Hey guys, here are my answers; please check and confirm/deny (I wasn't sure about #7). Hey, someone told me that even though we have to hand this in, it's not graded...just marked 1 or 0 for handed in or not. Is that true?:

1. 930/1465=.63

2. 1023/1465=.7

3. 815/1465= .56

4. P(D+,T+)=P(D+)P(T+|D+)=(1023/1465)(815/1023)=.56=probability for 1 person, so take this to the 5th power for 5 => .053

5. 1-P(all have T+); P(all have T+)=P(T+)to the fifth=> so, 1-.63e5=.9

6. .63e5=>.1

7. 1-P(none have T+)=??

8. P(T+|D+)=815/1023=.8; sensitivity of the test.

9. P(T-|D-)=322/442=.74; specificity of the test.

10. P(D+|T+)=815/930=.88; predictive value of a positive test.

11. P(D-|T-)=327/535=.61; predictive value of a negative test.

12. blah blah blah

-- Omar Akhtar (akhtarom@umdnj.edu), September 01, 1999

Answers

#5. Omar, your calculation for this one is wrong. There are two ways to find the probability that none are +: a) 1-P(1+4-)-P(2+3-)-P(3+2-)-P(4+1-)-P(5+)= b) P(5 negative) = (535/1465) to the 5th = 0.006 <-- EASIER

#7. P(at least one negative) = 1-P(all positive) = 1-0.1 = 0.9 P(all positive) from question #6

-- Irina Sigal (sigalir@umdnj.edu), September 01, 1999.

#5 it is (1-.63)e5

#7 5!/1!*5! * .63e4*.37e1

-- Mateusz Opyrchal (opyrchma@umdnj.edu), September 01, 1999.

i agree with irina on #5... i got .0065

-- Fakhra Chaudhry (chaudhfs@umdnj.edu), September 01, 1999.

Guys, let's move this over to the students-njms bulletin board because it's better if we keep everything student related on that site.

Go to students-njms.umdnj.edu, log in, click on "Class" right under "Announcements" on the left hand side and it'll kick you right into the board.

I already posted this entire thread there.

-- Omar Akhtar (foo@bar.com), September 02, 1999.

Omar, the "post reply" button on the student site does not work...

Matt,

for #5 you have the same thing I do, just you had 0.63 rounded off, so your answer came out a little different. I used (535/1465)x^y5.

#7, on the other hand... If you look carefully, your calculation is for the probability of 1 negative and 4 positive, but the question asks for AT LEAST 1 negative.

-- Irina Sigal (sigalir@umdnj.edu), September 03, 1999.