Page 497 - Design for Six Sigma a Roadmap for Product Development
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456 Chapter Twelve
2. If AB 0, then
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
3. If AC 0, then
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
4. If BC 0, then
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
5. If A 0, then
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
By simultaneously assuming that ABC 0 and AB 0, solve
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
We can get the solutions m 22.1 and x 24.9.
Note that many similar pairs of equations will not give us answers; for
example, by simultaneously assuming that ABC 0 and AC 0, we get
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
We cannot get a solution of m and x from above because the signs for m
and x in the above two equations are the same.
Another pair of equations that we can derive solutions for m and x are
ABC 0 and A 0:
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
21.1 16.3 m 17.3 16.1 14.5 x 23.3 0
We can get the solutions m 15.7 and x 18.5.
By putting solution (a): m 22.1, x 24.9, and solution (b): m 15.7,
x 18.5 back in Table 12.18, we can draw the following half-normal plots
illustrated in Fig, 12.20.
By merely looking at the half-normal plots in Fig. 12.20a and b,we
cannot judge which set of estimated missing values is better. So either
(1) m 22.1, x 24.9 or (2) m 15.7, x 18.5 could be accepted as the
estimated missing values. In Draper and Stoneman’s method, one of
these solutions will be put back in Table 12.19 and we will complete the
experimental data analysis.
