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Retired NBME 23 Answers

nbme23/Block 3/Question#37 (reveal difficulty score)
A 48-year-old man is referred for evaluation ...
108 to 118 ๐Ÿ” / ๐Ÿ“บ / ๐ŸŒณ / ๐Ÿ“–
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 +13  upvote downvote
submitted by โˆ—humble_station(85)
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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

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paulkarr  Glad I wasn't the only one to solve it this way...didn't even think to bother with the calculation. +2
cagri  I do not think 95% is more precise because CI is basically range of mean values that expected to exist in an experiment. Thus 99% CI should be greater because higher CI means that ''I m widening my expectations so my prediction is more likely to involve other higher or smaller numbers.'' +



 +7  upvote downvote
submitted by โˆ—sajaqua1(607)
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A standard deviation is a measure of probability in resembling the average. One standard deviation on a bell curve distribution creates a 67% chance that the answer will lie in there. Two standard deviations will create a 95% chance. Three standard deviations creates a 99.7% chance.

This patient has an average of 113, and a 95% confidence at 110-116 means that the SD is 1.5 . So one additional SD would give us a range of 108.5-117.5, rounded to 108-118.

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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) +



 +3  upvote downvote
submitted by โˆ—usmleuser007(464)
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NVM got it.

Just FYI: the CI was stated to be from 110-116 with 95% and mean of 113. So, on either there are two SD on either sides of 113 (the mean) that give the 95%.

116-113= 3 within 2SD above the mean 113-110= 3 within 2SD below the mean

3 divided by the 2 SD = 1.5 per SD.

to get from 95% to 99% you have to incorporate one more SD (3 SD) on either sides of the mean (113)

Therefore; at 99% CI 110-1.5= 108.5 CI 116+1.5= 117.5

Round these up and you get 108-118

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tyrionwill  95%CI = M ยฑ 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 ยฑ 2SD = 95%CI to know SD, then use Mean ยฑ 3SD to only know 99.7%CI, a bit larger than 99%CI +



 +3  upvote downvote
submitted by brainchild(3)
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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!

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lulumomovicky  I did exactly the sameeee +



 -1  upvote downvote
submitted by โˆ—drmohandes(193)
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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 โ†’ 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.

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