In acceptance sampling by variables, we make and record actual measurements instead of simply classifying items as good as or as bad as in attributes sampling. This difference in procedure changes the details of determining a plan that meets our specifications of AQL, α, LTPD, and β because the appropriate statistical distribution is now the normal distribution instead of distributions for proportions. Conceptually, however, the basic ideas on which the control of outgoing quality is maintained remain the same. The discriminating power of a plan is represented by an OC curve, which shows the probability of acceptance for different levels of actual quality presented to the plan. To specify a plan that gives the desired protection requires basically the same procedure as for sampling by attributes.

Kinds of Variables Sampling Plans:

There are two main categories, which depend on our knowledge of the population standard deviation, Sx: where Sx is known and constant and where Sx is known and constant and where Sx is unknown and may be variable. Furthermore, the classification may be extended to the nature of the decision criterion; that is, where the criterion is the average of measurements and where the criterion is percent defectives (PD). To summarize, the classification is as follows:

1. Sx is known and constant

i) The decision criterion is expressed as the average of measurements, xa

ii) The decision criterion is expressed as PD in the lot

2. Sx is unknown and may be variable

i) The decision criterion is expressed as the average of measurements, xa

ii) The decision criterion is expressed as PD in the lot.

Factors influencing the choice among types of sampling plans.

Factor Single Double Sequential

Protection against rejecting high

Quality lots and accepting low

quality lots Same Same Same

Total inspection cost Highest Intermediate Least

Amount of record keeping Least Intermediate Most

Variability of inspection load Constant Variable Variable

Sampling costs when all

samples can be taken as

needed Least Highest Intermediate

Accurate estimate of lot

Quality Best Intermediate Worst

Sampling costs when dependent

on the number of samples

drawn Least Intermediate Highest

Relationship with suppliers,

that is, give more than one chance Worst Intermediate Best

We will discuss plan design for which Sx is known and constant the decision criterion is expressed by the average of measurements. Procedures for the situation where Sx is unknown and may be variable may be found in Duncan (1974).

Variables Sampling Plans where Sx is known and Constant:

We will discuss these procedures in the context of an example in which steel bar stock is received in batches from a vendor. It has been determined that a tensile strength of 90,000 pounds per square inch (psi) is required, and we wish to specify that Pa = 10 percent for lots of this average tensile strength. Lots with an average tensile strength of 95,000 psi are regarded as good quality, and we wish to specify that Pa = 95 percent for lots of this average tensile strength. We have a long history with this supplier, so Sx is known to be 6000 psi, and the measurements are normally distributed. To summarize, our plan specifications are

AQL = 95,000 psi

Xt = 90,000 psi (equivalent to LTPD in attributes sampling)

α = 5 percent

β = 10 percent

We wish to determine a sampling plan that will indicate an acceptance average for sample tests, xa, and a sample size n that will accept lots according to our specifications. The acceptance average for sample tests xa is equivalent to acceptance number c, in attributes sampling plans. In other words, when xa is less than the value we determine as critical, the lot from which the sample was drawn will be rejected and returned to the supplier. Lots for which the sample average tensile strength is equal to or greater than xa will be accepted.

The standard deviation of the sampling distribution of means for samples of size n will be 6000√n. To be accepted 95 percent of the time, AQL=95,000 psi must be 1.654 S units above the grand means, x = xa since 5 percent of the area under a normal curve is beyond µ + 1.645S. Therefore, xa – 95,000 is 1.64Sx units. Then,

xa – 95,000 = – 1.645 x (6000/√n)

Also, to ensure that lots of average tensile strength xt = 90,000 have only a 10 percent chance of acceptance, which ensures that samples with the xi=90,000 psi must be 1.28 S units below the grand mean,

xa – 90,000 = + 1.282 x (6000/√n)

We now have two independent equations with two unknown, xa and n. They may be solved simultaneously to yield the following values:

xa = 92,000 psi

n=12

The relationships of the various elements of the problem can be drawn on a graph that answer the question, What is the grand mean, x, and the sample size, n, of a normal distribution with S = 6000psi and Sx = 6000/√n?’

The OC curve for the plan just described is determined by cumulating the areas under the normal curve for the sampling distribution of sample size n. The OC curve for this plan is shown.

—