NBME 24 5% (NBME Answers)

NBME 24 Answers

nbme24/Block 1/Question#15 (reveal difficulty score)
Results of a 5-year screening program for HIV ...
5% 🔍 / 📺 / 🌳 / 📖
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submitted by ∗youssefa(162)

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So number of cases with positive serology during "intake" represents prevalence and should not be accounted for when calculating population at risk. Population at risk = Total population - prevalence. Doing that, sum up the annual incidences left = 400 + 250 + 250 + 300 + 300 = 1500.

Applying incidence formula = 1500/(10,000-4000) = 0.25 over a total of 5 years.

Annual = 0.25/5 = 0.05%

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submitted by ∗karljeon(140)

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I don't know if there is an equation for this, but I basically pumped out every division across the table to get ~5% on average.

Here they are: 400 / 6,000 = 0.067 250 / 5,600 = 0.045 300 / 5,350 = 0.056 300 / 5,050 = 0.059 250 / 4,800 = 0.052

The average of these %s for all the years = 5.58%. So that's close enough to 5%.

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seagull good work. I found this question annoying and gave up doing those considering the amount of time we are given. +4

vshummy Well just don’t include the intake year... because that messed me up.. +17

_yeetmasterflex How would we have known not to include the intake year? From average **annual** incidence? +

lamhtu Do not include intake year because the question stem is asking average annual incidence. The 4000 positives at intake could have acquired HIV whenever, not just in the last year. +9

neels11 literally didn't think there was an actual way to figure this out. but my thought process was: okay incidence means NEW cases. so the annual average at the end of 5 years would be: (# of NEW people that tested positive at the end of year 5) / (# of people at that were at risk at the beginning of year 5) <--- aka at the end of year 4 250/5050 = 4.95% also if you look at year 5: you'll see that the at risk population is 4800 when 300 new cases were found the year before. 5050 at the end of year 4 MINUS the 300 new cases at the end of year 4 should give you 4750 as the new population at risk. but notice that end of year 5 we have 4800. idk if that means 50 people were false positives before or 50 people were added but in incidence births/death/etc don't matter it's kind of like UWORLD ID 1270. assuming average annual incidence is the same as cumulative incidence this was just a bunch of word vomit. sorry if it was unbearable to follow +

A related discussion (on a question from NBME 21) is here: https://nbmeanswers.com/exam/nbme21/714

+2/- csalib2(4)

The quickest way to do this is to add all the numerator ... = 1500. Then Divide by both year and susceptible population =1500 / (5*6000) = 5%.

+2/- peqmd(74)

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submitted by jotajota94(14)

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incidence = number of new cases/ new cases+ population at risk. new cases= 250. People at risk that year 5050 (including new cases. In one point they were a population at risk) 250/ 5050= 4.9%

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submitted by felxordigitorum(12)

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Exclude the intake year. Add all the positive results (400+250+300+300+250 = 1500). Divide 1500 by the total number of patients from the start of the incidence period, year 1, so 6000 ---> 1500/6000=0.25, this is the average across all 5 years. For annual, divide 0.25 by the number of years, 5 --> 0.25/5 = 0.05 --> 5%.

That's how I got it.

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submitted by sabistonsurgery(5)

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Incidence is typically calculated as a number diseased per exposed population per unit time, usually year. For each year, divide the people found, by the people exposed. You arrive at figures fr each year being 4%, 5% and 6%, avergae being 5%.

Q: Why not count intake year for which the incidence seems to be 40%?

Ans: So that's the trick of the question. IT seems to imply the intake year also has 4000 new cases, but this isn't incidence, this is prevalence! What happened here was that the population of 10,000 was not being screened for AIDS. Then you started the screening program.

Typically in any screening program, you will establish the baseline prevalence of the disease by testing all of the general population. You find 4000 new cases, but these were just undetected cases. You exclude them from your monitoring. For the next year, you keep a watch on the 6000 left, of which 400 develop AIDS. You calculate incidence for this year as 6.7% and exclude these 400 from the next year and so on.

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submitted by sweetsummerblowout(0)

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https://step1.medbullets.com/stats/101011/statistical-hypotheses-and-error

The chart on this site shows alpha/beta/power/True negative(correct). This question asks "likelihood of missing an association" - which basically means what is the false positive rate, which is the definition of alpha/Type I error. Question stem states alpha=0.05, which is 5%.

Beta is Type II error, and would be 10% (false negative). 90% would be Power (1-beta)

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submitted by ∗guillo12(58)

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I don't know the equation for this, but I did this: The sum of incidence per year/ the sum of number reamainig in population. So... 1,500/28,800 = .05208 or 5.2%

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