MSA Manual revision 3 Attribute Data - Signal Detection Formula

T

TommyT - 2003

#1
Rev. 3, p. 134:
"d" is an estimate of the width of region II areas and, thus, an estimate of the GRR=5.15xSDgrr"
d= 0.0237915
How do they make the jump to %GRR = 24% ?
(10/29/02 Errata changes it to 29%, I know, but I still can't figure the math here.)
GRR math experts available ?
Thankful Tommy.
:bigwave:
 
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T

TommyT - 2003

#2
Statistics - the answer

I got a very quick response on this by using the AIAG website.
Submitted question:
Rev. 3, p. 134:
"d" is an estimate of the width of region II areas and, thus, an estimate of the GRR=5.15xSDgrr" d= 0.0237915
How do they make the jump to %GRR = 24% ?
(10/29/02 Errata changes it to 29%, I know, but I still can't figure the math here.)

Answer:
The d=0.0237915 is the 5.15 * std dev(GRR). To determine the %GRR we need (1) to determine the std dev (GRR) and the process/target std dev. Since the process variation exceeds the tolerance we use the estimated std dev based on the tolerance = 6 * std dev. The tolerance range is .095 (this was not given in the
example). Putting the pieces together yields a GRR percent of 29% = (100%)*std dev (grr)/std dev (target std dev) = (100%)*(0.0237915/5.15)/(.095/6).

This worked out exactly.
 

Atul Khandekar

Quite Involved in Discussions
#3
TommyT,

Thanks a lot for that explanation.

I arrived at this figure in the following way:
Example says that the gage had a GRR of 25% of tolerance (p.126). If you actually calculate the mean of the tolerance range (from the given values), it comes out to 0.0956635. However what confused me was the figure 28 on p. 126 and the fact that the manual states that the process has Pp=Ppk=0.5: together this would make the tolerance=0.1

The manual further states (p.134) that "the goodness of this estimate depends on the sample size and how close the sample represents the process. The larger the sample, the better the estimate." I interprete this to mean that if you have a larger sample size, the calculated tolerance range would approximate to 0.1

Is this correct?

Rgds,

- Atul
 
Last edited:
T

TommyT - 2003

#4
Yes, I think you are correct.
However, I am not very used to these statistical methods. The MSA r.3 seems to be loaded with new interpretations that are different than what I am used to. It's much more complex.
 
K

kinwaiw

#5
Re: MSAr3 Attrib.-Signal Detection Formula

I have a question, MSA ver3, p134,
do you know how to calculate dLSL & dUSL, although it shows
dLSL=0.470832-0.446697=0.024135
dUSL=0.566152-0.542704=0.023448
why subtract 0.446697 from 0.470832 in dLSL?
and why subtract 0.542704 from 0.566152 in dUSL?

Thanks
 

Marc

Hunkered Down for the Duration
Staff member
Admin
#6
How to calculate dLSL & dUSL

Can anyone help on post 5 - How to calculate dLSL & dUSL?
 
G

Gagandeep S. Datta

#8
Atul,

How do we calculate the zone of ambiguity?

If LSL =.45, USL=.55, GRR% = 25% is there a formula to calculate zone of ambiguity?
 

Atul Khandekar

Quite Involved in Discussions
#10
Gagandeep, parts need to be ordered by their Reference Value. Then parts that are clearly bad or out of spec on both sides are all called "Bad" (Code -) by all appraisers, parts that are clearly good are called "Good" by all appraisers (Code +). Some ambiguity may exist for borderline parts (Code x) near USL as well as near LSL.

Please see Rameshwar's Excel file above that explains the calculations beautifully.
Thanks, Rameshwar!
 
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