Comparing Consistency of Inspection Among Inspectors
Six Sigma – iSixSigma › Forums › Old Forums › General › Comparing Consistency of Inspection Among Inspectors
- This topic has 6 replies, 7 voices, and was last updated 18 years, 7 months ago by
ABC.
-
AuthorPosts
-
January 7, 2004 at 3:43 am #34229
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?0January 8, 2004 at 6:28 am #94022
indreshParticipant@indreshInclude @indresh in your post and this person will
be notified via email.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
0January 8, 2004 at 9:15 am #94032
ALEK DEParticipant@ALEK-DEInclude @ALEK-DE in your post and this person will
be notified via email.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.
Alek0January 8, 2004 at 12:29 pm #94042I 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
0January 8, 2004 at 1:06 pm #94043This 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 consistency0January 10, 2004 at 9:15 am #94138Given 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?”
0January 12, 2004 at 9:12 am #94148Use 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
0 -
AuthorPosts
The forum ‘General’ is closed to new topics and replies.