# 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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