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Six Sigma for Electronics Design and Manufacturing
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                     dence; k is the number of boundaries, and DOF is the degrees of free-
                     dom. Selected values of the   table are given in Table 5.3.
                                     2
                     2.4.3 Example of   goodness of fit to normal
                     distribution test
                     Thirty parts were selected from a production line that was assumed to
                     be normally distributed, and lengths were measured in  m. The data
                     was sorted in ascending order and five boundaries were created for
                     the sorted data set.     2
                       Table  2.5  shows  the  original  as  well  as  the  sorted  data  set  of  30
                     measurements. The average is calculated at 8843.43 and the standard
                     deviation  is  743.  The  boundary  limits  are  a  minimum  of  less  then
                     8000 to a maximum of more than 9500 in 500 increments. The first
                     two boundary calculations are shown for illustration:
                     Boundary 1
                        P 1 = P(  < 8000) = P{(  –  )/ }
                        z = (8000 – 8843.43)/743 = –1.135
                        P 1 = f(z) = 0.128 from normal distribution tables
                        Expected frequency = NP i = 30 · 0.128 = 3.84
                                               2
                         2
                                                             2
                          contribution = (m 1 – nP 1 ) /nP 1 = (4 – 3.84) /3.84 = 0.0067
                     Boundary 2
                        P 2 = P(8000 <   < 8500)
                        z 2 = (8500 –  )/  = (8500 – 8843.43)/743 = –0.462
                        z 1 = –1.135 (from previous boundary)
                        P 2 = f(z 2 ) – f(z 1 ) = 0.3228 – 0.128 = 0.1948
                        Expected frequency = NP i = 30 · 0.1948 = 6.27
                         2
                          contribution = (m i – nP i ) /nP i = (8 – 5.844) /5.844 = 0.795
                                                             2
                                              2
                       Other results for the remaining boundaries are shown in Table 2.5.
                     It  can  be  seen  that  the  total  number  of  observed  and  expected  fre-
                     quencies in all of the boundaries should be equal to the data set total
                     of 30. The total probability P i should also equal to 1, and the total ex-
                     pected frequency should equal 30.
                       The total chi-square value for the data set is 2.36, which falls be-
                     tween the limits of   = 0.10 (90% confidence) of 1.064 and   = 0.5 (50%
                                               2
                     confidence)  of  3.357  for  the   distribution  with  degrees  of  freedom
                     DOF = 4 (5 boundaries – 1), from Table 5.3. That implies that the data
                     set is normal since it corresponds with the normal distribution expec-
                     tations. A plot is shown of the data set values versus their normal
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