20.1 – Area under the curve

Introduction
Physiology area under the curve
Machine Learning area under the curve
Area under the receiver operating carrier curve
Area under the precision recall curve
References and suggested readings
Questions
Chapter 20 contents

Introduction.

The classical physiological area under the curve, AUC, represents the total change in $y$ given change in $x$ . For example, if $x$ is time, and $y$ is oxygen consumption — eg, respirometry: recorded as an analog voltage output (millivolts or volts) from a gas analyzer to a data acquisition (DAQ) board — an AUC would be appropriate to quantify the total oxygen consumption following strenuous exercise (Excess post-exercise oxygen consumption, EPOC) or following a large meal (Specific Dynamic Action, SDA). The area represents the total, cumulative volume of oxygen consumed over that specific timeframe, eg, calculating an individual’s total EPOC.

In machince learning, the axes are probabilities and error rates. The area tells you how accurate a decision-making model is. We plot True Positive Rate (Sensitivity) against False Positive Rate (1 – Specificity) across every possible mathematical decision threshold (from 0% to 100% confidence). The curve now represents the shifting balance between catching true positives and committing false positives and the area summarizes how cleanly the model separates two distinct classes, eg, Sick vs. healthy. An area of 1.0 means perfect separation with zero overlap.

Area under the relative (receiver) operating carrier, AUROC, shows characteristics of a diagnostic model, a graphic used to show trade off between sensitivity and specificity. Classifier performance. Used to find the appropriate cut-off. Plot true positive rates against false positive rates as cumulative functions, shows the relationship between sensitivity and specificity for every possible cut off value. Can then calculate AUC to get a measure of the intervention’s ability to discriminate between true and false positive rates.

Related, area under precision-recall curve, AUPRC, is a single scalar metric that measures a binary classification model’s ability to balance precision (how many predicted positives are actually positive) and recall (how many actual positives are caught) across all possible decision thresholds. Here, precision and recall refer to both the accuracy of the model’s positive predictions and its ability to find all the positive cases.

From Chapter 7 we defined

$\begin{align*} Precision = \frac{TP}{TP+FP} \end{align*}$

where $TP$ is true positives and $FP$ is false positives.

We now introduce recall, which is also refered to as sensitivity

$\begin{align*} Recall = \frac{TP}{TP+FN} \end{align*}$

Precision measures the quality of your model’s positive guesses; Recall measures how many actual positive cases in the dataset were identified by the model.

AUC evaluates a model’s ability to distinguish between both $TP$ and $FP$ equally, while AUPRC focuses strictly on the model’s performance regarding the positive (rare) class.

AUC would be preferred when your dataset is balanced (equal positives and negatives) OR when you care equally about performance on both classes. For example, dataset of patients with flu or Covid from a clinic during winter months where there is roughly balanced 50/50 split of patients presenting with either Flu or Covid. You want a model that accurately classifies both which favors AUC approach.

AUPRC would be the choice when your dataset is highly imbalanced, ie, the positive class is rare. In contrast, AUPRC would be favored for scenarios where a researcher is screening 100,000 asymptomatic patients for a rare disease that only 10 people actually have.

Physiology Area under the curve

Download and install R package MESS; requires geepack, geeM, and Matrix packages

R code

x <- seq(1:10) 
y <- c(1,4,5,2,11,22,9,7,5,1) 
#length(x)==length(y)
#smooth the data 
loxy <- loess(y~x)
#Make a plot (Fig. 1)
plot(x,y, pch=19, cex=2, col="blue") 
lines(predict(loxy), type="l", col="red")

where == is an R comparison operator.

And R output

area under the curve, AOC, example

Figure 1. Area under the curve example.

library(MESS) 
auc(x,y,from=0,rule=2) 
auc(x,loxy$fitted,from=0,rule=2)

And R output

#area under curve for raw data
[1] 67
#area under curve for smoothed data
[1] 66.77616

Machine Learning area under the curve.

# Install and load the ROCR package if you haven't already
# install.packages("ROCR")
# modified from https://rviews.rstudio.com/2019/03/01/some-r-packages-for-roc-curves/

library(ROCR)
df <- data.frame(ROCR.simple)
pred <- prediction(df$predictions, df$labels)
perf <- performance(pred,"tpr","fpr")

# 1. Calculate the actual numeric AUC value
auc_perf <- performance(pred, "auc")
auc_value <- round(unlist(auc_perf@y.values), 4)

# 2. Plot the ROC curve with colorized thresholds and custom labels
plot(perf, 
colorize = TRUE, 
main = "ROC Curve",
xlab = "False Positive Rate (1 - Specificity)",
ylab = "True Positive Rate (Sensitivity)")

# 3. Add the AUC value as a subtitle
mtext(paste("AUC =", auc_value), side = 3, line = 0.5)

# 4. Add a diagonal 50/50 reference line
abline(a = 0, b = 1, lty = 2, col = "darkgray")

End R code.

Note 1. I mostly provide S3 R code examples in this eBook. In R, the @ symbol is used to extract a specific piece of data (called a slot) from an S4 object. It works almost exactly like the $ sign, but while $ is used for basic lists or standard data frames (S3 objects), @ is reserved strictly for the more structured, formal object system in R known as S4.

R output.

The calculated AUC was $0.8341875$ . The right-hand axes are color codes by AUC values (Fig 2): good tests AUC between 0.8 and 0.9, very good tests greater than 0.9.

ROC plot

Figure 2. Example ROC curve.

Area under the precision recall curve

AUPRC measures the overall performance of a binary classification model by summarizing how well it balances precision and recall across all possible decision thresholds. A score closer to 1.0 means the model is very good at finding real positive cases without making many mistakes, which is especially useful when your dataset has very few positive examples. A score closer to 1.0 means the model is very good at finding real positive cases without making many mistakes.

R code.

library(ROCR)

# 1. Prepare data and prediction object
df <- data.frame(ROCR.simple)
pred <- prediction(df$predictions, df$labels)

# 2. Extract Precision and Recall for the plot
perf_pr <- performance(pred, measure = "prec", x.measure = "rec")

# 3. Calculate the numeric AUPRC value
# In ROCR, AUPRC is calculated by finding the Area Under the Precision/Recall curve ("aucpr")
perf_auprc <- performance(pred, measure = "aucpr")
auprc_value <- round(unlist(perf_auprc@y.values), 4)

# Print the final AUPRC score to the console
cat("The calculated AUPRC is:", auprc_value, "\n")

# 4. Plot the Precision-Recall Curve using base R graphics
plot(perf_pr, 
colorize = TRUE, 
main = paste("Precision-Recall Curve (AUPRC =", auprc_value, ")"),
xlab = "Recall (Sensitivity)", 
ylab = "Precision (Positive Predictive Value)")

End R code.

R output.

The calculated AUPRC was $0.7875$ .

Figure 3. Precision recall curve.

The differences: AUC evaluates how well the model separates both classes equally, while AUPRC evaluates how reliably the model handles the positive class alone.

The Interpretive baselines:

Figure 2: AUC Baseline is always 0.5. A completely random model yields a diagonal line from (0,0) to (1,1), resulting in an AUC of 0.5. This baseline never changes, regardless of how rare your disease is.
Figure 3: AUPRC Baseline depends on prevalence. A random model yields a flat horizontal line across the plot. The height of this line equals the proportion of positive cases in your dataset ( $\text{Positives} / \text{Total}$ ). If only 1% of the patients have a rare condition, a random model’s AUPRC baseline is 0.01.

What a shifting threshold means:

On an AUC Curve: As you lower your decision threshold to catch more sick people, you track how fast your True Positive Rate rises compared to your False Positive Rate. You are looking at the overall trade-off between the entire healthy population and the entire sick population.
On an AUPRC Curve: As you lower your threshold to catch more sick people, you track Recall against Precision. This evaluates the quality of the “Positive” alerts. It directly measures how quickly the pool of patients you flagged as “high risk” becomes diluted with false alarms.

References and suggested readings

Çorbacıoğlu, Ş. K., & Aksel, G. (2023). Receiver operating characteristic curve analysis in diagnostic accuracy studies: A guide to interpreting the area under the curve value. Turkish Journal of Emergency Medicine, 23(4), 195–198.

Lobo, J. M., Jiménez-Valverde, A., & Real, R. (2008). AUC: A misleading measure of the performance of predictive distribution models. Global Ecology and Biogeography, 17(2), 145–151.

Martin, B., Bennett, T. D., DeWitt, P. E., Russell, S., & Sanchez-Pinto, L. N. (2025). Use of the Area Under the Precision-Recall Curve to Evaluate Prediction Models of Rare Critical Illness Events. Pediatric Critical Care Medicine, 26(6), e855–e859.

Ozenne, B., Subtil, F., & Maucort-Boulch, D. (2015). The precision–recall curve overcame the optimism of the receiver operating characteristic curve in rare diseases. Journal of Clinical Epidemiology, 68(8), 855–859. https://doi.org/10.1016/j.jclinepi.2015.02.010

Steyerberg, E. W., & Vergouwe, Y. (2014). Towards better clinical prediction models: Seven steps for development and an ABCD for validation. European Heart Journal, 35(29), 1925–1931.

Questions

Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help

Chapter 20 contents

/MD

7.3 – Conditional Probability and Evidence Based Medicine

Conditional probability and risk communication
Probability and independent events
Example of multiple, independent events
Probabilistic risk analysis
Conditional probability of non-independent events
Diagnosis from testing
Standard million
Per capita rate
Practice and introduce PPV and test decision criteria
Evidence Based Medicine
Software
Example
Questions
Quiz
Chapter 7 contents

Conditional probability and risk communication.

Conditional probabilities can be used to accurately and effectively convey the risk of specific health outcomes to patients and the public. Tools introduced in this chapter help move from an initial assessment of disease risk to a revised probability after a test result is returned. Although there are other interpretations, here we frame EBM — evidence-based-medicine — as a form of Bayesian decision-making.

This sub-chapter covers a lot of ground. We begin by addressing how probability of multiple events are calculated assuming each event is independent. The assumption of independence is then relaxed, and how to determine probability of an event happening given another event has already occurred, conditional probability, is introduced. Use of conditional probability to interpret results of a clinical test are also introduced, along with the concept of EBM. We also introduce Bayes’ rule and Bayesian probability.

Probability and independent events.

Probability distributions are mathematical descriptions of the probabilities of how often different possible outcomes occur. We also introduced basic concepts related to working with the probabilities involving more than one event.

For review, for independent events, you multiply the individual chance that each event occurs to get the overall probability.

Example of multiple, independent events.

Figure 1. Now that is a box full of kittens. Creative Commons License, source: https://www.flickr.com/photos/83014408@N00/160490011

Figure 1. Now that’s a box full of kittens. Creative Commons License, source: https://www.flickr.com/photos/83014408@N00/160490011.

What is the chance of five kittens in a litter of five to be of the same sex? In feral cat colonies, siblings in a litter share the same mother, but not necessarily the same father, superfecundation. Singleton births are independent events, thus the probability of the first kitten is female is 50%; the second kitten is female, also 50%; and so on. We can multiply the independent probabilities (hence, the multiplicative rule), to get our answer:

kittens <- c(0.5, 0.5, 0.5, 0.5, 0.5)
prod(kittens)
[1] 0.03125

Probabilistic risk analysis.

Risk analysis is the use of information to identify hazards and to estimate the risk. A more serious example. Consider the 1986 Space Shuttle Challenger Disaster (Hastings 2003). Among the crew killed was Ellison Onizuka, the first Asian American to fly in space (Fig 2, first on left back row). Onizuka was born and raised on Hawai`i and graduated from Konawaena High School in 1964.

Figure 2. STS-51-L crew: (front row) Michael J. Smith, Dick Scobee, Ronald McNair; (back row) Ellison Onizuka, Christa McAuliffe, Gregory Jarvis, Judith Resnik. Image by NASA - NASA Human Space Flight Gallery, Public Domain.

Figure 2. STS-51-L crew: (front row) Michael J. Smith, Dick Scobee, Ronald McNair; (back row) Ellison Onizuka, Christa McAuliffe, Gregory Jarvis, Judith Resnik. Image by NASA – NASA Human Space Flight Gallery, Public Domain.

The shuttle was destroyed just 73 seconds after lift off (Fig 3).

Figure 3. Space Shuttle Challenger launches from launchpad 39B Kennedy Space Center, FL, at the start of STS-51-L. Hundreds of shorebirds in flight. Image by NASA - NASA Human Space Flight Gallery, Public Domain.

Figure 3. Space Shuttle Challenger launches from launchpad 39B Kennedy Space Center, FL, at the start of STS-51-L. Hundreds of shorebirds in flight. Image by NASA – NASA Human Space Flight Gallery, Public Domain.

This next section relies on material and analysis presented in the Rogers Commission Report June 1986. NASA had estimated that the probability of one engine failure would be 1 in 100 or 0.01; two engine failures would mean the shuttle would be lost. Thus, the probability of two rockets failing at the same time was calculated as 0.01 X 0.01, which is 0.0001 or 0.01%.

NASA had planned to fly the fleet of shuttles 100 times per year, which would translate to a shuttle failure once in 100 years. The Challenger launch on January 28, 1986, represented only the 25th flight of the shuttle fleet.

One difference on launch day was that the air temperature at Cape Canaveral was quite low for that time of year, as low as 22 °F overnight.

Attention was pointed at the large O-rings in the boosters (engines). In all, there were six of these O-rings. Testing suggested that, at the colder air temperatures, the chance that one of the rings would fail was 0.023. Thus, the chance of success was only 0.977. Assuming independence, what is the chance that the shuttle would experience O-ring failure?

shuttle <- c(0.977, 0.977, 0.977, 0.977, 0.977, 0.977)
#probability of success then was
prod(shuttle)
[1] 0.869

#and therefore probability of failure was
1 - prod(shuttle)
[1] 0.1303042

Conditional probability of non-independent events.

But in many other cases, independence of events cannot be assumed. The probability of an event given that another event has occurred is referred to as conditional probability. Conditional probability is used extensively to convey risk. We’ve touched on some of these examples already:

the risk of subsequent coronary events given high cholesterol;
the risk of lung cancer given a person smokes tobacco;
the risk of mortality from breast cancer given that regular mammography screening was conducted.

There are many, many examples in medicine, insurance, you name it. It is even an important concept that judges and lawyers need to be able to handle (eg, discussion in Berry 2008).

A trigger warning: what follows is emotional, traumatic material, provided to emphasize the importance of these concepts, and the far-reaching implications understanding conditional probability can have for us.

A now famous example of conditional probability in the legal arena came from arguments over the chance that a husband or partner who beats his wife will subsequently murder her — this was an argument raised by the prosecution during pre-trial in the 1995 O.J. Simpson trial (The People of the State of California v. Orenthal James Simpson), and successfully argued by O.J. Simpson’s attorneys… judge ruled in favor of the defense and evidence of O.J. Simpson’s prior abuse were not included in trial. Gigerenzer (2002, following discussions by others, eg, Good 1995) and others (eg, Skorupski and Wainer 2015), have called this the Defendent’s Fallacy, the reverse of the Prosecutor’s Fallacy, where the more typical scenario is that the prosecution provides a list of probabilities about characteristics of the defendant, leaving the jury to conclude that no one else could have possibly fit the description (Thompson and Shumann 1987).

In the O.J. Simpson case, the argument went something like this. From the CDC we find that the chance of an average woman being murdered by a stranger or an unrelated party was about 1 in 20,000. CDC statistics indicate 1.3 million women are abused each year by their partner or former partner; each year about 1000 women are murdered. One thousand divided by 1.3 million is a small number, so even when there is abuse the argument goes, 99% of the time there is not murder. The Simpson judge ruled in favor of the defense and much of the evidence of prior abuse was excluded.

Something is missing from the defense’s argument. Nicole Brown Simpson did not belong to a population of battered women — she belonged to the population of murdered women. When we ask, if a woman is murdered, what is the chance that she knew her murderer, we find that more than 55% knew their murderer — and of that 55%, 93% were killed by a partner who had a history of prior abuse against the victim.

Instead of looking at all 1.3 million abused women, we isolate a representative cohort of 100,000 abused women to see what mathematically happens to them in a given year:

Group 1: Murdered by their abuser. According to the defense’s rate of 1 in 2,500:

$\begin{align*} \frac{100K}{2.5K} = 40 / women \end{align*}$

Group 2: Murdered by someone else:

$\begin{align*} \frac{100K}{20K} = 5 / women \end{align*}$

Given that an abused woman was murdered, what is the probability it was done by her abuser?

$\begin{align*} p(\text{abuser is killer} \mid \text{abused \& killed}) = \frac{40}{40+5} \end{align*}$

or 89%.

This is a huge difference from how the Defense portrayed the chance, where they wanted the judge to assume every single one of those 2500 abusers is equally likely to have committed the crime. In reality, we must condition the match on all other available evidence.

The key — Nicole Simpson (and Ron Goldman) were murdered; O.J. Simpson was an ex-partner who had been guilty of assault against Nicole Simpson. Now, it goes from an impossibly small chance, to a much greater chance. Conditional probability, and specifically Bayes’ rule, is used for these kinds of problems. The defense successfully masked a probability that was heavily stacked against their client.

Bayes’ rule provides guidance for how to update the probability of an event based on new evidence. Named after Thomas Bayes (1701- 1761), more generally, Bayes’ rule, described below, is used to quantify conditional probability. Bayesian probability, where probability is interpreted as reasonably expectation instead of frequency of some event. We discuss Bayesian and Frequentist approaches to hypothesis testing in Chapter 8.2.

Note 1: Following our epidemilogy definitions, women murdered by their abuser (Group 2), epidemiology terms, that would be $CER$ , the exposure rate in controls, whereas abused woman murdered by her abuser (Group 1), that would be $EER$ , the exposure rate in cases.

Diagnosis from testing.

Let’s turn our attention to medicine. A growing practice in medicine is to claim that decision making in medicine should be based on approaches that give us the best decisions. A search of PubMed texts for “evidence based medicine” found more than 91,944 (13 October 2021, and increase of thirteen thousand since last I looked (10 October, 2019). Evidence based medicine (EBM) is the “conscientious, explicit, judicious and reasonable use of modern, best evidence in making decisions about the care of individual patients” (Masic et al 2008). By evidence, we may mean results from quantitative, systematic reviews, meta-analysis, of research on a topic of medical interest, eg, Cochrane Reviews.

Note 2: Primary research refers to generating or collecting original data in pursuit of tests of hypotheses. Both systematic reviews and meta-analysis are secondary research or “research on research.” As opposed to a literature review, systematic review make explicit how studies were searched for and included; if enough reasonably similar quantitative data are obtained through this process, the reviewer can combine the data and conduct an analysis to assess whether a treatment is effective (De Vries 2018).

As you know, no diagnostic test is 100% fool-proof. For many reasons, test results come back positive when the person truly does not have the condition — this is a false positive result. Correctly identifying individuals who do not have the condition, 100% – false positive rate, is called the specificity of a test.

Think of specificity in this way — provide the test 100 true negative samples (eg, 100 samples from people who do not have cancer) — how many times out of 100 does the test correctly return a “negative”? If 99 times out of 100, then the specificity rate for this test is 99%. Which is pretty good. But the test results mean more if the condition/disease is common; for rare conditions, even 99% is not good enough. Incorrect assignments are rare, we believe, in part because the tests are generally quite accurate. However, what we don’t consider is that detection and diagnosis from tests also depend on the base rate — how frequent the incidence of the condition is in the population. Paradoxically, the lower the base rate, the poorer diagnostic value even a sensitive test may have.

Note 3: This concept, that the base rate effects clinical relevance of interpreting testing — has really important, real-world consequences. It directly bears on clinical testing of cancer and other relatively rare diseases; it also dramatically influences applicability of highly sensitive tests for drug use in clinical samples.

To summarize our jargon for interpreting a test or assay, so far we have

True positive (a), the person has a disease and the test correctly identifies the person as having the disease.
False positive (b), test result incorrectly identifies disease; the person does not have the disease, but the test classifies the person as having the disease.
False negative (c), test result incorrectly classifies person does not have disease, but the person actually has the disease.
True negative (d), the person does not have the disease and the test correctly categorizes the person as not having the disease.
Positive predictive value (PPV), the proportion of patients with a positive test result who actually have the disease, .
- Negative Predictive Value (NPV), the probability that a person who receives a negative test result really does not have the disease, $NPV = \frac{TN}{TN+FN}$ .
Sensitivity of test (Se), the proportion of persons who test positive and do have the disease (true positives): $Se = \frac{TP}{TP+FN}$
If a test has 75% sensitivity, then out of 100 individuals who do have the disease, then 75 will test positive (TP = true positive).
Specificity of a test (Sp), refers to the rate that a test correctly classifies a person that does not have the disease (TN = true negatives): $Sp = \frac{TN}{TN+FP}$
If a test has 90% specificity, then out of 100 individuals who truly do not have the disease, then 90 will test negative (= true negatives).

A worked example. A 50 year old male patient is in the doctor’s office. The doctor is reviewing results from a diagnostic test, eg, a FOBT — fecal occult blood test — a test used as a screening tool for colorectal cancer (CRC). The doctor knows that the test has a sensitivity of about 75% and specificity of about 90%. Prevalence of CRC in this age group is about 0.2% (cf NCI SEER). The prevalence is also the prior probability — the baseline rate of the condition in the population before a test is taken. Figure 4 shows our probability tree using our natural numbers approach (Fig 4).

Figure 4. Probability tree for FOBT test; Good test outcomes shown in green: TP stands for true positive and TN stands for true negative. Poor outcomes of a test shown in red: FN stands for false negative and FP stands for false positive.

Note 4: Since U.S. Preventative Task Force (2021), FOBT has largely been replaced by Fecal Immunochemical Test (FIT): sensitivity 74%, specificity 95%, and Multitarget Stool DNA Test: sensitivity 76%, specificity 90%. Cologuard combines both: sensitivity 92%, specificity 97% (cf discussion in Imperiale et al 2014).

The U.S. Supreme Court (Kennedy v. Braidwood Management, Inc) supported USPSTF’s authority to recommend preventive services covered by insurers (see 2025 article by R. Payerchin in Medical Economics).

The associated probabilities for the four possible outcomes of these kinds of tests (eg, what’s the probability of a person who has tested positive in screening tests actually has the disease?) are shown in Table 1.

Table 1. A 2 X 2 table of possible outcomes of a diagnostic test.

	Person really has the disease
Test Result	Yes	No
Positive	a TP	b FP
Negative	c FN	d TN

We want the posterior probability — the chance that a positive test result is a true positive. Bayes’ rule is often given in probabilities,

$\begin{align*} p\left ( D\mid \oplus \right )=\frac{p\left ( D \right )\cdot p\left (\oplus\mid D \right )}{[p\left ( D \right )\cdot p\left (\oplus \right )+p\left ( ND \right )\cdot p\left (\oplus \mid ND \right )]} \end{align*}$

Bayes’ rule, where truth is represented by either D (the person really does have the disease”) or ND (the person really does not have the disease”) and ⊕ is symbol for “exclusive or” and reads “not D” in this example.

An easier way to see this is to use frequencies and our table instead (Table 1, numbers in Figure 4).

Now, the formula is

$\begin{align*} p\left ( disease \mid positive \right )=\frac{a}{a+b} \end{align*}$

Simplified Bayes’ rule, where a is the number of people who test positive and DO HAVE the disease and b is the number of people who test positive and DO NOT have the disease. Thus, $p$ , the posterior probability was

$\begin{align*} p\left ( disease \mid positive \right )=\frac{15}{15+9998} \end{align*}$

or 0.15%. Because the condition is so rare (0.2% prior) and the test generates a very high number of False Positives (9,998), getting a positive result actually lowers oour confidence slightly from the base rate (from 0.2% down to 0.15%). This occurs because the test is highly unreliable in screening this specific population — it’s known as the base rate fallacy. Thus, the chance that a positive FOBT test for our 50 year old male patient is 99.85% likely wrong.

Standardized million.

Where did the 100,000 come from? We discussed this in chapter 2: it’s a simple trick to adjust rates to the same population size. We use this to work with natural numbers instead of percent or frequencies. You choose to correct to a standardized population based on the raw incidence rate. A rough rule of thumb:

Table 2. Relationship between standard population size and incidence rate.

Raw IR rate about	IR	Standard population
1/10	10%	1000
1/100	1%	10,000
1/1000	0.1%	100,000
1/10,000	0.01%	1,000,000
1/100,000	0.001%	10,000,000
1/1,000,000	0.0001%	100,000,000

The raw incident rate is simply the number of new cases divided by the total population.

Per capita rate.

Yet another standard manipulation is to consider the average incidence per person, or per capita rate. The Latin “per capita” translates to “by head” (Google translate), but in economics, epidemiology, and other fields it is used to reflect rates per person. Tuberculosis is a serious infectious disease of primarily the lungs. Incidence rates of tuberculosis in the United States have trended down since the mid 1900s: 52.6 per 100K in 1953 to 2.7 per 100K in 2019 (CDC). Corresponding per capita values are 5.26 x 10^-4 and 2.7 x 10^-5, respectively. Divide the rate by 100,000 to get per person rate.

Practice and introduce PPV and test decision criteria.

Let’s break these problems down, and in doing so, introduce some terminology common to the field of “risk analysis” as it pertains to evaluating assay and diagnotistc tests in biology and epidemiology. Our first example considers the fecal occult blood test, FOBT, test. Blood in the stool may (or may not) indicate polys or colon cancer (Park et al 2010).

A table like the one shown above (Table 1) will appear again and again throughout this and subsequent chapters, but in different forms. We start with Table 3.

Table 3. A 2 X 2 table of possible outcomes of FOBT test.

	Person really has the disease
	Yes	No
Positive	15	9998	PPV = 15/(15+9998) = 0.15%
Negative	5	89,982	NPV = 89982/(89982+5) = 99.99%

We want to know how good is the test, particularly if the goal is early detection? This is conveyed by the PPV, positive predictive value of the test. Unfortunately, the prevalence of a condition is also at play: the lower the prevalence, the lower the PPV must be, because most positive tests will be false when population prevalence is low.

Youden (1950) proposed a now widely adopted index that summarizes how effective a test is. Youden’s J is the sum of specificity and sensitivity minus one.

$\begin{align*} J = Se+Sp-1 \end{align*}$

where Se stands for sensitivity of the test and Sp stands for sensitivity of the test.

Youden’s J takes on values between 0, for a terrible test, and 1, for a perfect test. For our FOBT example, Youden’s J was 0.65. This statistic looks like it’s independent of prevalence, but it’s use as a decision criterion (eg, a cutoff value, above which test is positive, below test is considered negative), assumes that the cost of misclassification (false positives, false negatives) are equal. Prevalence affects number of false positives and false negatives for a given diagnostic test, so any decision criterion based on Youden’s J will also be influenced by prevalence (Smits 2010).

Youden’s J is insensitive to base rates and therefore may be of little practical use in the field. Unsurprisingly, other tools are available, including Matthew’s Correlation Coefficient (MCC),

$\begin{align*} MCC = \frac{(TP X TN) - (FP X FN)}{\sqrt{(TP X FP)(TP + FP)(TN + FP)(TN + FN)}} \end{align*}$

and the positive and negtive Clinical Utility Index (CUI+, CUI-),

$\begin{align*} Positive CUI+ = Sensitivity X PPV Negative CUI- = Sensitivity X NPV \end{align*}$

or the F1-score,

$\begin{align*} F1-score = 2 X \frac{PPV X Sensitivity}{PPV + Sensitivity} \end{align*}$

epiR will calculate Youden’s J, but the others are either calculated by hand or by way of additional packages. So, we’ll provide a function.

R code to calculate testing decision criteria — code genereated with Gemini AI assistance.

run_screening_wizard <- function() {
# 1. Sequential user prompts starting with cell a
cat("--- 2x2 Matrix Entry Wizard ---\n")
a <- as.numeric(readline(prompt = "Enter cell a (True Positives, TP): "))
b <- as.numeric(readline(prompt = "Enter cell b (False Positives, FP): "))
c <- as.numeric(readline(prompt = "Enter cell c (False Negatives, FN): "))
d <- as.numeric(readline(prompt = "Enter cell d (True Negatives, TN): "))

# 2. Map cells to short variables for calculation
tp <- a
fp <- b
fn <- c
tn <- d

# 3. Construct and print the visual 2x2 matrix
user_matrix <- matrix(
c(tp, fp, fn, tn), 
nrow = 2, 
byrow = TRUE,
dimnames = list(Test = c("+", "-"), Disease = c("+", "-"))
)
cat("\n[+] Captured Matrix:\n")
print(user_matrix)

# 4. Run background metrics math
sensitivity <- tp / (tp + fn)
specificity <- tn / (tn + fp)
ppv <- tp / (tp + fp)
if (is.nan(ppv)) ppv <- 0 

youden_j <- sensitivity + specificity - 1
cui_pos <- sensitivity * ppv

f1_score <- 2 * ((ppv * sensitivity) / (ppv + sensitivity))
if (is.nan(f1_score)) f1_score <- 0 

mcc_num <- (tp * tn) - (fp * fn)
mcc_den <- sqrt(as.numeric(tp + fp) * as.numeric(tp + fn) * 
as.numeric(tn + fp) * as.numeric(tn + fn))
mcc <- if (mcc_den == 0) 0 else mcc_num / mcc_den

# 5. Compile into the copy/paste ready console matrix
output_matrix <- matrix(
c(round(youden_j, 4), "No", "Good (Overly optimistic baseline)",
round(mcc, 4), "Yes", "Poor (Correctly captures imbalance)",
round(cui_pos, 4), "Yes", "Unacceptable clinical utility",
round(f1_score, 4), "Yes", "Near-total prediction failure"),
nrow = 4, byrow = TRUE
)

dimnames(output_matrix) <- list(
c("Youden's J Index", "MCC", "CUI+", "F1-Score"),
c("Value", "Base_Rate_Dep", "Console_Interpretation")
)

final_df <- as.data.frame(output_matrix)

# 6. Print and invisibly return the dataframe
cat("\n[+] Resulting Metrics Table (Copy/Paste Ready):\n")
print(final_df)
return(invisible(final_df))
}

# Run the function at the R command prompt.
run_screening_wizard()

End R code.

Thus, using a metric sensitive to low base rate, our FOBT tests are (modified from R output)

Table 4. Decision criteria.

	Value	Base rate dependent?	Interpretation
$Youden's \ J$	0.65	No	Good
$MCC$	0.0306	Yes	Poor
$Positive \ Clinical \ Utility \ Index$	0.0011	Yes	Unacceptable clinical utility
$F1-score$	0.0030	Yes	Near-total prediction failure

Another worked example. A study on cholesterol-lowering drugs (statins) reported a relative risk reduction of death from cardiac event by 22%. This does not mean that for every 1000 people fitting the population studied, 220 people would be spared from having a heart attack. In the study, the death rate per 1000 people was 32 for the statin versus 41 for the placebo — recall that a placebo is a control treatment offered to overcome potential patient psychological bias (see Chapter 5.4). The absolute risk reduction due to statin is only 41 – 32 or 9 in 1000 or 0.9%. By contrast, relative risk reduction is calculated as the ratio of the absolute risk reduction (9) divided by the proportion of patients who died without treatment (41), which is 22% (LIPID Study Group 1998).

Note that risk reduction is often conveyed as a relative rather than as an absolute number. The distinction is important for understanding arguments based in conditional probability. Thus, the question we want to ask about a test is summarized by absolute risk reduction (ARR) and number needed to treat (NNT), and for problems that include control subjects, relative risk reduction (RRR). We expand on these topics in the next section, 7.4 – Epidemiology: Relative risk and absolute risk, explained.

Evidence Based Medicine.

One culture change in medicine is the explicit intent to make decisions based on evidence (Masic et al 2008). Of course, the joke then is, well, what were doctors doing before, diagnosing without evidence? The comic strip xkcd offers one possible answer (Fig 5).

A summary of "evidence based medical" decisions, perhaps? https://xkcd.com/1619/

Figure 5. A summary of “evidence based medical” decisions, perhaps? “Watson Medical Algorithm,” https://xkcd.com/1619/.

As you can imagine, there’s considerable reflection about the EBM movement (see discussions in response to Accad and Francis 2018, eg, Goh 2018). More practically, our objective is for you to be able to work your way through word problems involving risk analysis. You can expect to be asked to calculate, or at least set up for calculation, any of the statistics listed above (eg, False negative, false positive, etc.). Practice problems are listed at the end of this section, and additional problems are provided to you (Homework 4). You’ll also want to check your work, and in any real analysis, you’d most likely want to use R.

Software.

R has several epidemiology packages, and with some effort, can save you time. Another option is to run your problems in OpenEpi, a browser-based set of tools. OpenEpi is discussed with examples in the next section, 7.4.

Here, we illustrate some capabilities of the epiR package, expanded more also in the next section, 7.4. We’ll use the example from Table 3.

R code

library(epiR) 
Table3 <- matrix(c(15, 5, 9998, 89982), nrow = 2, ncol = 2)
epi.tests(Table3)

R output

         Outcome + Outcome -   Total
Test +          15      9998   10013
Test -           5     89982   89987
Total           20     99980  100000

Point estimates and 95% CIs:
--------------------------------------------------------------
Apparent prevalence *                      0.10 (0.10, 0.10)
True prevalence *                          0.00 (0.00, 0.00)
Sensitivity *                              0.75 (0.51, 0.91)
Specificity *                              0.90 (0.90, 0.90)
Positive predictive value *                0.00 (0.00, 0.00)
Negative predictive value *                1.00 (1.00, 1.00)
Positive likelihood ratio                  7.50 (5.82, 9.67)
Negative likelihood ratio                  0.28 (0.13, 0.59)
False T+ proportion for true D- *          0.10 (0.10, 0.10)
False T- proportion for true D+ *          0.25 (0.09, 0.49)
False T+ proportion for T+ *               1.00 (1.00, 1.00)
False T- proportion for T- *               0.00 (0.00, 0.00)
Correctly classified proportion *          0.90 (0.90, 0.90)
--------------------------------------------------------------
* Exact CIs

Oops! I wanted PPV, which by hand calculation was 0.15%, but R reported “0.00?” This is a significant figure reporting issue. The simplest solution is to submit options(digits=6) before the command, then save the output from epi.tests() to an object and use summary(). For example

options(digits=6)
myEpi <- epi.tests(Table3)
summary(myEpi)

And R returns

     statistic           est        lower        upper
1           ap  0.1001300000 0.0982761568  0.102007072
2           tp  0.0002000000 0.0001221693  0.000308867
3           se  0.7500000000 0.5089541283  0.913428531
4           sp  0.9000000000 0.8981238085  0.901852950
5      diag.ac  0.8999700000 0.8980937508  0.901823014
6      diag.or 27.0000000000 9.8110071871 74.304297826
7         nndx  1.5384615385 1.2265702376  2.456532054
8       youden  0.6500000000 0.4070779368  0.815281481
9       pv.pos  0.0014980525 0.0008386834  0.002469608
10      pv.neg  0.9999444364 0.9998703379  0.999981958
11      lr.pos  7.5000000000 5.8193604069  9.666010707
12      lr.neg  0.2777777778 0.1300251423  0.593427490
13      p.rout  0.8998700000 0.8979929278  0.901723843
14       p.rin  0.1001300000 0.0982761568  0.102007072
15      p.tpdn  0.1000000000 0.0981470498  0.101876192
16      p.tndp  0.2500000000 0.0865714691  0.491045872
17      p.dntp  0.9985019475 0.9975303919  0.999161317
18      p.dptn  0.0000555636 0.0000180416  0.000129662

There we go — pv.pos reported as 0.0014980525, which, after turning to a percent and rounding, we have 0.15%. Note also the additional statistics provided — a good rule of thumb — always try to save the output to an object, then view the object, eg, with summary(). Refer to help pages for additional details of the output (?epi.tests).

What about R Commander menus?

Note 3: Confirmed Fall 2026 — I have not been able to run the EBM plugin successfully! Simply returns an error message — — on data sets which have in the past performed perfectly. Thus, until further notice, do not use the EBM plugin. Instead, use commands in the epiR package. I’m leaving the text here on the chance the error with the plugin is fixed.

Rcmdr has a plugin that will calculate ARR, RRR and NNT. The plugin is called RcmdrPlugin.EBM (Leucuta et al 2014) and it would be downloaded as for any other package via R.

Download the package from your selected R mirror site, then start R Commander.

install.packages("RcmdrPlugin.EBM")

From within R Commander (Fig 6), select

Tools → Load Rcmdr plug-in(s)…

To install an Rcmdr plugin, first go to Rcmdr → Tools → Load Rcmdr plug-in(s)...

Figure 6. To install an Rcmdr plugin, first go to Rcmdr → Tools → Load Rcmdr plug-in(s)…

Next, select from the list the plug-in you want to load into memory, in this case, RcmdrPlugin.EBM (Fig 7).

Select the Rcmdr plugin, then click the "OK" button to proceed.

Figure 7. Select the Rcmdr plugin, then click the “OK” button to proceed.

Restart Rcmdr again (Fig 8),

Select "Yes" to restart R Commander and finish installation of the plug-in.

Figure 8. Select “Yes” to restart R Commander and finish installation of the plug-in.

and the menu “EBM” should be visible in the menu bar (Fig 9).

After restart of R Commander the EBM plug-in is now visible in the menu.

Figure 9. After restart of R Commander the EBM plug-in is now visible in the menu.

Note that you will need to repeat these steps each time you wish to work with a plug-in, unless you modify your .RProfile file. See

Rcmdr → Tools → Save Rcmdr options…

Clicking on the EBM menu item brings up the template for the Evidence Based Medicine module. We’ll mostly work with 2 X 2 tables (eg, see Table 1) , so select the “Enter two-way table…” option to proceed (Fig 10).

Select "Enter two-way table...".

Figure 10. Select “Enter two-way table…”.

And finally, Figure 11 shows the two-way table entry cells along with options. We’ll try a problem by hand then use the EBM plugin to confirm and gain additional insight.

Two-way table Rcmdr EBM plug-in.

Figure 11. Two-way table Rcmdr EBM plug-in.

For assessing how good a test or assay is, use the Diagnosis option in the EBM plugin. For situations with treated and control groups, use Therapy option. For situations in which you are comparing exposure groups (eg, smokers vs non-smokers), use the Prognosis option.

Example.

Here’s a simple one (problem from Gigerenzer 2002).

About 0.01% of men in Germany with no known risk factors are currently infected with HIV. If a man from this population actually has the disease, there is a 99.9% chance the tests will be positive. If a man from this population is not infected, there is a 99.9% chance that the test will be negative. What is the chance that a man who tests positive actually has the disease?

Start with the reference, or base population (Figure 12). It’s easy to determine the rate of HIV infection in the population if you use numbers. For 10,000 men in this group, exactly one man is likely to have HIV (0.0001X10,000), whereas 9,999 would not be infected.

For the man who has the disease it’s virtually certain that his results will be positive for the virus (because the sensitivity rate = 99.9%). For the other 9,999 men, one will test positive (the false positive rate = 1 – specificity rate = 0.01%).

Thus, for this population of men, for every two who test positive, one has the disease and one does not, so the probability even given a positive test is only 100*1/2 = 50%. This would also be the test’s Positive Predictive Value.

Note that if the base rate changes, then the final answer changes! For example, if the base rate was 10%

It also helps to draw a tree to help you determine the numbers (Fig 12)

Draw a probability tree to help with the frequencies.

Figure 12. Draw a probability tree to help with the frequencies.

From our probability tree in Figure 12 it is straight-forward to collect the information we need.

Given this population, how many are expected to have HIV? Two.
Given the specificity and sensitivity of the assay for HIV, how many persons from this population will test positive? Two.
For every positive test result, how many men from this population will actually have HIV? One.

Thus, given this population with the known risk associated, the probability that a man testing positive actually has HIV is 50% (=1/(1+1)).

Use the EBM plugin. Select two-way table, then enter the values as shown in Fig 13.

EBM plugin with data entry

Figure 13. EBM plugin with data entry.

Select the “Diagnosis” option — we are answering the question: How probable is a positive result given information about sensitivity and specificity of a diagnosis test. The results from the EBM functions are given below

Rcmdr> .Table
      Yes No
   +  1    1
   -  0 9998

Rcmdr> fncEBMCrossTab(.table=.Table, .x='', .y='', .ylab='', .xlab='', 
Rcmdr+ .percents='none', .chisq='1', .expected='0', .chisqComp='0', .fisher='0', 
Rcmdr+ .indicators='dg', .decimals=2)

# Notations for calculations
        Disease + Disease -
Test +      "a"         "b" 
Test -      "c"         "d"

# Sensitivity (Se) = 100 (95% CI 2.5 - 100) %. Computed using formula: a / (a + c)
# Specificity (Sp) = 99.99 (95% CI 99.94 - 100) %. Computed using formula: d / (b + d)
# Diagnostic accuracy (% of all correct results) = 99.99 (95% CI 99.94 - 100) %. Computed using formula: (a + d) / (a + b + c + d)
# Youden's index = 1 (95% CI 0.02 - 1). Computed using formula: Se + Sp - 1
# Likelihood ratio of a positive test = 9999 (95% CI 1408.63 - 70976.66). Computed using formula: Se / (Sp - 1)
# Likelihood ratio of a negative test = 0 (95% CI 0 - NaN). Computed using formula: (1 - Se) / Sp
# Positive predictive value = 50 (95% CI 1.26 - 98.74) %. Computed using formula: a / (a + b)
# Negative predictive value = 100 (95% CI 99.96 - 100) %. Computed using formula: d / (c + d)
# Number needed to diagnose = 1 (95% CI 1 - 40.91). Computed using formula: 1 / [Se - (1 - Sp)]

Note that the formulas used to calculate Sensitivity, Specificity, etc., follow our Table 1 (compare to “Notations for calculations”). The use of EBM provides calculations of our confidence intervals.

Questions.

The sensitivity of the fecal occult blood test (FOBT) is reported to be 0.68. What is the False Negative Rate?
The specificity of the fecal occult blood test (FOBT) is reported to be 0.98. What is the False Positive Rate?
For men between 50 and 54 years of age, the rate of colon cancer is 61 per 100,000. If the false negative rate of the fecal occult blood test (FOBT) is 10%, how many persons who have colon cancer will test negative?
For men between 50 and 54 years of age, the rate of colon cancer is 61 per 100,000. If the false positive rate of the fecal occult blood test (FOBT) is 10%, how many persons who do not have colon cancer will test positive?
A study was conducted to see if mammograms reduced mortality

Mammogram Deaths/1000 women

No 4

Yes 3

data from Table 5-1 p. 60 Gigerenzer (2002)
What is the RRR?
A study was conducted to see if mammograms reduced mortality

Mammogram Deaths/1000 women

No 4

Yes 3

data from Table 5-1 p. 60 Gigerenzer (2002)
What is the NNT?
Does supplemental Vitamin C decrease risk of stroke in Type II diabetic women? A study conducted on 1923 women, a total of 57 women had a stroke, 14 in the normal Vitamin C level and 32 in the high Vitamin C level. What is the NNT between normal and high supplemental Vitamin C groups?
Sensitivity of a test is defined as
A. False Positive Rate
B. True Positive Rate
C. False Negative Rate
D. True Negative Rate
Specificity of a test is defined as
A. False Positive Rate
B. True Positive Rate
C. False Negative Rate
D. True Negative Rate
In thinking about the results of a test of a null hypothesis, Type I error rate is equivalent to
A. False Positive Rate
B. True Positive Rate
C. False Negative Rate
D. True Negative Rate
During the Covid-19 pandemic, number of reported cases each day were published. For example, 155 cases were reported for 9 October 2020 by Department of Health. What is the raw incident rate?

Mammogram	Deaths/1000 women
No	4
Yes	3

Mammogram	Deaths/1000 women
No	4
Yes	3

Quiz Chapter 7.3

Conditional Probability and Evidence Based Medicine

Chapter 7 contents

Introduction
Epidemiology definitions
Epidemiology basics
Conditional Probability and Evidence Based Medicine
Epidemiology: Relative risk and absolute risk, explained
Odds ratio
Confidence intervals
References and suggested readings

7.2 – Epidemiology basics

Introduction
Prevalence rate
Prevalence: R code
Incidence rate
Person-time
Incidence rate: R code
Age-specific rates
Age-adjusted rates
Questions
Quiz
Chapter 7 contents

Introduction

A principle aim of epidemiology is to identify frequency and causes of illnesses. Kinds of evidence gathered range from case studies to randomized controlled trials, to meta-analysis and systematic reviews. In turn, how these kinds of evidence are evaluated and turned into policy recommendations generally accounts for additional concerns eg, GRADE – Grading of Recommendations Development, Assessment and Evaluation Schünemann et al 2023). Learning about conditional probability in biostatistics is essential for discussing evidence because it provides a rigorous framework for updating strength of evidence based on new, specific information. It allows for the objective assessment of how new evidence — results from a clinical test — influences the likelihood of a hypothesis — presence/absence of disease for a patient. Conditional probability analysis allows interpretation of a positive test result in the context of disease prevalence, which may lead to the paradoxical conclusion that a test with high sensitivity applied to a person belonging to a low at risk group is more likely a false positive than evidence of a true positive.

Thus, to introduce conditional probability, I have elected to push you into epidemiology and risk analysis. We introduced epidemiology definitions and here, we build on basic terminology of epidemiology. Epidemiology is the study of the causes and distribution of health-related events in a population. It is has been called the basic science of public health (p. 16, Decker 2008).

Prevalence rate

Prevalence of a disease (condition), is defined as the proportion of the population that has the disease (condition) at a point or duration in time. The prevalence statistic is the ratio of the number of existing cases divided by the total population.

For example, prevalence of Type 2 diabetes in Hawai`i. The 2020 populations was about 1.4 million. A survey was conducted on a random of 1000 individuals and 80 were reported to have Type 2 diabetes. What is the estimated prevalence of Type 2 diabetes in Hawai`i?

Start from perspective — what if we actually have something close to a census count? For the actual estimates, see the report at the Hawaii State DOH website. With every estimate of a statistic, we need a confidence interval (CI). An approximate (good for large samples) formula for the 95% CI of prevalence is

$\begin{align*} 95\% \ CI=\hat{p}\pm z \sqrt{\frac{1}{N}\hat{p}\left (1-\hat{p} \right )} \end{align*}$

where $\hat{p}$ is the 2010 prevalence in the population (8.3%), N is the population size (1,360,301), and z is the standard normal probability. For a 95% CI, then we want z_95%.

If you recall, we can get this from our standard normal probability table, or directly from R. We want to know z that is in the ± 2.5% tails of the distribution (that’s 0.05 ÷ 2, see Ch 8.4 – Tails of a test). Our R code then is

qnorm(c(0.025), mean=0, sd=1, lower.tail=FALSE)

which returns

[1] 1.959964

You should confirm that setting lower-tail = TRUE yields -1.96 (rounded).

Alternately, if using Rcmdr: Distributions → Continuous distributions → Normal distribution → Normal quantiles… (Fig 1).

Figure 1. R Commander popup menu for Normal quantiles.

Figure 1. R Commander popup menu for Normal quantiles.

Thus, for our example, the 95% CI was ( $8.3 - 0.046, 8.3, + 0.046)$ ; our confidence in our estimate of the prevalence of diabetes in Hawaii is between 8.25% and 8.35%. Again, note that for our purposes it is OK to calculate the approximate confidence interval, replace $\pm 1.96$ with $2$ (for large N, the differences are observed in the 0.001 decimal).

Prevalence: R code

Rather than census counts, more likely we have results of smaller surveys. We use the epi.conf() function from the epiR package. Code adapted from example provided in epiR_descriptive vignette.

Note 1: Many R packages include vignettes, which, together with the package manual, is often helpful to understand what a function is intended to do and how to get the most from the function. R code to call all vignettes available for a package, eg, epiR:

vignette(package="epiR")

which will return names of available vignettes. For epiR, these are

epiR_descriptive Descriptive epidemiology (source, html)
epiR_surveillance Disease surveillance (source, html)
epiR_measures_of_association Measures of association (source, html)
epiR_sample_size Sample size calculations (source, html)

To call up the vignette we need for this example,

vignette("epiR_descriptive", package="epiR")

which brings up the page (assuming you installed the html help files during installation of base R).

library(epiR)
pop.Hawaii = 1.4e06
pop.Survey = 1000
type2.Survey = 80 
Table.2 <- as.matrix(cbind(type2.Survey, pop.Survey)) 
epi.conf(Table.2, ctype = "prevalence", method = "exact", N = pop.Hawaii, design = 1, conf.level = 0.95) * 100

R output

   est    lower    upper
1    8 6.394198 9.857978

Thus, estimated Type 2 diabetes prevalence from the survey was 8, with 95% confidence interval 6.4 to 9.9 cases per 100 individuals. From 2020 U.S. Census, Hawai`i population was 1,455,271.

Note 2: Type 2 diabetes is one of many conditions for which prevalence is greater among Native Hawaiian and Pacific Islander populations compared to other groups in Hawai’i (Galinski et al 2016); these are called health disparitiess.

Incidence rate

Incidence of a disease (or condition) is defined as the occurrence of new cases of a disease. The simplest way to view incidence is that if everyone was followed for the same period of time, then incidence rate is the number of new cases since the start of the study divided by the total population. This is too simplistic, so we define a better metric called person-time. Incidence rate (IR) is then the ratio of the number of new cases divided by total person-time.

Again, incidence rate is an estimate. Therefore, we need a confidence interval. Assuming the population is large, then confidence interval can be calculated as

$\begin{align*} IR\pm z\sqrt{\frac{N}{T^{2}}} \end{align*}$

where IR is the incidence rate, T is the person-time, N is the number of events or cases, and z again is the standard normal probability (z ± 1.96 for 95% confidence interval).

For our example, N = 3 cases, T = 236 p-d, so 95% CI is (11.3%, 14.1%).

Person-time

Person-time can be days, months, years. Person-time is best defined with an example.

Five men join a study that will last 70 days. These men were selected because they had suffered a myocardial infarction (MI) and at the start of the study, they receive the same treatment. The outcome of the study is whether or not the subjects suffer a second MI.

The results are shown below

Subject A, 53 days
Subject B, 70 days
Subject C, 24 days
Subject D, 70 days
Subject E, 19 days

Add up the person days,

$\begin{align*} 53+70+24+70+19=236 \ person-days\left (p-d \right ) \end{align*}$

Now calculate the incidence rate,

$\begin{align*} 1000\cdot \frac{3 \ cases}{236p-d}=12.7 \ person-days \end{align*}$

that’s 3 cases (A, C, E) divided by 236 p-d = 0.0127. Multiply this by 1000 and we get our final answer for the incidence rate, 12.7 per person-days.

Incidence rate: R code

Incidence rate of myocardial infarction (MI) study. Code adapted from example provided in epiR_descriptive vignette.

ncas = 3
ntar = 236
tmp <- as.matrix(cbind(ncas, ntar))
epi.conf(tmp, ctype = "inc.rate", method = "exact", N = 1000, design = 1, conf.level = 0.95) * 1000

R output

          est    lower    upper
ncas 12.71186 2.621492 37.14946

The incidence rate of myocardial infarction was 12.7 (95% CI 2.6 to 37.2) cases per 1000 person-days.

Age-specific rates

Incidence or prevalence rates may be reported for specific age groups. For example, we can distinguish between number of live births in Hawaii in 2017, and numbers of live births by age group of the mother.

Age-adjusted rates

If populations with different age demographics, then the convention is to adjust the populations to a standard reference population with known age and other demographic properties. For example, the CDC uses the 2000 census as the standard population (CDC definitions, age adjustment, retrieved January 2023).

Questions

Calculate the confidence interval of type 2 prevalence in Hawai`i with the 2020 census population value. How much did it change?
Recalculate the confidence interval for a 99% confidence interval. Which estimate communicates greater confidence in the estimate?

Quiz

Quiz Chapter 7.2

Epidemiology basics

Chapter 7 contents

6.5 – Discrete probability distributions

Binomial distribution
Hypergeometric distribution
Discrete uniform distribution
Poisson distribution
Next — Continuous distributions
Questions
Quiz
Chapter 6 contents

Binomial distribution

Discrete refers to particular outcomes — data that take on only specific separate values. Discrete data types include all of the categorical types we introduced earlier, including binary, ordinal, and nominal.

The binomial probability distribution is a discrete distribution for the number of successes, k, in a sequence of n independent trials, where the outcome of each trial can take on only one of two possible outcomes. For cases of 0 or 1, yes or no, “heads” or “tails,” male or female, pass or fail, we talk about the binomial distribution, because the outcomes are discrete and there can be only two possible (binary) outcomes.

Note 1: The biological definition of sex is binary — whether sexually reproducing organism produces male (sperm) or female (ovum) gametes (cf discussion in Goymann et al 2022). For the argument that sex is a continuum held by some biomedical and social scientists, see Ainsworth 2015.

The mathematical function of the binomial is written as

$\begin{align*} Pr\left [ X \ successes \right ] = \binom{n}{X} p^X \left ( 1-p \right )^{n-X} \end{align*}$

where the binomial coefficient is given by

$\begin{align*} \binom{n}{X} = \frac{n!}{X!\left (n-X \right )!} \end{align*}$

and X refers to the number of ways to choose “success” from n observations.

Consider an example.

We have to define what we mean by success. For coin toss, this might be the number of heads.

The mean for the binomial this is given simply as

$\begin{align*} \mu_{X} = np \end{align*}$

where X is “Heads” (the category of successes for our example), and p corresponds to the probability the selected event occurs, in this case, “Heads.”

The variance of the binomial distribution is given by

$\begin{align*} \sigma_{X}^{2} = np\left ( 1-p \right ) \end{align*}$

Here’s a density plot of two trials with success 2% with n (x) equal to 20 (Fig 1).

plot generated with KMggplot2 Rcmdr plugin

Figure 1. Plot generated with KMggplot2 Rcmdr plugin.

Here’s the R code

Create the trials, 1 through 20, then create an object to hold the number of trials

nSize=1:20
Size <- length(nSize); Size

R returns

[1] 20

Assign the probability value to an object

prob <- 0.02

Next, calculate the mean, mu, and the variance, var, for the binomial with prob = 0.02 and number of trials Size = 20

mu <- Size*prob
var <- Size*prob*(1-prob)

Print the mean and variance; lets assign them to an object then print the object

stats <- c(mu, var); stats

And R returns

[1] 0.400 0.392

And here’s a real-world example. Twinning in humans is rare. In Hawaiʻi in the 1990’s the rate of twin births (monozygotic and dizygotic) was about 20 for every 1000 births or 2%. “Success” here then is twin births.

Figure 2. Example of binomial-like distribution: reported twins born in Hawaiʻi.

Figure 2. Example of binomial-like distribution: reported twins born in Hawaiʻi.

Interestingly, rates of twins have since increased in Hawaiʻi (31 out of 1000 births) and in the United States overall (33 out of 1000 births) (Table 2, NCHS Data Brief No. 80, 2012). Data for Fig 2 were for year 2009.

Out of ten births, what is the probability of two twin births in Hawaiʻi?

$\begin{align*} Pr\left [ 2 \ twinbirths \right ] = \left ( \frac{10}{2} \right ) 0.031^2 \left ( 1 - 0.031 \right )^{10-2} \end{align*}$

You can solve this with your calculator (yikes!), or take advantage of online calculators (GraphPad QuickCalcs), or use R and Rcmdr.

In R, simply type at the prompt

dbinom(2,10,0.031)
[1] 0.03361446

Try in R Commander.

Rcmdr → Distributions → Discrete distributions → Binomial distribution → Binomial probabilities …

Rcmdr, enter Binomial probabilities

Figure 3. Rcmdr menu to get binomial probability.

Note I used p = 0.033 the rate for entire USA. Here’s the output

> .Table <- data.frame(Pr=dbinom(0:10, size=10, prob=0.033))
> rownames(.Table) <- 0:10
> .Table
Pr
0  7.149320e-01
1  2.439789e-01
2  3.746728e-02 ← Answer, 0.0375 or 3.75%
3  3.409639e-03
4  2.036263e-04
5  8.338782e-06
6  2.371422e-07
7  4.624430e-09
8  5.918028e-11
9  4.487991e-13
10 1.531579e-15

And here is the output for our example from Hawaiʻi (p = 0.031).

> .Table <- data.frame(Pr=dbinom(0:10, size=10, prob=0.031))
> rownames(.Table) <- 0:10
> .Table
Pr
0  7.298570e-01
1  2.334940e-01
2  3.361446e-02 ← Answer, 0.0336 or 3.36%
3  2.867694e-03
4  1.605494e-04
5  6.163507e-06
6  1.643178e-07
7  3.003893e-09
8  3.603741e-11
9  2.561999e-13
10 8.196283e-16

We use the binomial distribution as the foundation for the binomial test, ie, the test of an observed proportion against an expected population level proportion in a Bernoulli trial.

Hypergeometric distribution

The binomial distribution is used for cases of sampling with replacement from a population. When sampling without replacement is done, the hypergeometric distribution is used. It is the number of successes, k, in a sequence of n independent trials drawn from a fixed population.

Note 2: A fixed population in epidemiology refers to a group of individuals, defined by set of characteristics. Examples include individuals selected because they all share a common life event, eg, giving birth. The population is “fixed” because, once an individual is included in the population, they remain a member of the population even if their characteristics change during the duration of the study. The opposite of a fixed population is an dynamic or open population. Definitions from Chapter 2 in Aschengrau and Seage (2003).

This sampling scheme means that each draw is no longer independent — with each draw you decrease the remaining number of observations and thus change the proportion.

The P-value from the hypergeometric distribution is often used to study if genes from a list of Gene Ontology (eg, biological process functional terms like “cell cycle,” or “apoptosis”) are enriched in the sets of genes with expression differences across the treatment groups (cf, discussion in Cao and Zhang 2014). The R package gprofiler2, which links to a Cloud service at g:Profiler, the can be used to do gene enrichment analyses with stunning visualizations (eg, Manhattan plot) suitable for publication.

The mathematical function of the hypergeometric is written as

$\begin{align*} Pr\left [X = k \right ] = \frac{\left ( \frac{K}{k} \right )\left ( \frac{N - K}{n - k} \right )}{\left ( \frac{N}{n} \right )} \end{align*}$

where N is the population size, K is the number of successes in that population, and n and k are defined as above. Lets look apply this to the twinning problem.

In 2009, 2200 women gave birth in Hawaiʻi County, Hawaiʻi. Out of 10 births, what is the probability of 2 twin births in Hawaiʻi?

Assuming “risk” of twinning is the same rate as in rest of USA, then we have expected 72 successes in this population (0.033 * 2200).

Here’s the graph (Fig 4).

hypergeometric distribution twinning Hawaiʻi

Figure 4. Plot of hypergeometric distribution twinning Hawaiʻi.

where the X axis values shows the number of events with successes (twin births). Taking the bin 2 (we wanted to know about probability of 2 out of ten), we can draw a line back to the Y-axis to get our probability — looks like about 5% roughly. Plot drawn with KMggplot2

To get the actual probability,

Rcmdr → Distributions → Discrete distributions → Hypergeometric distribution → Hypergeometric probabilities …

Rcmdr menu get hypergeometric probability

Figure 5. Rcmdr menu to get hypergeometric probability.

where m is the number of successes, n is the number of “failures,” and k is the number of trials.

> .Table
Pr
0 0.716453457
1 0.243438645
2 0.036688041 ← Answer, 0.0367 or 3.67%
3 0.003228871

The reference to white and black balls and urns is a device described by Bernoulli himself and has been used by others ever since to discuss probability problems (called the urn problem) and so I apply it here to be consistent. The urn contains a number of white (x) and a number of black (y) balls mixed together. One ball is drawn randomly from the urn — what color is it? The ball is then is either returned into the urn (replacement) or it is left out (without replacement) as in the hypergeometric problem, and the selection process is repeated.

Besides applications in gambling and balls-in-urns problems, this distribution is the basis for many tests of gene enrichment from microarray analyses. The hypergeometric forms the basis of the Fisher Exact test (see Chapter 9.5).

Discrete uniform distribution

For discrete cases of “1,” “2,” “3,” “4,” “5,” or “6,” on the single toss of fair dice, we can talk about the discrete uniform distribution because all possible outcomes are equally likely. If you are branded as a “card-counter” in Las Vegas, all you’ve done is reached an understanding of the uniform distribution of card suits!

One biological example would be the fate of a random primary oocyte in the human (mammal) female — three out of four will become polar bodies, eventually reabsorbed, whereas one in four will develop into a secondary oocyte (egg); the uniform distribution has to do with the counts of the products — each of the four primary oocytes has the same (apparently) chance (25%) of becoming the egg (Edwards and Beard 1997).

Analyses in biological research often use a uniform distribution as a baseline or null model (Gotelli and Ulrich 2011). For DNA sequencing, the uniform distribution provides a baseline for assessing coverage bias, eg, GC bias (Ross et al 2013). The uniform distribution underlies the Jukes–Cantor evolutionary model. The Jukes–Cantor model assumes equal substitution rates among all four nucleotides (A, T, C, G) and equal base frequencies (Felsenstein 2003).

The uniform distribution exists also for continuous data types, where every value within a specified interval has equal probability density. For example, DNA shearing — by endonucleases, proteases, or tansposases — during initial library prep for sequencing reactions, breakpoints may be modeled as uniformly distributed along fragment length. Non-uniform distributions reveal cleavage bias. Sequence-dependent cleavage bias is concluded when certain DNA fragments or sequences are generated or detected more (or less) frequently than would be expected by chance (Meyer and Liu 2014). This generates many false positives — that is, identifying a biological signal or molecule that isn’t actually there (Nix et a; 2008).

Poisson distribution

An extension from the binomial case is that, rather than following success or failure, you may have the following scenario. Consider a wind-dispersed seed released from a plant. If we markup the area around the plant in grids, we could then count the number of seeds within each grid. Most grids will have no seeds, some grids will have one seed, a few grids may have two seeds, etc. Multiple seeds in grids is a rare event. The graph might look like

the number of wind-dispersed seeds within each grid

Figure 6. Example, poisson-like graph: the number of wind-dispersed seeds within each grid.

In molecular biology, use of Poisson distribution is common. In digital PCR for example, a DNA sample is diluted and divided into thousands of small reaction volumes or partitions. Thus, each partition contains either zero molecules, one molecule, or more than one molecule, where the loading of molecules into partitions is assumed to be independent, random, and rare per partition. In sequencing, Poisson models are used to describe random sampling of molecules and reads, especially when coverage is generated by non-deterministic processes like shotgun sequencing.

The Poisson has interesting properties, one being that the expected mean, λ, is equal to the variance. Lambda is, with respect to the Poisson distribution, the average number of times an event happens over a specific period or area.

An equation is

$\begin{align*} Pr\left [X\right ] = \frac{\mu^X e^{-\mu}}{X!} \end{align*}$

where μ is the mean (or could substitute with variance!), e is the natural logarithm, and X is number of successes you are interested in. For example, if μ = 1, what is the probability of observing a grid with five seeds? Simple enough to do this by hand, but let’s use Rcmdr instead. Here’s the graph (Fig 7) from Rcmdr (KMggplot2 plugin).

poisson plot, mean = 1

Figure 7. The probability of observing a grid with five seeds, poisson μ = 1 (ggplot2).

and for the actual probability we have from R

dpois(5, lambda = 1)
[1] 0.003065662

Rcmdr → Distributions → Discrete distributions → Poisson distribution → Poisson probabilities … (Fig 8)

The only thing to enter is the mean (some call μ lambda with symbol λ).

Rcmdr menu calculate poisson probability

Figure 8. Rcmdr menu, poisson probability.

Here’s the output from R. For intervals 0, 1, 2, 3, …, 6 (Rcmdr just enters this range for you!

> .Table <- data.fram(Pr=dpois(0:6, lambda = 1))
> rownames(.Table) <- 0:6
> .Table
Pr
0 0.3678794412
1 0.3678794412
2 0.1839397206
3 0.0613132402
4 0.0153283100
5 0.0030656620 ← Answer, 0.0307 or 3.07%
6 0.0005109437

Next — Continuous distributions

And finally, for ratio (continuous) scale data, which can take on any value, we can express the chance that probability of a given point as a continuous function, with the normal distribution being one of the most important examples (there are others, like the F-distribution). Many statistical procedures assume that the data we use can be viewed as having come from a “normally distributed population.” See Chapter 6.6.

Questions

1.2. Quarterback sacks by game for the NFL team Seahawks, years 2011 through 2022 are summarized below (data extracted from https://www.pro-football-reference.com/ ).

Sacks	How many games?
0	25
1	46
2	49
3	39
4	25
5	14
6	8
7	2
8	1
9	0

a) Assuming a Poisson distribution, what are the mean (lambda) and variance?

b) The table covers a total of 112 games. How many sacks (events) were observed?

c) What is the probability of the Seahawks getting zero sacks in a game (in 2022, a season was 17 games; prior years a season was 16 games)?

Quiz Chapter 6.5

Discrete probability

Chapter 6 contents

Introduction
Some preliminaries
Ratios and proportions
Combinations and permutations
Types of probability
Discrete probability distributions
Continuous distributions
Normal distribution and the normal deviate
Moments
Chi-square distribution
t distribution
F distribution
References and suggested readings