calculating standard deviation
- This topic has 3 replies, 2 voices, and was last updated 16 years, 8 months ago by Anonymous.
October 16, 2005 at 12:57 pm #41064
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WHEN MEASURING 5 DIAMETERS ON A CMM DIAMETER AND FORM ARE GIVIN. HOW WOULD I CALCULATE THE STANDARD DEVIATION? R BAR DEVIDED BY D2 OF 2.33?0October 16, 2005 at 1:42 pm #128375
You need to be careful because calculations of standard deviation assume random independence.
From your description it would appear you’re interested in ‘uniformity’ or rather cylindricity. This is course is measured relative to a datum and categorised by location.
I might be wrong about your intentoins, but this is one of the most common mistakes in quality engineering and in six sigma in particular. For example, some people actually plot R for X-bar and R charts using correlated data and wonder why the process is out-of-control!!!
Then they go back to the office and write a report to a senior manager stating the process is crap … and the process engineer tells the quality engineer if he can do the job better he is welcome to it; and goes off to find another job sell BMW cars. (Not that you would do this of course. I’m just trying to explain why statistical assumptions are important!)
Andy0October 16, 2005 at 2:02 pm #128376
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DEAR ANDY: THANKYOU FOR THE ANSWER. I THINK YOUR SAYING I SHOULD USE ONLY MY DIAMETERS AND THE NORMAL FORMULA OF CALCULATING STANDARD DEVIATION? I SHOULD NOT THEN CONSIDER THE FORM OF THE PART FOR SIGMA.0October 16, 2005 at 3:16 pm #128377
To be safe, if I were you I would expect to have a ‘part’ for each ‘n’ in the calculation of sigma. Clearly, if n < 15, then Range would be a better estimate of sigma, but the same rule applies: You should expect to choose the Range (Max-Min) from n parts.
When you have a single part and you take multiple measurements, it’s not always safe to assume random independence.
If you agree that your interest is the ‘Form’ of the part, then you could regard this as just another ‘response,’ so your process would comprise two ‘responses,’ the process mean and the process Form (uniformity) of the part. This of course assumes that the process mean bears no relationship to the ‘uniformity,’ which isn’t always the case.
So .. you option is to either use multivariate methods using multiple measurements taken from one part, or use two responses!
Also, you should recall for Shewhart Charts (X-bar and R charts) in order to assess stability, the elements of the subgroup should be taken in the order of production and from a single stream, as this is a condition known as rational subgrouping. If you agree then it is obvious that we should not form subroups from multiple measurements taken from a single part.
This is why Motorola waferfabs used Multi-vari charts in the mid-1980s, much to the chagrin of quality management, Later, when we started using IBM PCs to manufacture 6800 microprocessors for Macintosh, we incoporporated Hotelling’s T-squared into the Multi-vari Chart.
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