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JWBK119-13
Comparison with Bootstrap Confidence Limits 207
Substituting these data in (13.1)--(13.4) gives
20.8 − 10
ˆ
C p = = 1.5,
6 × 1.2
1
m = (20.8 + 10) = 15.4,
2
ˆ |15.4 − 17.2|
k = = 0.3
(20.8 − 10)/2
ˆ
C pk = (1 − 0.3) 1.5 = 1.05.
ˆ
Since k falls between 0.2 and 0.5, Table 13.5 recommends C pk2 as the lower AM confi-
dence limit for C pk .
¯
The expression for C pk2 in (13.15) requires the determination of k. From Figure 13.4
we can determine k from p and C p , which are easily calculated from (13.5) and (13.6)
respectively:
−4
p = [−3 (1 + 0.3) 1.5] + [−3 (1 − 0.3) 1.5] = 8.164 × 10 ,
2
χ − 1, 1 − 0.05/2
¯ 50
C p = √ × 1.5 = 1.796.
50 − 1
Then k can be obtained either by direct estimation from Figure 13.4 as 0.40 or by
solving equation (13.11) as below using some numerical solvers:
−4
k = k : L k : p = 8.164 × 10 , C p = 1.796 = 0 = 0.415.
From equation (13.15), using the exact k-value, C pk2 is determined as
C pk2 = (1 − 0.415) 1.5 = 0.878.
Clearly, we have little faith in the new process meeting the customer’s quality require-
ment.
We can also proceed to calculate the confidence limits for the fraction noncon-
forming for this particular example. It is probable that the customer will be most
interested in the value ¯p. From Figure 13.2 we can locate p from k and C p (recall that
C pk2 = (1 − k)C p ), which were determined earlier as 0.415 and 1.5, respectively. Then
p can be determined either by direct estimation from Figure 13.2 as approximately
4.0 × 10 −3 or by calculating it from (13.5) using k and C p as
−3
¯ p = [−3 (1 + 0.415) 1.5] + [−3 (1 − 0.415) 1.5] = 4.24 × 10 .
13.7 COMPARISON WITH BOOTSTRAP CONFIDENCE LIMITS
We have arrived at some approximate confidence limits that are computationally easy
toobtainandprovideadequateprotectionintermsoftheircoverageprobability.Inthis
section we compare the proposed approximate method (AM) confidence limits with
three nonparametric bootstrap confidence limits for C pk : the standard bootstrap (SB),