Page 1 of 1: choice-corrected agreement (kappa)

determine · Credits: 3564 My Scrapbook

Five physicians are deliberating over whether the patient they have examined is exhibiting the auscultatory signs of mitral valve prolapse or not. Assuming that there would be agreement by chance of 50%, but, in fact, four of the five physicians actually agree on the findings, what is the choice-corrected agreement (kappa) for the clinical findings?

1) 0.8

2) 0.75

3) 0.6

4) 0.7

5) 0.5

I have no idea. What's kappa? Actually, I didn`t study such thing could anybody help please?

mickey_p · Credits: 652 My Scrapbook

its a little complex...let me try and be very simple

kappa is the unit of measuring agreement on observations between two (or more) observers, it is a measure of reliability (as oppsosed to validity which is measured by sensitivity, specificity etc).

Usually two observers could agree on a finding purely by chance (lets call this C), and there is an observed agreement (lets call it O) that is actually seen.
Thus in your case C= 0.5, and O=0.75 (4 out of 5).

I will skip all the discussion in between and come to the formula for estimating kappa (reliability)

k = O - C / 100 - C x 100.

thus the answer is 50% i.e.0.5

usually kappa bellow 0.8 is considered to be a poor indicator of reliabilty.

You can find more details in any biostats book, but for exam purpose this is ok I guess

Dr Manish

determine · Credits: 3564 My Scrapbook

Thank you very much Dr. Manish

determine · Credits: 3564 My Scrapbook

Hi, I just wanted to correct the calculation. 4 out of 5 is 0.8(not 0.75). So the answer should be 0.6.
0.8-0.5/1-0.5=0.6

mickey_p · Credits: 652 My Scrapbook

yes right!! good job!

maths was my weakest subject anyway!!!

Dr Manish

ophtha · Credits: 10361 My Scrapbook

THANKS guys! finally I am clarified with this kappa statistics.

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