Calculating Cp and Cpk on a Non-Normal Distribution

A

Andrews

#1
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
:bigwave:
 
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D

D.Scott

#2
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
 
K

KenK - 2009

#3
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.
 

Attachments

A

Andrews

#4
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

:confused:
 

Atul Khandekar

Quite Involved in Discussions
#5
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.
 
A

Andrews

#6
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?
 
A

Al Dyer

#7
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?
 
J

jospe.at.liu

#9
I found your problem a bit puzzling. Based on your description of the process, the data should be normally distributed. I have taken a look at your data and found some interesting things.

I have made time series plots of the subgroup averages and variability (st.dev and range). The averages shows a stable, random pattern. The variability charts gives some indications of special causes of variation, with spikes in subgroups 4, 10, 19 and 23.

I have made a normal probability plot and found that the shape indicates a logarithmic distribution of the data. However, there are four points that do not fit the pattern. These are the ones smaller than 6.600, and are located in subgroups 10, 18, 19 an 23. Three of these subgroups also have spikes in the variability charts, increasing my suspicions of special causes of variation.

When excluding the four points below 6.600 the standard deviation for the whole dataset drops from 0.0033 to 0.0019, making the natural variability range (avg +/- 3sigma) span from 6.600 to 6.611. The R^2 value for the normplot increases from 0.65 to 0.94, and the data displays a pattern that can be approximated with a normal distribution.

My conclusion is that your data quite nicely resembles a normal distribution, but you have some individual points that deviate from the general pattern. These should be able to assign to some special cause.
 

harry

Super Moderator
#10
Welcome and thank you for the maiden post.

Even though the original post was made in the year 2001, your answers are still good for those searching for reference on this subject. We look forward to your active participation.
 
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