20.15 — Meta-analysis

Introduction

Meta-analysis is a type of systematic review (Greenhalgh 1997, Akobeng 2005), a statistical method used to combine the results from multiple independent studies that addressed the same topic. By combining studies, the goal is to calculate a pooled estimate as a measure of overall effectiveness of the treatment. The underlying logic is that multiple studies draw subjects from the same reference population, thus, combining data provides a more powerful test of the hypothesis.

Meta-analysis can increase the overall sample size, which improves the study’s ability to detect an effect if one exists. So, how to combine results from different studies ostensibly addressing the same hypothesis? Inevitably, some subjectivity will come into decisions about including or excluding studies — the goal is to identify quality studies, which may have conflicting conclusions and test whether a single conclusion of effectiveness is warranted based on a summary of evidence.

Meta-analysis is now a routine type of study, conducted in multiple disciplines, including biomedical field (eg, Cochrane Reviews, more than 15K since 1993, Anderson et al 2024), and environmental studies (eg, Resende et al 2021), and sports science (eg, Miyamoto-Mikami et al 2018).

In general, meta-analysis studies return the combined effect size –the single, pooled effect across all included studies — of an intervention and the heterogeneity among the studies — how consistent measured differences. Models may be fixed effects — that is, any observed differences among studies are due to chance or sampling error — or random effects — the true effect size varies from study to study and that these effects are drawn from a distribution.

The purpose of this page is to provide a basic overview and an introduction to the meta-analysis approach. We previously introduced the Cochran’s Q test aka heterogeneity test in Chapter 9.4. The chi-square test test asks whether results from combined studies are consistent. After discussing inclusion criteria, we provide plotting techniques and additional statistical tests for heterogeneity of studies.

In Chapter 12.4, we introduced a simple meta-analysis combining ten separate one sample t-tests of the hypothesis that inbred mice and genetically diverse mouse strains do not differ for average lifespan. P-values were combined and we used Fisher’s method to evaluate the hypothesis.

Selection criteria

Any project, whether an observational study or an experiment, starts with a question that is modified into one or more testable hypothesis. We don’t collect the data then speculate about causes for perceived patterns that may appear. Rather, pattern recognition process can be used to generate hypotheses subject to future testing. Our example was about lifespan and genetic background and genetic diversity: Longevity refers to the capacity to live a long life whereas lifespan is the maximum amount of time an individual can live. We ask if genetic diversity increases longevity across a population; specific genetic background is linked to an individual’s predisposition to long life.

Given the importance such work has in clinical research, as you can imagine, there is plenty of advice and suggested guidelines for how to proceed. Of note, review systematic review Population, Intervention, Comparison, Outcome (PICO) guidelines (Schardt et al 2007). Additionally, see the Cochrane Handbook (2024).

With a clear hypothesis in hand, for both systematic review and meta-analysis, the study begins with a search of the published literature. Search results from the query “inbred mice strain lifespan,” in Pubmed returned about 1134 publications; replacing “longevity” for “lifespan,” returned 654 publications. We want primary literature, not reviews, so we add NOT (Review[Publication Type]) to the query: 1080 and 617, respectively. Further restricting the pool by restricting range to last ten years returned a manageable 197 and 116 results, respectively. We can begin selecting our data from the studies. We now have to decide whether all or only some meet our standards for consideration. Of course, this is not the same thing as looking at the results and selecting only the studies that confirm our thinking! Ideally, reporting of both positive (large effect size, small p-value) and negative (small effect size, large p-value) studies can — and should — be included. Our goal, combine the results of multiple independent studies on the same topic to reach a more powerful and reliable conclusion than a single study could provide.

Inclusion criteria are the requirements a study must meet to be considered for the review; exclusion criteria are the specific reasons a study is disqualified. Decisions about which studies to include consider demographics, intervention type, and study design. For example, the review may focus only on RCT studies. Likely, the review is focused on experimental work and so, excluding  previous systematic reviews would be an example of an exclusion criteria. At the same time we need to be transparent about our approach; that includes defining a clear research question, establishing strict inclusion/exclusion criteria, conducting a comprehensive literature search across multiple databases, and performing a rigorous data extraction and quality assessment of included studies (Meline 2006, Paldam 2015, Cochrane Handbook). For our lifespan and genetic diversity example I found two studies that met my criteria: the studies needed multiple inbred strains or outbred strains and the raw data needed to be available. Two studies met this need: Yuan et al (2009) and Mullis et al (2025).

An imperfect example. To practice applying selection and exclusion criteria for a meta-analysis, I ran a PubMed search using the keywords botox migraine. I limited the results to clinical trials published within the past five years, which gave me 14 papers. I read the abstracts and decided to include studies only if they had a control group, focused on patients with chronic migraine, and reported an exact p-value instead of just saying “P < 0.05.” Table 1 shows what I found, including the PubMed link for each paper, whether it was included, the sample size, the p-value, and my comments on why any studies were excluded.

Table 1. Example data collection for meta-analysis, topic botox and treatment of chronic migraines.

That leaves only three papers, not enough for a meta-analysis. By changing years from five to ten, 40 papers were available to search. By removing all time restrictions, 109 papers were returned since year 2000. Medical use of botox dates to the 1970s (Scott 2023).

R packages

metafor package: provides extensive functions for calculating various effect sizes, fitting different statistical models (e.g., fixed- and random-effects, mixed-effects), conducting meta-regression, and generating numerous meta-analytical plots.

See also metaforGUI package which provides a Graphical User Interface (GUI) for the R metafor Package.

meta package: provides user-friendly and comprehensive collection of functions for conducting standard meta-analyses.

Much has been written about how to conduct meta-analysis and caveat emptor — readers should be on notice that I present here just a fraction of the “how to” and pitfalls of meta-analysis. Interested readers should see Doing meta-analysis with R: A hands-on guide by M. Harrer et al (2021).

I² or I-squared test

In a meta-analysis, the I-squared test quantifies the percentage of total variation in a set of study results that is due to differences between studies rather than random sampling error (chance). An of 0% means all variation is due to chance, while a higher percentage indicates that differences between studies are a significant source of variation. As a reminder, the test returns a p-value that can be used to interpret whether that heterogeneity is statistically significant. We run our life span of inbred and genetically diverse outbred mouse strains example in Chapter 12.4; the p-value was small — we used Fisher’s method p-value combination approach to test the main hypothesis.

Here, we calculate ; we first need the effect size; Cohen’s d (Chapter 11.4) will do.

Table 1. Updated Table 1 from Chapter 12.4.

The effect size ranged from practically no difference (Cohen’s d of 4.5%) to substantial effect on average lifespan of inbred vs outbred strains (Cohen’s d of 126%). The simple average of the Cohen’s d values was 0.541;  a weighted average accounting for differences in variance needs to be applied. Regardless, the statistic we want is the coefficient from meta-regression, which indicates how a change in that predictor relates to the effect size.

With effect size calculated, we proceed to get . We use rma() from metafor package. This function performs several tests: a fixed-effects model and therefore we test a model where we assume all studies found the exact same effect, a random-effects model where the assumption is that effects are similar but vary across studies, plus a mixed-effects model uses a combination of both, and finally, a meta-regression. The meta-regression uses the data from those studies to try and explain why the results differ.

R code

library(metafor)
strains <- 1:10
my_effsize <- c(0.0446092, 1.258065, 0.3891731, 0.8390583,
0.5354427, 0.3302039, 0.9431162, 0.192082, 0.6626776, 0.2062765)
my_sample <- c(32, 32, 32, 24, 29, 29, 30, 26, 32, 31)
my_varC <- (my_effsize^2)/(2*my_sample)
myData <- data.frame(strains, my_effsize, my_sample, my_varC)
rma(yi = my_effsize, vi = my_varC, data = myData)

R output

Random-Effects Model (k = 10; tau^2 estimator: REML)

tau^2 (estimated amount of total heterogeneity): 0.1299 (SE = 0.0643)
tau (square root of estimated tau^2 value): 0.3604
I^2 (total heterogeneity / total variability): 99.15%
H^2 (total variability / sampling variability): 117.11

Test for Heterogeneity:
Q(df = 9) = 390.9415, p-val < .0001

Model Results:

estimate     se   zval   pval  ci.lb  ci.ub
  0.5208 0.1169 4.4558 <.0001 0.2917 0.7499 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Interpretation.

With an of 99%, which suggests a very high degree of heterogeneity among the comparison between the DO strain and the ten inbred strains for life span. As before, the heterogeneity or Q test was highly statistically significant, suggests that the observed variability is larger than what would be expected due to chance alone. Thus, all of the variability in the observed effects comes from between-strain differences, not random chance — it also justifies use of a random effects model, not a fixed model. Strain A/J, with the greatest difference in lifespan compared to the DO strain, is recognized for its susceptibility to carcinogenic tumors

From the meta regression, the coefficient was 0.52 — inbred strain membership was associated with change in the size of the effect. We conclude that in general, the genetically inbred strains differ substantially for average lifespan compared to the genetically variable DO strain: The true average effect was somewhere between 0.2917 and 0.7499.

Our results conform to our expectations — longevity is likely affected by a combination of effects from many genes along with an individual’s genetic background and interaction with the environment, cf discussion in Mullis et al (2025).

Forest plot of effect size.

A forest plot visually summarizes results from multiple studies, showing each study’s effect size (eg, odds ratio, mean difference, Cohen’s d) and variance or confidence interval as squares and lines, plus the overall pooled estimate (diamond). It helps quickly assess study consistency (heterogeneity), see individual study contributions, and determine if the combined evidence points to a significant overall effect.

For example, our inbred strains vs the outbred DO strain for lifespan shows a clear, overall effect (Fig 1). The diverse genetic strain lifespan is significantly greater than the majority of inbred strains, and the combined effect is about 0.5.

R code

library(metafor)
myData <- data.frame(
Strains = paste("Strains", c("129S1/SvImJ", "A/J", "BALB/cByJ", "BUB/BnJ",
"C3H/HeJ", "C57BL/6J", "CBA/J", "FVB/NJ", "P/J", "SWR/J")),
cohenD = c(0.045, 1.26, 0.389, 0.839, 0.536, 0.33, 0.943, 0.192, 0.663, 0.206),
cohenV = c(0.1910, 0.0400, 0.0368, 0.0485, 0.0403, 0.0399, 0.0405, 0.0438, 0.0374, 0.0376)
)

res2 <- rma(cohenD, cohenV, data = myData, method="REML")
# str(res2)

forest(res2,
slab = myData$Strains,
xlab = "Cohen's d")

Figure 1. Forest plot, Cohen’s effect size lifespan differences among inbred strains of mice compared to outbred strain.

Questions

1. What’s the difference between a literature review and a systematic review?

2. For which kinds of research goals would a meta-analysis be more appropriate than a systematic review?

Quiz Chapter 20.15

Meta-analysis

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References and suggested readings

Akobeng, A. K. (2005). Understanding systematic reviews and meta-analysis. Arch Dis Child, 90:845-848.

Andersen, M. Z., Zeinert, P., Rosenberg, J., & Fonnes, S. (2024). Comparative analysis of Cochrane and non-Cochrane reviews over three decades. Systematic Reviews, 13, 120.

Cochrane reviews, Cochrane Library. (2025). https://www.cochranelibrary.com/

Greenhalgh, T. (1997). How to read a paper: Papers that summarise other papers (systematic reviews and meta-analyses). BMJ, 315(7109), 672–675.

Gurevitch, J., Koricheva, J., Nakagawa, S., & Stewart, G. (2018). Meta-analysis and the science of research synthesis. Nature, 555(7695), 175–182.

Hansen, C., Steinmetz, H., & Block, J. (2022). How to conduct a meta-analysis in eight steps: A practical guide. Management Review Quarterly, 72(1), 1–19.

Harrer, M., Cuijpers, P., Furukawa, T.A., & Ebert, D.D. (2021). Doing Meta-Analysis with R: A Hands-On Guide. Boca Raton, FL and London: Chapman & Hall/CRC Press. Link to website.

Ioannidis, J. P. A. (2008). Why Most Discovered True Associations Are Inflated. Epidemiology, 19(5), 640.

Higgins JPT, Thomas J, Chandler J, Cumpston M, Li T, Page MJ, Welch VA (editors). Cochrane Handbook for Systematic Reviews of Interventions version 6.5 (updated August 2024). Cochrane, 2024. Available from www.cochrane.org/handbook.

Meline, T. (2006). Selecting studies for systemic review: Inclusion and exclusion criteria. Contemporary Issues in Communication Science and Disorders, 33(Spring), 21–27.

Miyamoto-Mikami, E., Zempo, H., Fuku, N., Kikuchi, N., Miyachi, M., & Murakami, H. (2018). Heritability estimates of endurance-related phenotypes: A systematic review and meta-analysis. Scandinavian Journal of Medicine & Science in Sports, 28(3), 834–845.

Mullis, M. N., Wright, K. M., Raj, A., Gatti, D. M., Reifsnyder, P. C., Flurkey, K., Archer, J. R., Robinson, L., Di Francesco, A., Svenson, K. L., Korstanje, R., Harrison, D. E., Ruby, J. G., & Churchill, G. A. (2025). Analysis of lifespan across diversity outbred mouse studies identifies multiple longevity-associated loci. Genetics, 230(4), iyaf081.

Nakagawa, S., & Cuthill, I. C. (2007). Effect size, confidence interval and statistical significance: A practical guide for biologists. Biological Reviews, 82(4), 591–605.

Paldam, M. (2015). Meta-analysis in a nutshell: Techniques and general findings. Economics, 9(1), 20150011.

Resende, P. S., Viana-Junior, A. B., Young, R. J., & Azevedo, C. S. (2021). What is better for animal conservation translocation programmes: Soft- or hard-release? A phylogenetic meta-analytical approach. Journal of Applied Ecology, 58(6), 1122–1132.

Schardt, C., Adams, M. B., Owens, T., Keitz, S., & Fontelo, P. (2007). Utilization of the PICO framework to improve searching PubMed for clinical questions. BMC Medical Informatics and Decision Making, 7, 16.

Scott, A. B., Honeychurch, D., & Brin, M. F. (2023). Early development history of Botox (onabotulinumtoxinA). Medicine, 102(Suppl), e32371.

van Aert, R. C. M., Wicherts, J. M., & van Assen, M. A. L. M. (2016). Conducting Meta-Analyses Based on p Values. Perspectives on Psychological Science, 11(5), 713–729.

Yuan, R., Tsaih, S.-W., Petkova, S. B., de Evsikova, C. M., Xing, S., Marion, M. A., Bogue, M. A., Mills, K. D., Peters, L. L., Bult, C. J., Rosen, C. J., Sundberg, J. P., Harrison, D. E., Churchill, G. A., & Paigen, B. (2009). Aging in inbred strains of mice: Study design and interim report on median lifespans and circulating IGF1 levels. Aging Cell, 8(3), 277–287.

Chapter 20 contents

• Additional topics
• Area under the curve
• Peak detection
• Baseline correction
• Surveys
• Time series
• Dimensional analysis
• Estimating population size
• Diversity indexes
• Survival analysis
• Growth equations and dose response calculations
• Plot a Newick tree
• Phylogenetically independent contrasts
• How to get the distances from a distance tree
• Binary classification
• Meta-analysis

3.5 – Statistics of error

Making sense of error in data.

This page explains how statistics of error help quantify differences between measured and true values, beginning with classical methods like standard error and confidence intervals that rely on large-sample assumptions. Related, statisticians talk about uncertainty — the quantification of doubt about a measurement. We introduce permutation-based approaches, including jackknife and bootstrap resampling, which avoid those assumptions and are especially useful with smaller or complex data sets. Observer variation (intra- and interobserver error) is also highlighted as an important source of measurement error.

Missing values are another common source of uncertainty in biological data sets. Data may be missing because of equipment failure, recording mistakes, sample loss, or biological factors that prevent a measurement from being collected. Even a small number of missing observations can influence summaries, graphs, and statistical tests, especially in smaller studies. In this chapter, we treat missing values mainly as a practical issue to identify and report during exploratory data analysis. Later, in the chapter on experimental design, we examine missing data in greater depth, including differences between random and non-random missingness, as well as approaches such as data imputation and the consequences these choices may have for interpretation.

We conclude with a discussion about confidence intervals as a measure of accuracy and emphasize proper use of significant figures to avoid overstating precision.

Statistics of error.

In statistics, error doesn’t mean mistake, an avoidable human misjudgement or oversight, eg, a typo. An error in statistics means there was a difference between the measured value and the actual value for an object. Efforts by many authors of notable classical statistics developed a large body of calculated statistics, eg, standard error the mean, which allows the user to quantify how large the error of measurement is given assumptions about the distribution of the errors. Thus, classical statistics requires user to make assumptions about the error distribution, the subject of our Chapter 6. A critical issue to understand is that these methods assume large sample sizes are available; they are called asymptotic statistics or large-sample statistics; the properties of the statistical estimates are evaluated as sample size approaches infinity.

Jackknife sampling and bootstrap sampling are permutation approaches to working with data when the Central Limit Theorem — as sample size increases, the distribution of sample means will tend to a normal distribution (see Chapter 6.7) — is unlikely to apply or, rather, we don’t wish to make that assumption (Chapter 19). The jackknife is a sampling method involving repeatedly sampling from the original data set, but each time leaving one value out. The estimator, for example, the sample mean, is calculated for each sample. The repeated estimates from the jackknife approach yield many estimates which, collected, are used to calculate the sample variance. Jackknife estimators tend to be less biased than those from classical asymptotic statistics.

Bootstrapping, and not jackknife resampling, may now be the preferred permutation approach (eg, Google Scholar search “bootstrap statistics” 36K hits; “jackknife statistics” 17K hits), but which method is best depends on qualities of the data set. Bootstrapping involves large numbers of permutations of the original data, which, in short, means we repeatedly take many samples of our data and recalculate our statistics on these sets of sampled data. We obtain statistical significance by comparing our result from the original data against how often results from our permutations on the resampled data sets exceed the originally observed results. By permutation methods, the goal is to avoid the assumptions made by large-sample statistical inference, ie, reaching conclusions about the population based on samples from the population. Since its introduction, “bootstrapping” has been shown to be superior in many — but, importantly, not all — cases for statistics of error compared to the standard, classical approach (Efron 1979, Hesterberg 2015).

There are many advocates for the permutation approaches, and, because we have computers now instead of the hand calculators our statistics ancestors used, permutation methods may be the approach you will take in your own work. However, the classical approach has its strengths — when the conditions, that is, when the assumptions of asymptotic statistics are met by the data, then the classical approaches tend to be less conservative than the permutation methods. By conservative, statisticians mean that a test performs at the level we expect it to. Thus, if the assumptions of classical statistics are met they return the correct answer more often than do the permutation tests.

Error and the observer.

Individual researchers make observations, therefore, we can talk about observer variation as a kind of error measurement. For repeated measures of the same object by an individual, we would expect the individual to return the same results. To the extent repeated measures differ, this is intraobserver error. In contrast, measures of the same object from different individuals is interobserver error. For a new instrument or measurement system, one would need to establish the reliability of the measure: confronted with the same object do researchers get the same measurement? Accounting for interobserver error applies in many fields, eg, histopathology of putative carcinoma slides (Franc et al 2003), liver biopsies for cirrhosis (Rousselet et al 2005), blood cell counts (Bacus 1973).

Confidence in estimates.

A really useful concept in statistics is the idea that you can assign how confident you are to an estimate. This is another way to speak of the accuracy of an estimate. Clearly, we have more confidence in a sample estimate for a population parameter if many observations are made. Another factor in our ability to estimate is the magnitude of observation differences. In general, the larger the differences among values from repeated trials, the less confident we will be in out estimate, unless, again, we make our estimates from a large collection of observations. These two quantities, sample size and variability, along with our level of confidence, eg, 95%, are incorporated into a statistic called the confidence interval.

We will use this concept a lot throughout the course; for now, a simple but approximate confidence interval is to use the 2 x SEM rule (as long as sample size large): twice the standard error of the mean. Take your estimate of the mean, then add (upper limit) or subtract (lower limit) twice the value of the standard error of the mean (if you recall, that’s the standard deviation divided by the square-root of the sample size).

Example. Consider five magnetic darts thrown at a dart board (28 cm diameter, height of 1.68m from the floor) from a distance of 3.15 meters (Fig 1).

Figure 1. Magnetic dart board with 5 darts. Click image to view full sized image.

The distance in centimeters (cm) between where each of the five darts landed on the board compared to the bullseye is reported in (Table 1),

Table 1. Results of five darts thrown at a target (Figure 2)*

* distance in centimeters from the center. 

How precise was the dart thrower? We’ll use the coefficient of variation as a measure of precision. Second, how accurate were the throws? Use R to calculate

darts = c(7.5, 3.0, 1.0, 2.7, 7.4)
#use the coefficient of variation to describe precision of the throws
coefVar = 100*(sd(darts)/mean(darts)); coefVar
[1] 68.46141

Confidence Interval to describe accuracy.

For a game of darts, the true value would be a distance of zero — all bullseyes. We need to calculate the standard error of the mean (SEM), then, we calculate the confidence interval around the sample mean.

#Calculate the SEM
SEM <- sd(darts)/sqrt(length(darts)); SEM
[1] 1.322649
#now, get the lower and upper limit, subtract from the mean
confidence <- c(mean(darts)-2*SEM, mean(darts), mean(darts)+2*SEM); confidence
[1] 1.674702, 4.320000, 6.965298

The mean was 4.3 cm, therefore to get the lower limit of the interval subtract 2.65 (2*SEM = 2.645298) from the mean; for the upper limit add 2.65 to the mean. Thus, we report our approximate confidence interval as (1.7, 7.0), and we read this as saying we are about 95% confident the population value is between these two limits. Five is a very small sample number, so we shouldn’t be surprised to learn that our approximate confidence interval would be less than adequate. In statistical terms, we would use the t-distribution, and not the normal distribution, to make our confidence interval in cases like this.

We made a quick calculation of the confidence interval; we can get make this calculation by hand by incorporating the t distribution. We need to know the degrees of freedom, which in this case is 4 (n – 1, where n = 5). We look up critical value of t at 5% (to get our 95% confidence interval), t = 2.78. Subtract for lower limit t*SEM and add for upper limit t*SEM to the sample mean for the 95% confidence interval.

We can get help from R, by using the one-sample t-test with a test against the hypothesis that the true mean is equal to zero. Among a variety of output we will introduce later (see Chapter 8 and 10), the confidence interval is reported (yellow highlight),

#make an attach a data frame object
Ddarts <- data.frame(darts)
t.test(darts,mu=0)
One Sample t-test
data: darts
t = 3.2662, df = 4, p-value = 0.0309
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 0.6477381 7.9922619
sample estimates:
mean of x 
 4.32

t.test uses the function qt(), which provides the quantile function. To recreate the 95% CI without the additional baggage output from the t.test, we would simply write

#upper limit
mean(darts)+ qt(0.975,df=4)*sd(darts)/sqrt(5)

#lower limit
mean(darts)+ qt(0.025, df = 4, lower.tail=TRUE)*sd(darts)/sqrt(5)

where sd(darts)/sqrt(5) is the standard error of the mean.

Or, alternatively, download and take advantage of a small package called Rmisc (not to be confused with the RcmdrMisc package) and use the function CI

library(Rmisc)
CI(darts)
     upper     mean     lower 
7.9922619 4.3200000 0.6477381

The advantage of using the CI() command from the package Rmisc is pretty clear; I don’t have to specify the degrees of freedom or the standard error of the mean. By default, CI reports the 95% confidence interval. we can specify any interval simply by adding to the command. For example, 

CI(darts, ci=0.90)

reports upper and lower limits for the 90% confidence interval.

Try it. On your own, get the 90% confidence interval and the 99% confidence intervals for our dart example. Compare to the 95% confidence interval — what do you conclude about “confidence” in our estimate from 90% to 99%?

Confidence intervals from p-values.

And finally, it is not unusual for a report to provide summary statistics, differences between means (ie, effect size), and p-values (orange highlight), but not report confidence intervals. We discuss p-values and the concept of statistical significance beginning in Chapter 8. But for now, we show how to calculate approximate confidence intervals from the p-value. This may be of use for conducting meta-analysis (Chapter 20.15). See two helpful “how to” articles by Bland and Altman in 2011 BMJ (references listed in Chapter 3.6 – References). 

In brief, calculate the test statistic for a normal distribution (letZ), calculate a standard error, and then the lower and upper limits for the 95% confidence interval. We illustrate use of the calculator formulas with the dart example, the p-value was 0.0309; the estimate was 4.32.

R code

letZ = (-0.862 + sqrt(0.743 - 2.404*log(0.0309, base = exp(1)))) = 2.154903
letSE = 4.32/letZ = 2.00473
1.96*letSE = 3.929271 # Add for upper and subtract for lower

The lower limit is 0.39 (4.32 – 3.929271) and the upper limit is 8.25 (4.32 + 3.929271). Although in the ballpark, these calculated estimates are wider than the confidence interval estimated by the t.test command. Small sample size, n = 5, is to blame — the accuracy of these methods, based on asymptotic statistical estimation approaches, assumes the law of large numbers applies.

Significant figures.

And finally, we should respect significant figures, the number of digits which have meaning. Our data were measured to the nearest tenth of a centimeter, or two significant figures. Therefore, if we report the confidence interval as (0.6477381, 7.9922619), then we imply a false level of precision, unless we also report our random sampling error of measurement.

R has a number of ways to manage output. One option would be to set number of figures globally with the options() function — all values reported by R would hold for the entire session. For example, options(digits=3) would report all numbers to three significant figures. Instead, I prefer to use signif() function, which allows us to report just the values we wish and does not change reporting behavior for the entire session.

signif(CI(darts),2)
upper  mean lower 
 8.00  4.30  0.65

The R function signif() applies rounding rules. We apply rounding rules when required to report estimates to appropriate levels of precision. Rounding procedures are used to replace a number with a simplified approximation. Wikipedia provides a comprehensive list of rounding rules. Notable rules include 

• directed rounding to an integer, eg, rounding up or down
• rounding to nearest integer, eg, round half up (down) if the number ends with 5
• randomly rounding to an integer, eg, stochastic rounding.

With the exception of stochastic rounding, all rounding methods impose biases on the sets of numbers. For example, the round half up method applied for numbers above 5, round down for numbers below 5 will increase the variance of the sample. In R, use round() for most of your work. If you need one of the other approaches, for example, to round up, the command is ceiling(); to round down we use floor().

When to round?

No doubt your previous math classes have cautioned you about the problems of rounding error and their influence on calculation. So, as a reminder, if reporting calls for rounding, then always round after you’ve completed your calculations, never during the calculations themselves. 

A final note about significant figures and rounding. While the recommendations about reporting statistics are easy to come by (and often very proscriptive, eg, Table 1, Cole 2015), there are other concerns. Meta-analysis, which are done by collecting information from multiple studies, would benefit if more and not fewer numbers are reported, for the very same reason that we don’t round during calculations.

Questions

1. Calculate the correct 90% and 99% confidence intervals for the dart data using the t-distribution
   • by hand
   • by one alternative method in R, demonstrated with examples in this page
2. How many significant figures should be used for the volumetric pipettor p1000? The p200? The p20 (data at end of this page)?
3. Another function, round(), can be used. Try

round(CI(darts),2)

1. • and report the results: vary the significant figures from 1 to 10 (signif() will take digits up to 22).
   • See any output differences between signif() and round()? Don’t forget to take advantage of R help pages (eg, enter ?round at the R prompt) and see Wikipedia.
2. Compare rounding by signif() and round() for the number 0.12345. Can you tell which rounding method the two functions use? 
3. Calculate the coefficient of variation (CV) for each of the three volumetric pipettors from the data at end of this page. Rank the CV from smallest to largest. Which pipettor had the smallest CV and would therefore be judged the most precise? 
4. (data provided at end of this page, click here).
5. Standards distinguish between within run precision and between run precision of a measurement instrument (see Chapter 16.5 – Instrument reliability and validity). The data in Table 1 were all recorded within 15 minutes by one operator. What kind pf precision was measured?
6. Calculate the standard error of the means for each of the three pipettors from the data
7. (data provided at end of this page, click here).
8. Calculate the approximate confidence interval using the 2SE rule and judge which of the three pipettors is the most accurate (narrowest confidence interval)
   • Repeat, but this time apply your preferred R method for obtaining confidence intervals.
   • Compare approximate and R method confidence intervals. How well did the approximate method work?
9. Return to the dart example, but now assume the p-value was p = 0.003 and again for p = 0.0003. Estimate the 95% confidence interval for each case; the difference remains 4.32. How does the change in p-value, from 0.03 to 0.0003 — a two orders of magnitude difference — affect the estimates for the confidence interval (hint: describe the width of the CI)?

Quiz Chapter 3.5

Statistics of error

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Data sets

Pipette calibration

Chapter 3 continues

• Introduction
• Data types
• Exploring data
• Measures of Central Tendency
• Measures of dispersion
• Estimating parameters
• Statistics of error
• References and suggested readings

18.5 – Selecting the best model

Introduction

This is a long entry in our textbook, many topics to cover. We discuss aspects of model fitting, from why model fitting is done to how to do it and what statistics are available to help us decide on the best model. Model selection very much depends on what the intent of the study is. For example, if the purpose of model building is to provide the best description of the data, then in general one should prefer the full (also called the saturated) model. On the other hand, if the purpose of model building is to make a predictive statistical model, then a reduced model may prove to be a better choice. The text here deals mostly with the later context of model selection, finding a justified reduced model.

From Full model to best Subset model

Model building is an essential part of being a scientist. As scientists, we seek models that explain as much of the variability about a phenomenon as possible, but yet remain simple enough to be of practical use.

Having just completed the introduction to multiple regression, we now move to the idea of how to pick best models.

We distinguish between a full model, which includes as many variables (predictors, factors) as the regression function can work with, returning interpretable, if not always statistically significant output, and a saturated model.

The saturated model is the one that includes all possible predictors, factors, interactions in your experiment. In well-behaved data sets, the full model and the saturated model will be the same model. However, they need not be the same model. For example, if two predictor variables are highly collinear, then you may return an error in regression fitting.

For those of you working with meta-analysis problems, you are unlikely to be able to run a saturated model because some level of a key factor are not available in all or at least most of the papers. Thus, in order to get the model to run, you start dropping factors, or you start nesting factors. If you were unable to get more things in the model, then this is your “full” model. Technically we wouldn’t call it saturated because there were other factors, they just didn’t have enough data to work with or they were essentially the same as something else in the model.

Identify the model that does run to completion as your full model and proceed to assess model fit criteria for that model, and all reduced models thereafter.

In R (Rcmdr) you know you have found the full model when the output lacks “NA” strings (missing values) in the output. Use the full model to report the values for each coefficient, i.e., conducting the inferential statistics.

Get the estimates directly from the output from running the regression function. You can tell if the effect is positive (look at the estimate for sample — it is positive) so you can say — more samples, greater likelihood to see more cases of cancer.

Remember, the experimental units are the papers themselves, so studies with larger numbers of subjects are going to find more cases of diabetes. We would worry big time with your project if we did not see statistically significant and positive for sample size.

Example: A multiple linear model

For illustration, here’s an example output following a run with the linear model function on a portion of the Kaggle Cardiovascular disease data set.

The variables (and data types) were

SBP = systolic blood pressure (from ap_hi)

age.years = Independent variable, ratio scale (calculated from age in days)

BMI = Dependent variable, ratio scale (calculated from height and weight variables)

active = Independent variable, physical activity, binomial

alcohol = Independent variable, alcohol use, binomial

cholesterol = Independent variable, cholesterol levels (scored 1 = “normal,” 2 = “above normal, and 3 = “well above normal”)

gender = Independent variable, binomial

glucose = Independent variable, glucose levels (scored 1 = “normal,” 2 = “above normal, and 3 = “well above normal”)

smoke = Independent variable, binomial

R output

Call:
lm(formula = SBP ~ BMI + gender + age.years + active + alcohol + 
cholesterol + gluc + smoke, data = kaggle.train)

Residuals:
Min 1Q Median 3Q Max 
-33.359 -9.701 -1.065 8.703 45.242

Coefficients:
             Estimate Std. Error t value   Pr(>|t|) 
(Intercept)   76.7664    14.9311   5.141  0.00000155 ***
BMI            0.8395     0.2933   2.862  0.00522 ** 
gender        -2.0744     1.6102  -1.288  0.20092 
age.years      0.2742     0.2403   1.141  0.25672 
active         7.1411     3.4934   2.044  0.04383 * 
alcohol       -7.9272     9.1854  -0.863  0.39039 
cholesterol    7.4539     2.3200   3.213  0.00182 ** 
gluc          -1.2701     2.9533  -0.430  0.66817 
smoke         11.4132     6.7015   1.703  0.09197 . 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 14.96 on 91 degrees of freedom
Multiple R-squared: 0.2839, Adjusted R-squared: 0.221 
F-statistic: 4.51 on 8 and 91 DF, p-value: 0.0001214

Question. Write out the equation in symbol form.

We see from the modest R² (adjusted) that the model explains little variation in systolic blood pressure; only BMI, physical activity (yes, no), and serum cholesterol levels, were statistically significant at 5% level. Smoker (yes, no) approached significance (p-value = 0.092), but I wouldn’t go on and on about positive or negative coefficients and effects.

We report the value (I would round to 7.5) for cholesterol on SBP (Table 1), and note that those who had elevated serum cholesterol levels tended to have higher systolic blood pressure (e.g., the sign of the coefficient — and I related the coefficient back to the most important thing about your study — the biological interpretation). Repeat the same for the other significant coefficients.

Now’s a good time to be clear about HOW you report statistical results. DO NOT SIMPLY COPY AND PASTE EVERYTHING into your report. Now, for the estimates above, you would report everything, but not all of the figures. Here’s how the output should be reported in your Project paper.

Table 1. Coefficients from full model.

term     Coefficient Error  P-value* 
Intercept       76.8  14.93 < 0.00001
BMI              0.8   0.29   0.00522
gender          -2.1   1.61   0.20092 
age.years        0.3   0.24   0.25672 
active           7.1   3.49   0.04383
alcohol         -7.9   9.19   0.39039 
cholesterol      7.5   2.32   0.00182 
gluc            -1.3   2.95   0.66817 
smoke           11.4   6.70   0.09197

* t-test

Looks better, doesn’t it?

Once you have the full model, we should run simple diagnostics on the regression fit to the data (VIF, Cook’s Distance, basic diagnostic plots), then proceed to use this model for the inferential statistics.

> vif(model.1)
BMI gender age.years active alcohol cholesterol gluc smoke 
1.112549 1.091154 1.081641 1.048519 1.096382 1.178635 1.168442 1.131081

No VIF was greater than 2, suggesting little multicollinearity among the predictor variables.

> cooks.distance(model.1)[which.max(cooks.distance(model.1))]
60 
0.355763

This person had high SBP (180), was a smoker, but active and normal BMI.

Use the significance tests of each parameter in the model from the corresponding ANOVA table. Now, where is the ANOVA table? Remember, right after running the linear regression,

Rcmdr: Models → Hypothesis testing → ANOVA tables

Accept the default (partial marginality), and, Boom! … the ANOVA table you should be familiar with.

From the ANOVA table you will tell me whether a Factor is significant or not. You report the ANOVA table in your paper. You describe it.

Now, the next step is to decide what is the best model. It then guides you to the next step which is to decide whether a better model (fewer parameters, Occam’s razor) can be found. Identify the parameter from the ANOVA table with the highest P-value and remove it from the model when you run the regression again. Repeat the steps above, return the ANOVA table, checking the estimates and P-values, until you have a model with only statistically significant parameters.

Find the best model

We may be tempted to run the linear model again, but with non-significant predictors excluded. For example, re-run the model on SBP but with just BMI, active, cholesterol and excluding smoke.

We can calculate the AIC, Akaike Information Criterion, the lower value is considered the better model.

AIC(model.1)
[1] 835.4938

AIC(model.2)
[1] 834.0782

The full model had eight predictor variables, excluding the intercept, whereas the reduced model had just three predictor variables, again, not including the intercept. Unsurprisingly, the reduced model is better. But now, a stronger test — what about smoking, where the p-value approached 5%? We ran the linear model (model.3).

> AIC(model.3)
[1] 831.9495

We can use an anova test or the likelihood ratio test to compare the models. The anova test, the models must be nested.

stats::anova(model.2, model.3)
Analysis of Variance Table

Model 1: SBP ~ BMI + active + cholesterol
Model 2: SBP ~ BMI + active + cholesterol + smoke
  Res.Df   RSS Df Sum of Sq      F   Pr(>F) 
1     96 22205 
2     95 21307  1     898.1 4.0043 0.04824 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

and

lmtest::lrtest((model.2,model.3)
Likelihood ratio test

Model 1: SBP ~ BMI + active + cholesterol
Model 2: SBP ~ BMI + active + cholesterol + smoke
 #Df  LogLik Df  Chisq Pr(>Chisq) 
1  5 -412.04 
2  6 -409.97  1 4.1286 0.04216 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Both tests tell us the same thing — our model with four predictors, including smoking, is better.

2 December 2024 — Past this point, edits pending

But first, I want to take up an important point about your models that you may not have had a chance to think about. The order of entry of parameters in your model can effect the significance and value of the estimates themselves. The order of parameter model entry above can be read top to bottom. Age was first, followed in sequence by CalsPDay, CholPDay, and so on. By convention, enter the covariates first (the ratio-scale predictors), that’s what I did above.

Here’s the output from a model in which I used a different order of parameters.

Anova(LinearModel.2, type="II")
Anova Table (Type II tests)

Response: BMI
              Sum Sq    Df    F value     Pr(>F)
Sex             1.52     1     0.0478    0.82764
Smoke           8.84     1     0.2782    0.59964
Age             0.62     1     0.0196    0.88918
CalsPDay       10.79     1     0.3394    0.56215
CholPDay       92.37     1     2.9065    0.09286 .
Sex:Smoke       3.35     1     0.1055    0.74631
Residuals    2129.34    67

The output is the same!!! So why did I give you a warning about parameter order? Run the ANOVA table summary command again, but this time select Type III type of test, i.e., ignore marginality

> Anova(LinearModel.2, type="III")
Anova Table (Type III tests)

Response: BMI
               Sum Sq    Df    F value       Pr(>F)
(Intercept)   1544.34     1    48.5929   1.708e-09 ***
Sex              4.69     1     0.1475     0.70214
Smoke           11.82     1     0.3720     0.54400
Age              0.62     1     0.0196     0.88918
CalsPDay        10.79     1     0.3394     0.56215
CholPDay        92.37     1     2.9065     0.09286 .
Sex:Smoke        3.35     1     0.1055     0.74631
Residuals     2129.34    67

The output has changed — and in fact it now reports the significance test of the intercept. This output is the same as the output from the linear model. Try again, this time selecting Type I, sequential

> anova(LinearModel.2)
Analysis of Variance Table

Response: BMI
           Df   Sum Sq    Mean Sq  F value    Pr(>F)
Sex         1     0.68      0.681   0.0214   0.88402
Smoke       1     2.82      2.816   0.0886   0.76690
Age         1     3.44      3.436   0.1081   0.74333
CalsPDay    1     2.27      2.272   0.0715   0.78998
CholPDay    1    96.30     96.299   3.0301   0.08633
Sex:Smoke   1     3.35      3.354   0.1055   0.74631
Residuals  67  2129.34     31.781

Here, we see the effect of order. So, as we are working to learn all of the issues of statistics and in particular mode fitting, I have purposefully restricted you to Type II analyses — obeying marginality correctly handles most issues about order of entry.

Easy there…. Take a deep breath, and guess what? Your best model needs to have significant parameters in it, right? Your best fit model then is Model 5. And that model will be your candidate for best fit as we proceed to complete our model building.

Now we proceed to gain some support evidence for our candidate best model. We are going to use an information criterion approach.

Use a fit criterion for determining model fit

To help us evaluate evidence in favor of one model over another there are a number of statistics one may calculate to provide a single number for each model for comparison purposes. The criteria model evaluators available to us include Mallow’s Cp, adjusted R², Akaike Information Criterion (AIC) or the Bayesian Information Criterion (BIC) to select best model.

We already introduced the coefficient of determination R² as a measure of fit – in general we favor models with larger values of R². However, values of R² will always be larger for models with more parameters. Thus, the other evaluators attempt to adjust for the parameters in the model and how they contribute to increased model fit.  For illustrative purposes we will use Mallow’s Cp. The equation for Mallow’s Cp in linear regression is

where p is the number of parameters in the model. Mallow’s Cp is thus equal to the number of parameters in the model plus an additional amount due to lack of fit of the model (i.e., large residuals). All else being equal we favor the model in which the Cp is close to the number of parameters in the model.

In Rcmdr, select Models → Subset model selection … (Fig. 1)

Figure 1. Rcmdr popup menu, Subset model selection…

From the menu, select the criterion and how many models to return. The function returns a graph that can be used to interpret which model is best given the selection criterion used. Below is an example (although for a different data set!) for Mallow’s Cp (Fig. 2).

Figure 2. Mallow’s Cp plot

Lets break down the plot. First define the axes. The vertical axis is the range of values for the Cp calculated for each model. The horizontal axis is categorical and reads from left to right: Intercept, Inspection[T.Yes], etc., up to Suspended. Looking into the graph itself we see horizontal bars — the extent of shading indicates which model corresponds to the Cp value. For example, the lowest bar which is associated with the Cp value of 8 extends all the way to the right of the graph. This says that the model evaluated included all of the variables and therefore was the saturated or full model. The next bar from the bottom of the graph is missing only one block (lgCarIns), which tells us the Cp value 6 corresponds to a reduced model, and so forth.

Cross-validation

Once you have identified your Best Fit model, then, your proceed to run the diagnostics plots. For the rest of the discussion we return to our first example.

Rcmdr: Models → Graphs → Basic diagnostic plots.

We’ll just concern ourselves with the first row of plots (Fig. 3).

Figure 3. Diagnostic plots

The left one shows the residuals versus the predicted values — if you see a trend here, the assumption of linearity has been violated. The second plot is a test of the assumption of normality of the residuals. Interpret them (residuals OK, Residuals normally distributed? Yes/ No), and you’re done. Here, I would say I see no real trend in the residuals vs fitted plot, so assumption of linear fit is OK. For normality, there is a tailing off at the larger values of residuals, which might be of some concern (and I would start thinking about possible leverage problems), but nothing dramatic. I would conclude that our Model 6 is a good fitting model and one that could be used to make predictions.

Now, if you think a moment, you should identify a logical problem. We used the same data to “check” the model fit as we did to make the model in the first place. In particular if the model is intended to make predictions it would be advisable to check the performance of the model (e.g., does it make reasonable predictions?) by supplying new data, data not used to construct the model, into the model. If new data are not available, one acceptable practice is to divide the full data set into at least two subsets, one used to develop the model (sometimes called the calibration or training dataset) and the other used to test the model. The benefits of cross-validation include testing for influence points, over fitting of model parameters, and a reality check on the predictions generated from the model.

A note on best practices

In the 1980s, researchers often “removed” the effect of a covariate by first fitting a regression and then saving the residuals to use in separate analyses, such as t-tests, ANOVAs, or variance component estimates (eg, Dohm et al 2001). For example, in a study of plant growth, scientists might regress final plant height on initial size to remove the effect of size, then compare the residuals between fertilizer groups. While this approach adjusts for the covariate, it is problematic because residuals are not raw data: they are correlated, have constrained sums, and underestimate uncertainty. This can distort group differences, especially if the covariate distributions differ among groups, and it ignores the variability introduced by the first regression. This approach was common in ecology, evolutionary biology, physiology, and psychology before modern mixed modeling and generalized regression workflows became standard.

Modern best practices instead recommend fitting a single unified model that includes both the covariate and the grouping variable directly. Using the same plant growth example, a linear model such as estimates the effect of fertilizer while properly accounting for initial plant size. This approach provides valid standard errors, allows correct inference about group differences, and can accommodate interactions (eg, if the relationship between and differs by group) or hierarchical data structures using mixed-effects models. Additionally, modern workflows include diagnostics such as residual plots, Cook’s Distance, and variance inflation factors to check model assumptions and influential points. Use mixed models when data are hierarchical or repeated data. Random effects handle variation among groups, individuals, years, etc., in a single coherent model. When covariates need to be “controlled for,” we use partial regression. This happens inside the model, not by subtracting covariate effects first. By analyzing all relevant predictors in a single model, researchers obtain more accurate estimates, properly reflect uncertainty, and avoid the statistical pitfalls of the old residual-based two-stage approach.

Thus, there are several statistical reasons the older approach fails under modern scrutiny. First, Residuals are not independent observations. Residuals are model-dependent quantities, not raw data. Once you fit a regression, the residuals are correlated with each other, constrained to sum to zero, and, likely heteroscedastic if the model was imperfect. Using them in a new test violates standard assumptions of the second test.

Second,  you underestimate uncertainty. If you perform tests on residuals, the standard errors do not include uncertainty from the first regression, which may lead to inflated Type I error and an overly confident “adjusted” group comparisons.

Third, the old approach creates a “two-stage model” that breaks the likelihood framework. A two-stage analysis discards the correct joint likelihood and therefore produces biases p-values, and interferes with valid confidence intervals or predictions.

Under the modern one-stage models include ML/REML, generalized estimating equations, and fully integrated Bayesian models, which keeps the model coherent. Use of mixed models make residualization unnecessary. In studies with repeated measures, blocks, sites, subjects, or years, the old approach cannot partition sources of variation correctly.

In short, the advantages of the one-stage approach is that resulting models have valid uncertainty, correct group comparisons, better handling of interactions and hierarchy, full likelihood-based inference. They fit a single unified regression or mixed-effects model with all relevant predictors, examine diagnostics, and make inferences directly. This is one more benefit of the increased power and availability of more powerful computers and more complete development of mixed methods modeling.

Questions

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

Quiz Chapter 18.5

Selecting the best regression model

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Chapter 18 contents

• Introduction
• Multiple linear regression
• Nonlinear regression
• Logistic regression
• Generalized Linear Squares
• Selecting the best model
• Compare two linear models
• References and suggested readings (Ch17 & 18)

12.4 – ANOVA from “sufficient statistics”

Introduction

By now you should be able to run a one-way ANOVA using R (and R Commander) with ease. As a reminder, You should also be aware that, if you need to, you could use spreadsheet software like Microsoft Excel or LibreOffice Calc to run one way ANOVA on a small data set. Still, there are times when you may need to run a one-way ANOVA on a small data set, and doing so by hand calculator may be just as convenient. What are your available options?

Following the formulas I have given would be one way to calculate ANOVA by hand, but it would be tedious and subject to error. Instead of working with the standard formulas, calculator shortcuts can be derived with a little algebra and this is where I want to draw your attention now. This technique will come in handy in lab classes or other scenarios where you collect some data among a set number of groups and calculate means and standard deviations. The purpose of this posting is to show you how to obtain the necessary statistics to calculate a one-way ANOVA from the available descriptive statistics: means, standard deviations, and sample sizes. In other words, these are the sufficient statistics for one-way ANOVA.

As you recall, a one-way ANOVA yields a single F test of the null hypothesis that all group means are equal. To calculate the F test, you need

• Mean Square Between Groups,
• Mean Squares Within Groups or Error,

F is then calculated as

with degrees of freedom k – 1 for the numerator and N – 1 for the denominator. can also appear as

We can calculate  as

where k is the number of i ^th groups, n_i is the sample size of the ith group,  refers to the overall mean for all of the  sample means.

Next, for the Error Mean Square, , all we need is the average of the sample variances (the square of the sample standard deviation, s).

ANOVA from sufficient statistics

Consider an example data set (Table 1) for which only summary statistics are available (mean and standard deviation, sd). The data set is for life span(days) for several inbred strains of laboratory mice from a study (. Sample size for each group was seven mice.

Table 1. Descriptive statistics female life span (days) for ten different inbred strains of mice (Mus domesticus).

Data from Yuan et al (2021; https://phenome.jax.org).

Spreadsheet calculations.

You have several options at this point, ranging from using your calculator and the formulas above (don’t forget to square the standard deviation to get the variances!), or you could use Microsoft Excel or LibreOffice Calc and enter the necessary formulas by hand (Table 2). You’ll also find many online calculators for one-way ANOVA by sufficient statistics (eg, https://www.danielsoper.com/statcalc/calculator.aspx?id=43).

Table 2. Spreadsheet with formulas for calculating one-way ANOVA from means and standard deviations from statistics presented in Table 1.

For this example, you should get the following

MSB = 219899.5
MSE = 31634.1
F = 6.951
P-value = 4.45E-09

But, R can do it better.

Here’s how. Install the HH package (or RcmdrPlugin.HH for use in Rcmdr) and call the ?aovSufficient function.

Step 1. Install the HH package from a CRAN mirror, eg, cloud.r-project.org, in the usual way.

chooseCRANmirror()
install.packages("HH") 
library(HH)

Step 2. Enter the data. Do this in the usual way (eg, from a text file), or enter directly using the read.table command as follows.

MouseData <- read.table(header=TRUE, sep = ",", text= "
Strain, n.size, average, stdev 
129S1, 32, 787.4, 159.16
A.J, 32, 630.7, 130.20
BALB.c, 32, 734.4, 154.43
BUB.BnJ, 24, 611.3, 218.34
C3H.HeJ, 29, 724.1, 131.48
C57BL.6J, 29, 855.7, 185.34
CBA.J, 30, 622.9, 181.95
FVB.NJ, 26, 750.3, 230.11
P.J, 32, 676.0, 178.82
SWR.J, 31, 831.9, 181.31")
#Check import
head(MouseData)

End of R input

I know, a little hard to read, but from the MouseData to the end bracket ") before the comment line #Check import, that’s all one command.

Of course, you could copy the data and import the data from your computer’s clipboard in Rcmdr: Data → Import data → from text file, clipboard, or URL… 

Once the data set is loaded, proceed to Step 3.

Step 3. In our example, sample size was included for each group. Skip to step 4. If, however, the table lacked the sample size information, you can always add a new variable. For example, if we needed to add sample size to the data frame, we would use the repeat element function, rep().

MouseData$n <- rep(7, 10)

If you check the View data set button in Rcmdr, you will see that the command in Step 3 has added a new variable “n” for each of the eleven rows. The function rep() stands for “replicate elements in vectors” and what it did here was enter a value of 7 for each of the ten rows in the data set. Again, this step is not necessary for this example because sample size is already part of the data frame. Proceed to step 4.

Step 4. Run the one way ANOVA using the sufficient statistics and the HH function aovSufficient

MouseData.aov <- aovSufficient(average ~ Strain, data=MouseData)

Step 5. Get the ANOVA table

summary(MouseData.aov)

Here’s the R output

           Df  Sum Sq Mean Sq F value   Pr(>F) 
Strain      9 1977732  219748   7.123 2.54e-09 ***
Residuals 287 8853743   30849 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

End R output

To explore other features of the package, type ?aovSufficient at the R prompt (like all R functions, extensive help is generally available for each function in a package).

Meta-analysis and sufficient statistics.

Meta-analysis (see Chapter 20.15 — Meta-analysis), can be performed using sufficient statistics from different tests. Combining sufficient statistics allows for a more precise overall estimate by effectively merging the information from multiple studies, which can be especially useful when individual studies have small sample sizes or lack a unified outcome measure. Common methods include creating summary effect sizes like the standardized mean difference or using p-value combination methods, while ensuring the underlying statistical assumptions are met through goodness-of-fit tests.

Fisher’s method combines the p-values from multiple independent studies into a single overall p-value to test a joint null hypothesis. The method’s test statistic is calculated as the sum of the negative log-transformed p-values. P-values from number of tests are combined; each p-value, , contributes to a variable, , with degrees of freedom. The chi-square distribution is used to evaluate the significance of the Fisher’s method test statistic.

Fisher’s method is available in several R packages (eg,poolr package); here, we’ll just create a simple R code.

Limitations: Methods that only use p-values or test statistics may not adequately address the effects of heterogeneity and potential publication bias among studies. Because many older studies do not report effect sizes and confidence intervals, using test statistics or p-values is often the only available method to combine their results.

Example.

One hypothesis about longevity is that the greater the genetic diversity can lead to longer life. The mice in Table 1 were all from inbred mouse strains. By definition, genetic diversity is restricted in inbred mice, compared to outbred strains. Jackson Labs produced a “Diverse Outbred” line of mice, and a few individuals have lived nearly five years, well past the typical 1 – 2 year lifespan of mice (Cohen 2017, Mullis et al 2025). Thus, a natural comparison is to ask: is there a difference in average lifespan between the DO and inbred mouse strains? We’ll run a series of one-sample t-tests against the mean lifespan for the diverse outbred stock of mice, save the p-values, and apply Fisher’s method. While not strictly a meta-analysis, which would draw from multiple studies, this example outlines the basic steps to take given a series of hypothesis of the same kind.

do_avg= 794.5
# dataset MouseData
tsum.test(average[1], stdev[1], n.size[1], mu=do_avg)

t = -0.25235, df = 31, p-value = 0.8024

p_values <- numeric(length(average))

for (i in 1:length(average)) {
     suppressWarnings({
      test_result <- tsum.test(
      mean.x = average[i],
      s.x = stdev[i],
      n.x = n.size[i],
      mu = do_avg
      )
      p_values[i] <- test_result$p.value
      })
     }

print(p_values)

The p-values from all ten tests are sorted by strain in Table 2.

Table 2. Results of one sample t-tests inbred strains (Table 1) compared to average lifespan of the J:DO diversity outbred mouse strain from Jackson labs

R-code for Fisher’s method

my_pValues <- c(0.08623, 1.601e-05, 0.3368, 0.0007307, 0.2598)
X_squared <- -2 * sum(log(my_pValues),base = exp(1))
myDf <- 2 * length(my_pValues)
combinedP <- pchisq(X_squared, myDf, lower.tail = FALSE)
result_table <- as.data.frame(rbind(X_squared, myDf, combinedP))
rownames(result_table) <- c("chi_sqr", "df", "P-value")
colnames(result_table) <- c("Value")
print(result_table)

and the R output

                Value
chi_sqr 44.7017431931
     df 16.0000000000
P-value 0.0001541831

With the small p-value, we would reject the null hypothesis of no difference in lifespan among the inbred strains compared to the genetically diverse DO strain of mice.

Limitations of ANOVA from sufficient statistics.

This was pretty easy, so it is worth asking — Why go through the bother of analyzing the raw data, why not just go to the summary statistics and run the calculator formula? First, the chief reason against the calculator formula and use of only sufficient statistics loses information about the individual values and therefore you have no access to the residuals. The residual of an observation is the difference between the original observation and the model prediction. The residuals are important for determining whether the model fits the data well and are, therefore, part of the toolkit that statisticians need to do proper data analysis. We will spend considerable time looking at residual patterns and it is an important aspect of doing statistics correctly.

Secondly, while it is possible to extend this approach to more complicated ANOVA problems like the two-way ANOVA (Cohen 2002), the statistical significance of the interaction term(s) calculated in this way are only approximate (the main effects are OK to interpret). Thus, ANOVA from sufficient statistics has its place when all you have is access to descriptive statistics, but its use is limited and not at all the preferred option for data analysis when the original, raw observations are in hand.

Questions

1. Calculate the one-way ANOVA for body weight of 47 female (F) and 97 male (M) cats (kilograms, dataset cats in MASS R package) from the following summary statistics.

2. Bonus: Load the cats data set (package MASS, loaded with Rcmdr) and run a one-way ANOVA using the aov() function via Rcmdr. Are the ANOVA from sufficient statistics the same as results from the direct ANOVA calculation? If not, why not.

Quiz Chapter 12.4

ANOVA from “sufficient statistics”

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Chapter 12 contents

• Introduction
• The need for ANOVA
• One way ANOVA
• Fixed effects, random effects, and agreement
• ANOVA from “sufficient statistics”
• Effect size for ANOVA
• ANOVA posthoc tests
• Many tests one model
• References and suggested readings

12.3 – Fixed effects, random effects, and agreement

Introduction

Within discussions of one-way ANOVA models the distinction between two general classes of models needs to be made clear by the researcher. The distinction lies in how the levels of the factor are selected. If the researcher selects the levels, then the model is a Fixed Effects Model, also called a Model I ANOVA. On the other hand, if the levels of the factor were selected by random sampling from all possible levels of the factor, then the model is a Random Effects Model, also called a Model II ANOVA. This page is about such models and I’ll introduce the intraclass correlation coefficient, abbreviated as ICC, as a way to illustrate applications of Model II ANOVA.

Here’s an example to help the distinction. Consider an experiment to see if over the counter painkillers are as good as prescription pain relievers at reducing numbers of migraines over a six week period. The researcher selects Tylenol^®, Advil^®, Bayer^® Aspirin, and Sumatriptan (Imitrex^®), the latter an example of a medicine only available by prescription. This is clearly an example of fixed effects; the researcher selected the particular medicines for use.

Random effects, in contrast, implies that the researcher draws up a list of all over the counter pain relievers and draws at random three medicines; the researcher would also randomly select from a list of all available prescription medicines.

Fixed effects are probably the more common experimental approaches. To be complete, there is a third class of ANOVA called a Mixed Model or Model III ANOVA, but this type of model only applies to multidimensional ANOVA (eg, two-way ANOVA or higher), and we reserve our discussion of the Model III until we discuss multidimensional ANOVA (Table 1).

Table 1. ANOVA models

Although the calculations for the one-way ANOVA under Model I or Model II are the same, the interpretation of the statistical significance is different between the two.

In Model I ANOVA, any statistical difference applies to the differences among the levels selected, but cannot be generalized back to the population. In contrast, statistical significance of the Factor variable in Model II ANOVA cannot be interpreted as specific differences among the levels of the treatment factor, but instead, apply to the population of levels of the factor. In short, Model I ANOVA results apply only to the study, whereas Model II ANOVA results may be interpreted as general effects, applicable to the population.

This distinction between fixed effects and random effects can be confusing, but it has broad implications for how we interpret our results in the short-term. This conceptual distinction between how the levels of the factor are selected also has general implications for our ability to acquire generalizable knowledge by meta-analysis techniques (Hunter and Schmidt 2000). Often we wish to generalize our results: we can do so only if the levels of the factor were randomly selected in the first place from all possible levels of the factor. In reality, this may not often be the case. It is not difficult to find examples in the published literature in which the experimental design is clearly fixed effects (ie, the researcher selected the treatment levels for a reason), and yet in the discussion of the statistical results, the researcher will lapse into generalizations.

Random Effects Models and agreement

Model II ANOVA is common in settings in which individuals are measured more than once. For example, in behavioral science or in sports science, subjects are typically measured for the response variable more than once over a course of several trials. Another common setting of Model II ANOVA is where more than one raters are judging an event or even a science project. In all of these cases what we are asking is about whether or not the subjects are consistent, in other words, we are asking about the precision of the instrument or measure.

In the assessment of learning by students, for example, different approaches may be tried and the instructor may wish to investigate whether the interventions can explain changes in test scores. To assess instrument validity — an item score from a grading rubric (Goertzen and Klaus 2023), tumor budding as biomarker for colorectal cancer (Bokhorst et al 2020) — agreement or concordance among two or more raters or judges (inter-rater reliability) needs to be established (same citations and references therein). For binomial data, we would be tempted to use the phi coefficient, Is it actually reliable Cohen’s kappa (two judges) or Fleiss Kappa (two or more judges) is a better tool. Phi coefficient and both assessment scores were introduced in Chapter 9.2.

For ratio scale data, agreement among scores of  two or more scores from judges There is an enormous number of articles on reliability measures in the social sciences and you should be aware of a classical paper on reliability by Shrout and Fleiss (1979) (see also McGraw and Wong, 1996). Both the ICC and the Pearson product moment correlation, r, which we will introduce in Chapter 16, are measures of strength of linear association between two ratio scale variables (Jinyuan et al 2016). But ICC is more appropriate for association between repeat measures of the same thing, eg, repeat measures of running speed or agreement between judges. In contrast, the product moment correlation can be used to describe association between any two variables, eg, between repeat measures of running speed, but also between say running speed and maximum jumping height. The concept of repeatability of individual behavior or other characteristics is also a common theme in genetics and so you should not be surprised to learn that the concept actually traces to RA Fisher and his invention of ANOVA. Just as in the case of the sociology literature, there are many papers on the use and interpretation of repeatability in the evolutionary biology literature (eg, Lessels and Boag 1987; Boake 1989; Falconer and Mackay 1996; Dohm 2002; Wolak et al 2012).

Agreement and repeatability are Model II (random effects) because the subjects, items, or raters being studied are treated as a random sample from a larger population rather than specific, fixed levels of interest. In reliability studies, subjects are not chosen for their specific characteristics but as representatives of the population variability. The goal is to infer whether a measurement tool works for any future subject or rater, not just the ones tested in the study. These factors are designed to estimate variance components and generalize results beyond the specific sample used. Put another way, the goal of Model II is often to estimate how much variance is due to differences between subjects versus within subjects (repeatability), which requires treating the subject effect as random.

There are many ways to analyze these kinds of data but a good way is to treat this problem as a one-way ANOVA with Random Effects. Thus, the Random Effects model permits the partitioning of the variation in the study into two portions: the amount that is due to differences among the subjects or judges or intervention versus the amount that is due to variation within the subjects themselves. The Factor is the Subjects and the levels of the factor are how ever many subjects are measured twice or more for the response variable.

If the subjects performance is repeatable, then the Mean Square Between (Among) Subjects, , component will be greater than the Mean Square Error component, , of the model. There are many measures of repeatability or reliability, but the intraclass correlation coefficient or ICC is one of the most common. The ICC may be calculated from the Mean Squares gathered from a Random Effects one-way ANOVA. ICC can take any value between zero and one.

where

and

B and W refers to the among group (between or among groups mean square) and the within group components of variation (error mean square), respectively, from the ANOVA, MS refers to the Mean Squares, and k is the number of repeat measures for each experimental unit. In this formulation k is assumed to be the same for each subject.

By example, when a collection of sprinters run a race, if they ran it again, would the outcome be the same, or at least predictable? If the race is run over and over again and the runners cross the finish lines at different times each race, then much of the variation in performance times will be due to race differences largely independent of any performance abilities of the runners themselves and the Mean Square Error term will be large and the Between subjects Mean Square will be small. In contrast, if the race order is preserved race after race: Jenny is 1st, Ellen is 2nd, Michael is third, and so on, race after race, then differences in performance are largely due to individual differences. In this case, the Between-subjects Mean Square will be large as will the ICC, whereas the Mean Square for Error will be small, and so will the ICC.

Can the intraclass correlation be negative?

In theory, no. Values for ICC range between zero and one. The familiar Pearson product moment correlation, Chapter 16, takes any value between -1 and +1. However, in practice, negative values for ICC will result if .

In other words, if the within group variability is greater than the among group variability, then negative ICC are possible. The question is whether negative estimates are ever biologically relevant or, simply result of limitations of the estimation procedure. Small ICC values and few repeats increases the risk of negative ICC estimates. Thus, a negative ICC would be “simply a(n) “unfortunate” estimate (Liljequist et al 2019). However, there may be situations in which negative ICC is more than unfortunate.  For example, repeatability, a quantitative genetics concept (Falconer and Mackay 1996), can be negative (Nakagawa and Schielzeth 2010).

Repeatability reflects the consistency of individual traits over time (or spatially, eg, scales on on a snake), is defined as the ratio of within individual differences and among individuals differences. Thus, within individual variance is attributed to general environmental effects, , whereas variance among individuals is both genetic and environmental. Scenarios for which repeated measures on the same individual may occur under environmental conditions, termed specific environmental effects or by Falconer and Mackay (1996), that negatively affect performance are not hard to imagine — it remains to be determined how common these conditions are and whether or not estimates of repeatability are negative.

ICC Example

I extracted 15 data points from a figure about nitrogen metabolism in kidney patients following treatment with antibiotics (Figure 1, Mitch et al 1977). I used a web application called WebPlot Digitizer (https://apps.automeris.io/wpd/), but you can also accomplish this task within R via the digitize package. I was concerned about how steady my hand was using my laptop’s small touch screen, a problem that very much can be answered by thinking statistically, and taking advantage of the ICC. So, rather than taking just one estimate of each point I repeated the protocol for extracting the points from the figure three times generating a total of three points for each of the 15 data points or 45 points in all. How consistent was I?

Let’s look at the results just for three points, #1, 2, and 3.

In the R script window enter

points = c(1,2,3,1,2,3,1,2,3)

Change points to character so that the ANOVA command will treat the numbers as factor levels.

points = as.character(points)
extracts = c(2.0478, 12.2555, 16.0489, 2.0478, 11.9637, 16.0489, 2.0478, 12.2555, 16.0489)

Make a data frame, assign to an object, eg, “digitizer”

digitizer = data.frame(points, extracts)

The dataset “digitizer” should now be attached and available to you within Rcmdr. Select digitizer data set and proceed with the one-way ANOVA.

Output from oneway ANOVA command

Model.1 <- aov(extracts ~ points, data=digitizer)

 summary(Model.1)
            Df    Sum Sq   Mean Sq  F value     Pr(>F)    
points       2 313.38895 156.69448 16562.49 5.9395e-12 ***
Residuals    6   0.05676   0.00946                        
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

numSummary(extracts , groups = points, 
+   statistics=c("mean", "sd"))
         mean           sd data:n
1  2.04780000 0.0000000000      3
2 12.15823333 0.1684708085      3
3 16.04890000 0.0000000000      3

End R output. We need to calculate the ICC

I’d say that’s pretty repeatable and highly precise measurement!

But is it accurate? You should be able to disentangle accuracy from precision based on our previous discussion (Chapter 3.5), but now in the context of a practical way to quantify precision.

ICC calculations in R

We could continue to calculate the ICC by hand, but better to have a function. Here’s a crack at the function to calculate ICC along with a 95% confidence interval.

myICC <- function(m, k, dfN, dfD) {
 testMe <- anova(m) 
 MSB <- testMe$"Mean Sq"[1]
 MSE <- testMe$"Mean Sq"[2]
 varB <- MSB - MSE/k
 ICC <- varB/(varB+MSE)
 fval <- qf(c(.025), df1=dfN, df2=dfD, lower.tail=TRUE)
 CI = (k*MSE*fval)/(MSB+MSE*(k-1)*fval)
 LCIR = ICC-CI
 UCIR = ICC+CI
 myList = c(ICC, LCIR, UCIR)
 return(myList)
}

The user supplies the ANOVA model object (eg, Model.1 from our example), k, which is the number of repeats per unit, and degrees of freedom for the among groups comparison (2 in this example), and the error mean square (6 in this case). Our example, run the function

m2ICC = myICC(Model.1, 3, 2,6); m2ICC

and R returns

[1] 0.9999396 0.9999350 0.9999442

with the ICC reported first, 0.9999396, followed by the lower limit (0.9999350) and the upper limit (0.9999442) of the 95% confidence interval.

Instead of writing your own function, several R packages can calculate the intraclass correlation coefficient (ICC) and its variants. These packages include: irr, psy, psych, and rptR.  For complex experiments involving multiple predictor variables, these packages are helpful for obtaining the correct ICC calculation (cf Shrout and Fleiss 1979; McGraw and Wong 1996). For the one-way ANOVA it is easier to just extract the information you need from the ANOVA table and run the calculation directly. We do so for a couple of examples.

Example: Are marathon runners consistent more consistent than my commute times?

Figure 1.Honolulu Marathon 2024 participant. Image credit: Pdubs.94, licensed under CC BY 4.0 (Creative Commons Attribution 4.0 International License), via Wikipedia.

A marathon is 26 miles, 385 yards (42.195 kilometers) long (World Athletics.org). And yet, tens of thousands of people choose to run in these events. For many, running a marathon is a one-off, the culmination of a personal fitness goal. For others, it’s a passion and a few are simply extraordinary, elite runners who can complete the courses in 2 to 3 hours (Table 3). That’s about 12.5 miles per hour. For comparison, my 20 mile commute on the H1 freeway on Oahu typically takes about 40 minutes to complete, or 27 miles per hour (Table 2, yes, I keep track of my commute times, per Galton’s famous maxim: “Whenever you can, count”).

Figure 2. View west along Interstate H-1 (Lunalilo Freeway) at 6:30AM on a weekday, Honolulu, Oahu, Hawaii. Image used with permission, credit: S. Dohm.

Table 2. A sampling of average commute speeds, miles per hour (mph),  on the H1 freeway during DrD’s morning commute (distance about 19.3 miles).

Try on your own.

Calculate the ICC for my commute speeds.

Run the one-way ANOVA to get the necessary mean squares and input the values into our ICC function. We have

require(psych)
m2ICC = myICC(AnovaModel.1, 4, 4,11); m2ICC
[1] 0.7390535 0.6061784 0.8719286

Repeatability, as estimated by the ICC, was 0.74 (95% CI 0.606, 0.872), for repeat measures of commute times.

We can ask the same about marathon runners — how consistent from race to race are these runners? The following data are race times drawn from a sample of runners who completed the Honolulu Marathon in both 2016 and 2017 in 2 to 3 hours (times recorded in minutes). In other words, are elite runners consistent?

Table 3. Honolulu marathon running times (min) for eleven repeat, elite runners.

Stacked worksheet, P refers to person, therefore P1 is person ID 1, P2 is person ID 2, and so on.

After running a one-way ANOVA, here are results for the marathon runners

m2ICC = myICC(Model.1, 2, 10,11); m2ICC
[1] 0.9780464 0.9660059 0.9900868

Repeatability, as estimated by the ICC, was 0.98 (95% CI 0.966, 0.990), for repeat measures of marathon performance. Put more simply, knowing what a runner did in 2016 I would be able to predict their 2017 race performance with high confidence, 98%!

And now, we compare: the runners are more consistent!

Clearly this is an apples-to-oranges comparison, but it gives us a chance to think about how we might make such comparisons. The ICC will change because of differences among individuals. For example, if individuals are not variable, if everyone finishes the race at the same time, then the ICC will be low (or even zero).  Because there is no variance to be attributed to individual differences, the proportion of total variance explained by those differences will be being minimal.

Visualizing repeated measures.

Waterfall plot.

If just two measures, then a profile plot will do. We introduced profile plots in the paired t-test chapter (Ch10.3). For more than two repeated measures, a line graph will do (see question 9 below). Another graphic increasingly used is the waterfall plot (waterfall chart), particularly for before and after measures or other comparisons against a baseline. For example, it’s a natural question for our eleven marathon runners — how much did they improve on the second attempt of the same course? Figure 3 shows my effort to provide a visual.

Figure 3. Simple waterfall plot of race improvement for Table 3 data. Dashed horizontal line at zero.

R code

Data wrangling: Create a new variable, eg, race21, by subtracting the first time from the second time. Multiply by -1 so that improvement (faster time in second race) is positive. Finally, apply descending sort so that improvement is on the left and increasingly poorer times are on the right.

df <- data.frame(
ID = rep(paste0("P", 1:11), each = 2),
Time = c(179.9,192.0,
129.9,130.8,
128.5,129.6,
179.4,179.7,
174.3,181.7,
177.2,176.2,
169.0,173.4,
174.1,175.2,
175.1,174.2,
163.9,175.9,
179.3,179.8)
)

# Convert to wide format
df_wide <- reshape(df, idvar = "ID", timevar = "ID",
direction = "wide")

# Extract first and second times manually
time_matrix <- matrix(df$Time, ncol = 2, byrow = TRUE)
colnames(time_matrix) <- c("Time1", "Time2")

# Calculate improvement (second - first)
base21 <- time_matrix[,2] - time_matrix[,1]

# Sort for waterfall display
base21 <- sort(base21, decreasing = TRUE)

# Waterfall plot
barplot(base21, col="blue", border="blue", space=0.5,
ylim=c(-20,20),
ylab="Improvement from first race (min)")
abline(h=0, lty=2)

Interpretation.

From the graph (Fig 3) we see that few of the racers improved their times the second time around — a few had decreased times.

Spaghetti plots.

Spaghetti plots, which display individual trajectories over time or multiple measurements, is a specialized line plot effective for visualizing the consistency of performance. The vertical thickness or spread of a single subject’s line across measurements indicates their level of repeatability. A “flat” line indicates high repeatability, while a volatile, zig-zagging line indicates low repeatability. Spaghetti plots are useful to visualize growth, decline, or stability over time for a single performance measure.

Figure 4. A spaghetti plot of average commute speeds from Table 2 data. 

R code

# 1. Enter the data from Table 2.
commute <- data.frame(
Week = 1:4,
Monday = c(28.5, 25.8, 26.2, 23.3),
Tuesday = c(23.8, 22.4, 22.6, 26.9),
Wednesday = c(28.5, 29.3, 24.9, 31.3),
Thursday = c(30.2, 26.2, 24.2, 26.2),
Friday = c(26.9, 27.7, 34.3, 30.2)
)

head(commute)

# 2. Set the X-axis points
days <- 1:5
day_names <- c("Mon", "Tue", "Wed", "Thu", "Fri")

# 3. Create empty plot
# Find range of Y to ensure all data fits
ymin <- min(commute[, 2:6]) - 2
ymax <- max(commute[, 2:6]) + 2

plot(NA, xlim = c(1, 5), ylim = c(ymin, ymax),
xlab = "Day of Week", ylab = "Average Speed (mph)",
main = "Weekly Commute Speeds", xaxt = "n")

# Add X-axis labels
axis(1, at = 1:5, labels = day_names)

# 4. Add lines for each week (spaghetti)
colors <- c("red", "blue", "darkgreen", "purple")
for (i in 1:nrow(commute)) {
# Plot lines
lines(days, commute[i, 2:6], col = colors[i], lwd = 2, type = "b", pch = 19)
}

# 5. Add legend
legend("topleft", legend = paste("Week", 1:4), col = colors, lwd = 2, pch = 19)

Parallel coordinates plots.

Parallel coordinates plot allows each entity to be represented by a unique, continuous line traced across several vertical axes (each representing a different measurement or variable). High repeatability is indicated when a line remains roughly horizontal across all axes, while poor repeatability is visualized as lines that zig-zag or change direction between measurements.

Figure 5. A parallel coordinates plot of average commute speeds from Table 2 data.

Both spaghetti plots and parallel coordinate plots can be used to visualize consistency of performance over multiple measures, but they differ significantly in their handling of data, scale, and context. Spaghetti plots are best for tracking within-subject variability over time, but quickly become a mess — like pasta tossed and sticking to a wall — when subject count increases. Parallel coordinate plots are a multivariate generalization — each line represents an individual’s profile across different tests, eg, accuracy, speed, VO2max.

R code.

# 1. Create the data frame
# see above

# 2. Extract only the numerical commute data for plotting
commute_data <- commute[, 2:6]

# 3. Setup the plot area
# Define range for y-axis
y_range <- range(commute_data)
# Define number of days (5)
num_days <- ncol(commute_data)

# Create empty plot
plot(NA, xlim = c(1, num_days), ylim = y_range, 
xaxt = "n", xlab = "Day of Week", ylab = "Average Speed (mph)",
main = "Average Commute Speed: Weekly Comparison")

# Add x-axis labels
axis(1, at = 1:num_days, labels = colnames(commute_data))

# 4. Draw vertical lines for each day
for (i in 1:num_days) {
lines(c(i, i), y_range, col = "grey", lty = 2)
}

# 5. Add lines for each week (repeated measures)
colors <- c("red", "blue", "darkgreen", "orange")
for (i in 1:nrow(commute_data)) {
lines(1:num_days, commute_data[i, ], col = colors[i], lwd = 2, type = "b", pch = 19)
}

# 6. Add legend
legend("topleft", legend = paste("Week", commute$Week), col = colors, lwd = 2, pch = 19)

Heat map.

Heat map quickly identify patterns, such as an individual’s performance improving, degrading, or remaining stable over time. For example, uniform, solid colors across rows (individuals) or columns (time/trials) indicate high consistency, while varied colors indicate poor repeatability.

Figure 6. A heat map of the commute speeds data set.

R code.

# 1. Create the data
# Week is excluded from matrix to only include numerical commute times
commute_data <- matrix(c(
28.5, 23.8, 28.5, 30.2, 26.9,
25.8, 22.4, 29.3, 26.2, 27.7,
26.2, 22.6, 24.9, 24.2, 34.3,
23.3, 26.9, 31.3, 26.2, 30.2
), nrow = 4, byrow = TRUE)

# 2. Add row and column names
rownames(commute_data) <- c("Week 1", "Week 2", "Week 3", "Week 4")
colnames(commute_data) <- c("Mon", "Tue", "Wed", "Thu", "Fri")

# 3. Create the heatmap
# Use t() to transpose so weeks are rows and days are columns
# Use rev() on rows to have Week 1 at the top
image_data <- t(apply(commute_data, 2, rev))

# Set up the plotting area (margins for labels)
par(mar = c(5, 6, 4, 2))

# Plot the matrix
# Option 1
image(1:nrow(image_data), 1:ncol(image_data), image_data,
col = heat.colors(12),
axes = FALSE, xlab = "Days", ylab = "",
main = "Commute Times Heatmap")

# Add axes labels
axis(1, at = 1:5, labels = colnames(commute_data))
axis(2, at = 1:4, labels = rev(rownames(commute_data)), las = 1)
box()

Interpretation.

The graphs (Fig 4 – 6) show a lack of trends in consistency of drive speeds. The heat map suggests more consistency on Monday — low speeds (Fig 4 & Fig 5), but that’s a stretch.

An example for you to work, from our Measurement Day

If you recall, we had you calculate length and width measures on shells from samples of gastropod and bivalve species. In the table are repeated measures of shell length, by caliper in mm, for a sample of Conus shells (Fig 7 and Table 4).

Figure 7. Conus shells, image by M. Dohm.

Repeated measures of the same 12 Conus shells.

Table 4. Unstacked dataset of repeated length measures on 12 shells

Maximum length from from anterior end to apex with calipers. Repeated measures were conducted blind to shell id, sampling was random over three separate time periods.

Questions

1. Consider data in Table 2, Table 3, and Table 4. True or False: The arithmetic mean is an appropriate measure of central tendency. Explain your answer.
2. Enter the shell data into R; Best to copy and stack the data in your spreadsheet, then import into R or R Commander. Once imported, don’t forget to change Sample to character, otherwise R will treat Sample as ratio scale data type. Run your one-way ANOVA and calculate the intraclass correlation (ICC) for the dataset. Is the shell length measure repeatable?
3. The marathon results in Table 2 are paired data, results of two races run by the same individuals. Make a profile plot for the data (see Ch10.3 for help).
4. A profile plot is used to show paired data; for two or more repeat measures, use a line graph instead.
   A. Produce a line graph for data in Table 4. The line graph is an option in R Commander (Rcmdr: Graphs Line graph…), and can be generated with the following simple R code:

lineplot(Sample, Measure1, Measure2, Measure3)

B. What trends, if any, do you see? Report your observations.
C. Make another line plot for data in Table 2. Report any trend and include your observations.

Quiz Chapter 12.3

Fixed effects, random effects, and agreement

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Chapter 12 contents

• Introduction
• The need for ANOVA
• One way ANOVA
• Fixed effects, random effects, and agreement
• ANOVA from “sufficient statistics”
• Effect size for ANOVA
• ANOVA posthoc tests
• Many tests one model
• References and suggested readings

11.4 – Two sample effect size

Introduction

An effect size is a measure of the strength of the difference between two samples. Effect size tells us the practical importance of a result, say, the difference in outcome between a control group and the treatment group. A combined effect size is a common goal of any meta-analysis (Chapter 20.15).

By including effect sizes, this complement inferential statistics such as p-values. Importantly, effect size moves us beyond “significant” or “not significant.” Significance tells us a difference is likely between the groups, but offers no information about the practicality of the observed difference. After all, large studies inherently have more statistical power, which, by definition implicates even small differences can be statistically significant.

In statistics there are many estimates of effect size: for categorical data, odds ratio and relative risk between exposed and unexposed groups; for quantitative data, correlation, regression slope are effect size measures, in addition to the difference between group means. For differences in means between groups, Cohen’s d (and Hedge’s g) are common statistics. In ANOVA, eta-squared () and omega-squared (), summarizes the proportion of total variance in the dependent variable explained by the independent variables while the coefficient of determination, represents the proportion of variance in the outcome explained by the predictors. Advice and help for understanding “effect size” is an enormously popular topic — nearly 3 million hits in Google Scholar and about forty-five thousand papers in Pubmed. Two highly cited papers are those by Durlak (2009) and Ferguson (2009).

The effect size statistic, Cohen’s d, is calculated by subtracting one sample mean from the other and dividing by the pooled standard deviation.

where s_pooled is the pooled standard deviation for the two sample means.

Hedge’s d, which uses the s_pooled is the pooled standard deviation for the two sample means instead of the pooled population standard deviation, is written as

The used in d is from an equation for pooled standard deviation was provided in Chapter 3.3, but we’ll give it again here.

An alternative version of Cohen’s d is available for the t-test test statistic value

A d of one (1) indicates the effect size is equal to one standard deviation; a d of two (2) indicates the effect size between two sample means is equal to two standard deviations, and so on.

The variance of Cohen’s d, , can be estimated as

Hedges’ g (Hedges 1981), which includes a Bessel’s correction for bias in small samples, includes “ in the denominator of the pooled standard deviation calculation.

The “-2” forces the pooled variance to be larger — that makes g more conservative than d, which is of use for small sample studies.

A simple modification converts d to g. Multiple d by a small-sample size correction factor J (Durlak 2009):

What makes a large effect size?

Cohen cautiously suggested that values of d can be interpreted as

0.2 – small effect size

0.5 – medium effect size

0.8 – large effect size

That is, if the two group means don’t differ by much more than 0.2 standard deviations, than the magnitude of the treatment effect is small and unlikely to be biologically important, whereas a d = 0.8 or more would indicate a difference of 0.8 standard deviations between the sample means and, thus, likely to be an important treatment effect. Cohen (1992) provided these guidelines based on the following argument. The small effect 0.2 comes from the idea that it is much worse to conclude there is an effect when in fact there is no effect of the treatment rather than the converse (conclude no effect when there is an effect). The ratio of the Type II error (0.2) divided by the Type I error (0.05) gives us the penalty of 4. Similarly, for a moderate effect, equals 10. Clearly, these are only guidelines (see Lakens 2013).

Examples

The difference in average body size between six week old females of two strains of lab mice is 0.4 g (Table 1), and increases to 1.38 g by 16 weeks (Table 2).

Table 1. Average body weights of 6 week old female mice of two different inbred strains.†

Strain		s
C57BL/6J	18.5	0.9
CBA/J	18.1	1.27

†Source: Jackson Laboratories: C57BL/6J; CBA/J

Table 2. Average body weights of 16 week old female mice of two different inbred strains.†

Strain		s
C57BL/6J	23.9	2.3
CBA/J	25.38	3.76

†Source: Jackson Laboratories: C57BL/6J; CBA/J

The descriptive statistics are based on weights of 360 individuals in each strain (Jackson Labs).

The differences are both statistically significant from a independent t-test, i.e., p-value less than 0.05. I’ll show you how to calculate independent t-test given summary statistics (means, standard deviations), for Table 1 data, then will ask you to do this on your own in Questions.

Write an R script, example data from Table 1

sdd1 = 0.9
var1 = sdd1^2
sdd2 = 1.27
var2 = sdd2^2
mean1 = 18.5
mean2 = 18.1
n1 = 360
n2 = 360
dff = n1+n2-2
pooledSD <-sqrt((var1+var2)/2)
pooledSEM <-sqrt(var1/n1 + var2/n2); pooledSEM
tdiff<-(mean1-mean2)/pooledSEM; tdiff
pt(tdiff, df=dff, lower.tail=FALSE)
#get two-tailed p-value
2*0.0000006675956 
#get cohen's d
2*tdiff/sqrt(dff)

Results from the calculations we report (value of the test statistic, degrees of freedom, p-value), and the effect size, then are

t = 4.875773, df = 718, p-value = 0.0000006675956
cohen's d = 0.364

Now, I’m from the school of “don’t reinvent the wheel” or “someone has already solved your problems” (Freeman et al 2008), when it comes to coding problems. And, as you would expect, of course someone has written a function to calculate the t-test given summary statistics. In addition to base R and the pwr package (see Chapter 11.5), the package BSDA contains several nice functions for power calculations.

To follow this example, install BSDA, then run the following code

require(BSDA)
tsum.test(mean1, sdd1, n1, mean2, sdd2, n2, alternative = "two.sided", mu = 0, var.equal = TRUE, conf.level = 0.95)

R output

Standard Two-Sample t-Test

data: Summarized x and y
t = 4.8758, df = 718, p-value = 0.000001335
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.2389364 0.5610636
sample estimates:
mean of x mean of y 
18.5 18.1

Similarly, Cohen’s d is available from a package called effsize.

Questions

1. We needed an equation to calculate pooled standard error of the mean (pooledSEM in the R code). Read the code and write the equation used to calculate the pooled SEM.
2. Calculate the t-test and the effect size for the Table 1 data, but at three smaller sample sizes. Change from 360 for n₁ = n₂ = 20, repeat for n₁ = n₂ = 50, and finally, repeat for n₁ = n₂ = 100. Use your own code, or use the tsum.test function from the BSDA package.
3. Calculate Cohen’s effect size d for each new calculation based on different sample size.
4. Create a table to report the p-values from the t-tests, the effect size, for each of the four n₁ = n₂ = (20, 50, 100, 360).
5. True or false. The mean difference between sample means remains unaffected by sample size.
6. True or false. The effect size between sample means remains unaffected by sample size.
7. Based on comparisons in your table, what can you conclude about p-value and “statistical significance?” About effect size?
8. Repeat questions 2 – 7 for Table 2.

Quiz Chapter 11.4

Two sample effect size

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Chapter 11 contents

• Introduction
• What is Statistical Power?
• Prospective and retrospective power
• Factors influencing statistical power
• Two sample effect size
• Power analysis in R
• References and suggested readings

9.4 – Heterogeneity chi-square tests

Planned comparisons and meta-analysis.

After finding evidence to reject the statistical null hypothesis, it may be appropriate to proceed to test a number of data sets against the same theoretical distribution. Provided these are planned comparisons, part of the experiment and not an exercise in data dredging (Norman 2014), you may proceed to test a series of nested models where groups are combined to make new groups for comparison. This approach is analogous to the post-hoc tests one may conduct after a statistically significant ANOVA (see Chapter 12.4) or when identifying a best-fit regression model (see Chapter 18.4).

Combining groups to make new comparisons — this scenario is consistent with a meta-analysis. A type of systematic review (Greenhalgh 1997, Akobeng 2005), meta-analysis is a statistical method that combines the results from multiple independent studies on the same topic to provide a more precise and robust overall conclusion. By combining studies, the goal is to calculate a pooled estimate as a measure of overall effectiveness of the treatment. Meta-analysis can increase the overall sample size, which improves the study’s ability to detect an effect if one exists. So, how to combine results from different studies ostensibly addressing the same hypothesis? Inevitably, some subjectivity will come into decisions about including or excluding studies — the goal is to identify quality studies, which may have conflicting conclusions and test whether a single conclusion of effectiveness is warranted based on a summary of evidence.

Here we introduce the Cochran’s Q test (Cochran 1954, Hoaglin 2016), aka a special case of the more general heterogeneity chi-square test. The null hypothesis is no significant difference in the proportion of “successes” across all groups being compared — any observed differences between the groups are just due to chance. The test statistic, Q, has a chi-square distribution with degrees of freedom. A large Q value and small p-value suggest significant heterogeneity of results among the studies.

Heterogeneity tests ask whether results from combined studies are consistent; other tests exist, and are detailed in our special chapter on conducting meta-analysis. For example, I² or I-squared test provides a measure of how much variation in the combined studies is due to true differences  rather than by chance. In short, the I² test quantifies how much — the percentage — the different studies agree, while the chi-square test returns a p-value that can be used to interpret whether that heterogeneity is statistically significant. We outline the basics of meta-analysis in Chapter 20.15. Other tests include DerSimonian-Laird test and Mandel-Paule test — these use the Q statistic and its distribution to estimate the between-study variance and provide better estimators with small sample size. Between-study variance is used to determine if a fixed-effect or a random-effects model (Chapter 12.3 and Chapter 14.3), is appropriate for the analysis. A random-effects model is used when significant heterogeneity is present (Higgins et al 2009).

Example.

Imagine a scenario where a population geneticist has collected allele (gene) frequency and genotype frequency data on single nucleotide polymorphisms (SNP) for the BRCA1 locus for three USA groups, African Americans, Asian Americans, and European Americans. Data of this kind can be obtained from the SNP database at NCBI (data retrieved 14 & 15 July 2014), and I collated several SNP involving a Cytosine base for Thymine base change (Table 1). The BRCA1 locus is on chromosome 17 and some of the hundreds of mutations found for this gene are associated with high risk of breast and or ovarian cancer (Couch and Weber 1996, Tram et al 2013).

Table 1. SNP of BRCA1 locus

We may wish to test the combined data set to see if the large data set differs from Hardy Weinberg expectations before proceeding with a series of sample populations. By combining the data sets, we will be able to test whether one genotype is different from exception.

The goal then is to pool the data so that you have a more powerful test of the null hypothesis (remember our general discussion of statistical power and how increasing sample size increases your chances of correctly rejecting the null hypothesis).

For example, in our test of Hardy-Weinberg expectations (Example C), we calculated a test of 7.8955. This test has k – 1 = 3 – 1 = 2 degrees of freedom. At alpha 5%, the critical value was 5.991 We clearly would reject the null hypothesis. But do we reject because of a difference in the aa, ab, or bb genotypes? Simply combine categories. Let’s test the homozygotes (aa and bb) versus the heterozygotes (ab).

Table 2. Worksheet

	Categories
	aa + bb	ab	n
Observed	66	34	100
Expected	52	48	100

With one degree of freedom, we clearly reject the null hypothesis because the critical value at 5% and one degree of freedom is equal to 3.841. To get the p-value, use

pchisq(7.852,df=1,lower.tail=FALSE)

which will return

[1] 0.005076452

Oh, and what was the null hypothesis? That there was an equal number of homozygotes and heterozygotes.

Other tests would be possible here, but the point is that you can dissect an experiment to determine which group is causing you to reject the null original hypothesis. While this is an important tool, you should also consider issues of apriori and aposterori decisions in experiments.

Questions.

1. The SNP rs1799971 (A > G) in the μ-opioid receptor (MOR) is associated with opioid dependency. Allele frequencies for different populations are provided in the table. Global frequency for A at this SNP is 0.188 (therefore frequency of G is 0.812).

• Combine the data sets and test whether genotype frequency is different from the reported global exception.
• Test each population separately.

2. A behavioral ecologist studies the effect of three diets (high-protein, high-carbohydrate, and mixed) on the growth rate of juvenile frogs. Before collecting any data, she writes a protocol stating that she will compare only the high-protein vs mixed diet, based on a specific hypothesis about amino acid availability.

After analyzing the data, she also notices that frogs on the high-carbohydrate diet seem smaller, so she performs several additional pairwise comparisons that were not part of her original plan, reporting only the ones that are statistically significant.

Which statement best distinguishes the planned comparison from data dredging (p-hacking)?

A. The planned comparison is valid because it uses a larger sample size, while data dredging uses small samples.
B. The planned comparison was specified ahead of time based on a biological hypothesis, while data dredging involves unplanned tests added after seeing the data to search for significant results.
C. The planned comparison uses ANOVA, while data dredging uses regression.
D. The planned comparison must always produce a significant result, while data dredging never does.

Quiz Chapter 9.4

Heterogeneity chi-square tests

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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

9.2 – 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

7.3 – Conditional Probability and Evidence Based Medicine

Conditional probability and risk communication.

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

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

Probability and independent events.

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

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

Example of multiple, independent events.

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

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

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

Probabilistic risk analysis.

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

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

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

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

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

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

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

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

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

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

Conditional probability of non-independent events.

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

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

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

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

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

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

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

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

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

Group 2: Murdered by someone else:

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

or 89%.

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

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

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

Diagnosis from testing.

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

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

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

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

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

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

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

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

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

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

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

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

Now, the formula is

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

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

Standardized million.

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

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

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

Per capita rate.

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

Practice and introduce PPV and test decision criteria.

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

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

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

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

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

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

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

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

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

or the F1-score,

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

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

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

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

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

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

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

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

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

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

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

final_df <- as.data.frame(output_matrix)

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

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

End R code.

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

Table 4. Decision criteria.

	Value	Base rate dependent?	Interpretation
	0.65	No	Good
	0.0306	Yes	Poor
	0.0011	Yes	Unacceptable clinical utility
	0.0030	Yes	Near-total prediction failure

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

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

Evidence Based Medicine.

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

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

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

Software.

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

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

R code

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

R output

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

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

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

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

And R returns

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

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

What about R Commander menus?

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

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

install.packages("RcmdrPlugin.EBM")

From within R Commander (Fig 6), select

Tools → Load Rcmdr plug-in(s)…

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

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

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

Restart Rcmdr again (Fig 8),

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

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

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

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

Rcmdr → Tools → Save Rcmdr options…

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

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

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

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

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

Example.

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

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

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

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

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

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

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

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

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

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

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

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

Figure 13. EBM plugin with data entry.

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

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

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

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

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

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

Questions.

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

   Mammogram	Deaths/1000 women
   No	4
   Yes	3

   data from Table 5-1 p. 60 Gigerenzer (2002)
   What is the RRR?

6. A study was conducted to see if mammograms reduced mortality
   

   Mammogram	Deaths/1000 women
   No	4
   Yes	3

   data from Table 5-1 p. 60 Gigerenzer (2002)
   What is the NNT?

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

Quiz Chapter 7.3

Conditional Probability and Evidence Based Medicine

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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

7.2 – Epidemiology basics

Introduction

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

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

Prevalence rate

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

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

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

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

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

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

which returns

[1] 1.959964

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

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

Figure 1. R Commander popup menu for Normal quantiles.

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

Prevalence: R code

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

vignette(package="epiR")

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

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

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

R output

   est    lower    upper
1    8 6.394198 9.857978

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

Incidence rate

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

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

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

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

Person-time

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

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

The results are shown below

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

Add up the person days,

Now calculate the incidence rate,

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

Incidence rate: R code

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

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

R output

          est    lower    upper
ncas 12.71186 2.621492 37.14946

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

Age-specific rates

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

Age-adjusted rates

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

Questions

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

Quiz

Quiz Chapter 7.2

Epidemiology basics

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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