6.2 – Ratios and probabilities

Introduction

One common objective of a research project is comparison between quantities: one effective way to communicate a comparison is to construct a ratio. Ratios provide a standardized way to compare different data sets. The purpose of this page is introduce the wealth of types of ratios used for quantification of comparisons, from indexes to ratios.

Let’s review our definitions again. An event is some occurrence. As you know, a ratio is one number, the numerator, divided by another called the denominator. A proportion is a ratio where the numerator is a part of the whole. A rate is a ratio of the frequency of an event during a certain period of time. A rate may or may not be a proportion, and a ratio need not be a proportion, but proportions and rates are all kinds of ratios. If we combine ratios, proportions, and/or rates, we construct an index.

Ratios

Yes, data analysis can be complicated, but we start with this basic idea. Much of the statistics is based on frequency measures, eg, ratios, rates, proportions, indexes, and scales.

Ratios are the association between two numbers, one random variable divided by another. Ratios are used as descriptors and the numerator and denominator do not need to be of the same kind. Business and economics are full of ratios. For example, return on investment (ROI), eg, earnings equals net income divided by number of shares outstanding, the Price-Earnings, or P/E ratio, is the ratio of the price of a stock to the earnings per stock, as well as many others are used to summarize performance of a business, and to compare performance of one business against another.

Ratios are deceptively convenient way to standardize a variable for comparisons, ie, how many times one number contains another. For example, when estimating bird counts for different areas, or different birding effort (intensity, time searched), we may correct counts by accounting for area in which counts were made or the total time spent counting, for a per-unit ratio (Liermann et al 2004).

Practice: There were 1,326 day undergraduate students enrolled in 2014 at Chaminade University of Honolulu and the Sullivan Library added 8469 new items (ebooks, journals, etc,) to its collection during 2014. What is the ratio of new items per student?

Data collected from Chaminade University website at www.chaminade.edu on 3 July 2014.

Practice: For another example, what is ratio of annual institutional aid a student at Chaminade University may expect to receive compared to a student at Hawaii Pacific University?

Thus, in 2014, Chaminade offered more than twice the institutional aid to its students then does HPU (both institutions are private universities, data from retrieved on 3 July 2014 from www.cappex.com).

Relative change, Fold-change and orders of magnitude

It’s common practice to report on the amount of change between two quantities recorded over time. Relative change (often percentage) describes change relative to an initial value. Relative change in root length b

To make percent increase (decrease), simply multiple 100%.

(1)  

The ratio between two quantities, eg, to compare mRNA expression levels of genes from organisms exposed to different conditions, researchers may report fold-change. Fold change is often reported as log base 2 change.

Why log base 2? Symmetry and ease of reporting. A doubling becomes +1 and a halving becomes -1. Without log transformation, doubling is +2, but having is +0.5.

Relative change is useful for reporting percent changes, fold change for ratios, although if the change is small in amount, then relative change can be misleading, eg, value goes from five to ten, then the relative change is 100%, but effectively, a very small change in absolute units. Similarly, fold change becomes a large, but likely meaningless comparison, if the denominator is close to zero.

An example of calculation of fold change. Rates of the expression from cells exposed to heavy metal divided by expression under basal conditions. Gene expression under different treatments may be evaluated by calculating fold-change as the log base2 of the ratio of expression of a gene for one treatment divided by expression of same gene from control conditions. Copper is an essential trace element, but excess exposure to copper is known to damage human health, including chronic obstructive pulmonary disease. One proposed mechanism is that cell injury promotes an epithelial-to-mesenchymal shift. In a pilot study (N.I.H. INBRE P20RR016467), we investigated gene expression changes by quantitative real-time polymerase chain reaction (qPCR) in a rat lung Type II alveolar cell line exposed to copper sulfate compared to unexposed cells. Record cycle threshold values, C_T, for each gene, where C_T is number of cycles required for the fluorescent signal to exceed background levels; C_T is inversely proportional to amount of cDNA (mRNA) in the sample. Genes investigated were ECAD, FOXC2, NCAD, SMAD, SNAI1, TWIST, and VIM, with ATCB as reference gene. ECAD expression is marker of epithelial cells, whereas FOXC2, NCAD, SNAI1, TWIST, and VIM expression marker of mesenchymal cells. After calculating values, geometric means of normalized values of three replicates each are shown in Table 1.

myBits <- intToBits(8) 
paste(as.character(myBits),collapse="")

Table 1. Mean and fold change of gene expression values from qPCR for several genes from a rat lung cell line.*

* INBRE funding, P20RR016467.

At face value, following four hour exposure to copper sulfate in media, some evidence that the epithelial cell line adopted gene expression profile of mesenchyme-like cells. However, weakness of fold-change is clear from Table 1: the quantity is sensitive to small values. ECAD expression in the cell line is low, thus high numbers of PCR cycles (mean = 36) and treated cells not much fewer (mean = 34.4).

Orders of magnitude

In addition to fold change, order of magnitude is another common way to compare quantities. While fold change returns the multiplier between two numbers, order of magnitude generally refers to differences in multiples of ten, logarithmic.

Thus, a difference of one order of magnitude is the number multiplied by 10¹, three orders of magnitude is the number multiplied by 10³, and so on. A change from 10 to 100 is a one order of magnitude increase; the same change, 10 to 100, is a ten-fold change. It’s a difference of scale. If a person’s salary increases by 3%, from $49K to $50,470, reporting this as an “order of magnitude” increase is nonsensical as a communication device (magnitude of change is just 0.01). By contrast, fold change for the same increase would be 1.03 — also indicative of a negligible increase, but a cleaner representation.

Rates

Rates are a class of ratios in which the denominator is some measure of time. For example, the four year graduation rate of some Hawaii universities are shown in Table 2.

Table 2. Percent students graduation with bachelor’s degrees within four years or six years (cohort 2014, data source NCES.ed.gov)

Examples of rates

Rates are common in biology. To name just a few

• Basal metabolic rate (BMR), often measured by indirect calorimetry, reported in units kiloJoules per hour.
• Birth rates and mortality (death) rates, components of population growth rate.
• Extinction rate, how often over a given time species go extinct.
• Growth rate, which may refer to growth of the individual (somatic growth rate), or increase of number of individuals in a population per unit time.
• Molecular clock hypothesis, rate of amino acid (protein) or nucleotide (DNA) substitution is approximately constant per year over evolutionary time.
• Mutation rate, frequency of new mutations in a DNA sequence of an organism over time.
• Photosynthetic rate, convert light energy into chemical energy.
• Phred quality score, error rates of incorrectly called nucleotide bases by the sequencer.

Proportions

Proportions are also ratios, but they are used to describe one part to the whole. For example, 902 women (self-reported) day undergraduate students enrolled in 2014 at Chaminade University in Honolulu, Hawaii.

Practice: Given that the total enrollment for Chaminade in 2014 was of 1,326, calculate the proportion of female students to the total student body.

More practice:  A common statistic in agriculture is germination percentage, count the number of seeds that sprout over a period of time divided by total number of seeds planted multiplied by 100. Figure 1 shows an example of a germination trial for five tomato seeds.  Seed in position 3 (left to right 1 – 5) failed to germinate, whereas the other four seeds display range of seedling growth.  (A seed coat is observed next to plant 2.)

Figure 1. Example planting of five tomato seeds, day 5, on agar Petri dish (M. Dohm).

Observations conducted over 10 days. Germination defined as evidence of radicle by day 10. Data collected by student for biostatistics project. A random sample of results from three plates is presented in Table 3.

Table 3. Germination of tomato seeds on agar Petri dishes.

* Random sample of three plates drawn from among 30 plates.

We calculate the percent germination for the 15 seeds planted as

Thus, for the sample of plates the germination percent was 80%, which, coincidently, is the same result if germination percent was calculated for the five seeds displayed in Figure 1.

Beginning in Chapter 7 – Probability, Risk Analysis, we introduce and use a number of proportions, for example incidence and prevalence are proportions, as is PPV — positive predictive value. Sensitivity and specificity are both proportions and also qualified as indexes.

Comparing proportions

In some cases you may wish to compare two proportions or two ratios. The hypothesis tested is the difference between the two ratios, and the test is if the confidence interval of the difference includes zero. If it does, then we would conclude there is no statistical difference between the two proportions.  In R, use the prop.test function. For example, 63 women were on team sport rosters at Chaminade in 2014, a proportion of 59% of all student athletes (n = 106). Recall from the example above that women were 68% of all students at Chaminade University. Title IX compliance requires that a university “maintain policies, practices and programs that do not discriminate against anyone on the basis of gender” (NCAA, INCLUSION AND BELONGING Championing Women in Sports). In terms of athletic programs, then, universities are required to provide participation opportunities for women and men that are substantially proportionate to their respective rates of enrollment of full-time undergraduate students.

Consider Chaminade University: Is there a statistical difference between proportion of women athletes and their proportion of total enrollment? We introduce statistical inference in Chapter 8, but for now, this is a test of the null hypothesis that the difference between the two proportions is zero.

At the R prompt type (remember, anything after the # sign is a comment and ignored by R).

women = c(62,902)  #where 62 is the number of women athletes and 902 is the number of women students
students = c(106,1326)  #106 is the number of student athletes and 1326 is all students
prop.test(women,students)  #the default is a two-tailed test, ie, no group differences

And R returns

2-sample test for equality of proportions with continuity correction
 data: women out of students
X-squared = 3.6331, df = 1, p-value = 0.05664
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.197532407 0.006861073
sample estimates:
 prop 1 prop 2 
0.5849057 0.6802413

What is the conclusion of the test?

When you compare two groups, you’re asking whether the two groups are equal (the null hypothesis). Mathematically, that’s the same as saying the difference between the two groups is equal to zero.

First check the lower and upper limits of the confidence interval. A confidence interval is one way to report a range of plausible values for an estimate (see Ch 7.6 – Confidence intervals). It’s called a confidence interval because a probability is assigned to the range of values; a 95% confidence interval is interpreted as we’re 95% certain the true population value is somewhere between the reported limits. For our Chaminade University Title IX question, recall that we are asking whether the value of zero is included. The lower limit was -0.1975 and some change; the upper limit was 0.0068 and some change. Thus, zero is included and we would conclude that there was no statistical difference between the two proportions.

The second relevant output to look at is the p-value, or probability value. If the p-value is less than 5%, we typically reject the tested hypothesis. We will talk more about p-values and their relationship to inference testing in Chapter 8; for now, pay attention to the confidence interval (introduced in Chapter 3.4); if zero is included, then we conclude no substantial differences between the two proportions.

Indexes

Indexes are composite statistics that combine indicators. They condense complex diagnostic performance data into a single, universally understood numerical indicator.

Indexes are common in biology, and well-known business and economics indexes are reported on every day, eg, Dow Jones Industrial average combines stock prices from 30 companies listed on the New York Stock Exchange and the CPI, or Consumer Protection Index, constructed by the US Bureau of Labor Statistics from a “market basket of consumer goods and services, used by the government and other institutions to track inflation. The CPI itself is made up of numerous indexes to account for regional price differences and other factors.

Some indexes presented in this book include

• Grade point average
• Body Mass Index (BMI)
• Comet assay indexes (tail intensity, tail length, tail moment) are used to assess DNA damage among organisms exposed to environmental contaminants (eg, Mincarelli et al., 2019).
• Encephalization index, ratio of brain to body weight among species. Used to compare cognitive abilities.
• Sensitivity, specificity, NNT — number needed to treat, odds-ratio and RRR and related. See Chapter 7.1 – Probability, Risk Analysis.
• Species diversity indexes, measures of species richness and abundance. See Chapter 20.8.

Scales

Agreement scales for surveys, eg, Likert scale or sliding scale (Sullivan and Artino 2013). For example, after learning about Theranos, students were asked:

How serious is this violation in your opinion?

a 5-point scale.

Although an intuitive measure, how fast can an individual run, challenging to determine whether individual’s performance is at physiological maximum. Particularly for measures of performance capacity that involves behavior (motivation), may apply a race quality scale (eg., binary scale “good” or “bad” Husak et al 2006).

These examples reflect ordinal scales. Many of the nonparametric tests discussed in Chapter 15 are suitable for analysis of scales.

Limitations of ratios

Although the indexes may be easy to communicate, statistically, indexes have many drawbacks. Chief among these, variation in ratios may be due to change in numerator or denominator. Ratios and any index calculated by combining ratios seem simple enough, but have complicated statistical properties. Over the years, several authors have made critical suggestions for use of ratios and indexes. Some key references are Packard and Boardman (1988), Jasienski and Oikos (1999), Nee et al (2005), and Karp et al (2012).  For example, ratios, computing trait value by body weight, are often used to compare some trait among individuals or species that differ in body size. However, this normalization attempt only removes the covariation between size and the trait if there is a 1:1 relationship between size and the trait. More typically, relationship between the trait and body size is allometric, ie, the slope is not equal to one.  Thus, ratio will over-correct for large size and under-correct for small size. The proper solution is to conduct the comparison as part of an analysis of covariance (ANCOVA, see Chapter 17.6).

Example

Which is the safer mode of travel: car or airplane?

The following discussion covers travel safety in the United States of America for a typical year, 2000*.

Between 2000 and 2023, there were 779 deaths associated with accidents of major air carriers in the USA. Year 2009 was the last multiple-casualty crash of a major U.S. carrier (Colgan Air Flight 3407); between 2010 and 2021, two fatal accidents, two fatalities were reported.

We’ve all heard the claim that it’s much safer to fly with a major airline than it is to travel by car (eg, Flying is getting scary. But is it still safe? CNN June 20, 2024). There are a variety of arguments, but one statistical argument goes as follows. In 2000 in the United States, 638,902,993 persons traveled by major air carrier, whereas there were 190,625,023 licensed drivers. In 2000, 92 persons died in air travel (again, major carriers only), whereas 37,526 persons died in vehicle crashes (includes drivers and passengers). Thus, the risk of dying in air travel is given as the proportion , or 1.44e-07 (0.000014%), whereas the comparable proportion for death by motor vehicle is , or 1.97e-04 (0.0197%).

In other words, we can expect one death (actually 1.4) for every ten million airline passengers, but 20 deaths (actually 19.7) for every one hundred thousand licensed drivers. Thus, flying is a thousand times safer than driving (actual result 1,367 times; divide the rate of motor vehicle-caused deaths for licensed drivers by the rate for airlines). Proportions are hard to compare sometimes, especially when the per capita numbers differ (ten million vs. 100,000 in this case).

We can put the numbers onto a probability tree and get a sense of what we are looking at (Fig 2).

Figure 2. A probability tree to help visualize comparison of deaths (“yes”) by car travel and by airline travel in the United States for the year 2000.

Comparing rates and proportions

Without going into the details, we will do so in Chapter 9 Inferences on Categorical Data, comparing two rates is a chi-square, χ², contingency table type of problem. More specifically, however, it is a binomial problem (Chapter 3.1, Chapter 6.5); there are two outcomes, death or no death, and we can describe how likely the event is to occur as a proportion. Because the numbers are large, we can use rely on the normal distribution for comparing the two proportions. We’ll explain this more in the next chapters, but for now it may be enough to present the equation for the comparison of two proportions under the assumption of normality, proportion z test.

and the null hypothesis (see Chapter 8) tested as that the two proportions are equal. This may be written as

H0 : p₁-p₂=0

We can assign statistical significance to the differences in events for the two modes of travel under this set of assumptions. Rcmdr has a nice menu-driven system for comparing proportions, but for now I will simply list the R commands.

At the R prompt type each line then submit the command.

total = 100000000
prop.test(c(19700,14),c(total,total)))

And the R output

prop.test(c(19700,14),c(total,total))

    2-sample test for equality of proportions with continuity
    correction

data:  c(19700, 14) out of c(total, total)
X-squared = 19658, df = 1, p-value < 2.2e-16
alternative hypothesis: two.sided
95 percent confidence interval:
 0.0001940984 0.0001996216
sample estimates:
  prop 1   prop 2 
1.97e-04 1.40e-07

There’s a bit to unpack here. R is consistent; when it reports results of a statistical test, it typically returns the value of the test statistic (χ² = 19658), the degrees of freedom for the test (df = 1), and the p-value (< 2.2e-16).

By convention in statistics, if a p-value, where “p” stands for probability, is less than 5%, we would say that our results are statistically significant from the null hypothesis.  Looks pretty convincing to me, the difference of 19,700 deaths compared to 14 deaths is clearly different by any criterion and by the results of the statistical test, the p-value is several orders of magnitude smaller than 5%.

Safer to fly. By far, not even close. And similar conclusions would be reached if we compare different years, or averages over many years, or if we used a different way to express the amount of travel (eg, miles/year) by these modes of transportation.

Are you convinced, really? Is it safer to fly?

Let’s try a little statistical reasoning — what assumptions did I make to do these calculations? We recognize immediately that many more people travel by car: that there are way more cars being driven then there are airline planes being flown. The question then is, have we properly adjusted for this difference? Here are a few considerations. My source for the numbers is the NTS 2001 book published by the U.S. Department of Transportation (U.S. Department of Transportation). We are conducting a risk analysis, and the first step is to make sure that we are comparing “apples with apples.” Here are two alternative solutions that at least compare, “Red Delicious” apples with “Macintosh” apples.

Option 1. There are many, many more licensed drivers than there are licensed commercial airline pilots. The standard comparison offered in the background above compared deaths per licensed car driver, but a different metric for air travel, the rate per passenger. This isn’t as bad of a comparison as it may seem — after all, the majority of deaths in car accidents are of the driver themselves. But it isn’t that hard to make the direct comparison — just find out how many commercial pilots there are — a direct comparison with licensed car drivers (stated above as 190,625,023). From the FAA we see that in 2009 there were 125,738 persons with commercial certificates. Since there are only 20 major airline carriers in the United States now (a few more were active in 2000, but we’ll put this aside), the number of licenses is an over estimate of the actual number we want — how many pilots of commercial airlines — but lets use this number for starters — after all, just because a person has a drivers license doesn’t mean they drive or ride in a car.

Number of deaths/yr: Let’s use 2000 data, a typical year prior to 9/11 (and excluding the Covid-19 pandemic). Airlines: 92 deaths; motor vehicles (includes passenger cars, trucks, etc, but not motorcycles): 37,526 deaths (drivers = 25,567; passengers = 10,695; 86 others).

Which mode of travel is riskier? I get a rate a rate of 7.3 X 10^-4 deaths per commercial pilot, or

92125738

compared to a rate for car drivers of 1.97 X 10^-4 deaths.

To summarize what we have so far, I get a result that suggests car travel is almost four times safer

7.3 x 10-41.97 x 10-4→7.31.97 = 3.7

then traveling by commercial airliner. In whole numbers, these results translate to seven deaths for every 10,000 commercial pilots compared to two deaths for every 10,000 licensed car drivers (Fig 3).

Figure 3. Comparing totals of deaths adjusted by numbers of licensed drivers and by licensed commercial airline pilots in the United States.

R work follows. Enter and submit each command on a separate line in the script window

total = 10000
prop.test(c(2,7),c(total,total))

And the R output

prop.test(c(2,7),c(total,total))

    2-sample test for equality of proportions with continuity
    correction

data:  c(2, 7) out of c(total, total)
X-squared = 1.7786, df = 1, p-value = 0.1823
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.001187816  0.000187816
sample estimates:
prop 1 prop 2 
 2e-04  7e-04

What’s happened? The p-value (0.1823) is not less than 5% and so we would conclude under this scenario that there is no difference between the proportions of deaths between the two modes of travel. Let’s keep going.

Option 2. There are many, many more cars on the road then there are airplanes flying commercial passengers. The standard comparison offered in the background information above identified death rates per individual driver, but used a different metric for airline travelers (number of deaths per passenger), which confuses individuals with travelers: what we need is the number of individuals that traveled by airliner, not the total number of passengers (which is many times higher, because of repeat flyers). How can we make a fair comparison for the two modes of travel? Most people never fly whereas most people drive (or ride in a car) frequently in the United States. To me, risk of travel might be better expressed in terms of a per trip rate. I want to know, what are my chances of dying each time I get into my car versus each time I fly on a commercial jet in the United States?

Number of trips/yr. For airlines, I use the number of departures (2000 = 8,951,773). But for cars, we need to decide how to get a similar number. It’s not available directly from the DOT (and would be difficult to get — studies with randomly selected drivers can yield as many as 5 trips per day for licensed drivers). I took the number of licensed drivers and bound the problem — at the low end, let’s say that only 2 trips per week (eg, 50 weeks) are taken by licensed drivers (100 trips); at the upper end, let’s take 2 trips per day per week, or 500 trips/year. Thus, at the low end, we have 1.91 X10¹⁰ trips per year; at the upper end, 9.53 X 10¹⁰ trips per year.

Which mode of travel is riskier? Using the number of deaths/yr listed above in Option 1, I get a rate of 1.03 X 10^-5 deaths/trip for air carriers

928.95 x 10⁶

compared to a rate of 1.97 X 10^-6 deaths/trip for cars (lower bound) or 3.9 X 10^-7 deaths/trip for cars (upper bound). Here’s what the numbers look for in a tree (taking the lower number of trips per year for cars, Fig 4).

Figure 4. Comparing totals of deaths adjusted by numbers of car trips and by numbers of airline trips in the United States.

R work follow

total = 10000000
prop.test(c(20,103),c(total,total))

And the R output

prop.test(c(20,103),c(total,total))

    2-sample test for equality of proportions with continuity
    correction

data:  c(20, 103) out of c(total, total)
X-squared = 54.667, df = 1, p-value = 1.428e-13
alternative hypothesis: two.sided
95 percent confidence interval:
 -1.057370e-05 -6.026305e-06
sample estimates:
  prop 1   prop 2 
2.00e-06 1.03e-05

Now we have another really small p-value (1.428e-13), which suggests a statistically significant difference between the modes of air travel, but the difference in deaths is switched. I now have a result that suggests car travel is much safer than traveling with a commercial airliner! These calculations suggest that you are as much as 26 (upper bounds, five times for lower bounds) times more likely to die from a plane crash then you are behind the wheel. In whole numbers, these results indicate one death for every 100,000 airline flights compared to 1 death for every 500,000 (lower estimate) or 2,500,000 car trips!

Do I have it right and the standard answer is wrong? As Lee Corso used to say on ESPN’s College GameDay program, “Not so fast, my friend!” (Wikipedia). Mark Twain was right to hold the skeptic’s view. Begin by listing the assumptions and by checking the logic of the comparisons (there are still holes in my logic!!). For one, if I am considering my risk of dying by mode of travel, it is far more likely that I will be in a car accident than I will an airline accident, simply because I don’t travel by airline that much. When we consider lifetime risk, we can see why the assertion that it is “safer to fly than drive” is true — we’re far more likely to belong to one of the reference populations involving automobiles (eg, those who drive frequently, for many years) than we are to be among the frequent flyers reference populations.

Questions

1. Review and provide your own examples for

• index
• rate
• ratio
• proportion

3. Like travel safety, we are often confronted by risk comparisons like the following: Which animal is more deadly to humans, dogs or sharks? Between the two, which lead to more hospitalizations in the United States? Work through your assumptions and use results from the International Shark Attack file.

• If a person lives in Nebraska, and never visits the ocean, how does a “shark attack” risk analysis apply? Is it a fair comparison to make between dog attacks and shark attacks? Why or why not.

4. Go to cappex.com/colleges and update institutional (gift) aid offered by Chaminade and HPU. Compare to University of Hawaii-Manoa.

5. Calculate percent increase for

• Annual atmospheric CO2 1980 and 2024.

Quiz Chapter 6.2

Ratios and probabilities

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Chapter 6 content

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

9.2 – Chi-square contingency tables

Introduction

This page covers a lot of ground to extend our discussion of implementations of the chi-square test to experimental data and introduces several related statistics, including a correlation coefficient for categorical data and an introduction to agreement and reliability measurement statistics.

We begin with contingency tables.

We just completed a discussion about goodness of fit tests, inferences on categorical traits for which a theoretical distribution or expectation is available. We offered Mendelian ratios as an example in which theory provides clear-cut expectations for the distribution of phenotypes in the F₂ offspring generation. This is an extrinsic model — theory external to the study guides in the calculation of expected values — and it reflects a common analytical task in epidemiology.

But for many other kinds of categorical outcomes, no theory is available. Today, most of us accept the link between tobacco cigarette smoking and lung cancer. The evidence for this link was reported in many studies, some better designed than others.

Consider the famous Doll and Hill (1950) report on smoking and lung cancer. They reported (Table IV Doll and Hill 1950), for men with lung cancer, only two out of 649 were non-smokers. In comparison, 27 case-control patients (ie, patients in the hospital but with other ailments, not lung cancer) were nonsmokers, but 622 were smokers (Table 1).

Table 1. Doll and Hill (1950).

A more recent example from the study of the efficacy of St John’s Wort (Hypericum perforatum) as a treatment for major depression (Shelton et al 2001, Apaydin et al. 2016). Of patients who received St. John’s Wort over 8 weeks, 14 were deemed improved while 98 did not improve. In contrast 5 patients who received the placebo were deemed improved while 102 did not improve (Table 2).

Table 2. St. John’s Wort and depression

These kinds of problems are direct extensions of our risk analysis work in Chapter 7. Now, instead of simply describing the differences between case and control groups by Relative Risk Reduction or odds ratios, we instruct how to do inference on the risk analysis problems.

Thus, faced with absence of coherent theory as guide, the data themselves can be used (intrinsic model), and at least for now, we can employ the test.

In this lesson we will learn how to extend the analyses of categorical data to cases where we do not have a prior expectation — we use the data to generate the tests of hypotheses. This would be an intrinsic model. One of the most common two-variable categorical analysis involves 2X2 contingency tables. We may have one variable that is the treatment variable and the second variable is the outcome variable. Some examples provided in Table 3.

Table 3. Some examples of treatment and outcome variables suitable for contingency table analysis.

Table 4 were examples of published studies returned from a quick PubMed search; there are many examples (meta-analysis opportunities!). However, while they all can be analyzed by contingency tables analysis, they are not exactly the same. In some cases, the treatments are fixed effects, where the researcher selects the levels of the treatments (see Ch12.3 and Ch14.3 for discussion about fixed vs random effects in experimental design).

In other cases, each of these treatment variables need not be actual treatments (in the sense that an experiment was conducted), but it may be easier to think about these as types of experiments. These types of experiments can be distinguished by how the sampling from the reference population were conducted. Before we move on to our main purpose, to discuss how to calculate contingency tables, I wanted to provide some experimental design context by introducing two kinds of sampling (Chapter 5.5).

Our first kind of sampling scheme is called unrestricted sampling. In unrestricted sampling, you collect as many subjects (observations) as possible, then assign subjects to groups. A common approach would be to sample with a grand total in mind; for example, your grant is limited and so you only have enough money to make 1000 copies of your survey and you therefore approach 1000 people. If you categorize the subjects into just two categories (eg, Liberal, Conservative), then you have a binomial sample. If instead you classify the subjects a number of variables, eg, Liberal or Conservative, Education levels, income levels, homeowners or renters, etc., then this approach is called multinomial sampling. The point in either case is that you have just utilized a multinomial sampling approach. The aim is to classify your subjects to their appropriate groups after you have collected the sample.

Logically what must follow unrestricted sampling would be restricted sampling. Sampling would be conducted with one set of “marginal totals” fixed. Margins refers to either the row totals or to the column totals. The sampling scheme is referred to as compound multinomial. The important distinction from the other two types is that the number of individuals in one of the categories is fixed ahead of time. For example, in order to determine if smoking influences respiratory health, you approach as many people as possible to obtain 100 smokers.

The contingency table analysis

We introduced the 2×2 contingency table (Table 4) in Chapter 7.

Table 4. 2X2 contingency table

Contingency tables are all of the form (Table 5)

Table 5.  Basic format of a 2X2 table

Regardless of how the sampling occurred, the analysis of the contingency table is the same; mechanically, we have a chi-square type of problem. In both contingency table and the “goodness of fit” Chi-Square analyses the data types are discrete categories. The difference between GOF and contingency table problems is that we have no theory or external model to tell us about what to expect. Thus, we calculate expected values and the degrees of freedom differently in contingency table problems. To learn about how to perform contingency tables we will work through an example, first the long way, and then using a formula and the a, b, c, and d 2X2 table format. Of course, R can easily do contingency tables calculations for us.

Doctors noticed that a new form of Hepatitis (HG, now referred to as GB virus C), was common in some HIV+ populations (Xiang et al 2001). Accompanying the co-infection, doctors also observed an inverse relationship between HG loads and HIV viral loads: patients with high HG titers often had low HIV levels. Thus, the question was whether co-infection alters the outcome of patients with HIV — do HIV patients co-infected with this HG progress to AIDS and mortality at rates different from non-infected HIV patients? I’ve represented the Xiang et al (2001) data in Table 6.

Table 6. Progression of AIDS for patients co-infected with HG [GB virus C], Xiang et al data (2001)

Our question was: Do HIV patients co-infected with this HG progress to AIDS and mortality at rates different from non-infected HIV patients? A typical approach to analyze such data sets is to view the data as discrete categories and analyze with a contingency table. So we proceed.

Setup the table for analysis

Rules of contingency tables (see Kroonenberg and Verbeek 2018). What follows is a detailed walk through setting up and interpreting a 2X2 table. Note that I maintain our a, b, c, d cell order, with row 1 referencing subjects exposed or part of treatment group and row 2 referencing subjects not exposed (or part of the control group), as introduced in Chapter 7.

The Hepatitis G data from Xiang et al 2001 data, arranged in 2X2 format (Table 7).

Table 7.  Format of 2X2 table Xiang et al (2001) dataset.

The data are placed into the cells labeled a, b, c, and d.

Cell a : The number of HIV+ individuals infected with HG that lived beyond the end of the study.

Cell b : The number of HIV+ individuals infected with HG that died during the study.

Cell c : The number of HIV+ individuals not infected with HG that lived.

Cell d : The number of HIV+ individuals not infected with HG that died.

This is a contingency table because the probability of living or dying for these patients may have been contingent on coinfection with hepatitis G.

State the examined hypotheses

H_O: There is NO association between the probability of living and coinfection with hepatitis G.

For a contingency table this means that there should be the same proportion of individuals with and without hepatitis G that either lived or died (1:1:1:1).

Now, compute the Expected Values in the Four cells (a, b, c, d)

1. Calculate the Expected Proportion of Individuals that lived in Entire Sample:
   Column 1 Total / Total = Expected Proportion of those that lived beyond the study.
2. Calculate Expected Proportion of Individuals that died in Entire Sample:
   Column 2 Total / Total = Expected Proportion that died during the study.
3. Calculate the Expected Proportion of Individuals that lived and were HG+
   Expected Proportion of living (step 1 above) x Row 1 Total = Expected For Cell A.
4. Calculate the Expected Proportion of Individuals that died and were HG+
   Expected Proportion died x Row 1 Total = Expected For Cell B
5. Calculate the Expected Proportion of surviving individuals that were HG-
   Expected Proportion lived x Row 2 Total = Expected For Cell C
6. Calculate the Expected Proportion of individuals that died that were HG-
   Expected Proportion that died x Row 2 Total = Expected For Cell D

Yes, this can get a bit repetitive! But, we are now done — remember, we’re working through this to review how the intrinsic model is applied to obtain expected values.

To summarize what we’ve done so far, we now have

Observations arranged in a 2X2 table

Calculated the expected proportions, ie, the Expected values under the Null Hypothesis.

Now, we proceed to conduct the chi-square test of the null hypothesis — The observed values may differ from these expected values, meaning that the Null Hypothesis is False.

Table 8. Copy Table 6 data set

Get the Expected Values

1. Calculate the Expected Proportion of Individuals that lived in Entire Sample:
   Column 1 Total (198) / Total (362) = 0.5470
2. Calculate Expected Proportion of Individuals that died in Entire Sample:
   Column 2 Total (164) / Total (362) = 0.4530
3. Calculate the Expected Proportion of Individuals that lived and were HG+
   Expected Proportion of living (0.547) x Row 1 Total (218) = Cell A = 119.25
4. Calculate the Expected Proportion of Individuals that died and were HG+
   Expected Proportion died (0.453) x Row 1 Total (218) = Cell B = 98.75.
5. Calculate the Expected Proportion of surviving individuals that were HG-
   Expected Proportion lived (0.547) x Row 2 Total (144) = Cell C = 78.768
6. Calculate the Expected Proportion of individuals that died that were HG-
   Expected Proportion that died 0(.453) X Row 2 Total = Cell D = 65.232

Thus, we have (Table 9)

Table 9. Expected values for Xiang et al (2001) data set (Table 6 and Table 8)

Now we are ready to calculate the Chi-Square Value

Recall the formula for the chi-square test

Table 10. Worked contingency table for Xiang et al (2001) data set (Table 6 and Table 8)

Cell
a	(103 – 87.768)2 = 78.768	7.4547
b	(41 – 65.232)2 = 65.232	9.002
c	(95 – 119.25)2 = 119.25	4.93134
d	(123 – 98.75)2 = 98.75	5.95506
		27.3452

Adding all the parts we have the chi-square test statistic . To proceed with the inference, we test using the chi-square distribution.

Determine the Critical Value of test and evaluate the null hypothesis

Now recall that we can get the critical value of this test in one of two (related!) ways. One, we could run this through our statistical software and get the p-value, the probability of our result and the null hypothesis is true. Second, we look up the critical value from an appropriate statistical table. Here, we present option 2.

We need the Type I error rate α and calculate the degrees of freedom for the problem. By convention, we set α = 0.05. Degrees of freedom for contingency table is calculated as

Degrees of Freedom = (# rows – 1) x (# columns – 1)

and for our example → DF = (2 – 1) x (2 – 1) = 1.

Note. How did we get the 1 degree of freedom? Aren’t there 4 categories and shouldn’t we therefore have k – 1 = 3 df? We do have four categories, yes, but the categories are not independent. We need to take this into account when we evaluate the test and this lack of independence is accounted for by the loss of degrees of freedom.

What exactly is not independent about our study?

1. 1. The total number of observations is set in advance from the data collection.
   2. We also have the column and row totals set in advance. The number of HIV individuals with or without Hepatitis G infection was determined at the beginning of the experiment. The number of individuals that lived or died was also determined before we we conducted the test.
   3. Then the first cell (let us make that cell A) can still vary anywhere from 0 to N _i (sample size of the first drug).
   4. Once the first cell (A) is determined then the next cell in that row (B) will have to add up to N _i.
   5. Also the other cell in the same column as the first cell (C) must add up to the Column 1 Total.
   6. Lastly, the last cell (D) must have the Row 1 Total add up to the correct number.

All this translates into there being only one cell that is FREE TO VARY, hence, only one degree of freedom.

Get the Critical Value from a chi-square distribution table: For our example, look up the critical value for DF = 1, α = 0.05, you should get 3.841. The 3.841 is the value of the chi-square we would expect at 5% and the null hypothesis is in fact true condition in the population.

Once we obtain the critical value we simply use the previous statement regarding the probability of the Null Hypothesis being TRUE. We Reject the Null Hypothesis when the calculated test statistic is greater than the Critical Value. We Accept the Null Hypothesis when the calculated  is less than the Critical Value. In our example we got 27.3452 for our calculated test statistic, thus we reject the null hypothesis and conclude, that, at least for the sample in this study, there was an association between the probability of patients living past the study period and the presence of hepatitis G.

from a, b, c, and d 2X2 table format formula

If a hand calculation is required, a simpler formula to use is

where N is the sum of all observations in the table, r₁ and r₂ are the marginal row totals, and c₁ and c₂ are the marginal column totals from the 2×2 contingency table (eg, Table 4). If you are tempted to try this in a spreadsheet, and assuming your entries for a, b, c, d, and N look like Table 11, then a straightforward interpretation requires referencing five spreadsheet cells no less than 13 times! Not to mention a separate call to the distribution to calculate the p-value (one tailed test).

Table 11. Example spreadsheet with formulas for odds ratio (OR), Pearson’s , and p-value from distribution.

	A	B	C	D	E	F	G
1
2	a	103			OR	=(2*5)/(3*4)
3	b	41
4	c	95
5	d	123			chisq, p	=(((2*5-3*4)^2)*6)/ (SUM(2:3)*SUM(4:5)* SUM(2,4)*SUM(3,5))	=CHIDIST(F5,1)
6	N	632

Now, let’s see how easy this is to do in R.

effect size

Inference on hypotheses about association, there’s statistical significance — evaluate p-value compared to Type I error, eg, 5% — and then there’s biological importance, the strength of the association. The concept of effect size is an attempt to communicate the likely importance of a result. Several statistics are available to communicate effect size: for that’s φ, phi for the φ (phi) coefficient.

where N is the total. Phi for our example is

R code

sqrt(27.3452/362)

and R returns

[1] 0.274844

Effect size statistics typically range from 0 to 1; Cohen (1992) suggested the following interpretation

Table 12. Cohen’s interpretation of d values.

For our example, effect size of the association between co-infection with GB virus C and mortality was weak in the patients with HIV infection. The function phi() in psych package will calculate phi coefficient from a 2X2 table.

Agreement, kappa, and inter-rater reliability

I imagine by now you have made the leap — can we use the contingency table to assess agreement between (inter-) two judges (raters) who evaluate the same object? A naive approach would be to calculate simple agreement, the percent of objects in the list of items evaluated for which both judges agree (concur). For a binomial Yes or No, then agreement (concordance) occurs when both judges report “Yes,” or both report “No,” displayed on the main diagonal of the 2X2 table. The other possible combinations (Yes vs No and No vs Yes) are disagreement (discordance) and would be displayed on the off-diagonal of the 2X2 matrix.

Unfortunately, simple agreement, while intuitive, cannot discriminate between true agreement and agreement by chance. Cohen’s kappa (Cohen 1960, McHugh 2012) is used to assess agreement, aka inter-rater reliability, for binomial and nominal data types for two judges (raters). Cohen’s 1960 paper has been enormously impactful, with more than 53K citations (Google Scholar, Fall 2024), and spawned a number of additional measures of agreement, eg, weighted kappa for more than binomial levels, Fleiss kappa for three or more raters, ICC for ratio scale, discussed in Ch12.3). It’s beyond the scope of this text to review the context of such terms as agreement, and reliability, for just two terms used to evaluate results from two independent observers of the “same” event or object, and leave that to the reader to confirm how to report such endeavors.

Off we go. Cohen’s kappa is defined as

where p_o is the proportion in agreement and p_c is proportion of agreement expected by chance. Essentially, then, interpretable kappa values fall between 0 and 1 (0% to 100%); negative kappa estimates are possible.  Cohen (1960) noted that for kappa to equal 1, the off diagonal elements would all have to be zero.

Cohen provided an estimate of the standard error for kappa as

and 95% confidence intervals are then .

Here, we review calculation of simple agreement and Cohen’s kappa, with just the two levels. Both simple agreement and Cohen’s kappa (κ), are available in R via the icc package. Cohen’s kappa is also available in the psych package, with the cohen.kappa() function providing a confidence interval estimate directly.

Use of both is illustrated here with a simulated coin toss example modified from a nice 2018 summary of Cohen’s kappa at r-bloggers. The R code

library(irr)
# simulate some data
N = 24
flips1 <- rbinom(n = N, size = 1, prob = 0.5)
flips2 <- rbinom(n = N, size = 1, prob = 0.5)
coins<-cbind(flips1, flips2)

Confirm the data set.

head(coins)  
flips1 flips2 
[1,] 0 1 
[2,] 1 0 
[3,] 1 0 
[4,] 1 1 
[5,] 0 0 
[6,] 1 0

and create a 2X2 table.

I stacked the data

head(StackedCoin)
Side Coins    HT
1 0 flips1 tails
2 1 flips1 heads
3 1 flips1 heads
4 1 flips1 heads
5 0 flips1 tails
6 1 flips1 heads

then created a simple 2×2 table (see Note 6 below for a simple explanation for why “.Table” as a named object).

.Table <- xtabs(~Coins+HT, data=StackedCoin)

Table 13. coins as 2X2 table.

We can calculate the phi coefficient for Table 13.

phi(.Table, digits = 3)
[1] -0.167

and would conclude no correlation between the two sets of coin tosses.

Now, agreement statistics. Both agree() and kappa2() expect an unstacked (wide) data format. It’s not necessary, but we’ll call a value of 1 as “heads up” and a value of 0 as “tails up” of the tossed coin.

Next, simple agreement.

agree(coins, tolerance=0)

R returns

Subjects = 24 
Raters = 2 
%-agree = 41.7

Now, a proper agreement statistic.

kappa2(coins) 

and R returns

Subjects = 24 
Kappa = -0.135
z = -0.7 
p-value = 0.484

Results with psych:cohen.kappa, which provides the 95% confidence interval as default output.

Cohen Kappa and Weighted Kappa correlation coefficients and confidence boundaries 
                 lower estimate upper
unweighted kappa -0.51    -0.14  0.24
weighted kappa   -0.51    -0.14  0.24
Number of subjects = 24

By chance, the two independent toss of 24 coins had 10 rows of “agreement” (H:H or T:T), simply due to chance. However, results of kappa find no agreement between the two sets of coin tosses.

In summary, we see that simple agreement and the kappa statistic don’t return similar results. Percent agreement reported 41%, whereas kappa was zero. We know percent agreement can’t reflect anything in the coin toss case because each toss is independent from the next. We’ll have you explore repeated trials of coin tosses to help convince you that simple agreement is not a reliable measure of agreement (see question 6 below).

Interpreting agreement values is not straightforward, and some argue agreement statistics should be avoided in favor of disagreement measures (for example, see discussion in Pontius and Millones 2011). Cohen suggested (Table 14)

Table 14. Landis and Koch’s (1977) interpretation of Cohen’s kappa.

Even a cursory look at Table 14 should lead to questions about scenarios in which a range of agreement “substantial” would provide little comfort about the agreement between the judges or the reliability of the instrument used, for example forensic cases or histology of tissue biopsies.

Contingency table analyses in Rcmdr

Assuming you have already summarized the data, you can enter the data directly in the Rcmdr contingency table form. If your data are not summarized, then you would use Rcmdr's Two-way table… for this. We will proceed under the assumption that you have already obtained the frequencies for each cell in the table.

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

Here you can tell R how many rows and how many columns (Fig 1).

Figure 1. Screenshot R Commander menu for 2X2 data entry with counts.

The default is a 2X2 table. For larger tables, use the sliders to increase the number of rows, columns, or both.

Next, enter the counts. You can edit every cell in this table, including the headers. In the next panel, I will show you the data entry and options. The actual calculation of the test statistic is done when you select the “Chi-square test of independence” check-box (Fig 2).

Figure 2. Display of Xiang et al data entered into R Commander menu.

After entering the data, click on the Statistics tab (Fig 3).

Figure 3. Screenshot Statistics options for contingency table.

If you also select “Components of chi-square statistic” option, then R will show you the contributions of each cell (O-E) towards the chi-square test statistic value. This is helpful to determine if rejection of the null is due to a subset of the categories, and it also forms the basis of the heterogeneity tests, a subject we will pick up in the next section.

Here’s the R output from R Commander. Note the “2, 2, byrow=TRUE” instructions (check out R help to confirm what

.Table <- matrix(c(103,41,95,123), 2, 2, byrow=TRUE)
dimnames(.Table) <- list("rows"=c("HG+", "HG-"), "columns"=c("Lived", "Died"))

.Table # Counts
     columns
rows Lived Died
 HG+   103   41
 HG-   95   123

.Test <- chisq.test(.Table, correct=FALSE)
.Test

Pearson's Chi-squared test

data: .Table
X-squared = 27.339, df = 1, p-value = 0.0000001708

round(.Test$residuals^2, 2) # Chi-square Components
    columns
rows Lived Died
HG+   7.46 9.00
HG-   4.93 5.95

remove(.Test)
remove(.Table)

end R output.

Lots of output, take a deep breath and remember… The minimum output we need to look at is… (Table 15)?

Table 15. Minimally expected outcome for report from a statistical test.

We can see that the p-value 1.7^-7 is much less than Type 1 α = 0.05, thus, by our decision criterion we reject the null hypothesis (provisionally of course, as science goes).

myNumber = 0.0000001708
format(myNumber, scientific = TRUE, digits=2)

[1] "1.7e-07"

Questions

1. Instead of R Commander, try the contingency table problem in R directly.

   myData <- matrix(c(103,41,95,123))
   chisq.test(myData)

   Is this the correct contingency table analysis? Why or why not?

2. For many years National Football League games that ended in ties went to “sudden death,” where the winner was determined by the first score in the extra period of play, regardless of whether or not the other team got an opportunity to possess the ball on offense. Thus, in more than 100 games (140), the team that won the coin toss and therefore got the ball first in overtime, won the game following either a kicked field goal or after a touchdown was scored. In 337 other games under this system, the outcome was not determined by who got the ball first. Many complained that the “sudden death” format was unfair and in 2013 the NFL changed its overtime rules. Beginning 2014 season, both teams got a chance to possess the ball in overtime, unless the team that won the coin toss also went on to score a touchdown at which time that team would be declared the winner. In this new era of over-time rules ten teams that won the coin toss went on to score a touchdown in their first possession and therefore win the game, whereas in 54 other overtime games, the outcome was decided after both teams had a chance on offense (data as of 1 December 2015). These data may be summarized in the table
   

   	First possession win?
   	Yes	No
   Coin flip years	140	337
   New era	10	54

   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?

3. Return to the Doll and Hill (1950) data: 2 men with lung cancer were nonsmokers, 647 men with lung cancer were cigarette smokers. In comparison, 27 case-control patients (ie, patients in the hospital but with other ailments, not lung cancer) were nonsmokers, but 622 were cigarette smokers.
   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?
4. A more recent example from the study of the efficacy of St John’s Wort as a treatment for major depression (Shelton et al 2001). Of patients who received St. John’s Wort over 8 weeks, 14 were deemed improved while 98 did not improve. In contrast 5 patients who received the placebo were deemed improved while 102 did not improve.
   A. What is the null hypothesis?
   B. Which is more appropriate: to calculate an odds ratio or to calculate an RRR?
   C. This is a contingency table problem. Explain why
   D. Conduct the test of the null hypothesis.
   E. What is the value of the test statistic? Degrees of freedom? P-value?
   F. Evaluate the results of your analysis — do you accept or reject the null hypothesis?
5. Return to the coin toss simulation and repeat comparisons between simple agreement and Cohen’s kappa for different values of N. Do you see any trend(s) for the two measures as sample size increases or decreases?
6. Repeat the coin toss simulation for N = 24 up to five times, each time calculate simple agreement and Cohen’s kappa. Provide a table of your five results plus the example, and generalize simple agreement vs Cohen’s kappa as a reliable measure of agreement. Hint: you may want to use Descriptive statistics to help make your case.

Quiz Chapter 9.2

Chi-square contingency tables

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Chapter 9 contents

• Introduction
• Chi-square test: Goodness of fit
• Chi-square contingency tables
• Yates continuity correction
• Heterogeneity chi-square tests
• Fisher exact test
• McNemar’s test
• References and suggested readings

9 – Inferences, Categorical Data

Introduction

We introduced the chi-square distribution in 6.9 – Chi-square distribution. No doubt you have already been introduced to chi-square () tests (click here correct pronunciation), particularly if you’ve had a genetics class, but perhaps you were not told why you were using the   test, as opposed to some other test, for example t-test, ANOVA, or linear regression.

Chi-square analyses are used in situations of discrete (ie, categorical or qualitative) data types. When you can count the number of “yes” or “no” outcomes from an experiment, then you are talking about a problem. In contrast, continuous (ie, quantitative) data types for outcome variables would require you to use the t-test (for two groups) or the ANOVA-like procedures (for two or more groups). Chi-square tests can be applied when you have two or more treatment groups.

Two kinds of chi-square analyses

(1) We ask about the “fit” of our data against predictions from theory. This is the typical chi-square that student’s have been exposed to in biology lab. If outcomes of an experiment can be measured against predictions from some theory, then this is a goodness of fit (gof) . Goodness of fit is introduced in Ch09.1.

(2) We ask whether the outcomes of an experiment are associated with a treatment. These are called contingency table problems, and they will be the subject of the next lecture. The important distinction here is that there exists no outside source of information (“theory”) available to make predictions about what we would expect. Contingency tables are introduced in Ch09.2.

Suggested homework

Homework 6, includes BONUS

Quizzes in this chapter

A total of 45 questions among the several subchapters, a mix of true or false and multiple choice question format.

Chapter 9 contents

• Introduction
• Chi-square test: Goodness of fit
• Chi-square contingency tables
• Yates continuity correction
• Heterogeneity chi-square tests
• Fisher exact test
• McNemar’s test
• References and suggested readings

7.1 – Epidemiology definitions

Introduction

This subchapter may lack for drama, but let’s start by providing a list of key terms with definitions as you start our introduction to epidemiology. More definitions will follow in the sections as well.

Definitions

Absolute risk: The probability that a specified event will occur in a specified population. See Ch07.4 – Epidemiology: Relative risk and absolute risk, explained.

Absolute risk reduction (ARR): the decrease in risk of an event in an exposed (treatment) group compared to an unexposed (control) group. Also called the risk difference. , see Contingency table. See Ch07.4 – Epidemiology: Relative risk and absolute risk, explained.

: Control Event Rate and Experimental Event Rate. In observational studies, transition from “control” and “experimental” to exposure terms. For example, prospective or retrospective cohort studies, is Exposed Event Rate and its partner , Unexposed Event Rate. Equivalently, you will encounter for Incidence in the Exposed and forIncidence in the Unexposed. For Case-Control studies, where the direct event (incidence) rates over time are unavailable, then becomes Exposure Rate in Cases — the proportion of people with the condition who were exposed to the risk factor and is now Exposure Rate in Controls.

Contingency table, also called cross tabulation or crosstab, is a display of counts of variables in a matrix format. We briefly introduced the contingency table in Chapter 5.1. In epidemiology, rows of contingency table represent treatment or exposure groups, columns represent outcomes.

Table 1. A 2 X 2 contingency table

The 2X2 contingency table is referred to frequently in this chapter and again in Chapter 9.

R code

a = 4; b = 46; c = 5; d = 45

Table1 <- matrix(c(a,b,c,d), 2, 2, byrow=TRUE, dimnames = list(c("Treatment", "Control"), c("Yes", "No"))); Table1

R output

          Yes  No
Treatment   4  46
Control     5  45

In statistical classification, a field dedicated to establishing algorithms for correctly identifying and predicting classes of data, eg, correct disease diagnosis or , the 2X2 contingency table is often called a confusion or error matrix (Greene 2003). Machine learning methods are often built on developments from statistical classification findings (Michie et al 1995). We expand on contingency tables in Chapter 7.4 and Chapter 9.2.

Control event rate (CER): How often an event occurs in the control group.  , see Contingency table.

Diagnosis: identification of the nature of a disease or condition.

Event: From probability theory, an event is a set of one or more outcomes to which a probability is assigned to each outcome.

Experimental event rate (EER): How often an event occurs in the treatment group. , see Contingency table.

Exposure risk: the potential for harm from contact with or vulnerability to a hazard or risk factor.

Hazard: anything that can cause harm.

Incidence: the number of newly diagnosed individuals in a population having a condition, disease or other characteristic. Compare to prevalence.

Likelihood: In every-day English, the term likelihood and probability mean the same thing — the chance that an event will occur. In statistics, however, probability refers to possible outcomes in the future — a prediction for an event to occur, whereas likelihood is associated with the possible explanation (hypothesis) for the outcome that has already occurred, the past. Put another way, likelihood quantifies how likely is the data we observe, given the hypothesis (cf. excellent discussion by Gallistel 2015)? We will look at likelihood methods in practice in subsequent chapters.

Negative predictive value of a test (NPV), defined as the probability that a negative test result identifies a person who truly does not have the disease. Calculated as the total number of individuals without the disease divided by the total that tested negative. 

Number needed to treat (NNT): the inverse of the absolute risk reduction. . See Ch07.4 – Epidemiology: Relative risk and absolute risk, explained.

Odds, the ratio (OR) of two probabilities: the probability of getting a one on throwing a dice is , the probability of not getting a one is , therefore the odds of getting a one are 1 to 5.  . See Ch07.5 – Odds ratio.

Outcome, the single result from a random experiment. For example, result of a single coin toss is “heads.” Results of a birth, the outcome is the child born is a singleton, single birth.

Per capita rate, Latin phrase, for each head, meaning per person.

Positive predictive value of a test (PPV), defined as the probability that a positive test result identifies a person who truly has the disease. Calculated as the total number of individuals with the disease divided by the total that tested positive.

Posttest probability refers to the probability that the patient has the disease after the results of the test are known.

Pretest probability is the prevalence of the disease, ie, the chance that the a randomly selected person from the population has the disease.

Prevalence: The proportion of individuals in a population having a condition, disease, or characteristic. Compare to incidence.

Probability, defined in Chapter 6.1.

Prognosis, how a disease plays out.

Relative risk: Ratio of the risk of an event among those exposed to the risk factor to the risk among those not exposed to the risk factor. See Ch07.4 – Epidemiology: Relative risk and absolute risk, explained.

Relative risk reduction (RRR): is a measure calculated by dividing the absolute risk reduction by the control event rate. See Ch07.4 – Epidemiology: Relative risk and absolute risk, explained.

Risk: Probability of an event. Risk is not restricted to just bad events, but refers to the uncertainty of a particular event (eg, the risk that a child will be born male seems a melodramatic statement, but it is accurate as far as this definition goes).

Therapy, treatment intended to treat, relieve, or cure a disorder of condition.

Questions

1. Compare and contrast ARR and RRR.
2. What’s the difference between event, hazard, and risk?
3. What’s the difference between incidence and prevalence?
4. What’s the difference between diagnosis and prognosis?
5. What’s the implication of a NNT greater than 100 in terms of the utility of a proposed therapy or treatment?

Quiz

Quiz Chapter 7.1

Epidemiology definitions

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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