ms-95 mba assignment july dec 2010 Question 5

5) A sample of 48 tools produced by a machine shows the following sequence of good (G) and defective (D) tools

G

G

G

G

G

G

D

D

G

G

G

G

G

G

G

G

G

G

D

D

D

D

G

G

G

G

G

G

D

G

G

G

G

G

G

G

G

G

D

D

G

G

G

G

G

D

G

G

Test the randomness at the 0.05 significance level.

Solution:

       The numbers of D’s and G’s are N1 = 10 and N2 = 38, respectively, and the number of runs is V = 11.

Thus the mean and variance are given by

                        2 (10) (38)

           mv   =   _______       +1 = 16.83

                         10 + 38

        s2v   = 2 (10) (38)   [2 (10) (38) - 10 - 38]

                                        _________________

                           

                                        (10 + 38)2 (10 + 38 - 1)

                = 4.997

So that s v = 2.235

            For a two-tailed test at the 0.05 level, we would accept the Hypothesis Ho of randomness if -1.96 < z < 1.96 and would reject it otherwise since the z score corresponding to v = 11 is

         V - mv           11 - 16.83

Z =   ______                             =    ________              =  -2.61

            sv                2,235

           

      -2.61 < -1.96, we can reject H0 at the 0.05. The test shows that there are too few runs, level. indicating a clustering (or bunching) of defective tools. In other words - there seems to be a trend pattern in the production of defective tools. Further examination of the production process is warranted.