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24th October 2001, 06:05 AM
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Cp & CpK
We conducted Process Capability for a particular characteristic but found that the readings do not follow a normal distribution. The reading are given below. The specification is 6.75/6.45. Please let us know how the Cp and Cpk can be calculated for such a distribution.
26 sub groups are given below. Readings were taken once in an hour.
1 2 3 4
6.604 6.607 6.604 6.608
6.602 6.605 6.602 6.602
6.602 6.603 6.602 6.595
6.606 6.604 6.606 6.604
6.606 6.608 6.605 6.606
5 6 7 8
6.608 6.602 6.607 6.604
6.607 6.606 6.605 6.607
6.605 6.605 6.603 6.607
6.609 6.605 6.604 6.606
6.605 6.606 6.604 6.603
9 10 11 12
6.604 6.603 6.606 6.606
6.605 6.605 6.608 6.606
6.606 6.592 6.607 6.607
6.608 6.606 6.606 6.608
6.606 6.606 6.607 6.604
13 14 15 16
6.606 6.607 6.606 6.605
6.602 6.604 6.608 6.605
6.601 6.605 6.603 6.607
6.606 6.606 6.606 6.603
6.607 6.608 6.607 6.605
17 18 19 20
6.605 6.605 6.605 6.602
6.604 6.600 6.605 6.607
6.603 6.602 6.582 6.600
6.607 6.605 6.606 6.606
6.606 6.605 6.605 6.606
21 22 23 24
6.606 6.604 6.606 6.605
6.605 6.607 6.607 6.607
6.606 6.604 6.592 6.600
6.604 6.606 6.605 6.605
6.605 6.606 6.602 6.602
25 26
6.605 6.607
6.605 6.608
6.604 6.605
6.606 6.606
6.605 6.606
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24th October 2001, 09:31 AM
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Andrews - I am assuming your problem is not the data itself as the limits are so far outside the readings the capability of the process is not in question. I also assume you need a way to present a realistic Cpk/CP number for the process.
I calculate the Cpk on the data given as 4.59 with a CP of 8.67. The process shows skewness of -2.12.
To arrive at these numbers, you need to fit the population distribution with a Pearson curve rather than assuming a normal distribution.
Most SPC programs allow you the option of forcing a Pearson curve. The software I used recommends it.
Hope this helps.
Dave
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24th October 2001, 10:33 AM
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KenK Number 2
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Using MINITAB, I used probablity plots to look at the fit to several distributions - Normal, Lognormal, and Weibull.
The Weibull fit is really pretty good, although the lowest few data values do not fit well. I would think the fit is quite sufficient.
The Weibull-based capability indices are:
Pp = 19.21
Ppk = 2.40
Note: MINITAB does not estimate Cp or Cpk because the Weibull distribution does not behave like the normal distribution. I suppose we could fit a seperate Weibull distr. for each subgroup and then pool the slopes, but don't think it would add much value.
Clearly your process is adequately tight with respect to your tolerances. With these high values I wouldn't worry about the distributional fit too much.
One of advantages of the Weibull fit versus the Pearson curve fit is that we can easily assess the fit to Weibull using probablity plots. MINITAB uses the Weibull since it is a pretty "flexible" distribution.
I have yet to see a pearson curve tool (I've seen those in StatGraphics & Statistica) that provides an assessment of fit - they usually just give the best fit available, whatever that is. Maybe newer version do provide Pearson probability plots though.
Also note that subgroup 19 has an unusually large range - you might want to check that subgroup for hidden issues.
Ken K.
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26th October 2001, 05:25 AM
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Hello Mr.Scott and KenK
Thank you very much for the speedy response.
Sorry for my ignorance.What is a Pearson and Weibull curve ? In what way is it different from normal curve.Please also let me know what is the change in formula for calculating the control limits ,Cp,Cpk in cases where the readings do not follow a normal distribution.We would be grateful if you also tell me how you calculated the skewness value.Please help.
Andrews
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26th October 2001, 06:59 AM
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Andrews,
A little more information may help. What is the process, what is the measuring instrument? There are a lot of transforms available (like Pearson, etc...) Any particular reason why the distribution is not normal?
- Atul.
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27th October 2001, 05:30 AM
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Hello Mr. Atul,
The readings taken were from a machining operation on multi-spindle screw machine and the measuring instrument used was 0.001 Micrometer. The dia 6.75 / 6.45 is maintained by a roughing using dovetail form tool and finishing using circular size tool. We suspect that the sizing operation contributes to this non-normal distribution but the advantage of using this tool is that we can get a very good consistency in the job.
You had mentioned about some "transform" that could be used to solve this problem. Can you please elaborate and also tell me what effect it has on the formula used to calculate the control limits, Cp and Cpk. Please also note that the readings are not stable when X-R chart is plotted.Can you advice me what to do?
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27th October 2001, 09:35 AM
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Andrews,
Considering you have a non-normal distribution is it a problem with the gages or the machines?
Machines perform the way they are programmed, gages are a different matter.
Please tell me if I am wrong, but your machines are putting out product that doesn't meet spec? according to your gages?
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31st October 2001, 10:41 AM
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