9.5 – Fisher exact test

Introduction
R code
When to use the Fisher Exact Test?
Questions
Quiz
Chapter 9 contents

Introduction

We mentioned that chi-square tests for contingency tables are fine as long as two conditions are met. These are the assumptions of a $\chi^2$ test.

No cell should have expected values less than 5%.
The test performs poorly at DF = 1 because we are approximating an infinite distribution with an exact test.

You will note that any time you have a 2X2 table, the second condition is always an issue because 2X2 tables have DF=1. Thus, in biomedical research, it is common to have an experiment that may be appropriate for a contingency analysis but the data may suffer from one or both of these limitations. Fisher’s exact test is always an option for these types of problems, but with the advantage that it always returns the exact p-value.

A reminder, the 2×2 table looks like

Table 1. 2×2 table reporting numbers of subjects who have (Yes) or do not have (No) the event.

		Column 1	Column 2
	Subjects	Yes	No
Row 1	Treatment 1	a	b
Row 2	Treatment 2	c	d

where a is the count of Treatment 1 – treated subjects who have the event, b is the count of Treatment 1 – treated subjects who do not have the event, c is the count of Treatment 2 – treated subjects who have the event, d is the count of Treatment 2 – treated subjects who do not have the event. Note the row and column totals:

$\begin{align*} Row1 = a + b\\ Row2 = c + d\\ Column1 = a + c\\ Column2 = b + d \end{align*}$

For example, a fairly common, “Gee, that’s curious,” fact is the seven left-handed US Presidents since 1901 (21), is higher than the proportion of left-handers in the general population (about 10%). For comparison, we could ask the same question about Vice Presidents.†

Table 2. U.S. Presidents and Vice-Presidents handedness.

Subjects	Yes	No
Presidents	7	14
Vice presidents	5	20

†Seven Vice-Presidents went on to become President, four right-handers, three left-handers.

Ronald A. Fisher came up with a test that is now called “Fisher’s Exact test” that circumvents this problem. It is an extremely useful test to know about because it provides a way to get an exact probability of the outcome compared to all other possible outcomes. Thus, when asked for a possible alternate to the chi-square contingency test for a 2X2 table, you can respond “Fisher’s Exact test.”

Although tedious to calculate by hand, and resource demanding when done by computer because of the multiple factorial expressions, the major advantage of the test is that it does not rely on the assumption that an underlying distribution applies. The Fisher Exact test can be used to calculate the exact probability of the observed outcome (P).

The equation for the Fisher Exact test can be written as

$\begin{align*} P = \frac{{R_{1}! \cdot R_{2}! \cdot C_{1}! \cdot C_{2}!}}{a! \cdot b \cdot c! \cdot d! \cdot n!} \end{align*}$

where R stands for row total, C stands for column total, n is the sample size, ! is the factorial, and a, b, c, and d are defined as in Table 1.

How does Fisher’s Exact test work? The data are setup in the usual way for a contingency problem, but now, we calculate the probability for all possible outcomes that we COULD have seen from our experiment, and ask if the actual outcome is unusual (low p-value). The trick is recognizing that you have to keep the totals constrained (note row and column totals stay the same).

Table 3. Original 2X2 contingency table (bold), with the next two more extreme outcomes.

original data	$\longrightarrow$	more extreme	$\longrightarrow$	next more extreme still	$\longrightarrow$
Yes	No	Yes	No	Yes	No
10	5	11	4	12	3
4	12	3	13	2	14
	$p-value=0.0206$		$p-value=0.0029$		$p-value=0.0002$

I’ve shown just the one-tailed outcomes, so the p-values are for one-tailed tests of hypothesis. The essence of the test is to find all outcomes MORE extreme than the original, in one direction. The one-tailed P-value then is the sum of all probabilities from those more extreme tables of outcomes.

To get the two-tailed probability, remember that you multiply the one-tailed probability by two. More accurate methods are also available (Agresti 1992).

Calculation of Fisher’s test involves using all possible combinations and factorials. Rcmdr has Fisher’s 2X2 built in via the Contingency table and as part of some Rcmdr plugins (eg, RcmdrPlugin.EBM, the Evidence Based Medicine plugin). Here we illustrate Fisher Exact test from the context menu in the main Statistics menu.

Alternatively, there are many web sites out there that provide an online calculator for Fisher’s Exact test. Here’s a link to one such calculator on GraphPad’s web site, cookies must be enabled to run this calculator).

To get the Fisher Exact test, your data must already be summarized into a 2X2 table, in which case you can use R Commander via the Contingency tables… menu selection (Fig 1).

Table 4. 2X2 table smokers vs vitamin use.

	Smoker: No	Smoker: Yes
Vitamin use: No	14	26
Vitamin use: Yes	19	15

With the data summarized as in Table 4, we can proceed directly via R Commander to analyze the contingency table, either by Chi-square test of Fisher exact test. First, bring up menu to enter the data (Fig 1):

Rcmdr: Statistics → Contingency tables… → Enter and analyze two-way table…

Screenshot Rcmdr Contingency table

Figure 1. Screenshot Rcmdr menu, Contingency tables.

The image of the R worksheet below contains two columns: Smoker (No, Yes) and Vitamin use (No, Yes). Complete the form (Fig 2).

Screenshot Rcmdr menu, Enter two way table

Figure 2. Screenshot Rcmdr menu, Enter Two-Way Table.

Alternatively, if the original data are available, do not tally the counts, let R do the work for you. The worksheet would be stacked like so (Table 5).

Table 5. Portion of subject and row entries in stacked worksheet for smokers vs vitamin use.

Subject	Vitamin.use	Smoking
1	No	No
2	No	No
3	No	No
4	No	No
5	No	No
.	.	.
71	Yes	Yes
72	Yes	Yes
73	Yes	Yes
74	Yes	Yes

And from the R Commander menu select as before

Rcmdr: Statistics → Contingency tables… → Two way table … (Fig 3).

Screenshot Rcmdr two-way table menu, load the data from stacked worksheet.

Figure 3. Screenshot Rcmdr two-way table menu, load the data from stacked worksheet.

Data is now ready.

R code

To carry out contingency table analysis on the entered data (Fig 2 or Fig 3), select the Statistics option tab (Fig 4, green arrow).

Screenshot Rcmdr menu Statistics option Enter two way table

Figure 4. Screenshot Rcmdr menu Statistics option. Select Chi-square test of independence, Fisher’s exact test, or both.

Check the box next to the Fisher’s exact test (Fig 4, blue arrow).

Click OK, and here is the R output.

> fisher.test(.Table)
Fisher's Exact Test for Count Data

data: .Table
p-value = 0.1008
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.1496417 1.1985775
sample estimates:
odds ratio
0.4302094

We accepted the defaults Is this a one or two-tailed test of hypothesis?

What can we conclude about the null hypothesis? Do we accept or reject?

Want to know what the “odds ratio” is? Follow the link to the next subchapter.

When to use the Fisher Exact Test?

Here’s the take-home message: the Fisher exact test is an alternate and better choice over the contingency table chi-square for 2X2 tables if one or more of the cells has expected values less than 5%. It is also appropriate for cases in which you have only 1 degree of freedom (as do all 2X2 tables!), but it doesn’t make sense if each cell has more than 5% expected values (the calculation is too tedious), but rather, apply the Yate’s correction. As the sample sizes get larger, the different methods converge to virtually identical answers.

Some examples.

Is there an association between final grades and attendance on a randomly selected day?

Table 6. First 2X2 scenario.

cc	Yes	No
Letter grade A	2	3
Other letter grade	1	6

Table 7. Second 2X2 scenario.

cc	Yes	No
Letter grade A	5	6
Other letter grade	2	12

Table 8. Third 2X2 scenario.

cc	Yes	No
Letter grade A	10	12
Other letter grade	4	24

Code hints for tests with direct 2X2 entry with matrix() are

grades.Table <- matrix(c(2,3,1,6), 2, 2, byrow=TRUE)

Chi-square test of independence

chisq.test(grades.Table, correct=FALSE)

Fisher Exact test

fisher.test(grades.Table, alternative = "greater")

Questions

1. Apply Fisher exact test on the four contingency tables (a – d) introduced in section 9.3, question 2. Make note of the p-value from Fisher exact test and from analyses used to complete question 2. Note any trends. (Hint: make sure you are testing the same null hypothesis.)

(a)

	Yes	No
A	18	6
B	3	8

(b)

	Yes	No
A	10	12
B	3	14

(c)

	Yes	No
A	5	12
B	12	18

(d)

	Yes	No
A	8	12
B	3	3

Quiz Chapter 9.5

Fisher Exact Test

Chapter 9 contents

7.6 – Confidence intervals

Quantifying uncertainty
Confidence interval for proportions
Confidence interval for ARR
Confidence interval for NNT
Odds ratio Standard error and 95% confidence interval
P-value from confidence intervals
Questions
Quiz
Chapter 7 contents

Quantifying uncertainty.

Although I’ve already presented the concept (eg, Chapter 3.4), and equations for confidence intervals of risk estimates (Chapter 7.4, Chapter 7.5), here, we expand on the idea of confidence intervals. Confidence intervals are a central part of meeting one of the main objectives of statistics, that is, estimation and communicating uncertainty with any reported estimate.

There are three components of statistical analysis

Estimation
Inference
Modeling

Inference refers to statistical hypothesis testing — we ask questions of observations — do men (Rice et al 1999) and women (Fisher et al 2012) differ for blood glucose levels following a bout of aerobic exercise? T-tests, analysis of variance (ANOVA), chi-square, correlation, regression are types of statistical procedures used to do statistical inference. Modeling on the other hand refers to procedures used to relate cause and effect or for use in prediction of future values given new data. Many of the statistical procedures one uses for inference are also used to build statistical models (ANOVA, regression). Studies may intend to either test some hypothesis (inference) or to provide a predictive equation (modeling). But most studies that relate observations gathered from an experiment are obliged to also report statistics, and this is the realm of estimation. Estimates of the mean and standard deviation, for example, would be typical statistics one expects to find in a report. We call these descriptive statistics, and together with graphics, descriptive statistics are the chief way we describe our results.

Confidence intervals are crucial in statistics because they provide a range of values where the true population parameter is likely to fall, offering a more nuanced understanding of the estimate beyond just a single point value. Similarly, for statistical model model building, confidence intervals quantify the uncertainty inherent in data, providing a range of plausible values for model parameters rather than a single point estimate.

We will review how to calculate confidence intervals for proportions and for NNT. These intervals are available in epiR package and automatically returned in RcmdrPlugin.EBM.

Confidence interval for proportions

A proportion is the fraction of individuals in a population with some characteristic. The characteristic might be HIV positive for example. This would be called the population proportion and it would be a parameter of interest. In reality, we calculate a sample proportion and therefore estimate the population proportion with error. We can calculate the confidence interval (CI) of the proportion to communicate the precision of our estimate. For proportions, we use the binomial distribution — either a sample has the characteristic of interest or it does not — there are only two possibilities. There are a variety of ways to go here, and the simplest is to use a normal approximation. This will work well provided the sample size was reasonably large and the proportion is not close to zero or one, that is, we invoke the Central Limit Theorem here. Although the outcomes are binomial, the error is assumed to be normally distributed. The Wald confidence interval for p is

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

where $\hat{p}$ is the proportion of individuals with the characteristic (also called successes), z is the percentile from the normal distribution that corresponds to 1 – 1/2α. For 95% CI then α = 0.05 and z would be 1.965. (See standard normal table.) Of course, if making the normal approximation for the binomial is not appropriate, the CI is less than ideal. The binomial after all is a discrete distribution whereas the normal distribution is continuous, so errors will enter particularly for low sample numbers.

Other approaches may be used to get better estimates of CI for proportions, including Wilson score intervals and Jeffrey Intervals (Agresti and Coull 1998). See R package propCIs.

Because a statistic like the mean or a calculation of absolute or relative risk reduction are calculated from samples drawn from a population, the estimate comes with error. The error is basically this – if we calculate a statistic like number needed to treat (NNT) or its converse, the number needed to harm (NNH), we need to communicate to the reader how precise our estimate is. Estimation has to do with accuracy, error, and precision.

Confidence interval for ARR

The ARR is simply $\frac{c}{n_{2}}-\frac{a}{n_{1}}$

where $n_{1}$ is the number of treated or exposed individuals for which the event occurred and $n_{2}$ is the number of untreated or unexposed individuals which the event occurred.

	Event happened	Event did not happen
Treated or Exposed	a	b
Control or Not exposed	c	d

Our data from example (Ch 7.5), were a = 612, b = 2192, c = 375, and d = 2543.

$\begin{align*} SE_{ARR}=\sqrt{\left (\frac{a}{n_{1}} \cdot \frac{1-\frac{a}{n_{1}}}{n_{1}} \right )+\left (\frac{c}{n_{2}} \cdot \frac{1-\frac{c}{n_{2}}}{n_{2}} \right )} \end{align*}$

and the 95% confidence interval is then $ARR \pm 2 \cdot SE_{ARR}$

approximately. The “2” is only approximate; you need to use Z = 1.965, the value at probability value = 0.9725 (which comes from the Normal Table).

Confidence interval for NNT

For a sample of 100 people drawn at random from a population (which may number in the millions), then repeat the NNT calculation for a different sample of 100 people, do we expect the first and second NNT estimates to be EXACTLY the same number? No, but we do expect them to be close and we can define what we mean by close as we expect each estimate to be within certain limits. While we expect the second calculation to be close to the first estimate, we would be surprised if it was EXACTLY the same. And so, which is the correct estimate, the first or the second? They both are, in the sense that they both estimate the parameter NNT (a property of a population). But we can do better than two estimates. Confidence Intervals (CI) allow us to assign a probability to how certain we are about the statistic and whether it is likely to be close to the true value. We will

For CI of NNT, we need sample size for control and treatment groups; like all confidence intervals, we need to calculate the standard error of the statistic, in this, case, the standard error (SE) for NNT.

SE = sqrt(risk placebo * (1 - risk placebo) / (# in placebo group) + risk treatment * (1 - risk treatment) / (# in treatment group))

where SE is the standard error for NNT

The CI is approximately then $NNT \pm 2 X SE$

* – the “2” is only approximate; you need to use Z = 1.965, the value at probability value = 0.9725 (by multiplying this, which comes from the Standard Normal Table).

Odds ratio Standard error and 95% confidence interval

Like any statistic we can calculate, an estimate of odds ratio should be accompanied by the confidence limit. The standard error may be calculated with the following formula

$\begin{align*} SE\left \{ ln\left ( OR \right ) \right \}=\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}} \end{align*}$

R code

seOdds <- sqrt(sum(1/612, 1/2192,1/375,1/2543))

where ln refers to the natural logarithm. An estimate for the 95% confidence interval is

$\begin{align*} lower \ limit = exp\left (ln\left ( OR \right ) \right )-1.96\cdot SE\left \{ ln\left ( OR \right ) \right \} \end{align*}$

where exp is the exponential function

$e^{x}$

and ln is natural logarithm.

R code

 exp(log(1.89,base=exp(1)) - 1.96*seOdds)

Our example the lower limit was 1.64. For the upper limit

$\begin{align*} upper \ limit = exp\left (ln\left ( OR \right ) \right )+1.96\cdot SE\left \{ ln\left ( OR \right ) \right \} \end{align*}$

Our example the upper limit was 2.19.

R code

 exp(log(1.89,base=exp(1)) + 1.96*seOdds)

Thus, our estimate was 1.89 and the 95% confidence interval was $\left ( 1.64,2.19\right )$

which does not include one. Therefore, we conclude statistically different.

Note 1. A reminder, log() without specifying the base returns the natural logarithm. A better coding habit might suggest always specifying the base:

log(x, base = exp(1))

P-values from confidence intervals

While we expect certain reporting criteria for published, it is not uncommon that one or more elements are missing. For example, while we show how to obtain confidence intervals, the p-values were not reported. We discuss statistical inference and p-values in detail in Chapter 8, but for now, a low p-value (typically < 0.05) indicates that the odds ratio (OR) is statistically significant, meaning the observed relationship between a factor and an outcome is unlikely to be due to random chance. An OR of the difference may not be reported, or, an inexact p-value is reported like “< 0.05,” but we “need” an exact p-value for our meta-analysis. A little math, and these missing statistics can be calculated.

Note 2. We illustrated how to calculate confidence intervals from p-values in Chapter 3.4 – Estimating parameters. See two helpful “how to” articles by Bland and Altman in 2011 BMJ (references listed in Chapter 3.6 – References).

We can use the OR example. We need to take the logarithm of ratios like OR or NNT or RRR. From above, we have the confidence intervals.

We need

The OR estimate, $myEST$ , which was 1.89

lower and upper confidence intervals: 1.64 and 2.19, respectively.

the standard error, which can be calculated as $mySE = (upper − lower)/(2×1.96)=0.07377$

the test statistic, $myZ <- abs(myEST/mySE)$

and then the p-value, $exp(-.717*myZ-0.416*myZ^2) = 0.4044$

we would then report that the OR was not statistically different, p = 0.4044).

Questions

1. Instead of 95% confidence interval, obtain the 99% confidence interval for odds ratio 1.89.

Quiz Chapter 7.6

Confidence intervals

Chapter 7 contents

7.4 – Epidemiology: Relative risk and absolute risk, explained

Risk communication
An example of ARR and RRR risk calculations using natural numbers
Confidence interval for NNT is derived from the Confidence interval for ARR
Software: epiR
Improve output from epiR
Software: epitools
Software: OpenEpi
Software: RcmdrPlugin.EBM
Questions
Quiz
Chapter 7 contents

Risk communication.

Epidemiology is the study of patterns of health and illness of populations. An important task in an epidemiology study is to identify risks associated with disease. Epidemiology is a crucial discipline used to inform about possible effective treatment approaches, health policy, and about the etiology of disease.

Please review terms presented in section 7.1 before proceeding. RR and AR are appropriate for cohort-control and cross-sectional studies (see 2.4 and 5.4) where base rates of exposure and unexposed or numbers of affected and non-affected individuals (prevalence) are available. Calculations of relative risk (RR) and relative risk reduction (RRR) are specific to the sampled groups under study whereas absolute risk (AR) and absolute risk reduction (ARR) pertain to the reference population. Relative risks are specific to the study, absolute risks are generalized to the population. Number needed to treat (NNT) is a way to communicate absolute risk reductions.

An example of ARR and RRR risk calculations using natural numbers.

Clinical trials are perhaps the essential research approach (Sibbald and Roland 1998; Sylvester et al 2017); they are often characterized with a binary outcome. Subjects either get better or they do not. There are many ways to represent risk of a particular outcome, but where possible, using natural numbers is generally preferred as a means of communication, particularly to the general public. Consider the following example (pp 34-35, Gigerenzer 2015): What is the benefit of taking a cholesterol-lowering drug, Pravastatin, on the risk of deaths by heart attacks and other causes of mortality? Press releases (eg, Maugh 1995), from the study stated the following:

“… the drug pravastatin reduced … deaths from all causes 22%”

A subsequent report (Skolbekken 1998) presented the following numbers (Table 1).

Table 1. Reduction in total mortality (5 year study) for people who took Pravastatin compared to those who took placebo.

Deaths per 1000 people with high cholesterol
(> 240 mg/dL) No deaths Cumulative
incidence

Treatment

	Deaths per 1000 people with high cholesterol (> 240 mg/dL)	No deaths	Cumulative incidence
Treatment	Pravastatin (n = 3302)	`a` `= 32`	`b` `= 3270`	CIe
Placebo (n = 3293)	`c` `= 41`	`d` `= 3252`	CIu

Pravastatin

(n = 3302)

a
= 32

b
= 3270

CIe

Placebo

(n = 3293)

c
= 41

d
= 3252

CIu

where cumulative incidence refers to the number of new events or cases of disease divided by the total number of individuals in the population at risk.

Do the calculations of risk

The risk reduction (RR), or the number of people who die without treatment (placebo) minus those who die with treatment (Pravastatin), $41-32=9$

$\begin{align*} RR = \frac{\frac{a}{a+b}}{\frac{c}{c+d}} = 0.91 \end{align*}$

and the cumulative incidence in the exposed (treated) group, CIe, is $32/(32+3270) = 0.0097$ , and cumulative incidence in the unexposed (control) group, CIu, is $41/(41+3252) = 0.01245$ . We can calculate another statistic called the risk ratio,

$\begin{align*} RR = \frac{CI_{e}}{CI_{u}} = 0.78 \end{align*}$

because the risk ratio is less than one, we interpret that statins reduce the risk of mortality from heart attack. In other words, statins lowered the risk by 0.78.

But is this risk reduction meaningful?

Now, consider the absolute risk reduction (ARR) is $0.9 \% = 100 \% X \frac{9}{1000}$

Relative risk reduction, or the absolute risk reduction divided by the proportion of patients who die without treatment, is $22 \% = 100 \% X 9 \div 41$

Conclusion: high cholesterol may contribute to increased risk of mortality, but the rate is very low in the population as a whole (the ARR).

Another useful way to communicate benefit is to calculate the Number Needed to Treat (NNT), or the number of people who must receive the treatment to save (benefit) one person. The ideal NNT is a value of one (1), which would be interpreted as everyone improves who receives the treatment. By definition, NNT must be positive; however, a resulting negative NNT would suggest the treatment may cause harm, ie, number needed to harm (NNH).

For this example, the NNT is

$\begin{align*} \frac{1}{\frac{9}{1000}} = 111 \end{align*}$

therefore, to benefit one person, 111 need to be treated. The flip side of the implications of NNT, although one person may benefit by taking the treatment, 110 ( $=111 - 1$ ) will take the treatment, but will NOT RECEIVE THE BENEFIT, but do potentially get any side effect of the treatment.

Confidence interval for NNT is derived from the Confidence interval for ARR.

For a sample of 100 people drawn at random from a population (which may number in the millions), then repeat the NNT calculation for a different sample of 100 people, do we expect the first and second NNT estimates to be exactly the same number? No, but we do expect them to be close and we can define what we mean by close as we expect each estimate to be within certain limits. While we expect the second calculation to be close to the first estimate, we would be surprised if it was exactly the same. And so, which is the correct estimate, the first or the second? They both are, in the sense that they both estimate the parameter NNT (a property of a population).

We use confidence interval to communicate where we believe the true estimate for NNT to be. Confidence Intervals (CI) allow us to assign a probability to how certain we are about the statistic and whether it is likely to be close to the true value (Altman 1998, Bender 2001). We will calculate the 95% CI for the ARR using the Wald method, then take the inverse of these estimates for our 95% CI. The Wald method assumes normality.

For CI of ARR, we need sample size for control and treatment groups; like all confidence intervals, we need to calculate the standard error of the statistic, in this, case, the standard error (SE) for ARR is approximately

$\begin{align*} SE_{\left ( p_{1}-p_{2} \right )}=\sqrt{\frac{p_{1}\left ( 1-p_{1} \right )}{n_{1}}+\frac{p_{2}\left ( 1-p_{2} \right )}{n_{2}}} \end{align*}$

where SE is the standard error for ARR. For our example, we have

$\begin{align*} SE_{\left ( p_{1}-p_{2} \right )}=\sqrt{\frac{0.041\left ( 1-0.041 \right )}{1000}+\frac{0.032\left ( 1-0.032 \right )}{1000}} = 0.00838 \end{align*}$

The 95% CI for ARR is approximately $ARR\pm 2 X SE_{\left ( p_{1}-p_{2} \right )}$

For the Wald estimate, replace the 2 with $z = 1.965$ , which comes from the normal table for z at $\frac{0.95}{2}$

(why the 2 in the equation? Because it is plus or minus so we divide the frequency 0.95 in half) and for our example, we have $0.009 \pm 2 X SE_{\left ( p_{1}-p_{2} \right )} = \left (-0.0078,0.0258 \right )$

and the inverse for NNT CI is $\left (-128, 38 \right )$

Our example exemplifies the limitation of the Wald approach (cf. Altman 1998): our confidence interval includes zero, and doesn’t even include our best estimate of NNT (111).

Note 1: By now I trust you can see differences for results by direct input of the numbers into R and what you get by the natural numbers approach. In part this is because we round in our natural number calculations — remember, while it makes more sense to communicate about whole numbers (people) and not fractions (fractions of people!), rounding through the calculations adds error to the final value. As long as you know the difference and the relevance between approximate and exact solutions, this shouldn’t cause concern.

Software: epiR.

R has many epidemiology packages, epiR and epitools are two. Most of the code presented stems from epiR.

We need to know about our study design in order to tell the functions which statistics are appropriate to estimate. For 2X2 tables, we will use epi.2by2() command, which allows four kinds of studies (Table 2).

Table 2. The method argument in epi.2by2 chooses how to analyze your data based on your study design.

Method	Description	Notes
cohort.count	Use this when you follow a set number of people over time and count how many get sick.	This design directly calculates Relative Risk Reduction (RRR) and Absolute Risk Reduction (ARR) because you track baseline risk over time.
cohort.time	Use this when you follow people over time and track exactly how long each person was monitored before getting sick.	This calculates rate reductions instead of ARR or RRR because it deals with person-time data rather than raw counts.
case.control	Use this when you start with sick people and healthy people and look backwards at their past exposures.	You cannot calculate ARR or RRR here because you do not know the actual baseline risk of the population.
cross.sectional	Use this when you take a snapshot of a population at a single point in time to see who is currently exposed and sick.	This calculates prevalence reductions rather than true ARR or RRR because you are not tracking changes over time.

For our statin example, the design was prospective cohort (ie, cohort.count in epiR package language), not case-control or cross-sectional (review in Chapter 5.4).

library(epiR)
Table1 <- matrix(c(32,3270,41,3252), 2, 2, byrow=TRUE, dimnames = list(c("Statin", "Placebo"), c("Died", "Lived")))
Table1
        Died Lived
Statin    32  3270
Placebo   41  3252
epi.2by2(Table1, method="cohort.count", outcome = "as.columns")

R output

           Outcome + Outcome - Total           Inc risk *
Exposed +         32      3270  3302 0.97  (0.66 to 1.37)
Exposed -         41      3252  3293 1.25  (0.89 to 1.69)
Total             73      6522  6595 1.11  (0.87 to 1.39)

Point estimates and 95% CIs:
-------------------------------------------------------------------
Inc risk ratio                          0.78 (0.49, 1.23)
Inc odds ratio                          0.78 (0.49, 1.24)
Attrib risk in the exposed *           -0.28 (-0.78, 0.23)
Attrib fraction in the exposed (%)    -28.48 (-103.47, 18.88)
Attrib risk in the population *        -0.14 (-0.59, 0.32)
Attrib fraction in the population (%) -12.48 (-37.60, 8.05)
-------------------------------------------------------------------
Uncorrected chi2 test that OR = 1: chi2(1) = 1.147 Pr>chi2 = 0.284
Fisher exact test that OR = 1: Pr>chi2 = 0.292
Wald confidence limits
CI: confidence interval
* Outcomes per 100 population units

The risk ratio we calculated by hand is shown in green in the R output, along with other useful statistics (see ?epi2x2 for help with these additional terms) not defined in our presentation.

We explain results of chi-square goodness of fit (Ch 9.1) and Fisher exact (Ch 9.5) tests in Chapter 9. Suffice to say here, we interpret the p-value (Pr) = 0.284 and 0.292 to indicate that there is no association between mortality from heart attacks with or without the statin (ie, the Odds Ratio, OR, not statistically different from one).

Wait! Where’s NNT and other results?

Use another command in epiR package, epi.tests(), to determine the specificity, sensitivity, and positive (or negative) predictive value.

epi.tests(Table1)

and R returns

          Outcome + Outcome - Total
Test +           32      3270  3302
Test -           41      3252  3293
Total            73      6522  6595

Point estimates and 95% CIs:
--------------------------------------------------------------
Apparent prevalence *                    0.50 (0.49, 0.51)
True prevalence *                        0.01 (0.01, 0.01)
Sensitivity *                            0.44 (0.32, 0.56)
Specificity *                            0.50 (0.49, 0.51)
Positive predictive value *              0.01 (0.01, 0.01)
Negative predictive value *              0.99 (0.98, 0.99)
Positive likelihood ratio                0.87 (0.67, 1.13)
Negative likelihood ratio                1.13 (0.92, 1.38)
False T+ proportion for true D- *        0.50 (0.49, 0.51)
False T- proportion for true D+ *        0.56 (0.44, 0.68)
False T+ proportion for T+ *             0.99 (0.99, 0.99)
False T- proportion for T- *             0.01 (0.01, 0.02)
Correctly classified proportion *        0.50 (0.49, 0.51)
--------------------------------------------------------------
* Exact CIs

Additional statistics are available by saving the output from epi2x2() or epitests() to an object, then using summary(). For example save output from epi.2by2(Table1, method="cohort.count", outcome = "as.columns") to object myEpi, then

summary(myEpi)

look for NNT in the R output

$massoc.detail$NNT.strata.wald
       est    lower    upper
1 -362.377 -128.038  436.481

Thus, the NNT was 362 (compared to 111 we got by hand) with 95% Confidence interval between -436 and +128 (make it positive because it is a treatment improvement).

Note 2: Strata (L. layers), refer to subgroups, for example, sex or age categories (see discussion in Ch05.4). Our examples here are not presented as subgroup analysis, but epiR reports by name strata.

epiR reports a lot of additional statistics in the output and for clarity, I have not defined each one, just the basic terms we need for BI311. As always, see help pages (eg, ?epi.2x2 or ?epitests)for more information about structure of an R command and the output.

We’re good, but we can work the output to make it more useful to us.

Improve output from epiR.

For starters, if we set interpret=TRUE instead of the default, interpret=FALSE, epiR will return a richer response.

fit <- epi.2by2(dat = as.table(Table1), method = "cohort.count", conf.level = 0.95, units = 100, interpret = TRUE, outcome = "as.columns")
fit

R output. In addition to the table of coefficients (above), interpret=TRUE provides more context, shown below

Measures of association strength:
The outcome incidence risk among the exposed was 0.78 (95% CI 0.49 to 1.23) times less than the outcome incidence risk among the unexposed. 

The outcome incidence odds among the exposed was 0.78 (95% CI 0.49 to 1.24) times less than the outcome incidence odds among the unexposed.

Measures of effect in the exposed:
Exposure changed the outcome incidence risk in the exposed by -0.28 (95% CI -0.78 to 0.23) per 100 population units. -28.5% of outcomes in the exposed were attributable to exposure (95% CI -103.5% to 18.9%).

Number needed to treat for benefit (NNTB) and harm (NNTH):
The number needed to treat for one subject to be harmed (NNTH) is 362 (NNTH 128 to infinity to NNTB 436).

Measures of effect in the population:
Exposure changed the outcome incidence risk in the population by -0.14 (95% CI -0.59 to 0.32) per 100 population units. -12.5% of outcomes in the population were attributable to exposure (95% CI -37.6% to 8.1%).

That’s quite a bit. Another trick is to get at the table of results. We install a package called broom, which includes a number of ways to handle output from R functions, including those in the epiR package. Broom takes from the TidyVerse environment; tables are stored as tibbles.

Note 3: A tibble (package tibble) is a modified R data frame optimized for the tidyverse that enforces strict data handling and cleaner console printing. Unlike standard data frames, tibbles refuse to automatically simplify into vectors when sliced or guess column names using partial matching. Because of these intentional design choices, tibbles often don’t play well with R functions not written explicitly with the TidyVerse environment in mind.

To convert a data.frame to a tibble,

my_tibble <- as_tibble(my_dataframe)

To convert a tibble back to a standard data.frame

my_dataframe <- as.data.frame(my_tibble)

library(broom)

# Test statistics
tidy(fit, parameters = "stat")

R output

# A tibble: 3 × 4
term statistic df p.value
<chr> <dbl> <dbl> <dbl>
1 chi2.strata.uncor 1.15 1 0.284
2 chi2.strata.yates 0.909 1 0.340
3 chi2.strata.fisher NA NA 0.292

We can convert the tibbles into our familiar data.frame format, and then select only the statistics we want.

# Measures of association
fitD <- as.data.frame(tidy(fit, parameters = "moa")); fitD

R output, all 15 measures of association!

                  term     estimate     conf.low    conf.high
1       RR.strata.wald    0.7783605    0.4914679   1.23272564
2     RR.strata.taylor    0.7783605    0.4914679   1.23272564
3      RR.strata.score    0.8742994    0.6584540   1.10340173
4       OR.strata.wald    0.7761915    0.4876209   1.23553616
5     OR.strata.cfield    0.7761915           NA           NA
6      OR.strata.score    0.7761915    0.4894450   1.23093168
7        OR.strata.mle    0.7762234    0.4718655   1.26668220
8    ARisk.strata.wald   -0.2759557   -0.7810162   0.22910484
9   ARisk.strata.score   -0.2759557   -0.8000574   0.23482532
10     NNT.strata.wald -362.3770579 -128.0383246 436.48140194
11    NNT.strata.score -362.3770579 -124.9910314 425.84844829
12  AFRisk.strata.wald   -0.2847517   -1.0347210   0.18878949
13  PARisk.strata.wald   -0.1381661   -0.5933541   0.31702189
14  PARisk.strata.piri   -0.1381661   -0.3910629   0.11473067
15 PAFRisk.strata.wald   -0.1248227   -0.3760279   0.08052298

We can call out just the statistics we want from this table by calling to the specific elements in the data.frame (rows, columns).

fitD[c(1,4,7,9,12),]

R output

                 term   estimate   conf.low  conf.high
1      RR.strata.wald  0.7783605  0.4914679  1.2327256
4      OR.strata.wald  0.7761915  0.4876209  1.2355362
7       OR.strata.mle  0.7762234  0.4718655  1.2666822
9  ARisk.strata.score -0.2759557 -0.8000574  0.2348253
12 AFRisk.strata.wald -0.2847517 -1.0347210  0.1887895

Software: epitools.

Another useful R package for epidemiology is epitools, but it comes with it’s own idiosyncrasies. We have introduced the standard 2X2 format, with a, b, c, and d cells defined as in Table 1, above. However, epitools does it differently, and we need to update the matrix. By default, epitools has the unexposed group (control) in the first row and the non-outcome (no disease) is in the first column. To match our a,b,c, and d matrix, use the epitools command to change this arrangement with the rev() argument. Now, the analysis will use the contingency table on the right where the exposed group (treatment) is in the first row and the outcome (disease) is in the first column (h/t M. Bounthavong 2021). Once that’s accomplished, epitools returns what you would expect.

Calculate relative risk

risk1 <- 32 / (3270 + 32)
risk2 <- 41 / (3525 + 41)
risk1 - risk2

and R returns

-0.00180638

odds ratio

library(epitools)
oddsratio.wald(Table1, rev = c("both"))

and R returns

$data
              Outcome
Predictor Disease2 Disease1 Total
 Exposed2      517       36   553
 Exposed1      518       11   529
Total         1035       47  1082

$measure
           odds ratio with 95% C.I.
Predictor   estimate      lower      upper
 Exposed2  1.0000000        NA         NA
 Exposed1  0.3049657 0.1535563  0.6056675

$p.value
two-sided
Predictor  midp.exact  fisher.exact    chi.square
Exposed2           NA            NA            NA
Exposed1 0.0002954494  0.0003001641  0.0003517007

odds ratio highlighted in green.

Software: OpenEpi.

R is fully capable of delivering the calculations you need, but sometimes you just want a quick answer. Online, the OpenEpi tools at https://www.openepi.com/ can be used for homework problems. For example, working with count data in 2 X 2 format, select Counts > 2X2 table from the side menu to bring up the data form (Fig 1).

Data entry for 2X2 table at openepi.com.

Figure 1. Data entry for 2X2 table at openepi.com.

Once the data are entered, click on the Calculate button to return a suite of results (Fig 2).

Results for 2X2 table at openepi.com

Figure 2. Results for 2X2 table at openepi.com.

Software: RcmdrPlugin.EBM

Note 4: Fall 2023 — I have not been able to run the EBM plugin successfully! The package is still active at CRAN, but looks like it hasn’t been updated since 2015. 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.

This isn’t the place nor can I be the author to discuss what evidence based medicine (EBM) entails (cf. Masic et al. 2008), or what its shortcomings may be (Djulbegovic and Guyatt 2017). Rcmdr has a nice plugin, based on the epiR package, that will calculate ARR, RRR and NNT as well as other statistics. The plugin is called RcmdrPlugin.EBM

install.packages("RcmdrPlugin.EBM", dependencies=TRUE)

After acquiring the package, proceed to install the plug-in. Restart Rcmdr, then select Tools and Rcmdr Plugins (Fig 3).

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

Figure 3. Rcmdr: Tools → Load Rcmdr plugins…

Find the EBM plug-in, then proceed to load the package (Fig 4).

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

Figure 4. Rcmdr plug-ins available (after first download the files from an R mirror site).

Restart Rcmdr again and the menu “EBM” should be visible in the menu bar. We’re going to enter some data, so choose the Enter two-way table… option in the EBM plug-in (Fig 5)

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

Figure 5. R Commander EBM plug-in, enter 2X2 table menus.

To review, we have the following problem, illustrated with natural numbers and probability tree (Fig 6).

probability tree Illustration of the statin problem

Figure 6. Illustration of probability tree for the statin problem.

Now, let’s enter the data into the EBM plugin. For the data above I entered the counts as

	Lived	Died
Statin	468	32
Placebo	459	41

and selected the “Therapy” medical indicator (Fig 7).

EBM plugin with two-way table completed for the statin problem.

Figure 7. EBM plugin with two-way table completed for the statin problem.

The output from EBM plugin was as follows. I’ve added index numbers in brackets so that we can point to the output that is relevant for our worked example here.

(1) .Table <- matrix(c(468,32,459,41), 2, 2, byrow=TRUE, dimnames = list(c('Drug', 'Placebo'), c('Lived', 'Dead')))
(2) fncEBMCrossTab(.table=.Table, .x='', .y='', .ylab='', .xlab='', .percents='none', .chisq='1', .expected='0', .chisqComp='0', .fisher='0', .indicators='th', .decimals=2)

R output begins by repeating the commands used, here marked by lines (1) and (2). The statistics we want follow in the next several lines of output.

(3) Pearson's Chi-squared test data: .Table X-squared = 1.197, df = 1, p-value = 0.2739
(4) # Notations for calculations Event + Event -Treatment "a" "b" Control "c" "d"
(5)# Absolute risk reduction (ARR) = -1.8 (95% CI -5.02 - 1.42) %. Computed using formula: [c / (c + d)] - [a / (a + b)]
(6)# Relative risk = 1.02 (95% CI 0.98 - 1.06) %. Computed using formula: [c / (c + d)] / [a / (a + b)]
(7)# Odds ratio = 1.31 (95% CI 0.81 - 2.11). Computed using formula: (a / b) / (c / d)
(8) # Number needed to treat = -55.56 (95% CI 70.29 - Inf). Computed using formula: 1 / ARR
9)# Relative risk reduction = -1.96 (95% CI -5.57 - 1.53) %. Computed using formula: { [c / (c + d)] - [a / (a + b)] } / [c / (c + d)]
(10)# To find more about the results, and about how confidence intervals were computed, type ?epi.2by2 . The confidence limits for NNT were computed as 1/ARR confidence limits. The confidence limits for RRR were computed as 1 - RR confidence limits.

end R output

In summary, we found no difference between statin and placebo (P-value = 0.2739) and ARR of -1.8%

Questions.

Data from a case-control study on alcohol use and esophageal cancer (Tuyns et al (1977), example from Gerstman 2014). Cases were men diagnosed with esophageal cancer from a region in France. Controls were selected at random from electoral lists from the same geographical region.

	Esophageal cancer
Alcohol grams/day	Cases	Noncases	Total
> 80	96	109	205
< 80	104	666	770
Total	200	775	975

What was the null hypothesis? Be able to write the hypothesis in symbolic form and as a single sentence.
What was the alternative hypothesis? Be able to write the hypothesis in symbolic form and as a single sentence.
What was the observed frequency of subjects with esophageal cancer in this study? And the observed frequency of subjects without esophageal cancer?
Estimate Relative Risk, Absolute Risk, NNT, and Odds ratio?
1. Which is more appropriate, RR or OR? Justify your decision.
The American College of Obstetricians and Gynecologists recommends that women with an average risk of breast cancer (BC) over 40 get an annual mammogram. Nationally, the sensitivity of mammography is about 68% and specificity of mammography is about 75%. Moreover, mammography involves exposure of women to radiation, which is known to cause mutations. Given that the prevalence of BC in women between 40 and 49 is about 0.1%, please evaluate the value of this recommendation by completing your analysis.
A) In this age group, how many women are expected to develop BC?
B) How many False negative would we expect?
C) How many positive mammograms are likely to be true positives?
“Less than 5% of women with screen-detectable cancers have their lives saved,” (quote from BMC Med Inform Decis Mak. 2009 Apr 2;9:18. doi: 10.1186/1472-6947-9-18): Using the information from question 5, What is the Number Needed to Treat for mammography screening?

Quiz Chapter 7.4

Epidemiology: Relative risk and absolute risk, explained

Chapter 7 contents

Probability, Risk Analysis
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

7.3 – 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

4.4 – Mosaic plots

Introduction
Example data
mosaicplot() in R base
mosaic plot from EBM plugin
mosaic-like plot KMggplot2 plugin
Questions
Quiz
Depreciated material
Chapter 4 contents

Introduction

Mosaic plots are used to display associations among categorical variables, eg, from a contingency table analysis. Mosaic plots and tree plots (next chapter) are used to show part-to-whole associations or to show whether or not results are consistent across sub-groups (Ballarini et al 2020). Mosaic plots, a type of visual grid, are simple versions of heat maps (next chapter). Like stacked bar charts, the wdth of each column shows the size of each group, the height of any boxes within the columns shows the size of the sub-groups. Used appropriately, mosaic plots may be useful to show relationships. However, like pie-charts and bar charts, care needs to be taken to avoid their over use; the mosiac plot works for a few categories, but quickly loses clarity as numbers of categories increase. As an alternative, particularly if intended for a general audience, icon arrays, aka waffle plots, may be a better choice (see discussion in Garcia-Retamero et al 2013).

In addition to the function mosaicplot() in the base R package, there are a number of packages in R that will allow you to make these kinds of plots; depending on the analyses we are doing we may use any one of three Rcmdr plugins: RcmdrPlugin.mosaic (depreciated), RcmdrPlugin.KMggplot2, or RcmdrPlugin.EBM.

Example data

Table 1. Records of American and National Leagues baseball teams at home and away midway during 2016 season

	No	Yes
AL	10	5
NL	7	8

The configuration of major league baseball (MLB) parks differ from city to city. For example, Boston’s American League (AL) Fenway Park has the 30-feet tall “Green Monster” fence in left field and a short distance of only 302 feet along the foul line to right field fence. For comparison, Globe Life Park in Arlington, TX the distance along the foul lines left field (332 feet) and right field (325 feet). So, it suggests that teams may benefit from playing 81 games at their home stadium. To test this hypothesis I selected Win-Loss records of the 30 teams at the midway point of the 2016 season. Data are shown in the Table 1.

mosaicplot() in R base

The function mosaicplot() is included in the base install of R. The following code is one way to directly enter contingency table data like that from Table 1.

myMatrix <- matrix(c(10,  5,  7,  8),  nrow = 2, ncol = 2, byrow = TRUE)
dimnames(myMatrix) <- list(c("AL", "NL"), c("No","Yes"))
myTable <- as.table(myMatrix); myTable
mosaicplot(myTable, color=2:3)

The simple plot is shown in Figure 1. color = “2” is red, color = “3” is green.

moisaic plot made with function mosaicplot()

Figure 1. Mosaic plot made with basic function mosaicplot().

mosaic plot from EBM plugin

A good option in Rcmdr is to use the “evidence-based-medicine” or “EBM” plug-in for Rcmdr (RcmdrPlugin.EBM). This plugin generates a real nice mosaic plot for 2 X 2 tables.

After loading the EBM plugin, restart Rcmdr, then select EBM from the menu bar and choose to “Enter two-way table…”

Figure 2. First steps to make mosaic plot in R Commander EBM plug-in.

Complete the data entry for the table as shown in the image below. After entering the values, click the OK button.

Figure 3. Next steps to make mosaic plot in R Commander EBM plug-in.

Along with the requested statistics a mosaic plot will appear in a pop-up window.

Figure 4. Mosaic plot made from R Commander EBMplug-in

mosaic-like plot KMggplot2 plugin

The KMggplot2 plugin for Rcmdr will also generate a mosaic-like plot. After loading the KMggplot2 plugin, restart Rcmdr, then load a data set with the table (e.g., MLB data in Table 1). Next, from within the KMggplot2 menu select, “Bar chart for discrete variables…”

Figure 5. First steps to make mosaic plot in R Commander KMggplot2 plug-in.

From the bar chart context menu make your selections. Note that this function has many options for formatting, so play around with these to make the graph the way you prefer.

Figure 6. Next steps to make mosaic plot in R Commander KMggplot2 plug-in.

And here is the resulting mosaic-like plot from KMggplot2.

Figure 7. Mosaic-like plot made from R Commander KMggplot2 plug-in.

Questions

1. Most US states have laws that dictate pre-employment drug testing for job candidates; Interestingly, states are increasingly legalizing marijuana use. Data for states plus District of Columbia are presented in the table. Make a mosaic plot of the table.

Table 2. Marijuana use is US states, legal or not legal

	Marijuana-use legal	Marijuana-use not legal
Yes	19	12
No	14	6

Data adopted from https://www.paycor.com/resource-center/pre-employment-drug-testing-laws-by-state

Quiz Chapter 4.4

Mosaic plots

Depreciated material

As of summer 2020, Rcmdrplugin.mosaic is depreciated. While you can install the archived version, it is not recommended. Therefore, this material is left as is but for information purposes only. For a simple mosaic plot in Rcmdr I recommend working with the RcmdrPlugin.EBM.

Download the RcmdrPlugin.mosaic package, start Rcmdr, then navigate to Tools and choose Load Rcmdr plug-in(s).… Select Rcmdrplugin.mosaic (Fig. 8), then restart Rcmdr (Fig. 9). The plugin adds mosaic plot to the regular Graphics menu of Rcmdr.

Figure 8. Screenshot of popup menu from Rcmdr with mosaic plugin selected.

Figure 34. After clicking OK (Fig. 33), click Yes to restart Rcmdr. The plugin will then be available.

Figure 9. After clicking OK (Fig 8), click Yes to restart Rcmdr. The plugin will then be available.

Load a data set with 2X2 arranged data, or create the variables yourself (Yikes, 30 rows!). The mosaic plugin requires that you submit data in a table format. We can check whether our data are currently in that format. At the R prompt type

is.table(MLB)

And R will return

[1] FALSE

(To be complete, confirm that the data set is a data.frame: is.data.frame(MLB).)

You will need a table before proceeding with the mosaic plug-in. then create a table using a command like the one shown below.

MLBTable <- xtabs(~League+HomeWin, data=MLB)

Once the table is ready, select “mosaic or assoc plot” from the Rcmdr Graphics menu (Fig. 10)

Figure 35. How to access the mosaic plot in R Commander.

Figure 10. How to access the mosaic plot in R Commander.

A small window will pop up that will allow you to select the table of data you just created (Fig. 11). Note that you may need to hunt around your desktop to find this menu! Select the table (in this example, “MLBTable), then click on “Create plot” button.

Figure 11. Screenshot of popup menu in mosaic plugin in R Commander.

R Note: The popup from the mosaic menu shown in Fig. 11 will also display the data.frame MLB. If you mistakenly select the dataframe MLB, you’ll get an error message in Rcmdr (Fig. 12). The plugin behaves erratically if you select MLB: On my computer, the function hangs and requires restarting R.

Figure 37. Error message as result of selecting a dataframe for use in mosaic plugin.

Figure 12. Error message as result of selecting a dataframe for use in mosaic plugin.

After you select the table, two additional windows will pop up: on the left (Fig. 13) is the context menu to change characteristics of the mosaic plot; on the right (not shown) will be a mosaic plot itself in default grey scale colors.

Figure 13. Options for the mosaic plot

At a minimum, change the plot from grey scale to a colorized version by checking the box next to the “Colorize last variable” option. The new plot is shown in Figure 14.

Figure 14. Our new mosaic plot.

OK. Take a moment and look at the plot. What conclusions can be made about our hypothesis — are there any differences between the leagues for home versus road Wins-Loss records?

By default the mosaic command copies the command to the R window. You can change the graph by taking advantage of the options in the brewer palette. Here’s the command for the mosaic image above.

mosaic(structable(MLBTable), highlighting=2, highlighting_fill=brewer.pal.ext(2,"RdYlGn"))

Change the options in the brackets following “brewer.pal.ext.” For example, replace RdYlGn with Blues to make a plot that looks like the following

Figure 40. Mosaic plot with changed color scheme.

Figure 15. Mosaic plot with changed color scheme.

The colors are selected from the Rcolorbrewer package. For more, see this blog for starters.

Chapter 4 contents

Graphs and tables (How to report statistics)
Bar (column) charts
Histograms
Box plot
Mosaic plots
Scatter plots
Add a second Y axis
Q-Q plot
Ternary plots
Heat maps
Graph software
References and suggested readings