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

9.1 – Chi-square test: Goodness of fit

Introduction

We ask about the “fit” of our data against predictions from theory, or from rules that set our expectations for the frequency of particular outcomes drawn from outside the experiment. Three examples to illustrate goodness of fit, GOF, , A, B, and C follow.

A. For example, for a toss of a coin, we expect heads to show up 50% of the time. Out of 120 tosses of a fair coin, we expect 60 heads, 60 tails. Thus, our null hypothesis would be that heads would appear 50% of the time. If we observe 70 heads in an experiment of coin tossing, is this a significantly large enough discrepancy to reject the null hypothesis?

B. For example, simple Mendelian genetics makes predictions about how often we should expect particular combinations of phenotypes in the offspring when the phenotype is controlled by one gene, with 2 alleles and a particular kind of dominance.

For example, for a one locus, two allele system (one gene, two different copies like R and r) with complete dominance, we expect the phenotypic (what you see) ratio will be 3:1 (or round, wrinkle). Our null hypothesis would be that pea shape will obey Mendelian ratios (3:1). Mendel’s round versus wrinkled peas (RR or Rr genotypes give round peas, only rr results in wrinkled peas).

Thus, out of 100 individuals, we would expect 75 round and 25 wrinkled. If we observe 84 round and 16 wrinkled, is this a significantly large enough discrepancy to reject the null hypothesis?

C. For yet another example, in population genetics, we can ask whether genotypic frequencies (how often a particular copy of a gene appears in a population) follow expectations from Hardy-Weinberg model (the null hypothesis would be that they do).

This is a common test one might perform on DNA or protein data from electrophoresis analysis. Hardy-Weinberg is a simple quadratic expansion:

If p = allele frequency of the first copy, and q = allele frequency of the second copy, then p + q = 1,

Given the allele frequencies, then genotypic frequencies would be given by 1 = p² + 2pq + q².

Deviations from Hardy-Weinberg expectations may indicate a number of possible causes of allele change (including natural selection, genetic drift, migration).

Thus, if a gene has two alleles, and , with the frequency for , and for ,  (equivalently q = 1 – p) in the population, then we would expect 36 , 16 , and 48 individuals. (Nothing changes if we represent the alleles as A and a, or some other system, eg, dominance/recessive.)

Question. If we observe the following genotypes: 45  individuals, 34 individuals, and 21 individuals, is this a significantly large enough discrepancy to reject the null hypothesis?

Table 1. Summary of our Hardy Weinberg question

Recall from your genetics class that we can get the allele frequency values from the genotype values, eg,

We call these chi-square tests, tests of goodness of fit. Because we have some theory, in this case Mendelian genetics, or guidance, separate from the study itself, to help us calculate expected values in a chi-square test.

goodness of fit

For k groups, the equation for the chi-square test may be written as

where f_i is the frequency (count) observed (in class i) and f_i<hat> is the frequency (count) expected if the null hypothesis is true, sum over all k groups. Alternatively, here is a format for the same equation that may be more familiar to you… ?

where O_i is the frequency (count) observed (in class i) and E_i is the frequency (count) expected if the null hypothesis is true.

The degrees of freedom, df, for the GOF are simply the number of categories minus one, k – 1.

Explaining GOF

Why am I using the phrase “goodness of fit?” This concept has broad use in statistics, but in general it applies when we ask how well a statistical model fits the observed data.

At least for the chi-square test it is simple to see how the test statistic increases from zero as the agreement between observed data and expected data depart, where zero would be the case in which all observed values for the categories exactly match the expected values.

The goodness of fit test is designed to evaluate whether or not your data agree with a theoretical expectation (there are additional ways to think about this test, but this is a good place to start). Let’s take our time here and work with an example. The other type of chi-square problem or experiment is one for the many types of experiments in which the response variable is discrete, just like in the GOF case, but we have no theory to guide us in deciding how to obtain the expected values. We can use the data themselves to calculate expected values, and we say that the test is “contingent” upon the data, hence these types of chi-square tests are called contingency tables.

You may be a little concerned at this point that there are two kinds of chi-square problems, goodness of fit and contingency tables. We’ll deal directly with contingency tables in the next section, but for now, I wanted to make a few generalizations.

1. Both goodness of fit and contingency tables use the same chi-square equation and analysis. They differ in how the degrees of freedom are calculated.
2. Thus, what all chi-square problems have in common, whether goodness of fit or contingency table problems
   1. You must identify what types of data are appropriate for this statistical procedure? Categorical (nominal data type).
   2. As always, a clear description of the hypotheses being examined.

For goodness of fit chi-square test, the most important type of hypothesis is called a Null Hypothesis: In most cases the Null Hypothesis (H_O) is “no difference” “no effect”…. If H_O is concluded to be false (rejected), then an alternative hypothesis (H_A) will be assumed to be true. Both are specified before tests are conducted. All possible outcomes are accounted for by the two hypotheses.

From above, we have

A. H_O: Fifty out of 100 tosses will result in heads.

H_A: Heads will not appear 50 times out of 100.

B. H_O: Pea shape will equal Mendelian ratios (3:1).

H_A: Pea shape will not equal Mendelian ratios (3:1).

C. H_O: Genotypic frequencies will equal Hardy-Weinberg expectations.

H_A: Genotypic frequencies will not equal Hardy-Weinberg expectations

Assumptions: In order to use the chi-square, there must be two or more categories. Each observation must be in one and only one category. If some of the observations are truly halfway between two categories then you must make a new category (eg, low, middle, high) or use another statistical procedure. Additionally, your expected values are required to be integers, not ratio. The number of observed and the number of expected must sum to the same total.

The chi-square test is a good example of such tests, and we will encounter other examples too. Another common goodness of fit is the coefficient of determination, which will be introduced in linear regression sections (see Chapter 17.7 – Regression model fit). Still other examples are the likelihood ratio test (LRT), Akaike Information Criterion (AIC), and Bayesian Information Criterion (BIC), which are all used to assess fit of models to data. (See Graffelman and Weir [2018] for how to use AIC in the context of testing for Hardy Weinberg equilibrium.)

The likelihood ratio test is used to compare the goodness of fit of two hierarchically nested models to determine if a more complex model provides a significantly better fit to the data. It does this by forming a ratio of the likelihood of the data under the two models (the simpler “null” model and the more complex “alternative” model). A small ratio suggests the simpler model is a poor fit, while a value close to one indicates the added complexity is not statistically significant. AIC (and BIC) is a measure of relative fit — it balances goodness of fit and model complexity: A lower AIC suggests a better model by finding the optimal trade-off between fitting the data well and keeping the model simple. A model that is too complex will have a higher AIC than a simpler model with a nearly as good fit.

We use LRT, AIC and BIC when we talk about selecting best regression models (see Chapter 17.7 – Regression model fit). However, as we discuss the concept of comparing models, we need to distinguish between the concept of model choice — methods used to select a best model from competing options and estimating the “goodness of fit” of a single model.

How well does data fit the prediction?

Frequentist approach interprets the test as, how well does the data fit the null hypothesis, ? When you compare data against a theoretical distribution (eg, Mendel’s hypothesis predicts the distribution of progeny phenotypes for a particular genetic system), you test the fit of the data against the model’s predictions (expectations). Recall that the Bayesian approach asks how well does the model fit the data?

Table 2. A. 120 tosses of a coin, we count heads 70/120 tosses.

Table 3. B. A possible Mendelian system of inheritance for a one gene, two allele system with complete dominance, observe the phenotypes.

Table 4. C. A possible Mendelian system of inheritance for a one gene, two allele system with complete dominance, observe the phenotypes.

For completeness, instead of a goodness of fit test we can treat this problem as a test of independence, a contingency table problem. We’ll discuss contingency tables more in the next section, but or now, we can rearrange our table of observed genotypes for problem C, as a 2X2 table

Table 5. Problem C reported in 2X2 table format.

The contingency table is calculated the same way as the GOF version, but the degrees of freedom are calculated differently: df = number of rows – 1 multiplied by the number of columns – 1.

Thus, for a 2X2 table the df are always equal to 1.

Note that the chi-square value itself says nothing about how any discrepancy between expectation and observed genotype frequencies come about. Therefore, one can rearrange the  equation to make clear where deviance from equilibrium, D, occur for the heterozygote (het). We have

where D² is equal to

φ coefficient

The chi-square test statistic and its inference tells you about the significance of the association, but not the strength or effect size of the association. Not surprisingly, Pearson (1904) came up with a statistic to quantify the strength of association between two binary variables, now called the φ (phi) coefficient. Like the Pearson product moment correlation, the φ (phi) coefficient takes values from -1 to +1.

The formula for the absolute value of φ coefficient is

Thus, for A, B, and C examples, φ coefficient was 0.167, 0.190, and 0.771. Thus, only weak associations in examples A and B, but strong association in C. We’ll provide a formula to directly calculate φ coefficient from the cells of the table in 9.2 – Chi-square contingency tables.

Carry out the test and interpret results

What was just calculated? The chi-square, , test statistic.

Just like t-tests, we now want to compare our test statistic against a critical value — calculate degrees of freedom (df = k – 1, k equals the numbers of categories), and set a rejection level, Type I error rate. We typically set the Type I error rate at 5%. A table of critical values for the chi-square test is available in Appendix Table Chi-square critical values.

Obtaining Probability Values for the goodness-of-fit test of the null hypothesis:

As you can see from the equation of the chi-square, a perfect fit between the observed and the expected would be a chi-square of zero. Thus, asking about statistical significance in the chi-square test is the same as asking if your test statistic is significantly greater than zero.

The chi-square distribution is used and the critical values depend on the degrees of freedom. Fortunately for  and other statistical procedures we have Tables that will tell us what the probability is of obtaining our results when the null hypothesis is true (in the population).

Here is a portion of the chi-square critical values for probability that your chi-square test statistic is less than the critical value (Fig 1).

Figure 1. A portion of critical values of the chi-square at alpha 5% for degrees of freedom between 1 and 10. A more inclusive table is provided in the Appendix, Table of Chi-square critical values.

For the first example (A), we have df = 2 – 1 = 1 and we look up the critical value corresponding to the probability in which Type I = 5% are likely to be smaller iff (“if and only if”) the null hypothesis is true. That value is 3.841; our test statistic was 3.330, and therefore smaller than the critical value: so we do not reject the null hypothesis.

Interpolating p-values

How likely is our test statistic value of 3.333 and the null hypothesis was true? (Remember, “true” in this case is a shorthand for our data was sampled from a population in which the HW expectations hold). When I check the table of critical values of the chi-square test for the “exact” p-value, I find that our test statistic value falls between a p-value 0.10 and 0.05 (represented in the table below). We can interpolate

Table 6. Interpolated p-value for critical value not reported in chi-square table.

If we assume the change in probability between 2.706 and 3.841 for the chi-square distribution is linear (it’s not, but it’s close), then we can do so simple interpolation.

We set up what we know on the right hand side equal to what we don’t know on the left hand side of the equation,

and solve for x. Then, x is equal to 0.0724

R function pchisq() gives a value of p = 0.0679. Close, but not the same. Of course, you should go with the result from R over interpolation; we mention how to get the approximate p-value by interpolation for completeness, and, in some rare instances, you might need to make the calculation. Interpolating is also a skill used to provide estimates where the researcher needs to estimate (impute) a missing value.

Interpreting p-values

This is a pretty important topic, so much so that we devote an entire section to this very problem — see 8.2 – The controversy over proper hypothesis testing. If you skipped the chapter, but find yourself unsure how to interpret the p-value, then please go back to Ch 8.2. OK, that commercial message, what does it mean to “reject the chi-square null hypothesis?” These types of tests are called goodness of fit in the following sense — if your data agree with the theoretical distribution, then the difference between observed and expected should be very close to zero. If it is exactly zero, then you have a perfect fit. In our coin toss case, if we say that the ratio of heads:tails do not differ significantly from the 50:50 expectation, then we accept the null hypothesis.

You should try the other examples yourself! A hint, the degrees of freedom are one (1) for example B and two (2) for example C.

R code

Printed tables of the critical values from the chi-square distribution, or for any statistical test for that matter are fine, but with your statistical package R and Rcmdr, you have access to the critical value and the p-value of your test statistic simply by asking. Here’s how to get both.

First, let’s get the critical value.

Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Chi-squared quantiles (Fig 2).

Figure 2. R Commander menu for Chi-squared quantiles.

I entered “0.05” for the probability because that’s my Type I error rate α. Enter “1” for Degrees of freedom, then click “upper tail” because we are interested in obtaining the critical value for α. Here’s R’s response when I clicked “OK.”

qchisq(c(0.05), df=1, lower.tail=FALSE)
[1] 3.841459

Next, let’s get the exact P-value of our test statistic. We had three from three different tests:  for the coin-tossing example, for the pea example, and for the Hardy-Weinberg example.

Rcmdr: Distributions → Continuous distributions → Chi-squared distribution → Chi-squared probabilities… (Fig 3).

Figure 3. R Commander menu for Chi-squared probabilities.

I entered “3.333” because that is one of the test statistics I want to calculate for probability and “1” for Degrees of freedom because I had k – 1 = 1 df for this problem. Here’s R’s response when I clicked “OK.”

pchisq(c(3.333), df=1, lower.tail=FALSE)
[1] 0.06790291

I repeated this exercise for  I got ; for I got .

Nice, right? Saves you from having to interpolate probability values from the chi-square table.

3. How to get the goodness of fit  in Rcmdr.

R provides the goodness of fit  (the command is chisq.text()), but Rcmdr thus far does not provide a menu option to link to the function. Instead, R Commander provides a menu for contingency tables, which also is a chi-square test, but is used where no theory is available to calculate the expected values (see Chapter 9.2). Thus, for the goodness of fit chi-square, we will need to bypass Rcmdr in favor of the script window. Honestly, other options are as quick or quicker: calculate by hand, use a different software (eg, Microsoft Excel), or many online sites provide JavaScript tools.

So how to get the goodness of fit chi-square while in R? Here’s one way. At the command line, type

chisq.test (c(O1, O2, ... On), p = c(E1, E2, ... En))

where O1, O2, … On are observed counts for category 1, category 2, up to category n, and E1, E2, … En are the expected proportions for each category. For example, consider our Heads/Tails example above (problem A).

In R, we write and submit

chisq.test(c(70,30),p=c(1/2,1/2))

R returns

Chi-squared test for given probabilities.
data: c(70, 30) 
X-squared = 16, df = 1, p-value = 0.00006334

Easy enough. But not much detail — details are available with some additions to the R script. I’ll just link you to a nice website that shows how to add to the output so that it looks like the one below.

mike.chi <- chisq.test(c(70,30),p=c(1/2,1/2))

Let’s explore one at time the contents of the results from the chi square function.

names(mike.chi)   #The names function 
[1] "statistic" "parameter" "p.value" "method" "data.name" "observed"
[7] "expected" "residuals" "stdres"

Now, call each name in turn.

mike.chi$residuals
[1] 2.828427 -2.828427
mike.chi$obs
[1] 70 30
mike.chi$exp
[1] 50 50

And finally, let’s get the summary output of our statistical test.

mike.chi
Chi-squared test for given probabilities.
data: c(70, 30)
X-squared = 16, df = 1, p-value = 6.334e-05

GOF and spreadsheet apps

Easy enough with R, but it may even easier with other tools. I’ll show you how to do this with spreadsheet apps and with and online at graphpad.com.

Let’s take the pea example above. We had 16 wrinkled, 84 round. We expect 25% wrinkled, 75% round.

Now, with R, we would enter

chisq.test(c(16,80),p=c(1/4,3/4))

and the R output

Chi-squared test for given probabilities
data: c(16, 80)
X-squared = 3.5556, df = 1, p-value = 0.05935

Microsoft Excel and the other spreadsheet programs (Apple Numbers, Google Sheets, LibreOffice Calc) can calculate the goodness of fit directly; they return a P-value only. If the observed data are in cells A1 and A2, and the expected values are in B1 and B2, then use the procedure =CHITEST(A1:A2,B1:B2).

Table 7. A spreadsheet with formula visible.

The P-value (but not the Chi-square test statistic) is returned. Here’s the output from Calc.

Table 8. Spreadsheet example from Table 7 with calculated P-value.

You can get the critical value from MS Excel (=CHIINV(alpha, df), returns the critical value), and the exact probability for the test statistic =CHIDIST(x,df), where x is your test statistic. Putting it all together, a general spreadsheet template for goodness of fit calculations calculations of test statistic and p-value might look like

Table 9. Spreadsheet template with formula to calculate chi-square goodness of fit on two groups.

	A	B	C	D	E
1	f1	0.75
2	f2	0.25
3	N	=SUM(A5,A6)
4	Obs	Exp	Chi.value	Chi.sqr
5	80	=B1*3	=((A5-B5)^2)/B5	=SUM(C5,C6)
6	16	=B2*3	=((A6-B6)^2)/B6		=CHIDIST(D5, COUNT(A5:A6-1)
7
8

Microsoft Excel can be improved by writing macros, or by including available add-in programs, such as the free PopTools, which is available for Microsoft Windows 32-bit operating systems only.

Another option is to take advantage of the internet — again, many folks have provided java or JavaScript-based statistical routines for educational purposes. Here’s an easy one to use www.graphpad.com.

In most cases, I find the chi-square goodness-of-fit is so simple to calculate by hand that the computer is redundant.

Questions

1. A variety of p-values were reported on this page with no attempt to reflect significant figures or numbers of digits (see Chapter 8.2). Provide proper significant figures and numbers of digits as if these p-values were reported in a science journal.

1. 0.0724
2. 0.0679
3. 0.03766692
4. 0.004955794
5. 0.00006334
6. 6.334e-05
7. 0.05935
8. 0.058714340077662

2. For a mini bag of M&M candies, you count 4 blue, 2 brown, 1 green, 3 orange, 4 red, and 2 yellow candies.

1. What are the expected values for each color?
2. Calculate using your favorite spreadsheet app (eg, Numbers, Excel, Google Sheets, LibreOffice Calc)
3. Calculate using R (note R will reply with a warning message that the “Chi-squared approximation may be incorrect”; see 9.2 Yates continuity correction)
4. Calculate using Quickcalcs at graphpad.com
5. Construct a table and compare p-values obtained from the different applications

3. CYP1A2 enzyme involved with metabolism of caffeine. Folks with C at SNP rs762551 have higher enzyme activity than folks with A. Populations differ for the frequency of C. Using R or your favorite spreadsheet application, compare the following populations against global frequency of C is 33% (frequency of A is 67%).

1. 286 persons from Northern Sweden: f(C) = 26%, f(A) = 73%
2. 4532 Native Hawaiian persons: f(C) = 22%, f(A) = 78%
3. 1260 Native American persons: f(C) = 30%, f(A) = 70%
4. 8316 Native American persons: f(C) = 36%, f(A) = 64%
5. Construct a table and compare p-values obtained for the four populations.

Quiz Chapter 9.1

Chi-square test: Goodness of fit

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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

3 – Exploring data

Describing data: An introduction to summary statistics and basic visualization.

.

Scientists answer questions about natural phenomena using the scientific method. The questions in the form of hypotheses are tested by data, the raw information collected either by observing natural outcomes among groups (observational studies) or by manipulating variables (experimental studies) and recording the outcomes. 

Regardless, after a study has been completed, the collection of observations needs to be summarized or described, a process now referred to as data exploration. Interpretation of the data constitutes the evidence used to evaluate the hypothesis. How many mice grew tumors in the treated versus control group? Did all mice respond to treatment? What is the typical cost of a new home in Hilo, HI? How large of a difference is there between the most expensive home and the middle? These questions require simple, summary statistics also called descriptive statistics and informative data visualizations. Both are required of any data analysis report (students: see Making a report in Mike’s Workbook for Biostatistics) — to communicate the findings of the experiment or observational study. 

In Chapter 3 we introduce two aspects of summary statistics expected in any report about a data set (see Chapter 4 — How to report statistics). Summary statistics provide a brief overview of relevant characteristics of observations in the data set to provide the reader to get a quick but meaningful look at the data set. Data set refers to a collection of data, usually a collection of related, ordered observations, measurements, and related information.

Summary statistics vary according to the needs of the reporting vehicle, but may generally address two characteristics of the data set

1. Central tendency or the description of the middle of the data. See Chapter 3.2 – Measures of Central Tendency.
2. Data dispersion, the variability around the middle of the data set. See Chapter 3.3 – Measures of dispersion.
3. The other two sections in Chapter 3 introduce estimation, the role random sampling plays, and the kinds of error in statistics (Chapter 3.4 — Estimating parameters).
4. The final section, 3.5 – Statistics of error, introduces incorporating estimates of error in measurement calculations are used to report confidence — confidence intervals — in the estimates we make.

Before we can talk about descriptive statistics we need to introduce the concept of data types, which we do in Chapter 3.1 — Data types. Understanding data types is crucial for accurately calculating and interpreting measures of central tendency and dispersion, which are essential for effective data visualization.

We introduce statistical graphics, a specialized kind of data visualization, in Chapter 4 – How to report statistics . Statistical graphs are utilized to describe data sets, but also to communicate statistical inference, which we address formally beginning in Chapter 7 – Probability, Risk Analysis.

Formulas — a reminder.

Most of you have been asked at some point to calculate the average or the sample standard deviation. We will provide these again, but with additional statistics. Textbooks may present calculator formulas — nothing wrong with them, although we have to watch significant figures. But computer statistical packages generally do not use these formulations as the approach to calculate — the algorithms are typically much more involved. So why present formulas throughout the book as I do? Formulas often are the best definition of the statistical concept. I haven’t investigated this last point in any systematic way, but, because of access by scientists to increasingly powerful computers since the 1980s, I doubt anyone in the business of data analysis in the biological sciences has much use of the hand calculator or the formulas except as definitions.

Homework to go with this topic.

Mike’s Workbook for Biostatistics:

Homework 2A: Descriptive statistics

Quizzes in this chapter

A total of 56 questions among the several subchapters, a mix of true or false and multiple choice question format.

Chapter 3 contents

.

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