Remember that the CI= X +/- Z(SE) where SE=(SD/n^.5) and SE decreases as n increases, and vice versa.
In our question, n, the sample size, is decreasing because we will do fewer measurements of the BP.
This means SE will increase, so the CI will increase.
Basically, with fewer measurements, the CI gets wider because you need a larger sample size for more valuable results.
usmleuser007would require a a large sample size to see if there is a true difference+
claptainThis question is bogus. CI does not always increase with decreased sample size or vice versa. Four readings with small variation would give a narrower CI than 10 readings with greater variation. The only thing you can be certain about by adding more samples is that the CI will most likely change, but which direction is uncertain.+9
bartolomooseRecall the formula for 95%ci
Mean +/- 1.96* (SD/sqrt(samplesize))+1
the_enigma28@claptain The point you made is relevant in studies involving random data. But in case of this question, the data being collected is in fact the diastolic BP. We take several readings of BP to rule out white-coat hypertension and have as accurate reading as possible. In this case, taking more readings will actually narrow down the confidence interval. The readings here represent physiological parameter, which wouldn't vary veryyyy widely in an individual.+1
lowyield@claptain i was thinking the same thing but ended up choosing the increased because alot of NBME seems to reward the more simplistic answer than the overthinking answer+
submitted by โamirmullick3(76)
Remember that the CI= X +/- Z(SE) where SE=(SD/n^.5) and SE decreases as n increases, and vice versa. In our question, n, the sample size, is decreasing because we will do fewer measurements of the BP. This means SE will increase, so the CI will increase.
Basically, with fewer measurements, the CI gets wider because you need a larger sample size for more valuable results.
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