# Statistics Homework Question

I tried to got my my TA's office hours today but someone hogged him for 45 minutes and I didn't get a chance to ask my question. Hopefully someone can shed some light on this for me. Note that this is a question about diagnosing people with a disease $P(D)$ is the probability the person has the disease.

Assume that for a randomly selected person, $P(D) = 0.2$, $P(\text{Positive test}\mid D) = 1$. $P(\text{Positive test}\mid\overline{D}) = 0.05$, so that the inexpensive test only gives false positive, and not false negative results.

Suppose that this inexpensive test costs $\$10$. If a person tests positive then they are also given a more expensive test costing$\$100$, which correctly identifies all persons with the disease. What is the expected cost per person if a population is tested for the disease using the inexpensive test followed, if necessary, by the expensive test?

• You say $0.2$, but the answer in the back of your book is right if it's $0.02$. See below. – Michael Hardy Nov 12 '12 at 17:15

$$\begin{array}{cccccccccccc} & & & & & \bullet \\[15pt] & & & & \swarrow0.2 & & \searrow0.8 \\[15pt] & & & D & & & & \overline{D} \\[15pt] & & & & & & \swarrow & & \searrow \\[15pt] & & & & & & \text{positive} & & \text{negative} \\[15pt] & & & & & \swarrow & & \searrow \\[15pt] & & & & D & & & & \overline{D} \end{array}$$ Follow this tree. Everybody pays $\$10$for the first test. Then$5\%$of the aformentioned$80\%$of the population pays$\$100$ for the second test.
The $20\%$ who are ill also pay the $\$100$for the second test. So $$\10 + (0.2)\100 + (0.05)(0.8)\100 = \10 + \4 = \34.$$ Another p.o.v.: You have$100$people. Of those,$20$are ill. They take both tests, paying$\$110$ each. Of the other $80$ people, $5\%$ take both tests, paying $\$110$. That's$4$people. They other$76$test negative on the cheap test, and pay only$\$10$ each. So $$24\cdot\110 + 76\cdot\10 = \3400.$$ So the average is $\$34$. Later note: You say the answer at the end of the book is$\$16.90$. That would be correct if $P(D)=0.02$ rather than $0.2$. Just go through the above with those numbers instead of what we used.
• That makes a lot of sense to me, the only thing is that they are giving an answer of $16.90 at the end of the book... Do you have any idea how they are coming to that answer? – TopGunCpp Nov 12 '12 at 16:04 • I deleted this answer, then edited further, then un-deleted it. I initially omitted the fact that everyone who is ill also takes the second test. – Michael Hardy Nov 12 '12 at 16:50 • If$\$16.90$ is given as the answer, it's got to be a weighted average of $\$110$and$\$10$ for some weights, say $w$ and $1-w$. So $110w+10(1-w)=16.90$. That says $100w + 10=16.90$, so $100w=6.9$, and $w=0.069$. That would mean only $6.9\%$ of the population take the second test. To me that makes no sense, since all of the $20\%$ who are ill would take the second test. – Michael Hardy Nov 12 '12 at 16:55
• OK, here's the bottom line: if $P(D)=0.02$, i.e. $2\%$ rather than $20\%$ have the disease, then the answer is $\$16.90$. – Michael Hardy Nov 12 '12 at 17:14 • Michael Thank you soooo much for your help, I apologize for misreading the question, I even used 0.2 when I was working in my note book. As it turns out... P(D)=0.02 like you suspected. Sorry for the misunderstanding on my part and thank you for the helpful explanation!!! – TopGunCpp Nov 12 '12 at 17:24 Draw yourself a probability tree. On the first fork we have$P(D) = 0.2$and$P(D') = 0.8$. On the second set of forks we have$P(+|D) = 1$,$P(-|D) = 0$,$P(+|D')=0.05$and$P(-|D') = 0.95$. It follows that$P(+ \wedge D) = 0.2 \times 1 = 0.2$,$P(- \wedge D) = 0.2 \times 0 = 0$,$P(+ \wedge D') = 0.8 \times 0.05 = 0.04$, and$P(- \wedge D') = 0.8 \times 0.95 = 0.76\$.
$$E[X] = 0.2 \times 110 + 0 \times 10 + 0.04 \times 110 + 0.76 \times 10 = 34 \, .$$