Mixed Variables-Attributes Test Plans for Single and Double Acceptance Sampling under Exponential Distribution

Abstract

The mixed variables-attributes test plans for single acceptance sampling are proposed to protect “good lots” from attributes aspect and to optimize sample sizes from variables aspect. For the single and double mixed plans, exact formulas of the operating characteristic and average sample number are developed for the exponential distribution. Numerical illustrations show that the mixed sampling plans have some advantages over the variables plans or attributes plans alone.

1. Introduction

In reality, especially in industry, acceptance sampling is applied to audit the percent nonconforming of the incoming materials, semimanufactured goods, finished products, and so on. Among the available plans for testing the percent nonconforming, attributes sampling plans are the most widely used plans in practice, due to the fact that they only require easy-to-collect and low-cost attributes data and produce reasonable judgments about the number of nonconforming items, as pointed out in [1].

Compared with attributes plans, variables acceptance procedures require less data under the same error probabilities and produce more accurate quality information. Hence, for costly and/or destructive inspections, variables plans are more attractive for practitioners. However, the necessary preprocessing of translating the test of the percent nonconforming into that of a variables characteristic would have the following problems [1, 2]: (i) It is generally more difficult and time consuming to measure the variables quality characteristic. (ii) The variables quality characteristic and its distribution should be specified. Since testing and estimating the percent nonconforming highly depend on the assumed distribution, unreliable conclusions may be made from the inappropriate distribution especially when the percent nonconforming is extremely small or large and is greatly inflected by the tail feature of the distribution. (iii) Under the same requirements on error probabilities, the protection for the two types of error probabilities under variables plans may be weaker than that of attributes plans. (iv) For variables plans, a lot may be rejected even if all of the sampled items are conforming, which violates the rule of attributes procedures and is always questioned by practitioners.

In fact, the authors found the above problem (iv) when applying a variables plan to the following application. To test the percent nonconforming 𝑝(0<𝑝<1) of a batch of bullets, a bullets sample is inspected by shooting at a circular objective with the center (0,0) and the radius √𝑎(𝑎>0). A bullet is defined as conforming if it hits the objective under some circumstances, or as nonconforming otherwise. The inspection results include the variables data, that is, coordinates of the fall points, and the attributes data about whether the inspected bullets are nonconforming. According to specific needs, the test problem was set as 𝐻₀∶𝑝=𝑝₀=0.1versus𝐻₁∶𝑝=𝑝₁=0.4. (1.1)

Under the binomial distribution, the attributes plan for single acceptance sampling (single attributes plan) with the error probabilities 𝛼=𝛽=0.15 is (𝑛_𝑎,𝑐_𝑎)=(11,2), that is, to inspect 11 bullets and accept the null hypothesis in the test problem (1.1) when 2 or less bullets in the sample are nonconforming. The real error probabilities of the plan are 𝛼^′_𝑎=0.0896 and 𝛽^′_𝑎=0.1189.

The variables plan for single acceptance sampling (single variables plan) is also accessible based on locations of the fall points. Suppose the coordinate of the fall point is (𝑋,𝑌), where 𝑋 and 𝑌 are independent and identically distributed (i.i.d.) with 𝑁(0,𝜎²)(𝜎²>0) according to the practitioners’ experiences. Then, (𝑋²+𝑌²)/𝜎² conforms with 𝜒²(2) or Gamma(1,2). Equivalently, 𝑅=𝑋²+𝑌² conforms with the exponential distribution Exp(2𝜎²) with 2𝜎² as its mean. Hence, the percent nonconforming 𝑝 and the variables parameter 𝜎² have the following relationship: ()−𝑎𝑝=𝑃𝑅>𝑎=exp2𝜎², (1.2) or equivalently, 𝜎²𝑎=−2ln𝑝. (1.3)

Taking 𝑎=1 as an example, according to (1.3), the test problem (1.1) is equivalent to 𝐻₀∶𝜎²=𝜎²₀=0.2171versus𝐻₁∶𝜎²=𝜎²₁=0.5457. (1.4)

For the test problem (1.4), setting {𝑅_𝑛𝑣=∑^𝑛𝑣_𝑖=1𝑅_𝑖/𝑛_𝑣>𝑐_𝑣} as the rejection region, the single variables plan should be (𝑛_𝑣,𝑐_𝑣)=(6,0.6284) with the real error probabilities 𝛼^′_𝑣=𝛽^′_𝑣=0.1365.

To save the inspection costs, we chose the variables scheme and obtained coordinates of the 6 sampled bullets, that is, (−0.82,−0.51), (0.59,−0.06), (−0.74,0.39), (0.24,0.24), (−0.33,0.93), and (0.17,0.88). Since 𝑅_𝑛𝑣=∑⁶_𝑖=1𝑅_𝑖/6=0.6460>0.6284, the null hypothesis in (1.4) or in (1.1) was rejected. However, it is clear that all of the sampled bullets are conforming since 𝑅_𝑖≤1,𝑖=1,…,6. Thus, the judgement leads to some problems from attributes aspect as stated below.(i)The basic rule of attributes plans is to accept a lot when none of the sampled items is nonconforming. Obviously, our test result is contradictory to this rule.(ii)The producer can hardly be convinced about the decision of rejecting the lot when no nonconforming item is found. The producer may argue that the lot should be accepted even under the attributes plan (𝑛_𝑎,𝑐_𝑎̃)=(6,0), which guarantees the consumer’s risk since 𝛽^′=(1−0.4)⁶=0.0467<𝛽, although it employs the producer’s risk as large as 𝛼^′=1−(1−0.1)⁶=0.4686. Further, for the hypothesis “𝑝=𝑝₁=0.4” with {𝑑𝑛_𝑎𝑛=∑^𝑎_𝑖=1𝐼_(𝑎,+∞)(𝑅_𝑖)≤𝑐} (𝐼(⋅) is an indicator function) as its form of rejection region, the 𝑃-value of the test result is 𝑃_𝑝1(𝑑𝑛_𝑎≤0)=(1−0.4)⁶=0.0467. Therefore, the decision of rejecting the hypothesis “𝑝=𝑝₁=0.4” should be made with a quite sufficient evidence under the significance level 𝛾=0.05.

Motivated by this case and other similar problems in engineering (e.g., [3–8]), we try to find the answer about which one is more suitable for testing the percent nonconforming, the variables plan or the attributes plan. Scholars have different opinions and arguments in their papers. Some (e.g., [2]) support attributes plans due to their better protection for the two types of error probabilities. Some (e.g., [9, 10]) insist that variables plans be advantageous and acceptable according to the evaluation on some loss functions under superpopulations.

However, it is not necessarily a dichotomy scenario. Some scholars choose compromised schemes, that is, the mixed variables-attributes sampling plans, most of which (e.g., [11–15]) are double sampling procedures. For single acceptance sampling, Kao [16] proposes the mixed plan which permits to accept a lot when the number of nonconforming items is small and the conforming items in the sample are uniform under some variables characteristic. However, since it mainly aims at testing item variability and considers two independent quality characteristics for attributes and variables inspections, respectively, the mixed plan in [16] is inapplicable to our case. Anyway, the present mixed plans provide a way to find out a solution to our problem.

In Section 2, we propose the mixed variables-attributes plans for single acceptance sampling (single mixed plans). To make a comparison, we also obtain the existing mixed plans for double acceptance sampling (double mixed plans) with the derived performance formulas in Section 3. Summaries and further remarks are provided in Section 4. The appendix gives the related formula in evaluating performance of the mixed plans. Due to our interest in the bullets test problem, we only discuss the exponential distribution case, which is also vital for life testing according to [12].

2. Mixed Plans for Single Acceptance Sampling

To design the mixed variables-attributes schemes for single acceptance sampling, a key is to determine the form of the rejection region. For the bullets test problem in this paper, taking 𝐴={𝑅_𝑛=∑^𝑛_𝑖=1𝑅_𝑖/𝑛>𝑐_𝑟}(𝑐_𝑟≥0) as the rejection region for variables sampling and 𝐵={𝑑_𝑛=∑^𝑛_𝑖=1𝐼_(𝑎,+∞)(𝑅_𝑖)>𝑐_𝑑} (𝑐_𝑑 is an integer in [0,𝑛−1]) as the counterpart for attributes sampling, we set the rejection region of the single mixed plans as 𝐴𝐵={𝑅_𝑛>𝑐_𝑟}⋂{𝑑_𝑛>𝑐_𝑑}. Then, a single mixed plan contains 3 parameters, that is, the sample size 𝑛, the critical values 𝑐_𝑟 for variables measurement and 𝑐_𝑑 for attributes measurement.

Since 𝑐_𝑑≥0, the single mixed plans can guarantee the rule of accepting a lot when no nonconforming item is found in the sample. Moreover, variables and attributes information can be integrated in the single mixed plans to obtain more convincible and applicable results.

However, the feature of discrete distributions (e.g., the binomial distribution) for the number of nonconforming items 𝑑_𝑛 impairs the flexibility of designing the plans. Therefore, sample sizes of the single mixed plans are larger than or equal to those of the single variables plans with the same requirements on error probabilities.

Next, we discuss how to determine the operating characteristic (OC) formula of the single mixed plan (𝑛,𝑐_𝑟,𝑐_𝑑). With the rejection region 𝐴𝐵, the probability of accepting the null hypothesis under the percent nonconforming 𝑝 should be 𝐿𝑝=1−𝑃_𝑝(𝐴𝐵)=1−^𝑛_{𝑘=𝑐𝑑+1}𝑃_𝑝𝑅_𝑛>𝑐_𝑟,𝑑_𝑛=𝑘. (2.1)

As will be shown in the sequel, the OC formulas of the single mixed plans share the same difficulty as those of the double mixed plans in terms of the quantity 𝑃(𝑅_𝑛>𝑐_𝑟,𝑑_𝑛=𝑘) in (2.1). According to [12], finding the OC formulas of the double mixed plans involves “severe technical difficulties” in two aspects. First, the random variable 𝑅_𝑖 is truncated as 𝑅_𝑖>𝑎 or 𝑅_𝑖≤𝑎(𝑖=1,…,𝑛), which is specified by the number of nonconforming items 𝑑_𝑛. Commonly, it is not readily to obtain the distributions of algebraic functions of truncated random variables such as 𝑅_𝑛. Second, the random variables in 𝑅_𝑛 are not identically distributed when 0<𝑑_𝑛<𝑛.

Efforts have been made to derive the OC formulas of the double mixed plans. Savage [12] obtains the OC formulas for 𝑑_𝑛=0 and 1 based on the convolution for the exponential distribution. Although it is accessible in theory, the method is so complex that it is quite hard to derive the OC formulas for 𝑑_𝑛>1. For the double mixed plans under the normal distribution, Gregory and Resnikoff [11] provides the approximate OC formulas through numerical integration and asymptotic expansion, respectively. Thereafter, the exact OC formulas under the normal distribution are set up with a recursive form and tables about some related values are provided for applications in [13, 17–19].

In Theorem A.1 of the appendix, we derive an accurate formula for 𝑃(𝑅_𝑛>𝑐_𝑟,𝑑_𝑛=𝑘) to obtain the OC curve of the mixed plans under the exponential distribution. It is simple in form and easy in operation, which is believed to further increase attractions of the mixed plans.

According to Theorem A.1 in the appendix, the probability of accepting the null hypothesis under the single mixed plans is 𝐿𝑝=1−^𝑛_{𝑘=𝑐𝑑+1}𝑛𝑘^{𝑛−𝑘𝑖=0}(−1)^𝑖𝑖𝑛−𝑘1−𝐹_𝜎𝑝𝑛𝑐_𝑟−()𝑎−()𝑎𝑘+𝑖exp𝑘+𝑖2𝜎²_𝑝, (2.2) where 𝐹_𝜎𝑝(⋅) is the cumulative distribution function (c.d.f.) of Gamma (𝑛,2𝜎²_𝑝), and 𝜎_𝑝 is obtained from (1.3).

Finally, we determine the parameters (𝑛,𝑐_𝑟,𝑐_𝑑) with two proposed procedures here. One aims at protecting “good lots” from attributes aspect, and the other is to achieve optimization by minimizing the sample size.

2.1. Single Mixed Plan I: Protecting “Good Lots”

In Section 1, we point out the awkward problem of variables plans, that is, a “good lot” from the attributes aspect possibly being rejected. To solve the problem, the single mixed plan is designed to guarantee “good lots” to be accepted. This is called the single mixed plan I.

Then, the point is how to define a “good lot” from attributes aspect. According to the discussion in Section 1, we consider to find a proper region {𝑑_𝑛≤𝑐_𝑑} for rejecting the hypothesis “𝑝=𝑝₁” with a sufficient evidence under the significance level 𝛾. Then, when the number of nonconforming observations is not more than 𝑐_𝑑, we have a sufficient evidence to conclude a “good lot” from attributes aspect. This provides a way to determine the critical value 𝑐_𝑑 of the single mixed plan I.

The sample size 𝑛 of the single mixed plan I may be determined through searching in a computer program. In applications, the sample size 𝑛_𝑣 of the single variables scheme with the same error probabilities can be set as a starting value for searching 𝑛, since 𝑛 must be larger than or equal to 𝑛_𝑣 as stated above.

As the most flexible value with continuous measurement, the critical value 𝑐_𝑟 should be determined according to requirements on the two error probabilities under some provided 𝑛 and 𝑐_𝑑.

The following steps provide a guidance to find the single mixed plan I.

Step 1. Find the sample size 𝑛_𝑣 of the single variables plan with the required error probabilities and set the initial sample size 𝑛 as 𝑛=𝑛_𝑣.

Step 2. With the sample size 𝑛, determine the critical value 𝑐_𝑑 for “good lots” under the significance level 𝛾.

Step 3. For the given 𝑛 and 𝑐_𝑑, search 𝑐_𝑟 through dichotomy and try to find a single mixed plan with the required two error probabilities. If the plan is not found, set 𝑛=𝑛+1 and go to Step 2.

To illustrate the procedure, we reconsider the bullets test problem in Section 1.

Example 2.1. In this example, we set up the single mixed plan I for the bullets test problem (1.1) under the two error probabilities 𝛼=𝛽=0.15. Here, the significance level 𝛾 for defining “good lots” is set as 0.05. The steps are the following.(i)According to the results in Section 1, the single variables plan is (𝑛_𝑣,𝑐_𝑣)=(6,0.6284). Thus, we set 𝑛=6.(ii)For 𝑛=6, the region for rejecting “𝑝=𝑝₁=0.4” is {𝑑_𝑛≤0} under 𝛾=0.05. Therefore, we have 𝑐_𝑑=0.(iii) After searching 𝑐_𝑟 through dichotomy in a computer program, we find that when 𝑐_𝑟=0.6259, the single mixed plan (𝑛,𝑐_𝑟,𝑐_𝑑)=(6,0.6259,0) makes 𝛼^′=𝛽^′=0.1368<𝛼=𝛽=0.15. Thus, the determined single mixed plan I is (𝑛,𝑐_𝑟,𝑐_𝑑)=(6,0.6259,0).
With the data of the 6 inspected bullets in Section 1, since 𝑑_𝑛=0=𝑐_𝑑, the decision of accepting the null hypothesis in (1.1) should be made under the single mixed plan I (𝑛,𝑐_𝑟,𝑐_𝑑)=(6,0.6259,0).

More illustrations of the single mixed plan I are provided in Example 2.2.

Example 2.2. For the bullets test problem (1.1), let the two error probabilities be 𝛼=𝛽=0.10 and 0.05, respectively. Also take 𝛾=0.05. Tables 2 and 3 list the single mixed plans I and their performance.

From Examples 2.1 and 2.2, it can be found that the sample sizes of the single mixed plans I are equal to or a little more than those of the single variables plans with the same controlled error probabilities. In theory, sample sizes of some single mixed plans I can be improved by determining the parameter 𝑐_𝑑 more flexibly. With the principal objective of minimizing the sample size, we propose the single mixed plan II.

2.2. Single Mixed Plan II: Minimizing Sample Size

To minimize the sample size, we develop the single mixed plan II in this section. The main idea is to obtain a plan with minimized sample size by searching, and to optimize the critical value 𝑐_𝑑 (by maximizing 𝑐_𝑑) to protect “good lots”.

To make the sample size as small as possible under controlled error probabilities and to protect “good lots” at the same time, the single mixed plan II should be determined according to the following steps.

Step 1. Obtain the sample size 𝑛_𝑣 of the single variables plan and set 𝑛=𝑛_𝑣.

Step 2. Set 𝑐_𝑑=0.

Step 3. With the provided 𝑛 and 𝑐_𝑑, find the critical value 𝑐_𝑟 that maximizes 𝛼^′ under the constraint 𝛼^′≤𝛼. If no 𝑐_𝑟 satisfying 𝛼^′≤𝛼 is found, set 𝑛=𝑛+1 and go to Step 2.

Step 4. With the 𝑛, 𝑐_𝑑 and 𝑐_𝑟 from Step 3, calculate 𝛽^′. If 𝛽^′<𝛽, set 𝑐_𝑑=𝑐_𝑑+1 and then go to Step 3; If 𝛽^′=𝛽, go to Step 5; If 𝛽^′>𝛽 and 𝑐_𝑑>0, make 𝑐_𝑑=𝑐_𝑑−1 and go to Step 5; Otherwise, set 𝑛=𝑛+1 and go to Step 2.

Step 5. Readjust 𝑐_𝑟 under the determined 𝑛 and 𝑐_𝑑 to obtain a more appropriate plan when necessary.

Example 2.1 (continued). Continue the bullets test problem (1.1). The following indicates the process of finding the single mixed plan II.(i)According to the variables plan (𝑛_𝑣,𝑐_𝑣)=(6,0.6284), we make 𝑛=6.(ii)Set 𝑐_𝑑=0.(iii) Under 𝑛=6 and 𝑐_𝑑=0, the searched value is 𝑐_𝑟=0.6121 which makes 𝛼^′=0.15=𝛼.(iv) For the mixed plan (𝑛,𝑐_𝑟,𝑐_𝑑)=(6,0.6121,0), we have 𝛽^′=0.1277. Since 𝛽^′<𝛽, set 𝑐_𝑑=0+1=1.(v)Under 𝑛=6 and 𝑐_𝑑=1, search 𝑐_𝑟. The result shows that when 𝑐_𝑟=0, 𝛼^′ may reach its maximum value 0.1143.(vi) For the plan (𝑛,𝑐_𝑟,𝑐_𝑑)=(6,0,1), we have 𝛽^′=0.2333>𝛽. Therefore, make 𝑐_𝑑=1−1=0.(vii) With 𝑛=6 and 𝑐_𝑑=0, we search 𝑐_𝑟 again and find that when 𝑐_𝑟=0.6259, 𝛼^′=𝛽^′=0.1368<0.15. Thus, the determined single mixed plan II is (𝑛,𝑐_𝑟,𝑐_𝑑)=(6,0.6259,0), which is the same as the single mixed plan I.

Example 2.2 (continued). The single mixed plans II for 𝛼=𝛽=0.10 and 0.05 are also listed in Tables 2 and 3.

In Examples 2.1 and 2.2, all of the sample sizes of the single mixed plans II are the same as those of the single variables plans. It indicates that the single mixed plans II are attractive and practical due to the comparative costs with the single variables plans and the ability of protecting “good lots” to some degree.

To conduct further comparisons, we provide the performance formulas of the double mixed plans under the exponential distribution in the next section.

3. Mixed Plans for Double Acceptance Sampling

The double mixed plans were proposed decades ago and then were developed and applied in practice (e.g., [11–15, 17–19]). There are two forms of double mixed plans: the independent mixed plans and dependent mixed plans.

3.1. Double Independent Mixed Plans

Under a double independent mixed plan (𝑛₁,𝑛₂,𝑐_𝑟,𝑐₂), we should inspect 𝑛₁ items with variables measurement and accept the null hypothesis when 𝑅_𝑛1≤𝑐_𝑟. If 𝑅_𝑛1>𝑐_𝑟, another 𝑛₂ items should be inspected with attributes measurement. In the second stage, a decision is made based on the number of nonconforming items 𝑑_𝑛2 in the 𝑛₂ items, that is, accepting 𝐻₀ when 𝑑_𝑛2≤𝑐₂ or rejecting 𝐻₀ otherwise.

Obviously, the OC curve of the double independent mixed plan is 𝐿𝑝=𝑃_𝑝𝑅_𝑛1≤𝑐_𝑟+𝑃_𝑝𝑅_𝑛1>𝑐_𝑟𝑃_𝑝𝑑_𝑛2≤𝑐₂. (3.1) And its ASN formula is ASN𝑝=𝑛₁+𝑛₂𝑃_𝑝𝑅_𝑛1>𝑐_𝑟. (3.2)

The probabilities in (3.1) and (3.2) are easy to obtain because only one kind of measurement is considered in each of the stages, and the two stages are independent. Designing and selection of the double independent mixed plans for the normal distribution are discussed in [14, 15]. In the later examples, optimal double independent mixed plans, that is, plans with minimum ASN at some 𝑝, are obtained by searching in a computer program.

3.2. Double Dependent Mixed Plans

For a double dependent mixed plan (𝑛₁,𝑛₂,𝑐_𝑟,𝑐₁,𝑐₂), 𝑛₁ items should be tested with variables and attributes measurements. In the first stage, we can accept 𝐻₀ if 𝑅_𝑛1≤𝑐_𝑟 or reject 𝐻₀ if 𝑑_𝑛1>𝑐₁ where 𝑑_𝑛1 is the number of nonconforming items in the 𝑛₁ observations. Otherwise, another 𝑛₂ items should be inspected with attributes measurement. In the second stage, we should accept 𝐻₀ if 𝑑_𝑛1+𝑛2≤𝑐₂ or reject 𝐻₀ otherwise, where 𝑑_𝑛1+𝑛2 is the total number of nonconforming items in the 𝑛₁+𝑛₂ observations.

For the bullets test problem, according to Theorem A.1 in the appendix, the OC curve of the double dependent mixed plan is 𝐿𝑝=𝑃_𝑝𝑅_𝑛1≤𝑐_𝑟+𝑃_𝑝𝑅_𝑛1>𝑐_𝑟,𝑑_𝑛1≤𝑐₁,𝑑_𝑛1+𝑛2≤𝑐₂=𝑃_𝑝𝑅_𝑛1≤𝑐_𝑟+^𝑐1_𝑙=0𝑃_𝑝𝑅_𝑛1>𝑐_𝑟,𝑑_𝑛1=𝑙𝑃_𝑝𝑑_𝑛2≤𝑐₂−𝑙=𝐹_𝜎𝑝𝑛₁𝑐_𝑟+^𝑐1_𝑙=0𝑛₁𝑙^{𝑛1−𝑙𝑖=0}(−1)^𝑖𝑛₁𝑖−𝑙1−𝐹_𝜎𝑝𝑛₁𝑐_𝑟−()𝑎−()𝑎𝑙+𝑖exp𝑙+𝑖2𝜎²_𝑝×𝑃_𝑝𝑑_𝑛2≤𝑐₂−𝑙, (3.3) where 𝑑_𝑛2 is the number of nonconforming items in the second-stage 𝑛₂ observations.

The ASN formula of the double dependent mixed plan is ASN𝑝=𝑛₁+𝑛₂𝑃_𝑝𝑅_𝑛1>𝑐_𝑟,𝑑_𝑛1≤𝑐₁=𝑛₁+𝑛₂^𝑐1_𝑙=0𝑛₁𝑙^{𝑛1−𝑙𝑖=0}(−1)^𝑖𝑛₁𝑖−𝑙1−𝐹_𝜎𝑝𝑛₁𝑐_𝑟−()𝑎−()𝑎𝑙+𝑖exp𝑙+𝑖2𝜎²_𝑝. (3.4)

According to Schilling [20], a double dependent mixed plan in MIL-STD-414 is designed by combining the corresponding variables plan and attributes plan. In this procedure, 𝑛₁ should be much less than 𝑛₂, and 𝑐₁ is set equal to 𝑐₂. Although it can not find an exact double dependent mixed plan, this procedure is helpful to find an approximately matchable plan in a simple way. However, we find that such a design does not produce minimum ASN.

For the bullets test problems in Examples 2.1 and 2.2, we determine the double mixed plans with minimum ASN(𝑝₀) and then possibly small ASN(𝑝₁) by searching in a computer program. The plans and their performance are provided in Tables 1–3.

Table 1 

Plans for 𝑝₀=0.1,𝑝₁=0.4,𝛼=𝛽=0.15.

Table 2 

Plans for 𝑝₀=0.1,𝑝₁=0.4,𝛼=𝛽=0.10.

Table 3 

Plans for 𝑝₀=0.1,𝑝₁=0.4,𝛼=𝛽=0.05.

Further, the double variables plans and double attributes plans are found to make a comparison. Under the double variables plan, the null hypothesis is accepted when 𝑅_𝑛1≤𝑐_𝑟 or is rejected when 𝑅_𝑛1≥𝑐₁ at the first stage with a sample of size 𝑛₁. If 𝑐_𝑟<𝑅_𝑛1<𝑐₁, a second sample of size 𝑛₂ is needed and the null hypothesis is accepted when 𝑅_𝑛1+𝑛2≤𝑐₂ or is rejected otherwise. For the double attributes plan, the null hypothesis is accepted if 𝑑_𝑛1≤𝑐_𝑟 or is rejected if 𝑑_𝑛1≥𝑐₁ for the first-stage 𝑛₁ observations. If 𝑐_𝑟<𝑑_𝑛1<𝑐₁, a second sample of size 𝑛₂ is inspected and the null hypothesis is accepted if 𝑑_𝑛1+𝑛2≤𝑐₂ or is rejected otherwise.

To better analyze the performance of the mixed plans, the OC and ASN curves are illustrated in Figures 1, 2, 3, 4, 5, and 6. Interestingly, all of the plans have almost the same OC curves except the single attributes plans. Therefore, we provide the OC curves of the single mixed plans II (as a representative) and the single attributes plans in Figures 1–3. Figures 4-6 show the ASN curves of the double acceptance sampling plans.

Figure 1 

OC curves of plans for 𝑝0=0.1,𝑝1=0.4,𝛼=𝛽=0.15 (“—”: OC curve of the single mixed plan II; “- -": OC curve of the single attributes plan).

Figure 2 

OC curves of plans for 𝑝0=0.1,𝑝1=0.4,𝛼=𝛽=0.10 (“—": OC curve of the single mixed plan II; “- -”: OC curve of the single attributes plan).

Figure 3 

OC curves of plans for 𝑝0=0.1,𝑝1=0.4,𝛼=𝛽=0.05 (“—": OC curve of the single mixed plan II; “- -”: OC curve of the single attributes plan).

Figure 4 

ASN curves of double plans for 𝑝0=0.1,𝑝1=0.4,𝛼=𝛽=0.15 (“:”: ASN curve of the double independent mixed plan; “—”: ASN curve of the double dependent mixed plan; “−⋅”: ASN curve of the double variables plan; “- -”: ASN curve of the double attributes plan).

Figure 5 

ASN curves of double plans for 𝑝0=0.1,𝑝1=0.4,𝛼=𝛽=0.10 (“:”: ASN curve of the double independent mixed plan; “—”: ASN curve of the double dependent mixed plan; “−⋅”: ASN curve of the double variables plan; “- -”: ASN curve of the double attributes plan).

Figure 6 

ASN curves of double plans for 𝑝0=0.1,𝑝1=0.4,𝛼=𝛽=0.05 (“:": ASN curve of the double independent mixed plan; “—": ASN curve of the double dependent mixed plan; “−⋅": ASN curve of the double variables plan; “- -": ASN curve of the double attributes plan).

According to Figures 1–3, the single attributes plans perform a little stronger protection for producer and consumer's risks than other plans. Based on Figures 4–6, it is clear that the double dependent mixed plans are preferable in ASN than the double independent mixed plans. Especially, the ASN curves of the double independent mixed plans increase as 𝑝 becomes large, while the double dependent mixed plans have small ASNs beyond (𝑝₀,𝑝₁). Moreover, the ASN curves of the double dependent mixed plans lie between those of the double variables plans and the double attributes plans in (𝑝₀,𝑝₁).

4. Conclusions and Remarks

We have presented mixed plans for the single and double acceptance sampling problems. Our proposed methods combine certain strengths of the variables sampling plans and the attributes sampling plans. Numerical examples have shown that they perform well in various cases.

Compared to the variables sampling plans, the major advantages of our proposed mixed plans are as follows. (i) The mixed plans avoid the problem of rejecting a lot when no nonconforming item is inspected. (ii) Since attributes results are considered, the mixed plans can tolerate small quantity of outliers which depart from the “natural” form of the underlying distribution and then protect a generally good lot. Hence, the inference on the percent nonconforming does not depend on the assumed distribution so much like the variables sampling. (iii) For rejected lots, the actual number of nonconforming items is helpful to convince producers psychologically.

Compared to the attributes sampling plans, the major advantages of our proposed mixed plans are as follows. (i) The mixed plans can reduce sample sizes compared with the attributes plans with the same error probabilities. (ii) The mixed plans contain a far more careful inspection and analysis based on variables measurement. (iii) Due to the variables feature, the mixed plans introduce more flexibility to obtain desired plans with controlled risks.

With our derived formulas for evaluating their exact performance, our proposed mixed plans should be useful for various applications, especially for those in which the related population distributions are exponential. Comparing the proposed single mixed plans with the double mixed plans, the former may be more convenient to use, but the latter would require less average sample sizes. When the test cost is our major concern and the quality is assured near 𝑝₀ or 𝑝₁, the double dependent mixed plans are recommended. Otherwise, we recommend to use the single mixed plans for testing the percent nonconforming conveniently and cheaply.

In our opinion, much future research is required to study the possibility to optimize the mixed plans, especially the double mixed plans. Further, the development of mixed plans for the sequential sampling cases is still an open problem and requires much future research as well.

Appendix

A Related Formula for Mixed Plans’ Performance

Theorem A.1. Suppose 𝑅₁,…,𝑅_𝑛 are i.i.d. random variables from the exponential distribution Exp(2𝜎²) where 2𝜎² is the expectation of 𝑅_𝑖(𝑖=1,…,𝑛). Let 𝑅_𝑛=∑^𝑛_𝑖=1𝑅_𝑖/𝑛, 𝑑_𝑛=∑^𝑛_𝑖=1𝐼_(𝑎,+∞)(𝑅_𝑖). Then, for any integer 𝑙(0≤𝑙≤𝑛), 𝑃𝑅_𝑛>𝑐_𝑟,𝑑_𝑛=𝑛𝑙=𝑙^𝑛−𝑙_𝑖=0(−1)^𝑖𝑖𝑛−𝑙1−𝐹_𝜎𝑛𝑐_𝑟−()𝑎−()𝑎𝑙+𝑖exp𝑙+𝑖2𝜎², (A.1) where 𝐹_𝜎(⋅) is the c.d.f. of Gamma (𝑛,2𝜎²).

Proof. Let 𝐸_𝑖={0≤𝑅_𝑖≤𝑎},𝑖=1,…,𝑛; 𝐶_𝑘={0≤𝑅₁,…,𝑅_𝑘≤𝑎,𝑅_𝑘+1,…,𝑅_𝑛>𝑎,𝑅_𝑛>𝑐_𝑟},𝑘=0,…,𝑛; 𝐷_𝑘={𝑅_𝑘,…,𝑅_𝑛>𝑎,𝑅_𝑛>𝑐_𝑟},𝑘=1,…,𝑛, 𝐷_𝑛+1={𝑅_𝑛>𝑐_𝑟}.
Since 𝐷_𝑘+1=𝐶_𝑘+⋃^𝑘_𝑖=1(𝐷_𝑘+1𝐸^𝑐_𝑖)(𝑘=0,1,…,𝑛), we have 𝑃𝐶_𝑘=𝑃𝐷_𝑘+1−𝑃^𝑘_𝑖=1𝐷_𝑘+1𝐸^𝑐_𝑖=𝑃𝐷_𝑘+1−^𝑘_𝑖=1𝑃𝐷_𝑘+1𝐸^𝑐_𝑖−_{1≤𝑖<𝑗≤𝑘}𝑃𝐷_𝑘+1𝐸^𝑐_𝑖𝐸^𝑐_𝑗+⋯+(−1)^𝑘−1𝑃𝐷_𝑘+1𝐸^𝑐₁⋯𝐸^𝑐_𝑘=𝑃𝐷_𝑘+1−𝑘1𝑃𝐷_𝑘+𝑘2𝑃𝐷_𝑘−1+⋯+(−1)^𝑘𝑃𝐷₁=^𝑘_𝑖=0(−1)^𝑖𝑘𝑖𝑃𝐷_{𝑘+1−𝑖},𝑘=0,1,…,𝑛. (A.2)
As 𝑅_𝑖(𝑖=1,…,𝑛) are i.i.d. ~ Exp(2𝜎²), it is obvious that 𝑅_𝑖−𝑎∣𝑅_𝑖>𝑎(𝑖=1,…,𝑛) are i.i.d. ~ Exp(2𝜎²). Therefore, 𝑃𝐷_𝑘=𝑃𝑅_𝑘,…,𝑅_𝑛>𝑎,𝑅_𝑛>𝑐_𝑟=𝑃𝑅_𝑛>𝑐_𝑟∣𝑅_𝑘,…,𝑅_𝑛>𝑎𝑃𝑅_𝑘,…,𝑅_𝑛𝑅>𝑎=𝑃₁+⋯+𝑅_𝑘−1+𝑅_𝑘−𝑎+⋯+𝑅_𝑛−𝑎>𝑛𝑐_𝑟−(𝑛−𝑘+1)𝑎∣𝑅_𝑘,…,𝑅_𝑛𝑅>𝑎×𝑃_𝑘,…,𝑅_𝑛=>𝑎1−𝐹_𝜎𝑛𝑐_𝑟−()𝑎−()𝑎𝑛−𝑘+1exp𝑛−𝑘+12𝜎²,𝑘=1,…,𝑛+1. (A.3)
Then, for 0≤𝑙≤𝑛, 𝑃𝑅_𝑛>𝑐_𝑟,𝑑_𝑛=𝑛𝑙=𝑙𝑃𝐶_𝑛−𝑙=𝑛𝑙^{𝑛−𝑙𝑖=0}(−1)^𝑖𝑖𝑃𝐷𝑛−𝑙_{𝑛−𝑙+1−𝑖}=𝑛𝑙^{𝑛−𝑙𝑖=0}(−1)^𝑖𝑖𝑛−𝑙1−𝐹_𝜎𝑛𝑐_𝑟−()𝑎−()𝑎𝑙+𝑖exp𝑙+𝑖2𝜎². (A.4) The proof is completed.

Acknowledgment

This work was partly supported by the National Natural Science Foundation of China (NSFC) under the project Grant number 11001083. The authors cordially thank the editor and anonymous referees for their helpful reviews.