biostats nbme 3 - poojy36

[ArchivalUser]

1)if 1 in 10000 with a negative PPD have active TB and 1 in 100 with a positive PPD have active tB.which of the foll is teh estimated relative risk of active TB in persons with a positive PPD test?

a) 1
b)9.9
c)100
d)1000
e)10000

b) 2 drugs X and Y are evaluated for CHF...pts taking the drug X have cardiac index (CI) of 2.5 l /m2 with 95% confidence interval of 1.5-3.5.Patients taking drug Y have CI of 1.7 l/m2 with 95 % conf interval of 0.7-2.7.A test of significance of difference shows a p value of 0.1.which of the foll is the likelihood that the diff in the mean CI of pts receiving drug X and Y is due to chance?/

a)0%
b) 2.5 %
c) 5%
d) 7.5%
e)10%
f)66.7%
g)95%

[ArchivalUser]

pls try answering this

[ArchivalUser]

1. RR=(1/100)/(1/10,000)=100, answer is C.

2.e. 10%.

[ArchivalUser]

1....c

[ArchivalUser]

wud u plz explain the 2nd answer arya?

[ArchivalUser]

Agree with arya,

1. C.
2. E.

Catzung, for the 2nd Q, they are asking the type I error: you "saw" a difference that did not exist, that's the P value. P=0.1, so the likelihood that the diff in the mean CI of pts receiving drug X and Y is due to chance is 0.1 (10%)

[ArchivalUser]

hi arya i think for second q because drug y has a confidence of interval which includes 1--->it has not statistically significane effect--->and on the other hand confidence of interval of drug x &y have overlap--->these two drug are not different--->i think the answer is 95% ,what do you think?gl