q2! - sarim

[ArchivalUser]

NBME 3 Q.


Two experimental drugs, drug X and Y, are being evaluated for the treatment of congestive heart failure. Patients receiving drug X have a cardiac index of 2.5 L/m2 with a 95% confidence interval of 1.5 to 3.5. Patients receiving drug Y have a cardiac index of 1.7 L/m2 with a CI of 0.7 to 2.7. A test of the 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%

Kindly explain. Thanks

[ArchivalUser]

ee
actually there is no difference as p value more than .05
but if by chance difference is present that is 10%(type1 error)
difference should be more than 95% to reject null hypothesis.

[ArchivalUser]

why are we taking the "difference" between the 2 means for the p value...shouldnt we take the p values of each drug separately to see whether they are significant or not?

[ArchivalUser]

Is it e

[ArchivalUser]

If p value is not significant(more than 0.05),can we use it to deduce difference by chance???

Doesn't that mean as there is no difference???

[ArchivalUser]

0.1-->10% by chance

Thanks

[ArchivalUser]

saying it type 1 error is probably wrong.
because
if the null hypothesis is not rejected,there is no chance of a type 1 error (kaplan)
here null hypothesis is not rejected so it is not type 1 error.

[ArchivalUser]

E ,
It said how much we can't count on the study result.