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Comparing Consistency of Inspection Among Inspectors

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  • #34229

    weisusu
    Member

    Given the situation: Three inspectors performed the same inspection on the same batch of material (N=n1 + n2 + n3).
    Inspector A: n1= 100, rejected unit = 5, accepted unit = 95
    Inspector B: n2= 58, rejected unit = 3, accepted unit = 55
    Inspector C: n3= 350, rejected unit= 20, accepted unit = 330
    What kind of valid statistical analysis that I can perform to show that inspector C was inspecting the batch of material no difference than inspector A and B?

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    #94022

    indresh
    Participant

    Both x and y being discrete use a chi sqaure test to show that the population from where the sample is taken is same though the sample size is different
    any further assistance required get in touch
    rgds,
    indresh
     

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    #94032

    ALEK DE
    Participant

    What is the objective ? If consistency of judgements by different Inspectors is the objective I will suggest a Kappa or ICC test depending upon your response variable.
    Alek

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    #94042

    Bob J
    Participant

    I would approach the problem the same way as indresh has detailed using a chi-square contingency table.  This problem would result in a chi-square value of 0.091 with DF=2 resulting in a P of 0.956.  Any way you slice it (alpha value) this the result will be a failure to reject the null hypothesis. 
    Hope this helps…..
    BobJ
     

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    #94043

    Ron
    Member

    This is a classical gage R&R problem. Do a discrete gage R&R with the personnel involved.
    Most likely you will find that the training received and the margin of interpretation is sub-standard and will require a better methodology to assure consistency

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    #94138

    Srinivas
    Member

    Given the situation: Three inspectors performed the same inspection on the same batch of material (N=n1 + n2 + n3).Inspector A: n1= 100, rejected unit = 5, accepted unit = 95Inspector B: n2= 58, rejected unit = 3, accepted unit = 55Inspector C: n3= 350, rejected unit= 20, accepted unit = 330What kind of valid statistical analysis that I can perform to show that inspector C was inspecting the batch of material no difference than inspector A and B?”

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    #94148

    ABC
    Participant

    Use Chi-Square test
    Observed
                             A                        B                             C                        Total
    Accept          95                        55                           330                        480
    Reject             5                          3                              20                         28
    Total              100                     58                             350                       508
     
    Expected
                             A                        B                             C                        Total
    Accept          94.49                  54.80                       330.71                     480
    Reject             5.51                   3.20                          19.29                       28
    Total              100                     58                             350                         508
    Chi-square (calculation) = 0.09
    For 95% confidence interval with DOF = 2
    Chi-square = 5.99
    As 0.09 is much smaller than 5.99, so no operator/result is deffent to the others
     
     
     
     

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