HCOL 195 11/20/09

I’ve put the link to the NPR discussion I mentioned in class here and below.

And here is an interesting article from the New York Times that describes the history of breast cancer ideas, going back to the 19th century. Some of the debate today is very old.

Question 1

The distinguishing characteristic of Question 1 is that there are two independent cure rates; this means that the table of posterior probabilities must also be two-dimensional, that is, a square table. The posterior probability of each combination of cure rates r and s is the product of the posterior probability of r and that of s, and it is put into the appropriate square in the grid. Then the probability that the cure rates are equal is given by the sum along the main diagonal (red boxes in the picture), and the probability that B is better than A is the sum of all the numbers above the main diagonal, where s is greater than the corresponding r.

Question 2(A)

In the general picture, we see that according to the probabilities, the expected number of survivors is 200 in each case, but in the risky case, it’s possible that everyone will be killed. The risk-averse general would choose the sure thing rather than risk everyone being killed. What that really means is that the value of the 600 soldiers is less than three times the value of the 200 soldiers. That’s illustrated in red in the diagram, where we put 550 rather than 600 to reflect this assessment of this general.

Question 2(B)

In this problem, the objective is to win, and so there are just two gains, one for winning and the other for losing. To make the arithmetic come out nicely, 30 is a good gain for winning and 0 for losing. There are only two possible decisions, to go for the “sure thing” of 200 soldiers, versus the “risky” choice of possibly 600 soldiers, but possibly none. For the sure thing choice, the only chance node is the one that tells us whether we win or lose. But for the risky choice, the first thing that happens is that we take the soldiers to the scene of the upcoming battle, so that’s the first chance node. Once we get to the scene of the battle (if we do), then the battle takes place, so there is a second chance node for each of the two possible outcomes from the trip. We see that the expected gain is greater if we use the “sure thing” branch.

Question 3

Some background on this story can be found in the NPR story from All Things Considered, here, as I promised in class. This is a very interesting conversation, you can listen online, download an mp3 for your mp3 player, or read the transcript. I found the discussion very illuminating.

The New York Times today also had a very illuminating article about how the history of understanding cancer over the last 150+ years also influences the discussion today.

The diagram is very simple; you have the loss of one extra life ($5 M according to our class discussion) versus the cost of ten mammograms per person ($3M) plus whatever the loss is for the false positives (200, or 10% of the number in the group in the statement of the problem; but I found out that the actual number is closer to 1000). Also for simplicity I have drawn the diagram for 1 extra life saved out of 2000 women tested; the error is 5%, smaller than the error of other numbers that go into the calculation, so this is justified. We see that (in the diagram again) we will be indifferent if the additional cost C of the false positives is such that $5M=$3M + 200C. That makes C=$10,000. With the more accurate number of 1000 false positives, C=$2,000. Any C larger than that corresponds to a decision not to test.

Comments:

The actual discussion in the media indicates that they were not doing a cost-benefit analysis like the above, but rather trying to weigh the human cost of one extra life saved versus the human cost of false positives, which would include the anxiety, the pain of additional testing, the risk of additional testing (X-rays for example have a small association with cancers in and of themselves), the risk of unnecessary surgery that might cause a woman to be disfigured or even lose a breast, the risk of a woman dying due to unnecessary surgery. It’s not clear how they weighed the various costs.

Also, consider that money not spent for testing could be used for something else. Is the cost of testing in this age group the best way that we could spend the money? The amount of money available for health care is not unlimited, as Congress is finding out. It might be that more lives could be saved if the money were spent elsewhere.

Finally, as I mentioned in class, the 10% false positive rate is really for one mammogram. If you have ten mammograms in ten years, the probability is a lot closer to 50% that one of them will come up with a false positive. This explains the difference between the number I gave you in the question and the figure of 1000 false positives that the media is reporting. It also says that the probability of a false positive would be reduced if mammograms were only given every two years (five versus ten). This would reduce the negative outcomes from false positives. It would also cost less money, money that could be used in other, and possibly more productive ways. And (according to the discussions I’ve seen), the additional risk of a woman dying who could have been saved would be very small.

Question 4

This one is pretty straightforward and is similar to many that we’ve done during the course. The probability of observing 7 items of the first kind, 2 of the second, and 1 of the third, is (p_1)^7 (p_2)^2 (p_3)^1, so that’s the likelihood. The prior is given in the problem, and the rest is routine: Multiply likelihood and prior to get joint, add the joints to get the marginal, divide each joint by the marginal to get the corresponding posterior. Here we have only two states of nature. I gave full credit if you simply explained the calculation in detail but did not compute the numbers.

Question 5

This is similar to the problem we discussed on the study sheet, except that instead of H/T, there are three equally probable ways that the die can come up. So the probability of each of these ways is 1/3, and the calculation of the probability of hearing “yes”, given that p is the true proportion of people who did what the subject of the question is, is (p+1)/3=0.4. This gives p=0.2.

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