Capability analysis
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 This topic has 28 replies, 5 voices, and was last updated 16 years, 6 months ago by UFO.

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July 8, 2005 at 12:51 pm #39948
Hi all
It was my understanding that CP is always greater than Pp and CPk>Ppk. But today i got exactly oppositte results. Can anyone tell me , did i do anything wrong in my analysis? What is this results tells me?
Cp=1.91 Pp = 2.02 Cpk = 0.89 Ppk = 0.94
Thanks
Newbie0July 8, 2005 at 2:04 pm #122763Is your data abnormal or is there some sort of autocorrelation present in your data? This will cause the assumptions in calculations for Normally distributed process capabilitys to return incorrect results.
Kirk0July 8, 2005 at 2:24 pm #122764Kirk;
Thanks for your reply.
No i dont see any abnormal data. The data is normal. Below is my data if that help you.
0.991.230.981.411.230.780.841.420.861.341.041.201.581.030.781.230.890.820.881.720.411.281.320.891.120.860.610.780.681.80thanks
newbie0July 8, 2005 at 2:26 pm #122765Sorry i forgot to add the spec limit, It is +/ 2
0July 8, 2005 at 2:33 pm #122766What limits are you using to calculate the Capabilities? I agree, the data is normal and does not show any time dependence.
Kirk0July 8, 2005 at 2:41 pm #122767Newbie,
Sorry, I misread your original question. The issue you are seeing has nothing to do with the normality or time dependence. The reason that the reported value for Cp is less than Pp (and the same with Cpk and Ppk) has to do with how the standard deviations are approximated. Two different calculations are used to appoximate the within (short term) and overall (long term) standard deviation.
More than likely the short term stdev was calculated using the moving range method while the long term stdev is approximated using the standard formula. Because your data does not show any significant difference between short and long term, the errors in the approximations cause the effect you are seeing where the long term capability is “better” than the short term capability. This is just a figment of the assumptions used in approximating the different levels of variation.
Hope this helps.
Kirk0July 8, 2005 at 3:20 pm #122769Thanks once again Kirk;
My subgroup size is one. So i dont think minitab use Range method. Anyway the answer for my question is the approximation of two std deviations right?
Thanks once again. I have another doubt. Some time back when i showed a capabilty analysis sixpack(minitab output), one of my friends told me that my sample size is less. (I cannot find that minitab file now, otherwise i could have paste the data here.). My question is , how can we know that the sample size is less by looking at the sixpack (normal) results from minitab. Which graph tells me about my sample size. If i remember correctly the sample size was 30 pcs , subgroup one.
Thanks again
Newbie0July 8, 2005 at 3:36 pm #122770Newbie,
Right. It is the Average Moving Range method that Minitab may use for within standard deviation. Here is a link to a White Paper entitled “Formulas for Capability Analysis…” on the Minitab website that will describe this further… http://www.minitab.com/support/answers/answer.aspx?ID=294
As far as your second question is concerned, the sample size can be found on the xaxis of the IMR chart and the “Last 25 Observations” chart.
Kirk0July 8, 2005 at 4:58 pm #122775To Kirk:
Fm: UFO
You idiot, the counterintuitive disparity between Cp and Pp can also result from the difference in degrees of freedom when computing the shortterm and longterm standard deviations using the sumsofsquares method.
Using this method of calculation, the shortterm variance is based on g(n1) degrees of freedom while the longterm standard deviation is predicated on ng1 degrees of freedom. Given a stable process, it is possible for Cp > Pp owing to nothing more that a disparity in DF.
Get with the program you moron and seek a level of knowledge longer than your nose. You are only about two steps ahead of the person you are trying to offer advice to.0July 8, 2005 at 5:03 pm #122777So what you are saying is that the difference between the Cp and Pp numbers can be based on the degrees of freedom resulting from the different methods used to approximate long term and short term standard deviations?
0July 8, 2005 at 5:09 pm #122780No, I think he said you are an idiot. Let me check the previous post again.
0July 8, 2005 at 5:15 pm #122785Kirk,
Please excuse my previous post as I also have no clue what the things I cut and paste mean. Sorry about all that strut your stuff nonsense. I have no stuff to strut and am trying to compensate.0July 8, 2005 at 5:15 pm #122786To: Kirk
Fm: UFO
Go read about how to compute the short and longterm variance you idiot. This way you will discover what you obviously do not know.0July 8, 2005 at 5:18 pm #122787OK, I will review the theory behind calculating short and long term variance if you agree to go see someone about your obvious anger management issues.
0July 8, 2005 at 5:42 pm #122790UFO, kirk:
Thanks for your inputs. UFO said the disparity can also cause due to the denominator(df) correct? Along with the previously mentioned capability i also calculated other capability idicies for other characteristics. All other values are the same way i was expected(Cp&Cpk greater then Pp&Ppk).
So what is the bottom line, is that due df or the difference in round off std dev.
Kirk ; I did not put the second question in a good way. Let me try again. The six pack results from minitab was not good, basically it was hard to tell that the data was normal. When my friend see that results, he said that the non normality is basically because of small sample size. Do you know how can we tell that the non normality is caused due to small sample size, which grap tell us this?( sample size was 30)
UFO: i was reading a post regarding the calculation of sample size for ttest where std deviation is unknown. In that thread you replied ‘ do a Z test of delta to find the standard deviation. I did not understand this. Can you explain on this plese
Thanks once again for all the comments.
Newbie0July 8, 2005 at 5:48 pm #122792To: Kirk
Fm: UFO
Hey you stupid moron, take some of your own medicine. You, Chris, Stan and some others treat well intentioned newcomers in a very abrubt, abusive, impolite, and hostile way. Maybe you should treat others like you want to be treated. At least you recognize my anger. I doubt you see the anger you generate in others.
So yeah, I have a problem. My problem is with idots like you and Chris and Stan trying to pretend you are “experts” in the theory and practice of statistics. Get a glass of water, swallow the bitter pill and start being more humane to people. Maybe then I will quit dogging you and pointing out the severe limitations of your stupid advice.0July 8, 2005 at 6:00 pm #122793Newbie,
Although I am sure UFO will be predictable and call me an idiot, the issue you are seeing has nothing due to round off of the standard deviation.
Minitab defaults to an Average Moving Range Method to calculate the within standard deviation. This is given by the following equations:
MRbar = (MRw +…+MRn)/(nw+1) where
n = number of data points and w = number of points used in the moving range. This is 2 by default.
Stdev(within) = MRbar/d2(w)
As you change w (this is an option under the “Estimate” button in Minitab) the estimated value for Stdev(within) will change.
So to answer your question, the issue you are seeing with this data is a result of the assumptions / default settings used in the equation used to calculate the within standard deviation, which looking at a deeper view is based on degrees of freedom.
As far as your second question is concerned, if you are looking for one of the “magic” pvalues you won’t find one. The data will not change from a nonnormal distribution to a normal distribution as you increase your number of data points. One quick and dirty practical approach when just given a graphical output is to look at the histogram and normal probability plot. If the histogram “looks” normal and the points on the normal probability plot lie in a relatively straight line it is probably a safe assumption to say the analysis is OK based on a normal distribution. Does this answer your question?
Kirk0July 8, 2005 at 6:07 pm #122794Thanks once again Kirk:
Regarding the second question. Actually what i wanted to know is,
How can someone telll by looking at the six pack out put from minitab that the non normality of the data is caused due to small sample size. I know how to check if the data is not normal or not. But my question is how can i tell that the reason for non normality is due to low sample size. Which graph tells me this. Hope this time the question is clear.’ My sample size was 30 at that time. Thanks once again for your help
Thanks
Newbie0July 8, 2005 at 6:15 pm #122796
HF ChrisParticipant@HFChris Include @HFChris in your post and this person will
be notified via email.I don’t know about minitab, but there are programs that look at main effect and sample size. Find a sample size calculator (many on the web) and determine your power. It is also a way to choose an effective sample size. And no I’m not an expert, just someone who works with numbers and methods and tries to apply them in my current job. I don’t like best guesses using tribal knowledge, I like to attempt to see what is really happening.
Chris0July 8, 2005 at 6:15 pm #122797To: Newbie
Fm: UFO
Yes, I would be happy to explain. First, allow me to reiterate your question:
UFO: i was reading a post regarding the calculation of sample size for ttest where std deviation is unknown. In that thread you replied ‘ do a Z test of delta to find the standard deviation. I did not understand this. Can you explain on this plese
Since you do not know the standard deviation, it is possible to call upon the merits of the standard normal distribution. This means that you can assume a mean of 0.0 and standard deviation of 1.0. If you are willing to express the extent of benefit (mean difference) in the form of a Z value, then you can proceed with your sample size calculation.
For the following example, we assume a mean of 0 and standard deviation of 1.0:
Example: N = 2([ Z(a) + Z(b)]^2) / Z(diff)^2
Z(a) is the standard normal deviate associated with the selected alpha risk
Z(b) is the standard normal deviate associated with the selected beta risk
Z(diff) is the amount of beneficial change that must be detected in tems of mean difference. This is the amount of change that must be detected to be of practical concern.
There are many sources for this information. Do be aware that the example equation will change depending upon the specific circumstances surrounding your particular application. I am merely providing one such equation so as to illustrate how the sample size can be estimated without knowledge of the standard deviation.
Realizing the “proper” sample size often involves as much “art” as it does “science.” At best, any such calculation is merely an approximation of what is needed to satisfy various theoretical and practical requirements. In most applications, there is a distinctive tradeoff between risk, detection sensitivity and economic resources, not to mention the parameters that are unkown or can not be estimated.
Do an Internet search and you will discover much more. In fact, there are many sample size calculators that are distributed freely on the Internet.0July 8, 2005 at 6:18 pm #122798I guess I must be missing something in you question, but let me try again. Normality of individual data points has nothing to do with sample size. That being said, small sample sizes from normal distributions may appear to be non normal when looking at a histogram because of the small sample size, however if the underlying distribution is normal, then the normal probability plot should look “ok”.
Sorry if this still isn’t what you are looking for.
Kirk0July 8, 2005 at 6:23 pm #122799
HF ChrisParticipant@HFChris Include @HFChris in your post and this person will
be notified via email.Kirk,
Curious if he is worried about not meeting assumptions based on a small sample size. Pvalues and r are greatly exaggerated with small samples.
Chris0July 8, 2005 at 6:26 pm #122800That crossed my mind, but the question was asked from a pure graphical analysis point of view. I am still not clear what Newbie is looking for.
Kirk0July 8, 2005 at 6:30 pm #122801and he said the sample size was 30. There shouldn’t be an issue with sample size.
0July 8, 2005 at 6:46 pm #122802Thanks UFO for your detailed reply. I am “googling” to find some more materials.
Kirk: Sorry for the confusion.I was asking some stupid question. Please ignore that. i got the answer. basically that was a stupid question.
Thanks once again for all
Newbie0July 8, 2005 at 6:47 pm #122803Kirk,
Please excuse my previous post as I also have no clue what the things I cut and paste mean. Sorry about all that strut your stuff nonsense. I have no stuff to strut and am trying to compensate.0July 8, 2005 at 6:48 pm #122804Kirk,
Please excuse my previous post as I also have no clue what the things I cut and paste mean. Sorry about all that strut your stuff nonsense. I have no stuff to strut and am trying to compensate.0July 8, 2005 at 7:29 pm #122805To: Kirk
Fm: UFO
Of course I will respond to your post. Your advice to Newbie is not stupid, but it is somewhat misinformed. Let me point out the flaws in your advice:
Kirk: The data will not change from a nonnormal distribution to a normal distribution as you increase your number of data points.
UFO: This is generally false. When cummulative sampling is used, it is possible to initially observe a lack of fit to a normal distribution, due to sampling bias. But as the sequential sampling continues, the bias is overwhelmed by the unbiased data which does reside on or near “the line.”
Kirk: If the histogram “looks” normal and the points on the normal probability plot lie in a relatively straight line it is probably a safe assumption to say the analysis is OK based on a normal distribution.
UFO: Partially right. You will not be able to form a histogram when the sample size is extremely low, but can visually inspect the probability plot. Still, at best, it is a judgement call, not a statistical decision.
Overall, I give your advice a C+ grade. You can improve your “grade” by doing some research before posting your response. In this way, your target population gets better advice and you might learn something in the process. Until then, you will continue to look like a wannabe consultant and come across as a moron to those that do know such things.0July 8, 2005 at 7:36 pm #122806Sorry again Kirk, this last one was not as stupid as most of my previous posts, but give me time and I will return to my normal level of stupidity.
UFO
a.k.a 21st Century Schiziod Boy0 
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