Page 41 - Mechanical Engineers' Handbook (Volume 2)
P. 41

30   Instrument Statics

                                                             (O   E ) 2
                                                                   j
                                                               j
                                                        2 exp                                 (62)
                                                                E j
                          When using Eq. (62) at least five items per class interval are normally used, and a continuity
                          correction must be made if less than three class intervals are used.

                          Example 9 Use of the Chi-Square Distribution. In a test laboratory the following record
                          shows the number of failures of a certain type of manufactured part.

                                                                   Number of Parts
                                        Number of Failures    Observed      Expected a
                                              0                 364           325
                                              1                 376           396
                                              2                 218           243
                                              3                  89            97
                                              4                  33            30
                                              5                  13             7
                                              6                  4              1
                                              7                  3              0
                                              8                  2              0
                                              9                  1              0
                                                             1103
                                        a From an assumed statistical model that the failures are random.
                             At a 95% confidence level, determine whether the failures are attributable to chance
                          alone or to some real cause. For H , assume failures are random and not related to a cause
                                                     0
                          so that the expected distribution of failures follows the Poisson distribution. For H , assume
                                                                                          1

                                                                                              r
                          failures are not random and are cause related. The Poisson probability is P   e   /r !,
                                                                                       r
                          where      x P : i
                                     i
                               (0)        (1)         (2)        (3)        (4)        (5)
                                                                                            13
                                   364
                                              376
                                                          218
                                                                     89
                                                                                 33
                                   1103       1103       1103       1103        1103       1103
                                  24     21     16     9

                                  1103  1103   1103  1103
                               1346
                                      1.22
                               1103
                                   0.295 and E   P n   325. Similarly P   0.360, P   0.220, P   0.088,...
                          Then P 0          0    0                1         2         3
                          and, correspondingly, E   396, E   243, E   97,...are tabulated above for the expected
                                                            3
                                            1
                                                    2
                          number of parts to have the number of failures listed.
                             Goodness-of-fit test—H represents random failures and H represents real cause:
                                                                           1
                                                0
                                          P[  2      2  ]   	   0.05
                                             exp  Table
                                                    2      ( , 	)     (5,0.05)   11.070
                                                                   2
                                                          2
                                                   Table
                                                          (0   E) 2
                                                     2 exp
                                                            E
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