{𝕔}}_1},{\mathcal{A}}_{{\mathsf{𝕔}}_2},\dots {,\mathcal{A}}_{{\mathsf{𝕔}}_k}\right)$ is a sequence of integer acceptance levels.

$${\mathcal{A}}_{{\mathsf{𝕔}}_k}\ge {\mathcal{A}}_{{\mathsf{𝕔}}_{k-1}}\ge \dots \ge {\mathcal{A}}_{{\mathsf{𝕔}}_1}\ge 0, {\mathcal{A}}_{𝓇}=\left({\mathcal{A}}_{{𝓇}_1},\dots {\mathcal{A}}_{{𝓇}_k}\right)$$

is a sequence of integer rejection levels in which

$${\mathcal{A}}_{{𝓇}_k}\ge {\mathcal{A}}_{{𝓇}_{k-1}}\ge \dots \ge {\mathcal{A}}_{{𝓇}_1}\ge 1.$$

The $k$ parameters are related to the $k$ stages of MAS plans. Irrespective of the fact that MAS plans are known to be significantly more efficient in terms of the average sample number than single and double plans, research studies on MAS plans in the field of reliability testing are rare or nonexistent. In addition, no attempt has been taken to develop MAS plans for ML distribution. As a result, this section introduces multiple acceptance sample plans for the amputated life test at a pre-determined period, assuming that a product’s lifetime follows the ML distribution. To begin the algorithm, a sample of size ${𝓃}_1$ is taken at random from a batch, and the number of faulty items ${𝓇}_1$ in the sample is calculated and used as follows:

• If ${𝓇}_1\le {\mathcal{A}}_{{\mathsf{𝕔}}_1}$, the batch is accepted;
• If ${𝓇}_1\ge {\mathcal{A}}_{{\mathsf{𝕔}}_1}$, the batch is rejected;
• If ${\mathcal{A}}_{{\mathsf{𝕔}}_1}{<𝓇}_1<{\mathcal{A}}_{{𝓇}_1}$, another sample is drawn.

If further samples are required, the first sample technique is performed sample by sample. For each sample, the cumulative number of faulty units discovered at any stage (say, $j$th), $R_j=\sum _{i=1}^j{𝓇}_i$, is compared to the examination limit ${\mathcal{A}}_{{\mathsf{𝕔}}_k}$ and the rejection limit ${\mathcal{A}}_{{𝓇}_k}$ for that stage until a decision is made. Because ${\mathcal{A}}_{{𝓇}_k}={\mathcal{A}}_{{\mathsf{𝕔}}_k}+1$ for the last sample, a decision must be taken by the $k$th sample.

When ${\mathcal{A}}_{{\mathsf{𝕔}}_1}={\mathcal{A}}_{{\mathsf{𝕔}}_2}=\dots ={\mathcal{A}}_{{\mathsf{𝕔}}_k}$, the single accepting sampling plan is a special case from MAS plans. The proposed four-parameter multiple plan is established based on an amputate life test and determined with parameters $\left({𝓃}_k,{\mathcal{A}}_{{\mathsf{𝕔}}_k}.{\mathcal{A}}_{{𝓇}_k},\frac{𝓍}{{\psi }_0}\right)$ where ${\mathcal{A}}_{{\mathsf{𝕔}}_1}<{\mathcal{A}}_{{\mathsf{𝕔}}_2}<\dots <{\mathcal{A}}_{{\mathsf{𝕔}}_k}$ and ${\mathcal{A}}_{{\mathsf{𝕔}}_k}<{\mathcal{A}}_{{𝓇}_k}$. According to the operation of multiple acceptance sampling plans, the probability of accepting an inspection is given by the following equation:

$$\begin{array}{cc}\hfill {\mathrm{P}}_{i,i=\mathrm{1,2},\dots ,k}& =B\left({\mathcal{A}}_{{\mathsf{𝕔}}_1}|{𝓃}_1,𝓅\right)+{\displaystyle \sum _{{𝓇}_1={\mathcal{A}}_{{\mathsf{𝕔}}_1}+1}^{{\mathcal{A}}_{{𝓇}_1}-1}}b\left({𝓇}_1|{𝓃}_1,𝓅\right)B\left({\mathcal{A}}_{{\mathsf{𝕔}}_2}|{𝓃}_2,𝓅\right)\hfill \\ & +{\displaystyle \sum _{{𝓇}_1={\mathcal{A}}_{{\mathsf{𝕔}}_1}+1}^{{\mathcal{A}}_{{𝓇}_1}-1}}b\left({𝓇}_1|{𝓃}_1,𝓅\right){\displaystyle \sum _{{𝓇}_2={\mathcal{A}}_{{\mathsf{𝕔}}_2}-{𝓇}_1+1}^{{\mathcal{A}}_{{𝓇}_2}-1}}b\left({𝓇}_2|{𝓃}_2,𝓅\right)B\left({\mathcal{A}}_{{\mathsf{𝕔}}_3-R_1}|{𝓃}_3,𝓅\right)\hfill \\ & +{\displaystyle \sum _{{𝓇}_1={\mathcal{A}}_{{\mathsf{𝕔}}_1}+1}^{{\mathcal{A}}_{{𝓇}_1}-1}}b\left({𝓇}_1|{𝓃}_1,𝓅\right){\displaystyle \sum _{{𝓇}_2={\mathcal{A}}_{{\mathsf{𝕔}}_2}-{𝓇}_1+1}^{{\mathcal{A}}_{{𝓇}_2}-1}}b\left({𝓇}_2|{𝓃}_2,𝓅\right){\displaystyle \sum _{{𝓇}_3={\mathcal{A}}_{{\mathsf{𝕔}}_3-R_1}+1}^{{\mathcal{A}}_{{𝓇}_3}-1}}\left[\begin{array}{c}b\left({𝓇}_3|{𝓃}_3,𝓅\right)\\ \times B\left({\mathcal{A}}_{{\mathsf{𝕔}}_4-R_2}|{𝓃}_3,𝓅\right)\end{array}\right]\hfill \\ & ++\left\{\begin{array}{c}{\displaystyle \sum _{{𝓇}_1={\mathcal{A}}_{{\mathsf{𝕔}}_1}+1}^{{\mathcal{A}}_{{𝓇}_1}-1}}b\left({𝓇}_1|{𝓃}_1,𝓅\right){\displaystyle \sum _{{𝓇}_2={\mathcal{A}}_{{\mathsf{𝕔}}_2}-{𝓇}_1+1}^{{\mathcal{A}}_{{𝓇}_2}-1}}b\left({𝓇}_2|{𝓃}_2,𝓅\right){\displaystyle \sum _{{𝓇}_3={\mathcal{A}}_{{\mathsf{𝕔}}_3-R_1}+1}^{{\mathcal{A}}_{{𝓇}_3}-1}}b\left({𝓇}_3|{𝓃}_3,𝓅\right)\dots \\ {\displaystyle \sum _{{𝓇}_k={\mathcal{A}}_{{\mathsf{𝕔}}_k}-R_{k-1}+1}^{{\mathcal{A}}_{𝓇k}-1}}\left[\begin{array}{c}b\left({𝓇}_k|{𝓃}_k,𝓅\right)\\ \times B\left({\mathcal{A}}_{{\mathsf{𝕔}}_{m-1}}-R_{m-2}|{𝓃}_{m-1},𝓅\right)\end{array}\right]\end{array}\right\}\hfill \\ & +\left\{\begin{array}{c}{\displaystyle \sum _{{𝓇}_1={\mathcal{A}}_{{\mathsf{𝕔}}_1}+1}^{{\mathcal{A}}_{{𝓇}_1}-1}}b\left({𝓇}_1|{𝓃}_1,𝓅\right){\displaystyle \sum _{{𝓇}_2={\mathcal{A}}_{{\mathsf{𝕔}}_2}-{𝓇}_1+1}^{{\mathcal{A}}_{{𝓇}_2}-1}}b\left({𝓇}_2|{𝓃}_2,𝓅\right){\displaystyle \sum _{{𝓇}_3={\mathcal{A}}_{{\mathsf{𝕔}}_3-R_1}+1}^{{\mathcal{A}}_{{𝓇}_3}-1}}b\left({𝓇}_3|{𝓃}_3,𝓅\right)\dots \\ {\displaystyle \sum _{{𝓇}_k={\mathcal{A}}_{{\mathsf{𝕔}}_k}-R_{k-1}+1}^{{\mathcal{A}}_{{𝓇}_k}-1}}b\left({𝓇}_k|{𝓃}_k,𝓅\right){\displaystyle \sum _{{𝓇}_{m-1}={\mathcal{A}}_{{\mathsf{𝕔}}_{m-1}}-R_{m-2}+1}^{{\mathcal{A}}_{{𝓇}_{m-1}}-1}}\left[\begin{array}{c}b\left({𝓇}_{m-1}|{𝓃}_{m-1},𝓅\right)\\ \times B\left({\mathcal{A}}_{{\mathsf{𝕔}}_m}-R_{m-1}|{𝓃}_m,𝓅\right)\end{array}\right]\end{array}\right\},\hfill \end{array}$$

(18)

where $R_k=\sum _{i=1}^k{𝓇}_i$ is defined as the cumulative number of faulty units, and cumulative probabilities with negative arguments are taken to be zero. If ${\mathcal{A}}_{{\mathsf{𝕔}}_k}=-1$, that means acceptance is not allowed at the $k$ stage (Schilling and Nubauer (2009)). When acceptance is not allowed, the symbol ₡ is used for the examination limit. To illustrate how multiple inspection sampling plans perform, an example in which the third and fourth samples are plotted is given here. Multiple inspection sampling plans become more complex with the third sample. When this complexity is explained, the way this system works becomes evident. When the sampling plan is determined with parameters $\left({𝓃}_k,{\mathcal{A}}_{{\mathsf{𝕔}}_k}.{\mathcal{A}}_{{𝓇}_k},\frac{𝓍}{{\psi }_0}\right)$, where $k=$ 1, …, 4, the probabilities of acceptance are given as follows:

$$\begin{array}{c}{\mathrm{P}}_3={\mathrm{P}}_{{𝓃}_1,𝓅}+{\mathrm{P}}_{{𝓃}_1,{𝓃}_2,𝓅}+{\mathrm{P}}_{{𝓃}_1,{𝓃}_2,{𝓃}_3,𝓅},\\ {\mathrm{P}}_4={\mathrm{P}}_{{𝓃}_1,𝓅}+{\mathrm{P}}_{{𝓃}_1,{𝓃}_2,𝓅}+{\mathrm{P}}_{{𝓃}_1,{𝓃}_2,{𝓃}_3,𝓅}{+\mathrm{P}}_{{𝓃}_1,{𝓃}_2,{𝓃}_3,{𝓃}_4,𝓅}.\end{array}$$

(19)

The probabilities of ${\mathrm{P}}_{{𝓃}_1,𝓅}$, ${\mathrm{P}}_{{𝓃}_1,{𝓃}_2,𝓅}$, ${\mathrm{P}}_{{𝓃}_1,{𝓃}_2,{𝓃}_3,𝓅}$, and ${\mathrm{P}}_{{𝓃}_1,{𝓃}_2,{𝓃}_3{,𝓃}_4,𝓅}$ can be expressed as follows:

$$\begin{array}{c}{\mathrm{P}}_{{𝓃}_1,𝓅}={\sum }_{{𝓇}_1=0}^{{\mathcal{A}}_{{\mathsf{𝕔}}_1}}\left(\begin{array}{c}{𝓃}_1\\ {𝓇}_1\end{array}\right){𝓅}^{{𝓇}_1}{\left(1-𝓅\right)}^{{𝓃}_1-{𝓇}_1};\\ {\mathrm{P}}_{{𝓃}_1,{𝓃}_2,𝓅}={\sum }_{{𝓇}_1={\mathcal{A}}_{{\mathsf{𝕔}}_1}+1}^{{\mathcal{A}}_{{𝓇}_1}-1}\left(\begin{array}{c}{𝓃}_1\\ {𝓇}_1\end{array}\right){𝓅}^{{𝓇}_1}{\left(1-𝓅\right)}^{{𝓃}_1-{𝓇}_1}\left[{\sum }_{{𝓇}_2=0}^{{\mathcal{A}}_{{\mathsf{𝕔}}_2}-{𝓇}_1}\left(\begin{array}{c}{𝓃}_2\\ {𝓇}_2\end{array}\right){𝓅}^{{𝓇}_2}{\left(1-𝓅\right)}^{{𝓃}_2-{𝓇}_2}\right];\\ {\mathrm{P}}_{{𝓃}_1,{𝓃}_2,{𝓃}_3,𝓅}={\displaystyle \sum _{{𝓇}_1={\mathcal{A}}_{{\mathsf{𝕔}}_1}+1}^{{\mathcal{A}}_{{𝓇}_1}-1}}\left(\begin{array}{c}{𝓃}_1\\ {𝓇}_1\end{array}\right){𝓅}^{{𝓇}_1}{\left(1-𝓅\right)}^{{𝓃}_1-{𝓇}_1}\times {\displaystyle \sum _{{𝓇}_2={\mathcal{A}}_{{\mathsf{𝕔}}_2}-{𝓇}_1+1}^{{\mathcal{A}}_{{𝓇}_2}-1}}\left(\begin{array}{c}{𝓃}_2\\ {𝓇}_2\end{array}\right){𝓅}^{{𝓇}_2}{\left(1-𝓅\right)}^{{𝓃}_2-{𝓇}_2}\left[{\displaystyle \sum _{{𝓇}_3=0}^{{\mathcal{A}}_{{\mathsf{𝕔}}_3}-R_1}}\left(\begin{array}{c}{𝓃}_3\\ {𝓇}_3\end{array}\right){𝓅}^{{𝓇}_3}{\left(1-𝓅\right)}^{{𝓃}_3-{𝓇}_3}\right];\end{array}$$

and

$$\begin{array}{c}{\mathrm{P}}_{{𝓃}_1,{𝓃}_2,{𝓃}_3{,𝓃}_4,𝓅}={\sum }_{{𝓇}_1={\mathcal{A}}_{{\mathsf{𝕔}}_1}+1}^{{\mathcal{A}}_{{𝓇}_1}-1}\left(\begin{array}{c}{𝓃}_1\\ {𝓇}_1\end{array}\right){𝓅}^{{𝓇}_1}{\left(1-𝓅\right)}^{{𝓃}_1-{𝓇}_1}\times {\sum }_{{𝓇}_2={\mathcal{A}}_{{\mathsf{𝕔}}_2}-{𝓇}_1+1}^{{\mathcal{A}}_{{𝓇}_2}-1}\left(\begin{array}{c}{𝓃}_2\\ {𝓇}_2\end{array}\right){𝓅}^{{𝓇}_2}{\left(1-𝓅\right)}^{{𝓃}_2-{𝓇}_2}\\ \times {\sum }_{{𝓇}_3={\mathcal{A}}_{{\mathsf{𝕔}}_3}-R_1+1}^{{\mathcal{A}}_{{𝓇}_3}-1}\left(\begin{array}{c}{𝓃}_3\\ {𝓇}_3\end{array}\right){𝓅}^{{𝓇}_3}{\left(1-𝓅\right)}^{{𝓃}_3-{𝓇}_3}\left[{\sum }_{{𝓇}_4=0}^{{\mathcal{A}}_{{\mathsf{𝕔}}_3}-R_2}\left(\begin{array}{c}{𝓃}_4\\ {𝓇}_4\end{array}\right){𝓅}^{{𝓇}_4}{\left(1-𝓅\right)}^{{𝓃}_4-{𝓇}_4}\right].\end{array}$$

where $R_1={𝓇}_1+{𝓇}_2$ and $R_2={𝓇}_1+{𝓇}_2+{𝓇}_3$. Consider the following strategies as examples of how these formulas can be used:

	Strategy	First Strategy			Second Strategy
Stage		${\mathbf{𝓃}}_{\mathbf{1}}$	${\mathbf{𝓃}}_{\mathbf{2}}$	${\mathbf{𝓃}}_{\mathbf{3}}$	${\mathbf{𝓃}}_{\mathbf{1}}$	${\mathbf{𝓃}}_{\mathbf{2}}$	${\mathbf{𝓃}}_{\mathbf{3}}$	${\mathbf{𝓃}}_{\mathbf{4}}$
Examination limits		₡	1	2	₡	0	2	3
Rejection limits		2	2	2	2	3	4	4

In this table, ₡ means that acceptance is not allowed at the first stage. The OCs, ASN functions, and other measures of multiple acceptance sampling plans can also be computed using a control chart. The control charts for the first and second strategies, respectively, are shown in Figure 2 and Figure 3.

Here, $E\left(\varnothing |{𝓃}_{(.)}\right)=\left[1-B\left(\varnothing |{𝓃}_{(.)}\right)\right]$ and $B\left(\varnothing |{𝓃}_{(.)}\right)=\sum _{𝓇=0}^{\varnothing }\left(\begin{array}{c}{𝓃}_{(.)}\\ 𝓇\end{array}\right){𝓅}^{𝓇}{\left(1-𝓅\right)}^{{𝓃}_{(.)}-𝓇}$ are defined as the probability that the binomially distributed random variable is, respectively, larger than or equal to $\varnothing$. The results of Figure 2 give the following probabilities.

According to

Table 7. Convenient probabilities of three-stage inspection sampling plan.

Stages	$\mathbf{Accept} {\mathit{A}}_{\mathit{i}}$	$\mathbf{Indecision} {\mathit{I}}_{\mathit{i}}$	$\mathbf{Reject} {\mathit{R}}_{\mathit{i}}$
1	0	$b\left(0|{𝓃}_1\right)+b\left(1|{𝓃}_1\right)$	$E\left(1|{𝓃}_1\right)$
2	$p_{01}b\left(0|{𝓃}_2\right)$	$p_{01}b\left(1|{𝓃}_2\right)+p_{11}b\left(0|{𝓃}_2\right)$	$p_{01}E\left(1|{𝓃}_2\right)+p_{11}E\left(0|{𝓃}_2\right)$
3	$p_{01}b\left(1|{𝓃}_2\right)b\left(0|{𝓃}_3\right)+p_{11}b\left(0|{𝓃}_2\right)b\left(0|{𝓃}_3\right)$		$p_{01}b\left(1|{𝓃}_2\right)E\left(0|{𝓃}_3\right)+p_{11}b\left(0|{𝓃}_2\right)E\left(0|{𝓃}_3\right)$

, the probability of acceptance for the first strategy can be obtained as follows:

$${\mathrm{P}}_3={\left(1-𝓅\right)}^{{{𝓃}_1+𝓃}_2}+𝓅{\left(1-𝓅\right)}^{{{𝓃}_1+𝓃}_2+{𝓃}_3-1}\left[{𝓃}_2+{𝓃}_1\right]$$

(20)

It is noted from Figure 2 that the three-stage acceptance sampling plans are somewhat complicated. Thus, the four-stage acceptance sampling plans will be difficult to represent in the form of a flowchart and difficult to understand by readers. Thus, the four-stage acceptance sampling plans can be shown in the following simplified form.

The results of Figure 3 give the following probabilities.

According to

Table 8. Convenient probabilities of four-stage inspection sampling plan.

Stages	$\mathbf{Accept} {\mathit{A}}_{\mathit{i}}$	$\mathbf{No} \mathbf{Decision} {\mathit{I}}_{\mathit{i}}$	$\mathbf{Reject} {\mathit{R}}_{\mathit{i}}$
1	0	$p_{01}+p_{11}$	$p_{21}$
2	$p_{02}$	$p_{12}+p_{22}$	$p_{32}$
3	$p_{23}$	$p_{33}$	$p_{43}$
4	$p_{34}$	0	$p_{44}$

, the probability of acceptance for the second strategy can be obtained as follows:

$$\begin{array}{c}{\mathrm{P}}_4={\left(1-𝓅\right)}^{{{𝓃}_1+𝓃}_2}+{𝓅}^2{\left(1-𝓅\right)}^{{{𝓃}_1+𝓃}_2+{𝓃}_3-2}\left[{𝓃}_2{𝓃}_3+{𝓃}_1{𝓃}_3+\frac{{𝓃}_2\left({𝓃}_2-1\right)}{2}+{𝓃}_1{𝓃}_2\right]\\ +{𝓅}^3{\left(1-𝓅\right)}^{{{𝓃}_1+𝓃}_2+{𝓃}_3+{𝓃}_4-3}\left[{𝓃}_2\frac{{𝓃}_3({𝓃}_3-1)}{2}+{𝓃}_1\frac{{𝓃}_3({𝓃}_3-1)}{2}+{𝓃}_3\frac{{𝓃}_2({𝓃}_2-1)}{2}+{𝓃}_1{𝓃}_2{𝓃}_3\right]\end{array}$$

(21)

where $𝓅=F_{\psi }\left(𝓍\right)$ is the probability that the unit will not work out before the termination time ${𝓍}_0$, which is given in (2). The following inequalities can be used to find the optimal sample sizes for three-stage and four-stage sampling plans, respectively, that ensure that the veritable mean life is greater than or equal to the prescribed mean life (i.e., $\psi \ge {\psi }_0$) at the consumer’s confidence level $\varkappa *$:

$${\mathrm{P}}_3\le 1-\varkappa *,$$

(22)

and

$${\mathrm{P}}_4\le 1-\varkappa *$$

(23)

There may exist multiple solutions for a design parameter satisfying Equations (22) and (23); thus, we reduce the ASN to obtain them by using the condition ${𝓃}_4\le \dots \le {𝓃}_1$. The ASN can be acquired in the following manner:

$$ASN={𝓃}_1+{\psi }_1\left({𝓃}_2+{\psi }_2\left\{{𝓃}_3+{\psi }_3\left[\dots \left({𝓃}_m+0\right)\dots \right]\right\}\right),$$

(24)

where ${\psi }_k$ is the probability of making no decision at stage $k$. According to the first strategy, the ASN of (24) simplifies to

$$ASN={𝓃}_1+{\psi }_1\left[{𝓃}_2+{\psi }_2\left({𝓃}_3\right)\right]$$

(25)

The probabilities ${\psi }_1$ and ${\psi }_2$ can be found by using Table 7 with the following respective equations:

$${\psi }_1={\left(1-𝓅\right)}^{{𝓃}_1}+{𝓃}_1𝓅{\left(1-𝓅\right)}^{{𝓃}_1-1};$$

and

$${\psi }_2=𝓅{\left(1-𝓅\right)}^{{𝓃}_1+{𝓃}_2-1}\left[{𝓃}_2+{𝓃}_1\right].$$

Accordingly, in a three-stage sampling plan, the optimal sample sizes for the first strategy will be decided by solving the following non-linear optimization problem:

$$\mathrm{Minimize} ASN={𝓃}_1+{\psi }_1\left[{𝓃}_2+{\psi }_2\left({𝓃}_3\right)\right]$$

(26)

Subject it to

$${\mathrm{P}}_3\le 1-\varkappa *$$

(27)

$$1\le {𝓃}_3\le {𝓃}_2\le {𝓃}_1$$

(28)

$${𝓃}_1, {𝓃}_2 \mathrm{and} {𝓃}_3 \mathrm{positive} \mathrm{integer}$$

By altering the initial values of ${𝓃}_1$, ${𝓃}_2$, and ${𝓃}_3$, the optimal sample sizes of the first, second, and third samples meeting the inequality (22) can be discovered using a search process.

Table 9. Optimal sample size and ASN for ML distribution in three-stage sampling plan.

$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$	${\mathit{\varkappa }}^{\mathit{*}}$
	0.75				0.90				0.95				0.99
	${𝓃}_1$	${𝓃}_2$	${𝓃}_3$	ASN	${𝓃}_1$	${𝓃}_2$	${𝓃}_3$	ASN	${𝓃}_1$	${𝓃}_2$	${𝓃}_3$	ASN	${𝓃}_1$	${𝓃}_2$	${𝓃}_3$	ASN
0.650	5	5	2	7.6	7	5	2	8.9	7	7	3	9.3	9	9	3	10.9
0.910	5	2	2	5.5	6	2	2	5.9	6	2	2	6.6	6	6	1	7.0
1.000	4	2	2	4.8	5	2	2	5.3	6	2	2	5.8	5	5	1	6.1
1.585	3	1	1	3.1	3	1	1	3.3	4	1	1	3.8	4	2	1	4.4
2.000	2	1	1	2.6	3	1	1	3.1	3	1	1	2.8	3	1	1	3.5
3.500	2	1	1	2.1	2	1	1	2.2	2	1	1	2.5	3	1	1	3.0
4.000	2	1	1	1.8	2	1	1	2.1	2	1	1	2.3	3	1	1	2.8
6.000	1	1	1	1.7	2	1	1	1.9	2	1	1	2.0	2	1	1	2.3

shows the values of sample sizes ${𝓃}_1$, ${𝓃}_2$, ${𝓃}_3$, and the ASN for $\varkappa *$ = 0.75, 0.90, 0.95, and 0.99 and

$$\frac{𝓍}{{\psi }_0}=0.650, 0.910, 1.000, 1.585, 2.000, 3.500, 4.000 \mathrm{and} 6.000$$

From Table 9, it is seen that the sample sizes and the ASN required increases as the confidence level increases when the termination time ratio is constant. According to the second strategy, the ASN of (24) simplifies to

$$ASN={𝓃}_1+{\psi }_1\left\{{𝓃}_2+{\psi }_2\left[{𝓃}_3+{\psi }_3\left({𝓃}_4\right)\right]\right\},$$

(29)

where the probabilities ${\psi }_1$, ${\psi }_2$, and, ${\psi }_3$ can be found by using Table 8 and the following respective equations:

$$\begin{array}{c}{\psi }_1={\left(1-𝓅\right)}^{{𝓃}_1}+{𝓃}_1𝓅{\left(1-𝓅\right)}^{{𝓃}_1-1};\\ {\psi }_2=𝓅{\left(1-𝓅\right)}^{{𝓃}_1+{𝓃}_2-1}\left[{𝓃}_2+{𝓃}_1\right]+{𝓃}_2{𝓅}^2{\left(1-𝓅\right)}^{{𝓃}_1+{𝓃}_2-2}\left[\frac{\left({𝓃}_2-1\right)}{2}+{𝓃}_1\right];\end{array}$$

and

$${\psi }_3={𝓅}^3{\left(1-𝓅\right)}^{{𝓃}_1+{𝓃}_2+{𝓃}_3-3}\left\{{𝓃}_3\left[\frac{{𝓃}_2\left({𝓃}_3-1\right)}{2}+\frac{{𝓃}_1\left({𝓃}_3-1\right)}{2}+\frac{{𝓃}_2\left({𝓃}_2-1\right)}{2}+{𝓃}_1{𝓃}_2\right]\right\}.$$

Therefore, the design parameters for the proposed second strategy can be obtained by the solution from the following optimization:

$$\mathrm{Minimize} ASN={𝓃}_1+{\psi }_1\left\{{𝓃}_2+{\psi }_2\left[{𝓃}_3+{\psi }_3\left({𝓃}_4\right)\right]\right\}$$

(30)

Subject it to

$${\mathrm{P}}_4\le 1-\varkappa *$$

(31)

$$1\le {𝓃}_4\le {𝓃}_3\le {𝓃}_2\le {𝓃}_1$$

(32)

$${𝓃}_1, {𝓃}_2, {𝓃}_3 \mathrm{and} {𝓃}_4 \mathrm{positive} \mathrm{integer}.$$

A simple search can be used to tackle the above problem by altering the values of ${𝓃}_1$, ${𝓃}_2$, ${𝓃}_3$, and ${𝓃}_4$. The conditions of (28) and (32) are specified because it may not be convenient if the sample size in the $k$ stage is larger than that in the first stage.

Table 10. Optimal sample size and ASN for ML distribution in four-stage sampling plan.

$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$	${\mathit{\varkappa }}^{\mathit{*}}$
	0.75					0.90					0.95					0.99
	${𝓃}_1$	${𝓃}_2$	${𝓃}_3$	${𝓃}_4$	ASN	${𝓃}_1$	${𝓃}_2$	${𝓃}_3$	${𝓃}_4$	ASN	${𝓃}_1$	${𝓃}_2$	${𝓃}_3$	${𝓃}_4$	ASN	${𝓃}_1$	${𝓃}_2$	${𝓃}_3$	${𝓃}_4$	ASN
0.650	4	4	4	3	7.4	5	5	5	5	8.4	7	7	5	5	9.3	8	8	8	8	10.3
0.910	3	3	1	1	5.0	4	4	4	2	5.6	4	4	3	2	5.7	5	4	4	4	6.3
1.000	3	3	1	1	4.6	4	3	3	2	5.1	4	4	4	4	5.1	5	4	3	2	5.7
1.585	3	1	1	1	2.9	3	1	1	1	3.0	3	1	1	1	3.5	4	2	2	2	4.3
2.000	2	1	1	1	2.5	2	1	1	1	2.5	3	1	1	1	2.8	3	2	2	2	3.1
3.500	2	1	1	1	1.9	2	1	1	1	2.2	2	1	1	1	2.4	3	1	1	1	2.8
4.000	2	1	1	1	1.8	2	1	1	1	2.1	2	1	1	1	2.1	3	1	1	1	2.6
6.000	1	1	1	1	1.5	2	1	1	1	1.7	2	1	1	1	1.9	2	1	1	1	2.2

shows the values of sample sizes ${𝓃}_1$, ${𝓃}_2$, ${𝓃}_3$, and ${𝓃}_4$ and the ASN for $\varkappa *$ = 0.75, 0.90, 0.95, and 0.99 and $\frac{𝓍}{{\psi }_0}=$ 0.650, 0.910, 1.000, 1.585, 2.000, 3.500, 4.000, and 6.000.

Table 1, Table 4, Table 9 and Table 10 show that MAS plans provide even more flexibility and a lower ASN than double and single sampling plans, but they are sometimes harder to administer due to the complexity of handling and documenting all of the samples required.

The OC values for ML distribution according to (20) and (21) for diverse values of $\psi /{\psi }_0$ and $\varkappa *$ and for a given sampling plan $\left({𝓃}_k,{\mathcal{A}}_{{\mathsf{𝕔}}_k}.{\mathcal{A}}_{{𝓇}_k},\frac{𝓍}{{\psi }_0}\right)$ are shown in

Table 11. The OC values under three-stage acceptance sampling plan for a given $\varkappa *$ when ${\mathcal{A}}_{{\mathsf{𝕔}}_1}=\mathrm{₡}$, ${\mathcal{A}}_{{\mathsf{𝕔}}_2}=0$, ${\mathcal{A}}_{{\mathsf{𝕔}}_3}=1$, and ${\mathcal{A}}_{{𝓇}_1}={\mathcal{A}}_{{𝓇}_2}={\mathcal{A}}_{{𝓇}_3}=2$.

$\mathit{\varkappa }*$	$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$	$\mathit{\psi }/{\mathit{\psi }}_0$
		2	3	4	5	6	7	8	9
0.75	0.650	0.64	0.912	0.987	0.995	0.997	0.998	1	1
	0.910	0.528	0.848	0.955	0.988	0.996	0.997	0.999	1
	1.000	0.51	0.826	0.948	0.978	0.995	0.995	0.999	1
	1.585	0.346	0.668	0.831	0.918	0.967	0.98	0.989	0.999
	2.000	0.218	0.536	0.759	0.899	0.949	0.972	0.988	0.998
	3.500	0.032	0.239	0.499	0.706	0.824	0.891	0.932	0.981
	4.000	0.020	0.142	0.38	0.596	0.753	0.841	0.899	0.946
	6.000	0.0058	0.059	0.178	0.325	0.463	0.579	0.682	0.799
0.90	0.650	0.554	0.861	0.967	0.985	0.998	0.999	0.999	1
	0.910	0.482	0.835	0.952	0.979	0.993	0.994	0.997	0.999
	1.000	0.462	0.827	0.945	0.978	0.991	0.992	0.996	0.999
	1.585	0.295	0.695	0.884	0.951	0.979	0.981	0.995	0.999
	2.000	0.204	0.617	0.831	0.931	0.971	0.974	0.987	0.989
	3.500	0.019	0.196	0.452	0.682	0.809	0.831	0.89	0.991
	4.000	0.015	0.14	0.399	0.615	0.751	0.753	0.862	0.906
	6.000	3.85 × 10−4	0.019	0.114	0.361	0.411	0.4	0.568	0.699
0.95	0.650	0.432	0.819	0.952	0.987	0.991	0.999	0.999	0.999
	0.910	0.419	0.782	0.938	0.979	0.997	0.998	0.999	0.999
	1.000	0.407	0.781	0.923	0.97	0.994	0.997	0.997	0.999
	1.585	0.28	0.691	0.896	0.957	0.984	0.989	0.994	0.999
	2.000	0.175	0.586	0.792	0.905	0.953	0.954	0.99	0.989
	3.500	0.019	0.143	0.398	0.619	0.757	0.774	0.862	0.916
	4.000	0.0036	0.085	0.276	0.489	0.699	0.709	0.792	0.884
	6.000	2.37 × 10−4	0.091	0.093	0.199	0.395	0.413	0.518	0.674
0.99	0.650	0.401	0.805	0.945	0.981	0.99	0.997	0.999	0.999
	0.910	0.375	0.793	0.937	0.979	0.994	0.994	0.996	0.999
	1.000	0.363	0.786	0.927	0.97	0.987	0.993	0.996	0.998
	1.585	0.256	0.675	0.93	0.347	0.979	0.979	0.989	0.993
	2.000	0.144	0.544	0.771	0.893	0.959	0.961	0.968	0.984
	3.500	0.0094	0.145	0.373	0.623	0.751	0.764	0.86	0.901
	4.000	0.0034	0.075	0.292	0.468	0.648	0.674	0.786	0.862
	6.000	1.18 × 10−4	0.0091	0.091	0.197	0.353	0.369	0.496	0.641

and

Table 12. The OC values under four-stage acceptance sampling plan for a given $\varkappa *$ when ${\mathcal{A}}_{{\mathsf{𝕔}}_1}=\mathrm{₡}$, ${\mathcal{A}}_{{\mathsf{𝕔}}_2}=0$, ${\mathcal{A}}_{{\mathsf{𝕔}}_3}={\mathcal{A}}_{{\mathsf{𝕔}}_4}=3$, ${\mathcal{A}}_{{𝓇}_1}=2$, ${\mathcal{A}}_{{𝓇}_2}=3$, and ${\mathcal{A}}_{{𝓇}_3}={\mathcal{A}}_{{𝓇}_4}=4$.

$\mathit{\varkappa }*$	$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$	$\frac{\mathit{\psi }}{{\mathit{\psi }}_0}$
		2	3	4	5	6	7	8	9
0.75	0.650	0.645	0.929	0.989	0.998	0.998	0.999	1	1
	0.910	0.543	0.868	0.96	0.991	0.997	0.998	1	1
	1.000	0.523	0.827	0.952	0.979	0.995	0.996	0.999	1
	1.585	0.491	0.691	0.865	0.929	0.972	0.986	0.999	1
	2.000	0.464	0.586	0.791	0.903	0.951	0.982	0.989	0.999
	3.500	0.292	0.514	0.584	0.713	0.843	0.901	0.952	0.998
	4.000	0.204	0.451	0.557	0.624	0.787	0.852	0.903	0.958
	6.000	0.071	0.282	0.471	0.573	0.615	0.633	0.699	0.854
0.90	0.650	0.572	0.899	0.971	0.995	0.998	0.999	0.999	1
	0.910	0.538	0.871	0.965	0.989	0.995	0.998	0.999	1
	1.000	0.519	0.852	0.957	0.979	0.993	0.995	0.996	0.999
	1.585	0.482	0.721	0.892	0.974	0.98	0.991	0.995	0.999
	2.000	0.46	0.661	0.843	0.948	0.975	0.986	0.991	0.999
	3.500	0.162	0.403	0.518	0.708	0.852	0.855	0.909	0.995
	4.000	0.113	0.342	0.48	0.634	0.802	0.829	0.899	0.907
	6.000	0.027	0.171	0.357	0.487	0.557	0.592	0.612	0.727
0.95	0.650	0.503	0.837	0.972	0.989	0.994	0.999	0.999	1
	0.910	0.494	0.818	0.948	0.982	0.999	0.999	0.999	0.999
	1.000	0.483	0.805	0.937	0.979	0.994	0.999	0.999	0.999
	1.585	0.391	0.791	0.902	0.961	0.988	0.999	0.998	0.999
	2.000	0.388	0.608	0.809	0.915	0.959	0.964	0.991	0.999
	3.500	0.136	0.358	0.475	0.644	0.762	0.793	0.889	0.936
	4.000	0.102	0.331	0.477	0.541	0.719	0.732	0.805	0.904
	6.000	0.015	0.124	0.296	0.435	0.519	0.564	0.59	0.693
0.99	0.650	0.407	0.855	0.951	0.985	0.992	0.998	0.999	0.999
	0.910	0.387	0.817	0.949	0.981	0.997	0.996	0.999	0.999
	1.000	0.371	0.799	0.931	0.977	0.989	0.994	0.997	0.999
	1.585	0.278	0.699	0.939	0.635	0.981	0.982	0.99	0.995
	2.000	0.262	0.604	0.777	0.894	0.963	0.969	0.972	0.994
	3.500	0.077	0.277	0.415	0.637	0.775	0.779	0.892	0.933
	4.000	0.057	0.243	0.394	0.475	0.658	0.681	0.805	0.890
	6.000	0.013	0.108	0.262	0.389	0.468	0.999	0.543	0.673

, respectively.

Finally, the smallest values of ratio $\psi /{\psi }_0$ for the three-stage and four-stage sampling plans can be found by solving the following inequalities at the producer’s risk of $\tau =$ 0.05:

$${\mathrm{P}}_3\ge 1-\tau ,$$

(33)

and

$${\mathrm{P}}_4\ge 1-\tau .$$

(34)

For a given $\varkappa *$ and sampling plan $\left({𝓃}_k,{\mathcal{A}}_{{\mathsf{𝕔}}_k}.{\mathcal{A}}_{{𝓇}_k},\frac{𝓍}{{\psi }_0}\right)$, where $k=$ 1, …, 4, the smallest ratio of $\psi /{\psi }_0$, satisfying inequalities (33) and (34), is calculated and display in

Table 13. Minimum ratio for the multiple acceptance plans with producer’s risk of 0.05.

$\frac{\mathbf{𝓍}}{{\mathit{\psi }}_0}$	Three-Stage (First Strategy)				Four-Stage (Second Strategy)
	$\mathit{\varkappa }*$				$\mathit{\varkappa }*$
	0.75	0.90	0.95	0.99	0.75	0.90	0.95	0.99
0.650	3.072	3.493	4.094	4.19	2.674	3.358	4.036	4.008
0.910	3.744	3.978	4.312	4.255	3.669	3.873	4.202	4.101
1.000	3.879	4.13	4.427	4.394	3.65	4.024	4.406	4.286
1.585	5.115	4.961	4.563	5.036	4.32	4.734	4.514	4.969
2.000	5.036	5.649	6.746	6.088	4.933	5.061	5.872	5.72
3.500	7.744	8.479	8.522	9.44	7.631	8.393	7.476	8.784
4.000	8.916	9.357	10.277	9.959	8.815	8.513	9.756	8.806
6.000	10.785	13.466	14.065	14.875	10.306	12.723	13.788	14.029

.

From Table 13, it is noted that the minimum ratio increases as the time ratio increases and the confidence level remains constant. Additionally, the minimum ratio increases as the confidence level increases and the the time ratio remains constant.

6. Assessment Inspection Sampling Plans

Consider the proportion of defects as a function of $\psi /{\psi }_0$ for ML distribution with a time ratio of $𝓍/{\psi }_0$ = 0.910, a confidence level of $\varkappa *$ = 0.99, $\psi /{\psi }_0$ = 3, and a batch size of $N$ = 1000.

Table 14. Sampling plans assessment.

Plan	Measures of Performance
	ASN	OC Values	Mean Ratio
Single	20	0.710	4.791
Double	12.2	0.771	4.458
Three-stage	6.9	0.793	4.255
Four-stage	6.3	0.817	4.101

depicts the assessment of sampling plans for the ML distribution.

These statistics are helpful in evaluating types of sampling plans by attribute at a time ratio equal to 910 h. The probability of acceptance is the most significant criterion for evaluating sampling plans, specifically for the producer. A good plan is one that gives a value close to one. The ASN is the measure utilized to assess inspection sampling plans. When comparing multiple inspection sampling plans with single and double sampling plans, it is found that the multiple inspection sampling plans employ a smaller sample size on average than the other plans when the plans have the same time ratio and consumer risk. Additionally, the MAS plans use a smaller mean ratio compared to the mean ratio in single and double sampling plans when the plans have the same consumer risk. This proves that the MAS plans are better than single and double sampling plans, and that they save time, effort, and cost without increasing the risk. Although multiple-acceptance sampling plans are significantly more efficient in terms of the ASN, OCs, and mean ratio than single and double sampling plans, their administrative costs are higher due to their complexity. As a result, the single sample strategy is the easiest to apply and the least expensive.

7. Numerical Illustration

The numerical outputs for $X~$ML $\left(\psi \right)$ are given in Table 1, Table 2, Table 3, Table 4, Table 5, Table 6, Table 7, Table 8, Table 9, Table 10, Table 11, Table 12, Table 13 and Table 14, except for Table 7 and Table 8, as they show the convenient probabilities of three- and four-stage inspection sampling plans, respectively. Table 1, Table 4, Table 9 and Table 10 show the values of sample sizes to be examined for a time $𝓍$ in order to confirm that the mean life overrides a given value, ${\psi }_0$, with a consumer’s risk of $1-\varkappa *$ and a corresponding examination limit for single, double, three-stage, and four-stage sampling plans, respectively. From Table 1, Table 4, Table 9 and Table 10, it is seen that the sample sizes required and the ASN increase as the confidence level increases. Additionally, the sample sizes and the ASN decrease as the termination time ratio increases. Notice that, for a single sampling plan, $ASN=𝓃$. Additionally, the four-stage sampling plans provide even more elasticity and a smaller ASN than the single, double, and three-stage sampling plans but are frequently considered to be hard to administer because of the intricacy of handling and documenting all of the samples required. Table 2, Table 5, Table 11 and Table 12 present the operating characteristic values for different combinations of values of probability $\varkappa *$, quality parameter $\psi /{\psi }_0$, and time ratio $𝓍/{\psi }_0$ for single, double, three-stage, and four-stage sampling plans, respectively. Table 2, Table 5, Table 11 and Table 12 explain that the OC values increase when the quantity ratio $\psi /{\psi }_0$ increases. Additionally, the OC values decrease as the confidence level increases.

Table 3, Table 6 and Table 13 present the minimum ratios of veritable mean life to prescribed mean life for the acceptance of a batch with a producer’s risk of 0.05. In Table 3, Table 6 and Table 13, it is seen that the value of $\psi /{\psi }_0$ increases as the termination time ratio and the confidence limits increase. Table 14 present the assessment of the sampling plans.

7.1. First Experiment

Suppose that the examiner is interested in implementing a single sampling plan to confirm that the veritable mean lifetime for the mobile components is at least 1000 h with a confidence level of 0.99. The examiner intends to have the test amputated at ${𝓍}_0$ = 910 h. Then, for an examination limit of ${\mathcal{A}}_{\mathsf{𝕔}}=2$, the optimal sample size $𝓃$, which corresponds to the value of $\varkappa *$, is 20 (Table 1). If no more than two failures out of 20 are seen during 910 h, the experimenter can emphasize with a confidence level of 99% that the mean life is at least 1000 h. It is noted that the ASN for a plan with parameters $(𝓃,{\mathcal{A}}_{\mathsf{𝕔}}$ and $𝓍/{\psi }_0)$ is $𝓃$; thus, the ASN for this experiment is 20. For the sampling plan in which $\left(𝓃,{\mathcal{A}}_{\mathsf{𝕔}},\frac{𝓍}{{\psi }_0}\right)$ = (20, 2, and 0.910) and $\varkappa *$ = 0.99, the OC values from Table 2 are as follows:

$\mathit{\psi }\mathbf{/}{\mathit{\psi }}_{\mathbf{0}}$	2	3	4	5	6	7	8	9
OC values	0.281	0.710	0.894	0.959	0.982	0.992	0.996	0.998

This suggests that the producer’s risk is 0.29 if $\left(\psi /{\psi }_0=3\right)$ and 0.002 if the veritable mean life is nine times the prescribed mean life $\left(\psi /{\psi }_0=9\right)$. As a result, the producer’s risk tends to reduce as the quality parameter $\psi /{\psi }_0$ is increased. Table 3 can be employed to calculate the value of $\psi /{\psi }_0$ for different values of ${\mathcal{A}}_{\mathsf{𝕔}}$ and $𝓍/{\psi }_0$, so that the producer’s risk does not surpass 0.05. To illustrate, if ${\mathcal{A}}_{\mathsf{𝕔}}$ = 2, $𝓍/{\psi }_0$ = 0.910, and $\varkappa *$ = 0.99, the value of $\psi /{\psi }_0$ is 4.791; this implies that the product must have a mean lifetime of at least 4.791 times the prescribed life of 1000 h in order for the batch to be approved with a probability of 0.99.

7.2. Second Experiment

Assume that the examiner wishes to implement a double sampling plan with $\varkappa *$ = 99% to confirm that the veritable mean lifetime of the mobile components is at least 1000 h. The examiner wishes to amputate the experiment at ${𝓍}_0$ = 910 h. From Table 4, the optimal sample sizes ${𝓃}_1$ and ${𝓃}_2$ are 12 and 5, respectively, which are equivalent to the values of ${\mathcal{A}}_{{\mathsf{𝕔}}_1}$ = 0 and ${\mathcal{A}}_{{\mathsf{𝕔}}_2}$ = 2 where $\varkappa *$ = 99%. This sampling mechanism is explained as follows: the first sample size of 12 units is drawn from a batch and tested for 910 h. The lot is accepted if no defective units recorded during the test. The lot is rejected and terminated if more than two faulty units are registered. If the number of faulty units is 2 or fewer, the second sample of size 5 is drawn from the same batch and tested for another 910 h. The batch is accepted if the cumulative number of faulty units from the first and second samples is less than or equal to two. Otherwise, the test is terminated, and the batch is rejected. The ASN for this experiment is 12.2, so it would be beneficial to contrast the double plan with the single plan with an ASN of 20. When a double sampling plan is compared to a single sampling plan, it is discovered that the double plan uses a smaller sample size on average than the single plan when the same consumer risk is present. The OC values for the sampling plan in which $\left({𝓃}_1,{𝓃}_2,{\mathcal{A}}_{{\mathsf{𝕔}}_1},{\mathcal{A}}_{{\mathsf{𝕔}}_2},\frac{𝓍}{{\psi }_0}\right)$ = (12, 5, 0, 2, and 0.910) and $\varkappa *$ = 0.99 under ML distribution are provided in Table 5.

$\mathit{\psi }\mathbf{/}{\mathit{\psi }}_{\mathbf{0}}$	2	3	4	5	6	7	8	9
OC values	0.366	0.771	0.921	0.97	0.987	0.993	0.997	0.998

This signifies that if $\left(\psi /{\psi }_0=3\right)$, the batch is accepted with a probability of 77.1%, and the producer’s risk is 22.9%. The producer’s risk is 0.002 when $\psi /{\psi }_0$ = 9 times. This means there is an inverse relationship between the probability of acceptance and the producer’s risk: the higher the probability of acceptance, the lower the producer’s risk. The manufacturer may be interested in the acceptance probability as productivity increases in order to reduce the producer’s risk. Assume the manufacturer wants to inform what quality level will result in the producer’s risk being less than or equal to 0.05. Table 6 contains the answer: 4.458 is the mean ratio for $𝓍/{\psi }_0$ = 0.910 and $\varkappa *$ = 0.99. As a result, the product’s veritable mean life should be at least 4458 h.

7.3. Third Experiment

Assume the examiner receives large batches of mobile components and wishes to establish MAS plans in three stages (first strategy) to confirm that the veritable mean life is at least 1000 h. With a confidence level of 0.99, the examiner decides to amputate the experiment at time ${𝓍}_0$ = 910 h. According to the first strategy, the optimal sample sizes ${𝓃}_1$, ${𝓃}_2$, and ${𝓃}_3$ equivalent to the value of $\varkappa *$ = 0.99 are 6, 6, and 1 for examination limits ${\mathcal{A}}_{{\mathsf{𝕔}}_1}=₡$, ${\mathcal{A}}_{{\mathsf{𝕔}}_2}=0,$ and ${\mathcal{A}}_{{\mathsf{𝕔}}_3}=1$ and rejection levels ${\mathcal{A}}_{{𝓇}_1}={\mathcal{A}}_{{𝓇}_2}={\mathcal{A}}_{{𝓇}_3}=2$ (Table 9). The first strategy mechanism is described as follows: the first sample size of six units is drawn to be set to test for duration ${𝓍}_0$ = 910 h. No acceptance is allowed at the first stage, but the batch is rejected if the number of faulty units found is more than or equal to two. If one or fewer faulty units are found, a second sample size of six is drawn. The batch is accepted if no faulty units are found from the first and second samples and is rejected if two or more faulty units are found. If exactly one faulty units is found from the first and second samples, a third sample of size one is drawn. The batch is accepted if a total numbers of one faulty unit from the second and third samples are found. Otherwise, terminate the test and reject the batch. The ASN for the first strategy is 7.0. The OC values for the MAS plans with three stages equivalent to the quality parameter $\psi /{\psi }_0$ are shown in Table 11.

$\mathit{\psi }\mathbf{/}{\mathit{\psi }}_{\mathbf{0}}$	2	3	4	5	6	7	8	9
OC values	0.375	0.793	0.937	0.979	0.994	0.994	0.996	0.999

This indicates that if $\left(\psi /{\psi }_0=3\right)$, the batch is accepted with a probability of 79.3%, and the producer’s risk is 20.7%. When we need to determine the ratio that is equivalent to the producer’s risk of 0.05, we can search for it in Table 13. For instance, the ratio $\psi /{\psi }_0$ for the first strategy (the multiple sampling plan with three stages) is 4.255. This implies that the product must have a mean lifetime of at least 4.255 times the prescribed mean life of 1000 h for the batch to be accepted with 99% confidence level.

7.4. Fourth Experiment

Suppose the examiner receives large batches of mobile components and needs to create MAS plans with four stages (second strategy) to validate that the veritable mean life is at least 1000 h. The examiner decides to amputate the test at time ${𝓍}_0$ = 910 h with a confidence level of 0.99. According to the second strategy, the optimal sample sizes ${𝓃}_1$, ${𝓃}_2$, ${𝓃}_3$, and ${𝓃}_4$ corresponding to the value of $\varkappa *$ = 0.99 are 5, 4, 4, and 4 for examination limits ${\mathcal{A}}_{{\mathsf{𝕔}}_1}$ = $₡$, ${\mathcal{A}}_{{\mathsf{𝕔}}_2}$ = 0, ${\mathcal{A}}_{{\mathsf{𝕔}}_3}$ = 2, and ${\mathcal{A}}_{{\mathsf{𝕔}}_4}$ = 3 and rejection limits ${\mathcal{A}}_{{𝓇}_1}$ = 2, ${\mathcal{A}}_{{𝓇}_2}$ = 3, and ${\mathcal{A}}_{{𝓇}_3}$ = ${\mathcal{A}}_{{𝓇}_4}$ = 4, (Table 10). The following example shows how the sampling mechanism works: take the first sample size of 5 from the submitted batch and test it for 910 h. Acceptance is not permitted in the first stage, but the batch is refused if two or more faulty units are discovered. If one or fewer faulty units are discovered, a second sample size of four is taken. If no faulty units are recorded from the first and second samples, the batch is accepted; if three or more total faulty units are discovered, the lot is rejected. If two or fewer faulty units are found, a third sample size of 4 is taken. The lot is accepted if two faulty units are found in the second and third samples, and the lot is rejected if four or more total number of faulty units are discovered. If exactly three total number of faulty units are found from the second and third samples, a fourth sample size of four is taken. The lot is accepted if three faulty units are found from the third and fourth samples. Otherwise, terminate the test and reject the lot if four or more faulty units are found. The ASN for the second strategy is 6.3. The OC values for the MAS plans with four stages equivalent to the quality parameter $\psi /{\psi }_0$ are shown in Table 12.

$\mathit{\psi }\mathbf{/}{\mathit{\psi }}_{\mathbf{0}}$	2	3	4	5	6	7	8	9
OC values	0.387	0.817	0.949	0.981	0.997	0.996	0.999	0.999

This indicates that if $\left(\psi /{\psi }_0=3\right)$, the batch is accepted with a probability of 81.7%, and the producer’s risk is 18.3%. When we need to determine the ratio that correlates to the producer’s risk of 0.05, we can search for it in Table 13. For instance, the ratio $\psi /{\psi }_0$ for the second strategy (multiple sampling plan with four stages) is 4.101. This implies that the product must have a mean lifetime of at least 4.101 times the prescribed mean life of 1000 h for the batch to be accepted with a 99% confidence level.

8. Conclusions

In this study, single, double, three-stage, and four-stage acceptance sampling procedures are discussed based on the modified Lindley distribution. The problem of a single sample plan is taken into consideration when a lifetime test is cut short after a specific amount of time. To guarantee the estimated fixed mean life, the confidence limits and values ratio of the time and the sample size are desired for different levels of acceptability. With a certain probability, the results of the lowest actual mean life to fixed mean life ratio that confirms acceptance are presented. For this, a case study is offered. To provide a chance to draw a second sample, the double sampling plan is taken into consideration. For the purpose of demonstrating the possibilities of the multiple acceptance sampling schemes, four numerical examples are provided. The numerical example demonstrates how the single, double, and multiple acceptance sampling plans may be utilized to maintain product quality in terms of its mean life according to the consumer’s criterion at a set producer’s risk. A single acceptance sampling plan is the most widely used and simplest to use; however, they are not the most effective in terms of the average number of samples needed. It is feasible that, by taking into account the modified Lindley distribution, more authors will become interested in other cutting-edge models.

Some future point can be addressed as follows:

• The ML model could be useful in analyzing and assessing the actuarial risks.
• For analyzing the mortality rates, it is suggested to employ and apply the ML model.
• The ML model could be useful under other acceptance sampling plans.
• The ML distribution could be used for acceptance sampling plans from truncated life tests.
• Validation testing for modeling correctly censored data may be a good topic under the ML model.