# Total variation in MSA

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• #39469

Hello,     Is total variation in MSA equal to the meansurement variance? Below is a sample:Total variation in MSA is 2.3458£¬ but variance in “Stat > Basic Statistics > Display Descriptive Statistics” is 2.03062£¬waht is the diffence between them?
Tks!
TonySource           VarComp  StdDev   5.15*Sigma                                             Total Gage R&R   0.1721   0.41490  2.13672   Repeatability    0.1721   0.41490  2.13672   Reproducibility  0.0000   0.00000  0.00000   operator         0.0000   0.00000  0.00000   Part-To-Part     2.1737   1.47435  7.59290   Total Variation  2.3458   1.53162  7.88782   Part        operator        measure1        1        20.272        1        18.823        1        21.904        1        17.975        1        18.496        1        17.831        2        19.892        2        18.863        2        21.104        2        18.385        2        17.786        2        18.471        1        19.932        1        19.273        1        21.444        1        18.845        1        17.866        1        18.231        2        19.182        2        18.663        2        22.344        2        17.565        2        17.786        2        18.551        1        19.462        1        18.853        1        22.264        1        18.395        1        17.926        1        17.811        2        19.812        2        19.893        2        22.264        2        18.145        2        17.806        2        18.56

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#120104

Experts.
Need help!
Thanks very much!
Tony

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#120150

Tony –
Difference between the two numbers is (2.3458-2.03062 = 0.31524) LOL
Forgive me my wise-acre attempt at humor.
Really, the ANOVA method uses sums of squares as you see, but these have been adjusted according to some rules in the var comp analysis.
Return to your ANOVA table. I copied your data into Minitab so I would know the actual values. Your actual TOTAL SS is 70.018. But thast is total sum of squares. To get to the variance of the data we need to divide by the degrees of freedom (35) giving us 2.03062, exacly the same as the squared std deviation in discriptive statistics because it IS the same formula as the RMS estimate used there.
So the var comps calculated are correct and ratiometric to one another as displayed but NOT numerically the same as the RMS variance estimate.

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#120207

DaveS,
But I do not clearly understand still, do your mean the formular in total variation in MSA and variance in Stat > Basic Statistics > Display Descriptive Statistics is the same? What is difference is only df number? I copied all the data here, can you explain it more detail?
Total variance in MSA is 2.3458 but   variance in Stat > Basic Statistics > Display Descriptive Statistics is 2.03062
Tks!
Tony
Two-Way ANOVA Table Without Interaction
Source         DF  SS       MS       F        P                                                         Part            5  66.0719  13.2144  76.7655  0.0000operator        1   0.0078   0.0078   0.0453  0.8329Repeatability  29   4.9920   0.1721                 Total          35  71.0718
Gage R&R
Source           VarComp  StdDev   5.15*Sigma                                             Total Gage R&R   0.1721   0.41490  2.13672   Repeatability    0.1721   0.41490  2.13672   Reproducibility  0.0000   0.00000  0.00000   operator         0.0000   0.00000  0.00000   Part-To-Part     2.1737   1.47435  7.59290   Total Variation  2.3458   1.53162  7.88782

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#120302

Hi,
Can someone who has experience on it help me?
BR
Tony

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#120307

Tony:I will start with the understanding that the variance you get in descriptive stats is the total variation in the data set without attributing any part of it to a known effect. This comes from the SS of 71.0718 you see in the ANOVA table. The variance in descriptive stats is 71.0718/(36-1) = 2.0306. ANOVA and descriptive stats both agree here. The formulae are the same and the results are the same.
When you do a Gauge R&R you are attempting to place different components of the total SS into different ‘bins’ by manipulating the components of variation. This is done by1) decomposing the SS into different components2) calculating a mean square for each (MS,parts) = (SS,parts)/(df,Parts)3) combining the MSs in different ways to caluculate the proportion of variation assigned to each factor (VarComp). Consult the help function for VarComp calculation in Gauge R&R.Details of each of the above:1) The Gauge R&R will do the calculation of SS for each component of parts, operators, and repeatability (pure error left over after subtracting the SSparts and SSoperator from SStotal). The analysis is repeated with a parts*operator interaction. You can calculate these SS by hand by taking the difference between the the individual part dimensions and the mean part dimensions, squaring the numbers and summing. This is the same calculation as ‘within subgroup’ standard deviation as part of the Zst calculation.
2) The Gauge R&R will print the table where all these components have been calculated. Each SS is divided by its corresponding d.f. to get the mean square (MS). Think of this as the average variation within the subgroup. At this point we have taken the total SS apart into smaller pieces (SSparts, etc) and divided each by the d.f. to get the MSparts, MSoperator, etc.
3) The next step is to combine these MS numbers in different ways to assign the VarComp into the different ‘bins’. The formulae are shown in the ‘help’ file. Some of these formulae will give negative results (true in your case for ‘operator’) where the VarComps are assigned a value of zero. This manipulation is just an attempt to ‘bin’ the total variation into the different contributing factors. The ‘Total Variation’ is just the sum of these different components. It is not the variance, but is just a total of these estimates of the contribution of variation within each type of subgroup. The total value of 2.34585 is just the number from the manipulation of MSs. It will not be the same as the true variance of the data.
The StdDev (SD) entries are just the square roots of these VarComps and not the same as the true standard deviations within and between the subgroups.
The use of the VarComps is most valuable when it is normalized to give the division of the components in terms of 100% of the total variation, and when the ‘StdDev (SD)’ entries are multiplied by 5.15 to get the ‘width’ of the variation in terms of the tolerance window.
The alternate method is to use an X bar R chart to estimate the individual components of variation and different formulae to assign the VarComp to the different factors.
Hope this helps,
BTDT

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#120308

Tony:By the way, your part-to-part variation is extremely large and even though the Gauge R&R at ~25% is greater than the desired 10%, the number of categories (5) indicate that the gauge can still tell the difference between all your parts.The operators are not the problem either, every uses the (poor) gauge in the same way to measure your (very poor) parts.BTDT

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#120310

BTDT
Thanks very much! It do great help to me!
my email address is [email protected], wish be friend with you!
Tony

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#120381

Tony:Keep posting here and we’ll all pitch in and helpBTDT

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#120382

BTDT and Daves:
Nice to see when you guys comment. Try to do GLM with the data in this thread. You will get the same results. Make sure both operator and parts are random. I got it, so i thought i would let you guys know about it. Also i did send an email to you guys asking help to find a good book.
Tony: really sorry for the oftopic
Thanks and have a nice weekend.

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#120388 Ken Feldman
Participant

Hey BTDT, back off.  Only Vinny is allowed to be everyone’s Virtual Buddy.  The guy asked for a friend and that position is already taken by Dr. Vigorous Ph.D. himself.  By the way, you haven’t told us about your doctoral program.  Was it rigorous like his or a purchased one like some of our other posters?

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#120391

Darth:Consider me backed off. I invited everyone to join.Sally Struthers’ signature is the only proof of rigor anyone needs for a degree. They look pretty real to me, AND come free with an attractive frame and a cassette tape of “Pomp and Circumstance”.An additional \$60 value FOR FREE! Now how much would you pay? And if no one accepts the validity of your degree, return it for a full refund, but keep the frame and cassette as our gift.BTDT

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