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#1
08-16-2012
 USMLE Forums Master Steps History: --- Posts: 1,365 Threads: 648 Thanked 591 Times in 355 Posts Reputation: 601
Estimating the P value?

A ten-year prospective study is conducted to assess the effect of regular supplementary folic acid consumption on the risk of developing Alzheimer's dementia. The investigators report a relative risk of 0.77 {95% confidence interval 0.59 -0.98) in those who consume folic acid supplements compared to those who do not. Which of the following p values most likely corresponds to the results reported by the investigators?

A)0.03
B)0.05
C)0.07
D)0.09
E)0.15

#2
08-16-2012
 USMLE Forums Scout Steps History: Not yet Posts: 96 Threads: 15 Thanked 19 Times in 18 Posts Reputation: 29

is it A????????
#3
08-16-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,186 Threads: 119 Thanked 822 Times in 434 Posts Reputation: 832
a

it's a) 0.03
i think for calculating the p value we have to see the confidence interval(<1) not the relative risk ...and if it does not lies in between with 1,then stastical significance exist....then we have to calculate it with computed p value....so if statistical significance exist the p value should be <0.05 and there is only one option below it.........i hope it's true it's in kaplan book,plz post the answer soon

Last edited by venky2600; 08-16-2012 at 08:19 PM.

#4
08-17-2012
 USMLE Forums Master Steps History: Step 1 Only Posts: 864 Threads: 62 Thanked 374 Times in 225 Posts Reputation: 384
A

relative risk of 0.77 {95% confidence interval 0.59 -0.98)

According to this information, confidence interval is < 1. so patient is at low risk.
which should be less then alpha criteria of <0.05.
and the only answer which suits this description is 0.03
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 The above post was thanked by: sruthi (08-17-2012)

 Tags Biostatistics-Epidemiology, Step-1-Questions

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