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

7.4 – Epidemiology: Relative risk and absolute risk, explained

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.

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),

and the cumulative incidence in the exposed (treated) group, CIe, is , and cumulative incidence in the unexposed (control) group, CIu, is . We can calculate another statistic called the risk ratio,

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

Relative risk reduction, or the absolute risk reduction divided by the proportion of patients who die without treatment, is

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

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 () 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

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

The 95% CI for ARR is approximately

For the Wald estimate, replace the 2 with , which comes from the normal table for z at

(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

and the inverse for NNT CI is

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).

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. 

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).

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.

my_tibble <- as_tibble(my_dataframe)

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).

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).

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

Software: RcmdrPlugin.EBM

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).

Figure 3. Rcmdr: Tools → Load Rcmdr plugins…

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

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)

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).

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

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

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.

1. What was the null hypothesis? Be able to write the hypothesis in symbolic form and as a single sentence.
2. What was the alternative hypothesis? Be able to write the hypothesis in symbolic form and as a single sentence.
3. What was the observed frequency of subjects with esophageal cancer in this study? And the observed frequency of subjects without esophageal cancer?
4. Estimate Relative Risk, Absolute Risk, NNT, and Odds ratio?
   1. Which is more appropriate, RR or OR? Justify your decision.
5. 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?
6. “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

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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