Problem set #5

Problem set #5

by Elan Guterman -

Hi,

A few questions about the problem set:

For question 1(f) - are you expecting a particular number calculation as part of this answer?

For question 2 - I initially understood the phrase "willing to admit 25 patients to the hospital for 2 days unnecessarily in order to avoid discharging one from the emergency department" to mean that the threshold probability is 1/25, not that it is 25 times worse to have a stroke than admit somebody to the ED (C = X, B = 25X) but I am now confused about how to interpret questions 2(e) and 2(f).

Are we expected to use different thresholds for the two question parts? In other words: 2(e) C/B = 1/25 and 2(f) C/(C+B) = 1/25 ?

Thanks for any clarification


Re: Problem set #5

by Charles Murphy Iv -

For question 1(f) - are you expecting a particular number calculation as part of this answer? 
 No

For question 2 - I initially understood the phrase "willing to admit 25 patients to the hospital for 2 days unnecessarily in order to avoid discharging one from the emergency department" to mean that the threshold probability is 1/25, not that it is 25 times worse to have a stroke than admit somebody to the ED (C = X, B = 25X) but I am now confused about how to interpret questions 2(e) and 2(f).

It is 25 times worse, aka more costly, to send someone home who has a stroke in the next 2 days than it is to admit someone who will not have a stroke in the next 2 days.
C/C+B = treatment threshold

Are we expected to use different thresholds for the two question parts? In other words: 2(e) C/B = 1/25 and 2(f) C/(C+B) = 1/25 ? 
No. C/B = 1/25


Re: Problem set #5

by Michael Kohn -

This C/B vs. C/(C+B) is just odds/probability all over again.

If you are willing to admit 25 patients UNNECESSARILY to avoid failing to admit 1 who needs it, then C/B = 1/25.

That means you are willing to admit 26 TOTAL patients to avoid failing to admit 1, because you have to count the one who you admitted correctly.  Ptt = 1/26.  Note that it is just C/(C+B).

1/25 = C/B = threshold ODDS

1/26 = C/(C+B) = threshold PROBABILITY.


MAK


Re: Problem set #5

by Jerrine Morris -

Also, for problem 1 parts B and C, I am conceptualizing carrying the umbrella when the treatment threshold is less than the predicted risk. If so, I would plan to bring the umbrella based on both predictions which seems odd. 

For problem 1 parts d and e, I am calculating the predicted risk for rain for channel 3 and weighted the predictions based on the number of days the risk applied to. When I do this my numbers look strange.  Am I overthinking this question?