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NBME 23 Answers

nbme23/Block 3/Question#37 (26.0 difficulty score)
A 48-year-old man is referred for evaluation ...
108 to 118🔍
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So a simpler way than all the math being done is understanding what CI means.

CI - range of values w/in which the true mean of the population is expected fall

So a CI of 95% will be more precise and have a narrow range compared to a CI of 99% will be less precise because its including more values in and result in a wider range.

So if CI of 95% is 110 to 116 then a CI of 99% has to be a range that is wider... 108 to 118

paulkarr  Glad I wasn't the only one to solve it this way...didn't even think to bother with the calculation. +1  



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submitted by sajaqua1(461),
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A dandatrs oidneatvi si a emreuas fo labyoibript in mbnrisgeel teh ag.varee One nsatdard itievdnoa no a lble curev odunrtsibiti aesertc a 7%6 cnchea tath eth wrnsae lwli eli ni eetrh. wTo ndadtasr itadsvineo lilw eetcar a 95% cnhca.e hereT atndsadr eodsainvit easrcte a %.979 caecn.h

hTsi eitpant sah an veeraag fo 1,13 nda a %95 nodenccife ta 1011-61 amens ahtt the DS si 15. . oS one niddaatilo SD dlowu evig us a gnare of 5.,10.5181-7 rnuoded ot -.101188

usmleuser007  How did you get the SD to be 1.5? +  
usmleuser007  NVM Got it +1  
jesusisking  You wouldn't use Standard error with Confidence Interval? (pg. 262 FA 2019) +  



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VMN ogt it.

sJut YtF:eIh CI swa ttaeds ot be mrfo 116-101 iwth 9%5 nda anem of 3.S o11, no erihte teher era owt DS on ethier issed fo 311 et(h anme) taht gevi eht %.59

1=11-631 3 hwinti S2D obeav hte neam -011 11=3 3 nwihti SD2 wlboe hte aemn

3 ddvdiie by the 2 DS = 1.5 rep .SD

to get frmo 95% to %99 uoy ahve ot rcionepaotr neo oerm DS 3( DS) no rteeih idess fo teh mnea 1)31(

Teof;eehrr at %99 CI 10=15-1. 015.8 IC 11651+=. .7511

udRno ethse up dna yuo egt 811108-

tyrionwill  95%CI = M ± Z(SE) instead of SD 116-113 = 3 within 2SE, not 2SD SE = SD/extract the square root of n = SD/2 and SD = 2SE +  
tyrionwill  Sorry I made a mistake, neglect the abobe +  
tyrionwill  if you use Mean ± 2SD = 95%CI to know SD, then use Mean ± 3SD to only know 99.7%CI, a bit larger than 99%CI +  



For all the ~math~ people out here:

CI = mean + Z(SE)

For a 95% CI, Z = 1.96. For a 99% CI, Z = 2.58.

The CI is +3/-3 from the mean, so 3 = 1.96(SE) with the 95% CI. Solve for SE (which doesn't change if you change the CI), which comes out to about 1.53.

Now switch up the Z value for the 99% CI, with the SE you just calculated. CI = 2.58 * 1.53 = 3.95. Add this to both sides of the mean (113), and you get the answer!




I tried to calculate it more precise, and messed up the answer...

Here is why:

  • 99.7% CI = 3 SD
  • However: 99.0% CI is actually 2.5 SD (or 2.57 if you want to be more precise)

1 SD = 1.5 mmHg → 2.5 SD = 3.75 mmHG

This results in a 99% CI of 109.25 (113-3.75) to 116.75 (113+3.75)

Closer to answer C than B.