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          August 31, 2006
 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),
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