I am having a hard time with one of the chapter questions regarding process performance: I have followed the formula for Pp and I am getting it wrong: The problem is:
A process produces a part that is required to measure between 79.75 and 80.25 inches wide. As part of the supplier qualification process, samples were collected and yielded a process mean of 79.5 inches, and the long-term standard deviation was 0.625 inches. What is the Pp for this process?
I am using the formula Pp= USL-LSL/ 6(sigma). I am assuming the USL= 80.25 and LSL =79.75. The answer is .333 but I am not getting that with following this formula.
80.25 - 79.75 = 0.5
6 x .625 = 3.75 (You multiple the six by your standard deviation)
0.5 / 3.75 = .133
That suppliers process is not meeting expectations, as demonstrated by the sample size average being below the LSL.
Thank you so much! That is what I did, but the book answer is giving me an answer of .333; so I thought I was doing it wrong?
There is also a question in the study guide similar this where I would think you would use the same equation as above to find Cp but it says to use this equation instead:
Cp=USL-LSL/ 6(R bar/d2)
Are you familiar with this equation and why we would use it to find Pp or Cp?
With members and customers in over 130 countries, ASQ brings together the people, ideas and tools that make our world work better.
ASQ celebrates the unique perspectives of our community of members, staff and those served by our society. Collectively, we are the voice of quality, and we increase the use and impact of quality in response to the diverse needs in the world.
© American Society for Quality. All rights reserved.