Saturday, October 19, 2002

The exam is available on the class website. I was unable to post the file on the blog in a download-able format. The link to the class website is on the left. Good luck!

_karolina

Tuesday, October 15, 2002

I was reviewing problem set #7 online. For 2b, you calculated .00495/.0547 as .095 or 9.5%. According to my calculator, the answer is .0904936015 (to be painfully exact). Perhaps this is a mistake? Could you please check this when you have a moment? It may have been marked wrongon the set.

yes, i'm sorry about that. #2b on probelm set #7 should be 9.05%, rounded, not 9.5%.

Sunday, October 13, 2002

Answer key to problem set #7

1a) The sensitivity of the test (.99) times the prevalence of the condition (.50) = .495

One minus the specificity (1 - .95) of the test (.05) times the prevalence of the condition (.50) = .025

The sum of these two is (.52)

(.52) x (10,000) burglars = 5200 positive results.

b) Of all positive results (.52), the ones that are actual users are (.495). .495/.52 = .952 or 95.2%

2a) The sensitivity of the test (.99) times the prevalence of the condition (.005) = .00495

One minus the specificity (1 - .95) of the test (.05) times the prevalence of the condition (.995) = .04975

The sum of these two is (.0547)

(.0547) x (10,000) burglars = 547 positive results.

b) Of all positive results (.0547), the ones that are actual users are (.00495). .00495/.0547 = .095 or 9.5%
Note about the first exam: I will do a review for it, and will distribute the exams afterward. If you have a class conflict and cannot attend the discussion, Mark will distribute the exam after class. As always, once you have the exam, there is to be no more collaboration.

_k