Confusing Interpretation of Cpk
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 This topic has 6 replies, 4 voices, and was last updated 8 years, 2 months ago by Sven.

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September 17, 2014 at 6:45 am #54843
SvenParticipant@beginnersven Include @beginnersven in your post and this person will
be notified via email.Hello,
I´m about to study for my green belt certification. Sorry in advance, I have studied all day and I guess that I´m just doing a very simple mistake. Here it is:
I made up my own simple example so let´s assume the following data
Values: 1,2,2,3
LSL:0
USL:6
Mean:2
S: ~0.7If I calculate the cpk, I come up with min[(62)/(3*0.7);(20)/(3*0.7)]=min[1.9;0.9]
As 0.9 is the smaller value, cpk is 0.9. Various sources say, that if cpk<1, you have values outside the limits. But neither 1,2,2 nor 3 is outside 0 or 6.
Thank you in advance for opening my eyes
Sven0September 17, 2014 at 7:02 am #197386
Chris SeiderParticipant@cseider Include @cseider in your post and this person will
be notified via email.You are forgetting that if you created a normal distribution with the standard deviation and mean you stated, you would have values less than 0–assuming these are continuous measurements. This is why you have a Cpk less than one….you only have a sample with 4 readings in your example which HAPPEN to be all within your specifications.
Just think that 2 – (3)*(0.7) you’d get some values outside of zero.
0September 18, 2014 at 5:21 am #197389
SvenParticipant@beginnersven Include @beginnersven in your post and this person will
be notified via email.Hello Chris,
thank you for your quick help, much appreciated!
Have a lovely day
0September 19, 2014 at 11:59 pm #197395
agParticipant@amarghatak Include @amarghatak in your post and this person will
be notified via email.@beginnersven at least you should collect 30 data then you make histogram and see how many of the data point going out of specification.
0September 20, 2014 at 6:26 am #197396
Robert ButlerParticipant@rbutler Include @rbutler in your post and this person will
be notified via email.Ok, here’s 30 samples: seven 1’s, seven 3’s and sixteen 2’s. The mean is 2 and the standard deviation is .694 and you get the same results as before: min(1.92,.96) therefore Cpk = .96. If you run a distribution test on these 30 data points you will find, not surprisingly, that it fails various and sundry tests for normality. Thus one could argue the reason for the failure of the test was the “lack” of normality in the data (this, in spite of the roughly normal shape of the histogram).
To check this let’s take the range of 13 and the frequency of 1’s,2’s and 3’s in the OP and use Monte Carlo methods to generate a 30 point distribution with these parameters. What we get is a perfectly normal distribution with a mean of 1.90 a standard deviation of .64, a minimum of .65, and a maximum of 3.31. The Cpk calculations become min[((61.90)/(3*.64)),((1.900)/(3*.64))] = min[2.13,.989].
Therefore Cpk = .989 and you are still less than 1 and which means you will eventually have product outside of the spec and yet, no matter how you slice it, none of your sample data points are outside the spec limits either for the expansion of the OP distribution or the Monte Carlo simulation of normal data.
The key, as was noted by Chris, is not a matter of sample size rather it is that the standard deviation of the sample, regardless of which group you use, guarantees some product will exceed the lower bound. The overarching issue is that the mean of the sample is significantly different from the value of the center of the spec limits (1.9 vs. 3) and that, coupled with the sample spread, tells you that you are going to have problems with your process and its ability to meet the spec requirements.
0September 22, 2014 at 6:29 am #197399Cpk less than 1.0 does not mean you do have data out of spec, but rather you should expect to have data out of spec in the long run.
I get with the four data values:
Mean = 2.0
Estimated standard deviation (using mrbar/d2)= 0.591 <== This is the standard deviation used for Cpk
Therefore Cpk = 1.128
Ppk = 0.81650September 29, 2014 at 2:17 pm #197421
SvenParticipant@beginnersven Include @beginnersven in your post and this person will
be notified via email.Thank you very much @cseider, @amarghatak, @rbutler and @Matt for further clarification. It´s really nice to see such an active forum and people like you who take the effort to explain even such “simple” things to those who just started.
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