ANOVA Example (Continue From Last Post)

Step 1 : Normality Test


Step 3 : Process stability assessment


Step 4 : Process capability assessment


Step 5 : Test for equal variance


Step 6 : ANOVA


Conclusion :

  1. ANOVA analysis shows no significant difference of mean between M/C1 M/C2 and M/C3. However data of M/C 1 and M/C 3 show mean at the lower side to the control dimension.
  2. Test for equal variance shows data from 3 machine having equal variance, but M/C1 produce highest variation among the 3.
  3. Process capability analysis show Cpk value at 0.79. The process in general not capable.
  4. Process capability analysis show process is stable and no special cause happen in the process.
  5. Normality test show all data collected are normal data

Advice to my colleague :

Quite risky to mix the parts into one bin due to concern of machine to machine variation.

He need to do something to minimize machine to machine variation, bring the process mean to the target

He can consider to mix the parts once the line stabilized.

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