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MSA with only 1 replica, why varcomp for repeatibility is not zero?

Discussion in 'Gage R&R and MSA - Measurement Systems Analysis' started by laffranchinis, Jan 30, 2020.

  1. laffranchinis

    laffranchinis New Member

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    Hi all,

    Quick question:

    Theoretically, if for a MSA all the samples will be measured by the operators only once, we should not have information about the repeatability, right?
    Why also Minitab give me varcomp and all the other information about repeatability?

    thanks in advance.
    Samuele
     
  2. Miner

    Miner Moderator Staff Member

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    Please attach your data and analysis.
     
  3. laffranchinis

    laffranchinis New Member

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    Hi Miner,

    thanks for the quick answer,
    Here below the data and analysis:
    upload_2020-1-30_15-43-14.png

    Thanks in advance for the help
    Samuele
     
  4. Miner

    Miner Moderator Staff Member

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    This has to do with how ANOVA works. ANOVA will calculate the total variation, then the variation due to the operators and the variation due to the parts. It then subtracts the operator and part variation from the total variation as unexplained variation. In a normal study with replicates this would be the repeatibility. In this case it would be experimental error.
     
  5. laffranchinis

    laffranchinis New Member

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    Hi Miner,

    Thanks for the answer.
    Maybe I am missing somethings, but when calculating the variations following formula are used (I found them on support.minitab.com) :
    upload_2020-1-31_10-17-41.png with upload_2020-1-31_10-19-55.png

    Your answer is then referred to the calculation of SS, right.
    But the degree of freedom should be: a x b x (n-1) where n is the number of replicates, also 1 in this special case. So from the mathematical point of view, DF=0 and it should not be possible to calculate the variance of the repeatability.

    any other comment/explanation will be welcome !
    Samuele
     

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  6. Miner

    Miner Moderator Staff Member

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    Yes, I was referring to the SS. Degrees of freedom work the same way. Take the total df then subtract the df for operators and the df for parts. The remainder is the df for experimental error (repeatability). You had n=27 measurements, so dftotal = 27 - 1 = 26. There are 3 operators, so dfoperator = 2, 9 parts, so dfparts = 8. Therefore dferror(repeatability) = 26 - 2 - 8 = 16. You cannot directly calculate SS or df for error. It is always what is left over once you subtract the factors.
     

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