NBME 3..blck 1...quesn17 - dr_trepinator

[ArchivalUser]

Two experimental drugs X and Y are being evaluated for treatment of congestive heart failure.Pts receiving drug x have a cardiac index of 2.5L/m2 with a 95% CI of 1.5 to 3.5 . Pts receiving drug Y have a cardiac index of 1.7 L/m2 with a 95%confidence interval of 0.7 to 2.7 . A test of a significance of difference shows a p value of 0.1 . Which of the following is the likelihood that the difference in mean cardiac index of patients receiving drug X and drug Y is due to chance?

A)0%
B)2.5%
C)5%
D)7.5%
E)10%
F)66.7%
G)95%

thanks

[ArchivalUser]

some one plz...@samideb,kindmd,thein7thein,maryam2009,sarim????

i dont have a clue..abt this quesn

[ArchivalUser]

is anyone there?

[ArchivalUser]

Hi dr_trepinator

Please look at this explanation. In the answer key it is also given as "E"


http://www.usmleforum.com/showthread.php?tid=520076

I hope it helps.

[ArchivalUser]

thanks a lot psychmledr...

[ArchivalUser]

"E"

-The stem is simply saying that, we did a study with "Drug X" and "Drug Y"
and we drew a "Mean Cardiac Index" for both drugs.

-and we found that "There is a Difference b/w Mean Cardiac Index of Drug X and Y"

-now the stem is asking that "What is the likelihood that the difference that we found is not true and we got that difference "By Chance" ?

-and for that they gave us the "p-value" = 0.1 (10%)

Conclusion:

*There is a 0.1(10%) chance that The difference in Mean Cardiac Index of these two drugs in NOT TRUE.

[ArchivalUser]

thanks sarim...it looks very simple aftr ur explanation...i just got confused bcoz in the stem he mentions two p values...but indeed its easy if we break the questn down like u did..

[ArchivalUser]

i mean the confidnce intrvals...not p value sorry

[ArchivalUser]

[ArchivalUser]

yeah we didn't have calculate anything from those CI.