This is just an Excerpt from a larger document, click here to view the entire document.Double Sampling (Two Stage) Testing Procedures
In hypothesis testing, we often define a Null (H0) that expresses the desired value for the parameter under test (e.g., acceptable quality). Then, we define the Alternative (H1) as the unacceptable value for such parameter. We take a sample of pre-determined size "n" and, based upon the result obtained from drawing such a single fixed-size sample, we make a decision regarding these two hypotheses. This process is a single stage sampling procedure.
In a two-stage testing procedure (double sampling) one first draws a sample of size n1 and compares the number "X" of items "of interest" (e.g., "non compliant") with two integers: c1, c2. If X < c1, we accept the null hypotheses (H0) that the batch is of acceptable quality; and if X > c2, we accept the alternative hypotheses (H1) that the batch quality is unacceptable. However, if c1 ≤ X ≤ c2, we draw a second sample of size n2 and compare the total number X' (of items "of interest" found in the combined sample of size n1 + n2) with a third integer c3 (which may be the same as c2). We take the final decision based upon whether such combined X' is greater than, or equal to c3. We thus describe double sampling plans via the integers n1, n2, c1, c2, and c3, and denote them S (n1, n2, c1, c2, c3).
The logic behind double sampling schemes is that, if things are clearly good or bad (X < c1 or X > c2) then we make a decision based on the first sample only. If there are some doubts (c1 ≤ X ≤ c2), then we draw a second sample, collecting additional information before reaching a decision. This method lowers the risk of making the wrong decision, when initially things do not appear to be so clear, at the cost of a longer and more expensive process (i.e., drawing the second sample).
Consider the following example. Assume that we have a device with mission time "T" (say, 20 hours), which requires (H0) a reliability of say, 0.9, with confidence 0.95. To test this hypothesis, we place "n" (say 20) items on test for the "T" hours, and then count how many items survive. The number of survivals "X" is distributed as a Binomial (n = 20, p), where "p" is the reliability of the device; that is, the probability that such device survives beyond mission time "T." We then express such a Binomial probability model as:
Now, assume that a reliability of 0.8 or less is unacceptable (H1) and define the double sampling Plan S (n1 = 20, n2 = 20, c1 = 14, c2 = 15, c3 = 33), as described. We draw a first sample of size n1
= 20 and count the number of survivals "X." If X > 15, we don't reject H0. If X < 14 we reject H0 (reliability is unacceptable). If X is 14 or 15, we draw a second sample of size n2 = 20 and count the number of survivals (Y). Then, only if X + Y < 33 we reject H0 (and decide that the device reliability is unacceptable).
Plan S (n1 = 20, n2 = 20, c1 = 14, c2 = 15, c3 = 33) is constructed in the following way. For the first sample (n1) we selected, from the Binomial (n = 20, p = 0.9) tables, number "c2" = 15, because the probability of acceptance (survivals X > 15) of a "good" batch (reliability > 0.9) is P0.9 {X ≥ 16} = 0.957. We then selected number "c1" = 14 because the probability of rejecting a "good" batch P0.9 {X ≤ 13} = 0.002. Then, from the Binomial (n = 20, p = 0.8) table, the probability of "accepting a bad batch" (reliability 0.8 and X ≥ 16) is 1- P0.8 {X ≤ 15} = 1 - 0.3704 = 0.630, and that of "rejecting a bad batch" is P0.8{X ≤ 13} = 0.087. If the number of survivals is X = 14 or 15, results are considered inconclusive. The probability of an inconclusive result, when the batch is good, is 0.041, and when the batch is bad, is 0.284. If so, we take a second sample of n2 = 20 and define number c3 = 33. We accept the batch if the total survivals are 33 or more; and reject it if the survivals are less than 33. We don't claim that plan S is optimal; but it provides a good illustration of the construction approach.
The probability of acceptance for such double sampling plan S, for any p, is given by the following equation.
P{Accepting Batch} = P{Accepting Initially}
+ P{Initially Inconclusive Then Accept at 2nd}
= P{First Successes ≥ 16} + P{First Successes = 14
or 15 and Combined Successes ≥ 33} 20
= Σ Bin(x; n = 20, p) + Bin(x = 14; n = 20, p) x = 16
Table 1. Individual and Cum. Probs. for Binomial (n = 20; p = 0.9 and 0.8)
Surv.
P = 0.9
P = 0.8
Cum. For P = 9
Cum. For P = 8
10
0.000006
0.002031
0.00001
0.00259
11
0.000053
0.007387
0.00006
0.00998
12
0.000356
0.022161
0.00042
0.03214
13
0.001970
0.054550
0.00239
0.08669
14
0.008867
0.109100
0.01125
0.19579
15
0.031921
0.174560
0.04317
0.37035
16
0.089779
0.218199
0.13295
0.58855
17
0.190120
0.205364
0.32307
0.79392
18
0.285180
0.136909
0.60825
0.93082
19
0.270170
0.057646
0.87842
0.98847
20
0.121577
0.011529
1.00000
1.00000
x [Bin(x = 19; n = 20, p) + Bin(x = 20; n = 20, p)]
+ Bin(x = 15; n = 20, p) x [Bin(x = 18; n = 20, p)]
+ Bin(x = 19; n = 20, p) + Bin(x = 20; n = 20, p)
Notice that, to obtain a result of 33 survivals or more, in the combined first and second samples, we have to obtain 14 or 15 successes (inconclusive results) in the first sample and then, enough successes in the second (say 18, 19, or 20) to add up to 33 or more. For the case where reliability p = 0.9, the double sampling plan acceptance is 0.982, instead of just 0.957 for a fixed sample test with n = 20 and c = 16:
P{Accepting Batch} = P{Accepting Initially}
+ P{Initially Inconclusive Then Accept at 2nd
= 0.957 + 0.0089 x (0.2701 + 0.1215) + 0.0319
x (0.2851 + 0.2701 + 0.1215) = 0.982
The probability of rejection of our double sampling plan S is obtained just by substituting "acceptance" for "rejection," in the equations. For a true reliability p = 0.09, the probability of incorrect rejection is 0.018, instead of just P{X < 16} = 1 - 0.957 =
0.043, which is the corresponding probability for a fixed sample plan with n = 20 and c = 16:
P{Rejecting Batch} = P{Rejecting Initially}
+ P{Initially Inconclusive Then Reject at 2nd}
= P{First Successes < 13} + P{First Successes
= 14 or 15 and Combined Successes < 33} 13
= Σ Bin(x; n = 20, p) + Bin(x = 14; n = 20, p) x = 0
x {1 - [Bin(x = 19; n = 20, p) + Bin(x = 20; n = 20, p)]}
+ Bin(x = 15; n = 20, p) x {1 - [Bin(x = 18; n = 20, p)
+ Bin(x = 19; n = 20, p) + Bin(x = 20; n = 20, p)]}
= 0.0024 + 0.0089 x (1 - (0.2701 + 0.1215))
+ 0.0319 x (1 - (0.2851 + 0.2701 + 0.1215)) = 0.018
The preceding shows how the double sampling scheme, when compared with a single sample test, not only increases the probability of accepting a good batch but also reduces the probability of rejecting a good one, even if the initial test results are inconclusive. This characteristic is their strongest advantage that, in many cases, far outweighs the extra cost and effort involved in implementing such double sampling schemes.
