Chapter 8: Two Dichotomous Variables

Transcription

1 Chapter 8: Two Dichotomous Variables On the surface, the topic of this chapter seems similar to what we studied in Chapter 7. There are some subtle, yet important, differences. As in Chapter 5, we have one population box, which can represent a finite population or trials (but not BT s, as we will see). Each card has the value of two dichotomous variables. Being dichotomies, it is natural to think of each variable giving either an S or an F, but that would be confusing (think of keeping straight: S on first and F on second versus the reverse). Instead, we denote the possible values for the first variable as A and A c (read A-complement); and we denote the possible values for the second variable as B and B c. Suppose the population is all students enrolled this term at UW-Madison. Below are some examples of the two variables one might determine: + 311

2 1. GPA and anxiety about the future (both with values high/low). 2. Gender and residency. 3. GPA and activity (yes/no) in recent presidential campaign. For the second of these, if we had population boxes for females and males, then we could proceed as in Chapter 7. Chapter 8 is useful whenever: We can easily sample one box with two variables, but it might be difficult/impossible to sample two boxes. There is no obvious direction of influence as in 1 above. We might want to predict one variable based on the value of the other, even in cases like 2 above where the direction of influence is one-way. In Chapter 5, we had several ways to characterize the contents of a population box

3 The first was to specify the numerical values of s, f and N. We take a similar approach now, but the notation will be messier b/c we have two variables. We represent the population with the following table of population counts: B B c Total A A c N AB N A c B N AB c N A c B c N A A c Total N B N B c N We summarize this table by dividing each of its entries by N, giving the table of population proportions: B B c Total A A c p AB p A c B p AB c p A c B c p A A c Total p B p B c

4 Our goals in Chapter 8 are rather modest. We will cover only Sections 8.1 and 8.2. We will learn neither estimation (8.3) nor testing (8.4). Instead, we will study the structure of the population proportions; they sometimes reveal very interesting features of the population. In order to save time in presentation, I will focus on one example. It is a very important example in medicine and society. We will study screening tests for a disease. (In the text I used the softer term condition rather than disease, but b/c our main new idea involves something called conditional probabilities it s confusing to talk about conditional probability of a condition. I have no idea why I never realized this until after the book went to print. Mea culpa.) The population consists of some well-defined collection of people, a disease of interest and + 314

5 a screening test for the disease. Examples include: TB and skin test; breast cancer and mammogram; colon cancer and colonoscopy; prostate cancer and PSA measure; HIV infection and an ELISA test. The two variables are: presence (A) or absence (A c ) of the disease; and positive (B) or negative (B c ) result on the screening test. Thus, each person s card contains the actual disease status and the outcome of the screening test. This is clearly an idealization: Often disease status can be determined only by an autopsy; and even an autopsy might be inconclusive. It is unlikely that everyone in a population has had a screening test. A positive screening test is taken as an indication that the person has the disease

6 Thus, a positive might lead to: a more expensive test; treatment; quarantine; loss of health insurance. A negative screening test is taken as an indication that the person does not have the disease, usually resulting in no further attention from the medical profession. It is important to remember that screening tests make mistakes. It is useful to consider the following table: B A Correct Positive False Negative A c False Positive Correct Negative B c Before considering numbers, it is important to examine the consequences of errors. To a large extent, this examination is different for every disease/screening test combination. For example, what are the consequences of a false positive (false negative)? + 316

7 First, we realize that nobody knows there is an error. Thus, we begin with the question: what are the consequences of a positive (negative)? Consider the TB skin test. A negative means the person is sent back into the world believing that he/she is uninfected. Thus, the consequences of a false negative are twofold: no treatment received and others might be exposed. A positive could mean that the person begins treatment, but I have been told by past students that a positive means a more expensive/invasive screening test: a chest x-ray. Assuming that the x-ray detects the error (i.e. assuming that the chest x-ray does not yield another false positive; this is a very long road to follow) then the consequences of a false positive include: cost of unnecessary x-ray; patient anxiety; patient exposure to unneeded x-ray

8 Before our next example, let s consider the PSA test for prostate cancer. The PSA measures the concentration of an antigen in a man s blood (I don t know the units). I have been told that there are two ways for the test to be positive: PSA > 5; or PSA 5 but much larger than the previous year s value. It is also possible that a man s age is factored into the formula for positive. But here is the point: For many screening tests, there is a certain arbitrariness in deciding which values give positive and which give negative. In the early days of AIDS in the US, many scientists realized that it was caused by a virus that was present in blood. Thus, there was great interest in finding a screening test for infected blood. (My source for this information is the book The Band Played On by Randy Shilts.) + 318

9 The Abbott Corporation in Chicago developed a test for the HIV antibody in blood and this was proclaimed as a way to Make the blood supply safe. (Of course, b/c all tests make mistakes, one could debate the use of the word safe, but we won t do that.) Soon after, many civic leaders called for using the test to identify persons infected with HIV. The scientific community responded that it is a test for blood, not people. This seemed to confuse many civic leaders. I will discuss why a test for blood is not a test for people. Suppose a test measures a concentration, C, of something in the blood. For concreteness, let s suppose that C can take on values between 0 and 1 and that the higher the value of C, the stronger the indication of the presence of the disease. With these conditions, the screening test comes down to specifying a threshold d. If the concentration is larger than d, the test is positive; smaller than d it is negative

10 The picture below presents two possibilities for d: d 1 < d 2. Neg. d 2 Pos. 0 Neg. Pos. 1 d 1 If C < d 1 both thresholds agree; the test is negative. If C > d 2 both thresholds agree; the test is positive. But if d 1 < C < d 2 the thresholds disagree; d 1 says positive and d 2 says negative. Thus, comparing thresholds, d 1 gives more positives, and, hence, more false positives (and more correct positives too). And d 2 gives more false negatives. Now, let s return to the screening test for HIV in blood. First, as a test of blood. We begin with the consequences of a positive: the blood is discarded; of a negative: the blood is used in a transfusion

11 The consequence of a false positive is that a unit of safe blood is discarded. The consequence of a false negative is that a healthy person gets infected. Clearly, an FN is much more serious than an FP. Thus, when choosing a threshold, we would want to avoid FNs. We do this by making d very close to 0. But now let us consider testing people. A negative means a person is sent back into the world having been told he/she is not infected. A positive means the person is told he/she is infected. The consequences of a positive are impossible to specify exactly. There was talk of testing everyone and then quarantining those who test positive. There was fear that a positive test would lead to loss of health insurance, job and family relationships. For most diseases, a consequence of an FN is that the person does not receive treatment, + 321

12 but for HIV infection in the early 1980 s there was no treatment. Thus, I would argue that the only consequence of a person receiving an FN is that it might lead to more infections. (Discuss.) The consequences of an FP is major unwarranted anxiety and possibly unwarranted serious disruptions of life, liberty and the pursuit of happiness. Now everyone is entitled to a personal opinion about how serious these various consequences are. But I think that all would agree that the relative seriousness of FP versus FN for testing people is hugely different than for testing blood. Thus, as the scientists said, a test for blood is not a test for people b/c the different tests should have different thresholds. We will now return to numbers and formulas

13 The tables below appear on p. 260 of the text and is referred to as the first screening test for a population. Screening Test Disease B B c Total A A c Total Screening Test Disease B B c Total A A c Total Suppose we select a person at random from this population. The probability we select a person who has the disease is: P(A) = 100/1000 = 0.10, as shown in the second table as p A. Note: Henceforth, I will write P(A) instead of p A ; P(AB) instead of p AB ; and so on. I apologize for any confusion

14 I believe that one of the basic questions in science is how to compare two things. As a result, we studied this issue in Chapters 1 3, 7 and 16. Another basic question is: How do we make use of partial information? This is our current topic. In particular, suppose that we select a person at random and we learn (this is our partial information) that the person would test positive; Does this change the earlier probability? We proceed by reasoning from first principles. Look at the table of population counts again. Given that the person would test positive, we know that we have selected one of the 120 persons in the B column. Of these 120 persons, reading up the column we see that 12 of them have the disease. Thus, the probability we want is: P(A B) = 12/120 = We note that P(A B) = P(A) for this example; let s do one more computation before we interpret this equality

15 Suppose that we select a person at random and we learn that the person would test negative. The probability of having the disease given a negative test result is: P(A B c ) = 88/880 = To summarize, for this example, P(A) = P(A B) = P(A B c ) = In words, the screening test is perfectly worthless! Let s consider another possibility, a second screening test for this disease. The following table appears on p. 262 of the text. Screening Test Disease B B c Total A A c Total As above, P(A) = But now, P(A B) = 95/104 = and P(A B c ) = 5/896 =

16 This second screening test, while not perfect, is informative. The conditional probability of A given B is given by the following formula: P(A B) = N AB N B = P(AB) P(B). There are eight conditional probabilities of interest: In P(A B), A can be replaced by A c ; B can be replaced by B c and the positions of A and B can be reversed. Clearly, we do not want to derive (and remember) eight different formulas. We don t need to, provided we read the above formula creatively and not literally. Creatively, the formula says: the probability of one guy given another guy is the probability of both guys divided by the probability of the second guy. We can apply this interpretation to any situation. For example, P(B c A) = P(ABc ) P(A)

17 Note that in the numerator of this last fraction I write P(AB c ) instead of the perhaps more natural P(B c A). But they are the same and by convention mathematicians like to write these expressions alphabetically; i.e. A before B. We can rewrite the formula for P(A B) as P(AB) = P(B)P(A B) or, equivalently, P(AB) = P(A)P(B A). Either of these is referred to as the multiplication rule for conditional probabilities. Let s revisit our two screening tests. For the first screening test, P(AB) = P(B)P(A B) = P(B)P(A), which is our multiplication rule from Chapter 5. Thus, for the first screening test we say that the two variables are statistically independent. For the second screening test, P(AB) = P(B)P(A B) P(B)P(A), making the variables statistically dependent

18 Note the following. In Chapter 5, independence was good. Now, in Chapter 8 it is bad; a sign of a worthless screening test. Context is everything. We can now see a quick way to check for independence. Suppose we have the following table of population proportions: B B c Total A A c Total If we take the total for the first row, 0.30, and multiply by the total for the first column, 0.40, we have independence if, and only if, the product equals the number in the upper left entry, In symbols, we check to see whether P(A)P(B) equals P(AB). In this table it does and we have independence and a worthless screening test

19 Be careful when you hear somebody talk about the false positive (negative) rate. There are actually three of each and they have very different meanings. I will illustrate with FP s and the second screening test. An FP means that B is matched with A c. But there are three ways to combine them: P(A c B), P(A c B) and P(B A c ). P(A c B) = is always the smallest of these three. Next, P(B A c ) = P(A c B)/P(A c ) = 0.009/0.900 = 0.01 is larger b/c we are dividing P(A c B) by a proportion. If, however, the disease is rare, then P(A c ) is close to 1 and the division has little effect. Finally, P(A c B) = P(A c B)/P(B) = 0.009/0.104 = is considerably larger. If the disease is rare, it will be considerably larger b/c P(B) is close to 0 and dividing by a number close to 0 inflates the numerator

20 Especially if a disease is rare, conditional probabilities can give surprising insights. B B c Total A 24, ,000 A c 2,499, ,475, ,975,000 Total 2,524, ,475, ,000,000 P(A) = 25,000/250,000,000 = In words, 0.01% of the population has the disease; it is very rare. P(B c A) = 250/25,000 = 0.01 Of those with the disease, only 1% receive an FN. P(B A c ) = 2,499750/249,975,000 = Of those who are disease free, only 1% receive an FP. Sounds like a good test, but: P(A c B) = 2,499,750/2,524,500 = Of those who test positive, 99% are disease free

21 There is a reference at the end of Chapter 8 to a 1987 paper in the NEJM whose authors argue that P(A c B) was approximately one-third for testing for HIV infection. Thus, for example, if everyone was tested and those who test positive were quarantined, fully onethird of the people quarantined would be uninfected. Conditional probabilities can help us understand the effectiveness of screening tests and this can be useful for citizens as well as health care professionals. The big weakness in the above analyses, however, is that the tables of population counts are all hypothetical. Can we do better in practice? Well, first let me point out that what usually works, does not work here. Namely, suppose we take a random sample from the population and use our data to estimate the various proportions. Here are some of the difficulties: + 331

22 For many diseases, there is simply no way to determine whether a person has it. It is problematic to try to force people to take a screening test; especially, if the full consequences of a positive result are not known. Even if the previous two items do not apply, for a rare disease we would need a huge sample size to get a useful estimate of P(A), not to mention the even smaller P(AB) and P(AB c ). People at higher risk for a disease (e.g. IV users of illegal drugs) might be particularly resistant to appearing in a random sample, perhaps greatly biasing the study. Instead, medical researchers proceed as follows. Make an educated guess as to the value of P(A). (And repeat the analysis below for different choices of P(A) to obtain a range of answers.) Obtain a sample of people who definitely have the disease; give them the screening test and use these data to estimate P(B A) and P(B c A). Obtain a sample of people who appear to not have the disease. Give them the screening test and use these data to estimate P(B A c ) and P(B c A c )

23 The above ideas lead us to the last topic of Chapter 8: Building a table of probabilities. Here is the idea. Suppose that we know P(A), P(B A) and P(B A c ). We can then build the table of probabilities (population proportions). Here is an example. I call it a screening test for squirrels. The population is a model for trials. A trial consists of a 30 minute block of time in my backyard in summer. Below are the variables: A means that at least one squirrel entered my yard during the trial. B means that my dog Casey barked at least once. From past observation, I believe P(A) = 0.30, P(B A) = 0.80 and P(B A c ) = I can begin to complete the following table: Bark No Bark Total Squirrel 0.30 No Squirrel 0.70 Total

24 Applying the multiplication rule for conditional probabilities: P(AB) = P(A)P(B A) = 0.30(0.80) = 0.24 and P(A c B) = P(A c )P(B A c ) = 0.70(0.20) = Next, we plug these numbers into the table above, add and subtract to get: Bark No Bark Total Squirrel No Squirrel Total We can now use this table, as before, to answer questions about the population. For example, P(B) = 0.38 and P(A B) = 0.24/0.38 = In the above we have been given P(A B) s and been able to compute P(B A) s. This reversal is called Bayes rule or Bayes formula

25 Bayes rule is needed for anything with DNA testing, for example, paternity testing. Let A denote that Ralph is the father and let B denote that the DNA test is positive for Ralph being the father. Suppose that P(A) = 0.001, P(B A) = 1 and P(B A c ) = The table is: B B c Total A A c Total Thus, P(A B) = 0.001/ = 0.910, not the usually reported 9999 out of 10,000. Of course, the legal arguments usually surround the choice of P(A), which I have made very small