07-21-2015, 04:21 PM
sorry ,HCV, chronic carrier.
@ All forum members , Biostatistic help needed . - bugguy19
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07-21-2015, 04:21 PM
sorry ,HCV, chronic carrier.
07-21-2015, 04:25 PM
http://www.usmleforum.com/showthread.php?tid=811119 so tell me why Fomite then?
07-21-2015, 04:59 PM
07-22-2015, 01:23 AM
Answer is E. regard less of the infection type. I think the researcher. Here will be able to calculate the incidence ,
Right ??? Thank you cardio
07-22-2015, 02:06 PM
Rt *E*
@bug my pleasure. Can you tell me if post helping you?
07-22-2015, 05:28 PM
@cardio , I can't thank you enough for your great questions and your precious time , please keep it up , this is my only Bs review.
07-22-2015, 09:52 PM
Hello bug! NP just wanna make sure ATP not evaporating here.
___________________________________________________________________________________ http://i.imgur.com/7OFwbOW.png?1
07-23-2015, 06:36 AM
Incidence among smoker = 0.12
Incidence among non smoker 0.01 Attributable risk = incidence among exposed-incidence among non exposed Attributable risk = 0.12-0.01 = 0.11 Thank you
07-23-2015, 05:10 PM
Bug, let’s just try to understand How much of the dz that occurs can be attributed to a certain exposure? This is answered by another measure of “risk”, the Attributable Risk(AR). We def AR amount or proportion of dz incidence (or disease risk) that can be attributed to a specific EXP.
Incidence in “E”xposed group - Incidence in NON-exposed group = AR let plug it in -> (120/100,000) − (10/100,000) = 110/100,000 ->0.0012 -0.0001 = 0.0011 here you AR. Im asking percentage of lung cancer that can be attributed to smoking. So beside math you gone have to take one more formula to get home and what I ask ; Incidence in “E”xposed group - Incidence in NON-exposed group / Incidence in “E”xposed group X 100 lets plug it in (120/100,000) − (10/100,000) /120/100,000 = 0.0011/0.0012 = 0.9166 X 100 -> 91.6 ~ AR% of the incidence of a disease in the “exposED” that is due to the “exposURE” -> 92% *B* Cheers!
07-23-2015, 05:29 PM
http://i.imgur.com/kscN8wI.png?1 ? Exp your reasoning.
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