In clinical medicine, we are taught the important principle of using disease prevalence (how many people have the disease in question during a specific point in time) to frame the way we think about what is likely to be wrong with the patient in front of us. For example, the prevalence of malaria is high in sub-Saharan Africa, and quite low in Maine. If a patient presents to a doctor in Maine with a high fever, the doctor may know that a cardinal symptom of malaria is fever, but would also know that the risk of his average patient in Maine of having malaria is about 0.00000001% since the prevalence of malaria in Maine is virtually zero.
Determining the pre-test odds and the post-test odds of your patient having a disease is something we clinicians do intuitively hundreds of times per day when taking care of patients. It is also one of the places where patients can get into trouble when doing their own “Google research” or trying to interpret their own blood test results, where every rare disease and outcome seems just as likely as the next.
There are mathematical and statistical methods that are used to derive the chance of someone actually having the disease tested for when the test is positive. This situation is called a “true positive” (in contrast to a “false positive” which is when the test comes back positive but the patient does not actually have the disease tested for). But where do these numbers come from?
The Cologuard test is a non-invasive stool test designed to predict the chance of a patient having colorectal cancer. Overall, Cologuard is a decent test but is plagued by many false positive results (the basics of the test have been covered here previously). In the following original article, Dr. Douglas Cooper takes us on a deep-dive into the statistics and derives the mathematical odds of actually having colorectal cancer if you have a positive Cologuard test.
As you may know intuitively, when the prevalence of cancer increases in the community, so do the odds of having cancer when the Cologuard test is positive. Dr. Cooper has shown below that the actual risk of cancer with a positive Cologuard result may be quite a bit higher that the usual cited risk of 4%. The ultimate message remains the same however: A positive Cologuard test needs to be followed by a colonoscopy (the gold-standard test) to either diagnose or disprove the presence of colon cancer.
-Frederick Gandolfo, MD
A Bayesian Interpretation of a Colon Cancer Screening Test (Cologuard)
Douglas Winslow Cooper, PhD
August 12, 2019
The result of a screening test for colon cancer is often interpreted from the sensitivity of the test, the probability of a positive result given that the patient has cancer. Bayesian analysis, performed here, answers the more pertinent question, the probability that the patient has cancer, given the positive result. The analysis is performed for a range of prevalence of colon cancer and analogously for a negative result. For example, the Bayesian estimate of the probability the patient has colon cancer, from a positive test result on a Cologuard test with a sensitivity of 92%, for an assumed 4% prevalence, is 23%, quite different from the both sensitivity and the prevalence. The Bayesian analysis is shown to be consistent with a particular interpretation of a standard two-by-two table.
Introduction and Methods
As noted by Anderson and Ahlquist (2016), “Globally, gastrointestinal (GI) malignancies account for roughly 40% of all cancer deaths.” They make a case for improved screening techniques: “effective early detection methods are needed desperately.” We would add that correct interpretation of those screening techniques’ results is also important.
If one takes a screening test, like the Cologuard test, that has a 92% probability of correctly identifying colon cancer, and the test result is positive, does that mean the probability is 92% of having colon cancer? No, one needs a statistical technique, Bayes theorem, to find the right answer, which in the case of colon cancer is much smaller than 92%, but larger than the few percent prevalence in the general population.
Bayesian statistical analysis allows us to update prior probabilities using new information. Dan Morris and Mark Koning (2016) illustrated the fundamental Bayes formula for analyzing the degree to which a positive result on a test (+), such as exhibiting certain symptoms, indicates one has the condition (F); for example, whether one has the flu (or colon cancer). Here are the equations for one of their examples, testing positive for the flu:
P(F|+) = P(F) P(+|F) / [P(+)]
P(F) = probability of having the flu, in the general population
P(F|+) = probability of having the flu, given the test is positive
P(+|F) = probability of having a positive on the test, given the flu
P(+) = probability of positive on the test, in the general population
P(+) = P(F) P(+|F) + [1-P(F)] P(+|F’),
A formula involving the true positives and false positives and the population fractions having the flu and not having it.
We have put P(F) in bold-face, as it is our prior (naïve) estimate.
The ratio multiplying it, P(+|F) / P(+), might be considered a magnifier (not always greater than 1, however) that adjusts our “prior” P(F) probability (likelihood) that we have the flu by incorporating the information that we have tested positive (+), producing a “posterior” probability we seek, P(F|+).
Recently, we had to decide whether to have a patient undergo a colonoscopy to check for cancer. Because she is quadriplegic and on a ventilator, it is awkward to do such a test. We used the Cologuard screening test to get a preliminary assessment, and the test was positive. We took the advice of our gastroenterologist and had a conventional colonoscopy done. In an article by Frederick J. Gandolfo, MD, “What does a positive Cologuard test mean?”, he too recommended the conventional colonoscopy when the Cologuard test is positive.
We have now used Bayesian statistics to analyze this situation, and the analysis should be appropriate for others, as well. In the Thirteenth Edition of their book Statistics, James T. McClave and Terry Sincich provide the following Bayes formula (we have made some changes in notation):
P(F|+) = P(F) P(+|F) / [P(F) P(+|F) + P(F’) P(+|F’)]
Here, F stands for having cancer and F’ for not having cancer.
Although the lifetime risk of colon cancer is given variously as 4-6%, an extensive 10,000-patient study by Imperiale et al. (2014) found their cohort to have 0.7% prevalence; they determined the sensitivity of the Cologuard test as P(+|F) = 92% = true positive rate, and thus the false negative is P(-|F) = 1-0.92 = 0.080); the specificity is reported as P(-|F’) = 87% = the true negative rate, and thus a false positive rate of P(+|F’) = 1-0.87=0.13 (see Table 1). Using these values, including the 0.007 prevalence,
P(F|+) = (0.007) (0.92) / [(0.007) (0.92) + (0.993) (0.13)]
P(F|+) = (0.007) (0.92) / [ 0.136]
P(F|+) = (0.007) (6.8) = 0.048
The Bayesian estimate from the positive test result is almost seven times the assumed 0.7% prevalence.
The positive test result Bayesian multiplier (here, 6.8) raised our estimate of the probability of colon cancer from 0.7% to 5%, supporting getting the colonoscopy, which turned out to be negative (no cancer detected).
Given the false negative rates, one can carry out a similar analysis for the probability of having colon cancer, given a negative screening test result (-):
P(F|-) = P(F) P(-|F) / [P(F) P(-|F) + P(F’) P(-|F’)]
An estimated probability of having cancer given a negative test result would be
P(F|-) = (0.007) (0.08) / [(0.007) (0.08) + (0.993) (0.87)],
by assuming a true negative rate of P(-|F’)=0.87 and a false negative rate of P(-|F)=0.08, so
P(F|-) = (0.007) (0.08) / [0.864],
P(F|-) = (0.007) (0.09),
P(F|-) = 0.0006.
Thus, the negative screening result would decrease one’s expected probability of a colon cancer by a factor of about ten, under these assumptions.
These calculations were later confirmed using the site strattrek.com Bayesian analysis calculator.
Gandolfo (2016) notes the added value of a colonoscopy for detecting precancerous polyps, which are much more frequent than cancers (51% vs. 4%) and often warrant removal, so even a negative screening result might still lead to choosing a colonoscopy.
Table 1. Sensitivity, specificity, false positive, and false negative
|ILL (F)||NOT ILL (F’)|
|TEST||+||sensitivity = 0.92||false positive = 0.13|
|TEST||–||false negative = 0.08||specificity = 0.87|
The screening test gives added evidence of whether a colon cancer is likely to be present. It is notable that the “snapshot” prevalence of colon cancer was only 0.7% in the 10,000-patient Imperiale et al. test group. Presumably, a significant number of the cancer-free patients will later develop cancer, as indicated by the 4-6% lifetime incidence rates normally cited for colon cancer.
Those who do not have colon cancer (here, 99.3%) but who do test positive (false positive=13%) will likely go on to have a colonoscopy, which will almost certainly indicate no cancer. Those who do not have colon cancer (99.3%) and who do test negative (true negative=87%) will likely not go on to have a colonoscopy, which would have almost certainly (99+%) indicated no cancer.
Those who do have colon cancer (0.7%) and test negative (false negative=8%) will likely skip the colonoscopy and have their cancer missed (0.06%). Those who do have colon cancer (0.7%) and test positive (true positive=92%) will likely have the colonoscopy and have their cancer found (5%).
Because the prevalence of colon cancer depends on many factors, especially on age, we have calculated the corrected probabilities of having colon cancer or not, given positive or negative results on the test, and placed these results in Table 2, for populations with colon cancer probabilities (prevalence) of 0.001 to 0.10. Positive test results increased the calculated (a posteriori) probabilities of having colon cancer by factors of 7.0x to 4.4x over the prevalence (a priori) range studied. Negative test results, on the other hand, indicated 1% or less probability of cancer over the range studied and reduced the estimated probability of cancer by a factor of 1/10x.
Table 2. Corrected (Bayes Theorem) probabilities of having colon cancer (F) or not (F’), given positive (+) or negative (-) test results
|Population’s cancer probability (prevalance)||Patient’s Bayesian probabilities|
|Not cancer |
Two-by-two tables are more familiar than Bayes theorem analysis to many, and they are explored here for a hypothetical screening test having the same sensitivity (0.92) and specificity (0.87) (see Table 1) as the Cologuard test, operating on some hypothetical cases of a condition F or its absence, F’.
No illnesses, P(F) = 0.0, 1000 people tested
The 130/1000 = 13% false-positive (FP) rate (1-specificity) may lead to treating the “condition” unnecessarily, as none actually have it. The true negative (TN) rate was 87%.
All are sick, P(F) = 1.00, 1000 people tested
The 80/1000 = 8% false-negative (FN) rate (1-sensitivity) may lead to not treating the “condition” for these people, though all actually have it. The 92% true positive (TP) rate means these patients are likely to be identified and treated.
Next, we investigate a case of a “rare” illness, with a prevalence similar to that of colon cancer.
Rare illness (5% prevalence), P(F) = 0.05, 1000 people tested
There were 50 sick, and 46 of the 50 who were sick (sensitivity=46/50=92%), were identified and probably treated, but 124 of the 950 (124/950=13%) who were well and yet tested positive (false positive) were also probably treated also.
A fraction, 4 of the 50 (=4/50=8%) who were hypothesized here as sick were misdiagnosed by the test as being well and probably would not get treatment. 826 of the 1000 tested would have correctly been spared treatment.
Only 5% of those tested were sick, but 17% received a positive test result, an amplification of the apparent prevalence by 17/5=3.4. Of the 95% who were well, 826/950=87% received a true negative test result.
The values in this last two-by-two table can be converted to probabilities by dividing each by 1000. Doing so shows how to prepare such a table with fractions rather than counts.
Using the two-by-two table, and the values at the margins, one can find the desired result of the probability that one is sick, given a positive test result, P(F|+)=46/170=0.271. This is exactly the result achieved previously using the Bayesian analysis for 5% prevalence and the Cologuard sensitivity and specificity values (see Table 2).
Similarly, the probability that one is not sick, given a negative test result, is P(F|-)=(826/830)=0.995, again the same result found for P(F)=0.05 in Table 2, using the Bayesian approach.
Bayesian analysis allows combining information on the sensitivity and specificity of screening tests, such as the Cologuard test, with the known prevalence of the condition to produce substantially improved estimates of the presence or absence of the condition. For the characteristics of the Cologuard test, for assumed prevalence values from 0.1% to 10%, positive test results raised the estimated probability of having colon cancer by a factor of from 7.0x to 4.4x times the prevalence, and negative test results lowered the estimated probability of having colon cancer by a factor of 1/10x times the prevalence.
The same result can be obtained using a traditional two-by-two table when analyzed as indicated here.
Anderson BW, Ahlquist DA. Molecular Detection of Gastrointestinal Neoplasia: Innovations in Early Detection and Screening. Gastroenterology Clinics 2016; 43:529-542.
Anonymous, Cologuard Physician Brochure: https://cdn2.hubspot.net/hub/377740/file-1412311339-pdf/Document__LBL-0260.pdf?t=1541693758764
Imperiale TF, Ransohoff DF, Itzkowitz SH, et al. Multitarget stool DNA testing for colorectal-cancer screening. N Engl J Med 2014; 370:1287-1297.
McClave JT, Sincich T. Statistics, 13th Edition. Pearson Education, 2016.
Morris D, Koning M. Bayes’ Theorem Examples. Blue Windmill Media, 2016.
Rex D, Johnson D, Anderson J, et al. American College of Gastroenterology Guidelines for Colorectal Cancer Screening 2009. Am J Gastroenterol 2009; 104:739–750.