18.3 – Logistic regression

Introduction

We briefly introduced logistic regression in the previous chapter on nonlinear regression. We expand our discussion of logistic regression here.

Logistic regression is a statistical method for modeling the dependence of a categorical (binomial) outcome variable on one or more categorical and continuous predictor variables (Bewick et al 2005).

The logistic function may be used to transform a sigmoidal curve to a more or less straight line while also changing the range of the data from binary (0 to 1) to infinity (-∞,+∞).  For event with probability of occurring p, the logistic function is written as

where ln refers to the natural logarithm.

This is an odds ratio. It represents the effect of the predictor variable on the chance that the event will occur.

The logistic regression model then very much resembles the same general linear models we have seen before.

In R and Rcmdr we use the glm() function to model the logistic function. Logistic regression is used to model a binary outcome variable. What is a binary outcome variable? It is categorical! Examples include: Living or Dead; Diabetes Yes or No; Coronary artery disease Yes or No. Male or Female. One of the categories could be scored 0, the other scored 1. For example, living might be 0 and dead might be scored as 1. (By the way, for a binomial variable, the mean for the variable is simply the number of experimental units with “1” divided by the total sample size.)

With the addition of a binary response variable, we are now really close to the Generalized Linear Model. Now we can handle statistical models in which our predictor variables are either categorical or ratio scale. All of the rules of crossed, balanced, nested, blocked designs still apply because our model is still of a linear form.

We write our generalized linear model

    \begin{align*} G \sim Model \end{align*}

just to distinguish it from a general linear model with the ratio-scale Y as the response variable.

Think of the logistic regression as modeling a threshold of change between the 0 and the 1 value. In another way, think of all of the processes in nature in which there is a slow increase, followed by a rapid increase once a transition point is met, only to see the rate of change slow down again. Growth is like that (see Chapter 20.10 for related growth and related models). We start small, stay relatively small until birth, then as we reach our early teen years, a rapid change in growth (height, weight) is typically seed (well, not in my case … at least for the height). The curve I described is a logistic one (other models exist too). Where the linear regression function was used to minimize the squared residuals as the definition of the best fitting line, now we use the logistic as one possible way to describe or best fit this type of a curved relationship between an outcome and one or more predictor variables. We then set out to describe a model which captures when an event is unlikely to occur (the probability of dying is close to zero) AND to also describe when the event is highly likely to occur (the probability is close to one).

A simple way to view this is to think of time being the predictor (X) variable and risk of dying. If we’re talking about the lifetime of a mouse (lifespan typically about 18-36 months), then the risk of dying at one months is very low, and remains low through adulthood until the mouse begins the aging process. Here’s what the plot might look like, with the probability of dying at age X on the Y axis (probability = 0 to 1) (Fig 1).

lifespan mice, Yuan et al 2012 PNAS 109:8224

Figure 1. Lifespan of 1881 mice from 31 inbred strains (Data from Yuan et al [2012] available at https://phenome.jax.org/projects/Yuan2).

Note 1: I labeled Y axis labeled “Survival Probability”; “Inverse Survival Probability” would be more accurate.

We ask — of all the possible models we could draw — which model best fits the data? The curve fitting process is called the logistic regression. The sample data set is listed at end of this page (scroll down of click here). Create data.frame called yuan.

With some minor, but important differences, running the logistic regression is the same as what you have been doing so far for ANOVA and for linear regression. In Rcmdr, access the logistic regression function by calling the Generalized Linear Model (Fig 2).

Screenshot Rcmdr, select Statistics, Fit models, Generalized linear model...

Figure 2. Access Generalized Linear Model via R Commander

R results

GLM.1 <- glm(cumFreq ~ Months, family=gaussian(identity), data=yuan)

> summary(GLM.1)

Call:
glm(formula = cumFreq ~ Months, family = gaussian(identity), 
data = yuan)

Deviance Residuals: 
     Min       1Q   Median      3Q     Max 
-0.11070 -0.07799 -0.01728 0.06982 0.13345

Coefficients:
             Estimate Std. Error t value Pr(>|t|) 
(Intercept) -0.132709   0.045757   -2.90 0.0124 * 
Months       0.029605   0.001854   15.97 6.37e-10 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for gaussian family taken to be 0.008663679)

Null deviance: 2.32129 on 14 degrees of freedom
Residual deviance: 0.11263 on 13 degrees of freedom
AIC: -24.808

Number of Fisher Scoring iterations: 2

Rcmdr: Statistics → Fit models → Generalized linear model.

Screenshot Rcmdr GLM

Figure 3. Screenshot Rcmdr GLM menu. For logistic on ratio-scale dependent variable, select gaussian family and identity link function.

Select the model as before. The box to the left accepts your binomial dependent variable; the box at right accepts your factors, your interactions, and your covariates. It permits you to inform R how to handle the factors: Crossed? Just enter the factors and follow each with a plus. If fully crossed, then the interactions may be specified with “:” to explicitly call for a two-way interaction between two (A:B) or a three-way interaction between three (A:B:C) variables. In the later case, if all of the two way interactions are of interest, simply typing A*B*C would have done it. If nested, then use %in% to specify the nesting factor.

R output

GLM.1 <- glm(cumFreq ~ Months, family=gaussian(identity), data=yuan)

summary(GLM.1)

Call:
glm(formula = cumFreq ~ Months, family = gaussian(identity),
data = yuan)

Deviance Residuals:
     Min       1Q   Median      3Q     Max
-0.11070 -0.07799 -0.01728 0.06982 0.13345

Coefficients:
               Estimate   Std. Error   t value   Pr(>|t|)
(Intercept)   -0.132709     0.045757   -2.90     0.0124 *
Months         0.029605     0.001854   15.97     6.37e-10 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for gaussian family taken to be 0.008663679)

Null deviance: 2.32129 on 14 degrees of freedom
Residual deviance: 0.11263 on 13 degrees of freedom
AIC: -24.808

Number of Fisher Scoring iterations: 2

Note 2: The %in% operator is an infix function used to determine if elements of a vector or object on the left-hand side are present within another vector or object on the right-hand side. Quietly, we introduced use of %in% in our code to subset data set in a BONUS problem.

Assessing fit of the logistic regression model

Some of the differences you will see with the logistic regression is the term “deviance.” deviance in statistics simply means compare one model to another and calculate some test statistic we’ll call “the deviance.” We then evaluate the size of the deviance like a chi-square goodness of fit. If the model fits the data poorly (residuals large relative to the predicted curve), then the deviance will be small and the probability will also be high — the model explains little of the data variation. On the other hand, if the deviance is large, then the probability will be small — the model explains the data, and the probability associated with the deviance will be small (significantly so? You guessed it! P < 0.05).

The Wald statistic

where n and β refers to any of the n coefficient from the logistic regression equation and SE refers to the standard error if the coefficient. The Wald test is used to test the statistical significance of the coefficients. It is distributed approximately as a chi-squared probability distribution with one degree of freedom. The Wald test is reasonable, but has been found to give values that are not possible for the parameter (eg, negative probability).

Likelihood ratio tests are generally preferred over the Wald test. For a coefficient, the likelihood test is written as

where L0 is the likelihood of the data when the coefficient is removed from the model (ie, set to zero value), whereas L1 is the likelihood of the data when the coefficient is the estimated value of the coefficient. It is also distributed approximately as a chi-squared probability distribution with one degree of freedom.

Nonlinear regression

Nonlinear regression, nls() function, may be a better choice. The following

attach(yuan)
logisticModel <-nls(cumFreq~DD/(1+exp(-(CC+bb*Months))), start=list(DD=1,CC=0.2,bb=.5),data=yuan,trace=TRUE)
summary(logisticModel)

Formula: yuan$cumFreq ~ DD/(1 + exp(-(CC + bb * yuan$Months)))

Parameters:
      Estimate   Std. Error    t value    Pr(>|t|)
DD    1.038504     0.014471      71.77    < 2e-16 ***
CC   -4.626982     0.175109     -26.42   5.29e-12 ***
bb    0.206899     0.008777      23.57   2.03e-11 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.01908 on 12 degrees of freedom

Number of iterations to convergence: 11
Achieved convergence tolerance: 0.000006909

Get fit statistics

AIC(logisticModel)
[1] -71.54679

Because AIC for the nonlinear model much smaller (more negative) than AIC for logistic model, we may be tempted to judge fit of the nonlinear regression as best. However, this comparison of models is not valid because the Y variables are different between the two models and the fit families are different. One option, evaluate fit of models by plots of residuals (see 17.7 – Regression model fit).

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 18.3

Logistic regression

Data set this page

MonthsfreqcumFreq
000
30.0106330.010633
60.0170120.027645
90.0451890.072834
120.0643270.137161
150.0648590.20202
180.097820.299841
210.1185540.418394
240.1711860.58958
270.1621480.751728
300.1371610.888889
330.0696440.958533
360.0244550.982988
390.0116960.994684
420.0053161

Chapter 18 contents

18.2 – Nonlinear regression

Introduction

The linear model is incredibly relevant in so many cases. A quick look “linear model” in PUBMED returns about 22 thousand hits; 3.7 million in Google Scholar; 3 thousand hits in ERIC database. These results compare to search of “statistics” in the same databases: 2.7 million (PUBMED), 7.8 million (Google Scholar), 61.4 thousand (ERIC). But all models are not the same.

Fit of a model to the data can be evaluated by looking at the plots of residuals (Fig 1), where we expect to find random distribution of residuals across the range of predictor variable.

A good residual plot, no obvious trends.

Figure 1. Ideal plot of residuals against values of X, the predictor variable, for a well-supported linear model fit to the data.

However, clearly, there are problems for which assumption of fit to line is not appropriate. We see this, again, in patterns of residuals, eg, Figure 2.

Residual plot shows systematic trend.

Figure 2. Example of residual plot; pattern suggests nonlinear fit.

Fitting of polynomial linear model

Fit simple linear regression, data set Yuan et al lifespan, listed below.

R code

LinearModel.1 <- lm(cumFreq~Months, data=yuan)

summary(LinearModel.1)

Call:
lm(formula = cumFreq ~ Months, data = yuan)

Residuals:
     Min       1Q   Median      3Q     Max 
-0.11070 -0.07799 -0.01728 0.06982 0.13345

Coefficients:
                 Estimate Std. Error t value Pr(>|t|) 
(Intercept)     -0.132709   0.045757   -2.90  0.0124 * 
Months           0.029605   0.001854   15.97  6.37e-10 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.09308 on 13 degrees of freedom
Multiple R-squared: 0.9515, Adjusted R-squared: 0.9477 
F-statistic: 254.9 on 1 and 13 DF, p-value: 6.374e-10

We see from the R2 (95%), a high degree of fit to the data. However, residual plot reveals obvious trend (Fig 3)

Residual plot of nonlinear data on simple linear model

Figure 3. Residual plot, residuals against fitted values.

We can fit a polynomial regression.

First, a second order polynomial

LinearModel.2 <- lm(cumFreq ~ poly( Months, degree=2), data=yuan)

summary(LinearModel.2)

Call:
lm(formula = cumFreq ~ poly(Months, degree = 2), data = yuan)

Residuals:
     Min       1Q   Median      3Q     Max 
-0.13996 -0.06720 -0.02338 0.07153 0.14277

Coefficients:
                           Estimate Std. Error  t value   Pr(>|t|) 
(Intercept)                 0.48900    0.02458   19.891   1.49e-10 ***
poly(Months, degree = 2)1   1.48616    0.09521   15.609   2.46e-09 ***
poly(Months, degree = 2)2   0.06195    0.09521    0.651   0.528 
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.09521 on 12 degrees of freedom
Multiple R-squared: 0.9531, Adjusted R-squared: 0.9453 
F-statistic: 122 on 2 and 12 DF, p-value: 0.0000000106

Second, try a third order polynomial

LinearModel.3 <- lm(cumFreq ~ poly(Months, degree = 3), data=yuan)

summary(LinearModel.3)

Call:
lm(formula = cumFreq ~ poly(Months, degree = 3), data = yuan)

Residuals:
      Min        1Q   Median       3Q      Max 
-0.052595 -0.021533 0.001023 0.025166 0.048270

Coefficients:
                           Estimate Std. Error t value   Pr(>|t|) 
(Intercept)                0.488995   0.008982  54.442   9.90e-15 ***
poly(Months, degree = 3)1  1.486157   0.034787  42.722   1.41e-13 ***
poly(Months, degree = 3)2  0.061955   0.034787   1.781   0.103 
poly(Months, degree = 3)3 -0.308996   0.034787  -8.883   2.38e-06 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.03479 on 11 degrees of freedom
Multiple R-squared: 0.9943, Adjusted R-squared: 0.9927 
F-statistic: 635.7 on 3 and 11 DF, p-value: 1.322e-12

Which model is best? We are tempted to compare R-squared among the models, but R2 turn out to be untrustworthy here. Instead, we compare using Akaike Information Criterion, AIC

R code/results

AIC(LinearModel.1,LinearModel.2, LinearModel.3)
              df       AIC
LinearModel.1  3 -24.80759
LinearModel.2  4 -23.32771
LinearModel.3  5 -52.83981

Smaller the AIC, better fit.

anova(RegModel.5,LinearModel.3, LinearModel.4)
Analysis of Variance Table

Model 1: cumFreq ~ Months
Model 2: cumFreq ~ poly(Months, degree = 2)
Model 3: cumFreq ~ poly(Months, degree = 3)
  Res.Df   RSS    Df Sum of Sq       F      Pr(>F)
1     13 0.112628
2     12 0.108789  1  0.003838  3.1719 0.1025
3     11 0.013311  1  0.095478 78.9004 0.000002383 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Logistic regression

The Logistic regression is a classic example of nonlinear model.

R code

logisticModel <-nls(yuan$cumFreq~DD/(1+exp(-(CC+bb*yuan$Months))), start=list(DD=1,CC=0.2,bb=.5),data=yuan,trace=TRUE)
5.163059 : 1.0 0.2 0.5
2.293604 : 0.90564552 -0.07274945 0.11721201
1.109135 : 0.96341283 -0.60471162 0.05066694
0.429202 : 1.29060000 -2.09743525 0.06785993
0.3863037 : 1.10392723 -2.14457296 0.08133307
0.2848133 : 0.9785669 -2.4341333 0.1058674
0.1080423 : 0.9646295 -3.1918526 0.1462331
0.005888491 : 1.0297915 -4.3908114 0.1982491
0.004374918 : 1.0386521 -4.6096564 0.2062024
0.004370212 : 1.0384803 -4.6264657 0.2068853
0.004370201 : 1.0385065 -4.6269276 0.2068962
0.004370201 : 1.0385041 -4.6269822 0.2068989
summary(logisticModel)

Formula: yuan$cumFreq ~ DD/(1 + exp(-(CC + bb * yuan$Months)))

Parameters:
      Estimate  Std. Error  t value      Pr(>|t|)
DD    1.038504    0.014471    71.77   < 2e-16 ***
CC   -4.626982    0.175109   -26.42  5.29e-12 ***
bb    0.206899    0.008777    23.57  2.03e-11 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.01908 on 12 degrees of freedom

Number of iterations to convergence: 11
Achieved convergence tolerance: 0.000006909

> AIC(logisticModel)
[1] -71.54679

Logistic regression is a statistical method for modeling the dependence of a categorical (binomial) outcome variable on one or more categorical and continuous predictor variables (Bewick et al 2005).

The logistic function is used to transform a sigmoidal curve to a more or less straight line while also changing the range of the data from binary (0 to 1) to infinity (-∞,+∞). For event with probability of occurring p, the logistic function is written as

    \begin{align*} logit(p) = ln\left( \frac{p}{1-p} \right) \end{align*}

where ln refers to the natural logarithm.

The logit link function is used in models to transform a probability, which is restricted between 0 and 1, into the log-odds scale, which ranges from negative infinity to positive infinity. This transformation makes the relationship between predictors and the outcome linear. In other words, this is an odds ratio. It represents the effect of the predictor variable on the chance that the event will occur. Much of our contingency table work in Chapter 7 would be best analyzed with the logistic regression.

The logistic regression model then very much resembles the same as we have seen before.

    \begin{align*} logit(p) = logit(p) = \beta_{0}+\beta_{1}X_{1}+\beta_{2}X_{2}+\cdots + \beta_{n}X_{n} \end{align*}

In R and Rcmdr we use the glm() function to model the logistic function. Logistic regression is used to model a binary outcome variable. What is a binary outcome variable? It is categorical! Examples include: Living or Dead; Diabetes Yes or No; Coronary artery disease Yes or No. Male or Female. One of the categories could be scored 0, the other scored 1. For example, living might be 0 and dead might be scored as 1. (By the way, for a binomial variable, the mean for the variable is simply the number of experimental units with “1” divided by the total sample size.)

With the addition of a binary response variable, we are now really close to the Generalized Linear Model. Now we can handle statistical models in which our predictor variables are either categorical or ratio scale. All of the rules of crossed, balanced, nested, blocked designs still apply because our model is still of a linear form.

We write our generalized linear model

    \begin{align*} G \sim Model \end{align*}

just to distinguish it from a general linear model with the ratio-scale Y as the response variable.

Think of the logistic regression as modeling a threshold of change between the 0 and the 1 value. In another way, think of all of the processes in nature in which there is a slow increase, followed by a rapid increase once a transition point is met, only to see the rate of change slow down again. Growth is like that. We start small, stay relatively small until birth, then as we reach our early teen years, a rapid change in growth (height, weight) is typically seed (well, not in my case … at least for the height). The fitted curve I described is a logistic one (other models exist too). Where the linear regression function was used to minimize the squared residuals as the definition of the best fitting line, now we use the logistic as one possible way to describe or best fit this type of a curved relationship between an outcome and one or more predictor variables. We then set out to describe a model which captures when an event is unlikely to occur (the probability of dying is close to zero) AND to also describe when the event is highly likely to occur (the probability is close to one).

A simple way to view this is to think of time being the predictor (X) variable and risk of dying. If we’re talking about the lifetime of a mouse (lifespan typically about 18-36 months), then the risk of dying at one months is very low, and remains low through adulthood until the mouse begins the aging process. Here’s what the plot might look like, with the probability of dying at age X on the Y axis (probability = 0 to 1) (Fig 4).

lifespan mice, Yuan et al 2012 PNAS 109:8224

Figure 4. Lifespan of 1881 mice from 31 inbred strains (Data from Yuan et al (2012) available at https://phenome.jax.org/projects/Yuan2).

We ask — of all the possible models we could draw, which best fits the data? The curve fitting process is called the logistic regression.

With some minor, but important differences, running the logistic regression is the same as what you have been doing so far for ANOVA and for linear regression. In Rcmdr, access the logistic regression function by invoking the Generalized Linear Model (Fig 5).

Rcmdr: Statistics → Fit models → Generalized linear model.

Screenshot Rcmdr GLM menu.

Figure 5. Screenshot Rcmdr GLM menu. For logistic on ratio-scale dependent variable, select gaussian family and identity link function.

Select the model as before. The box to the left accepts your binomial dependent variable; the box at right accepts your factors, your interactions, and your covariates. It permits you to inform R how to handle the factors: Crossed? Just enter the factors and follow each with a plus. If fully crossed, then the interactions may be specified with “:” to explicitly call for a two-way interaction between two (A:B) or a three-way interaction between three (A:B:C) variables. In the later case, if all of the two way interactions are of interest, simply typing A*B*C would have done it. If nested, then use %in% to specify the nesting factor.

R output

> GLM.1 <- glm(cumFreq ~ Months, family=gaussian(identity), data=yuan)

> summary(GLM.1)

Call:
glm(formula = cumFreq ~ Months, family = gaussian(identity),
data = yuan)

Deviance Residuals:
Min 1Q Median 3Q Max
-0.11070 -0.07799 -0.01728 0.06982 0.13345

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.132709 0.045757 -2.90 0.0124 *
Months 0.029605 0.001854 15.97 6.37e-10 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for gaussian family taken to be 0.008663679)

Null deviance: 2.32129 on 14 degrees of freedom
Residual deviance: 0.11263 on 13 degrees of freedom
AIC: -24.808

Number of Fisher Scoring iterations: 2

 

Assessing fit of the logistic regression model

Some of the differences you will see with the logistic regression is the term deviance. Deviance in statistics simply means compare one model to another and calculate some test statistic we’ll call “the deviance.” We then evaluate the size of the deviance like a chi-square goodness of fit. If the model fits the data poorly (residuals large relative to the predicted curve), then the deviance will be small and the probability will also be high — the model explains little of the data variation. On the other hand, if the deviance is large, then the probability will be small — the model explains the data, and the probability associated with the deviance will be small (significantly so? You guessed it! P < 0.05).

The Wald test statistic

where n and β refers to any of the n coefficient from the logistic regression equation and SE refers to the standard error if the coefficient. The Wald test is used to test the statistical significance of the coefficients. It is distributed approximately as a chi-squared probability distribution with one degree of freedom. The Wald test is reasonable, but has been found to give values that are not possible for the parameter (eg, negative probability).

Likelihood ratio tests are generally preferred over the Wald test. For a coefficient, the likelihood test is written as

where L0 is the likelihood of the data when the coefficient is removed from the model (ie, set to zero value), whereas L1 is the likelihood of the data when the coefficient is the estimated value of the coefficient. It is also distributed approximately as a chi-squared probability distribution with one degree of freedom.

Questions

  1. Write up three learning outcomes for this page. Hint: Point your favorite generative AI to this page and ask for help.
  2. What is the primary difference between linear regression and logistic regression?
  3. What type of dependent variable is used in logistic regression?
  4. What is the purpose of the sigmoid (or logistic) function in this model?
  5. What is the role of a significance test (eg, a Wald test) in logistic regression?
  6. Why is the LRT generally considered more reliable than the Wald test for logistic regression, especially with smaller sample sizes?

Quiz Chapter 18.2

Nonlinear regression

Data set

Yuan data

MonthsfreqcumFreq
000
30.0106330.010633
60.0170120.027645
90.0451890.072834
120.0643270.137161
150.0648590.20202
180.097820.299841
210.1185540.418394
240.1711860.58958
270.1621480.751728
300.1371610.888889
330.0696440.958533
360.0244550.982988
390.0116960.994684
420.0053161

Chapter 18 content

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).

Smokers Non-smokers
Lung cancer 647 2
Case controls, no lung cancer 622 27

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

Improved Not improved
St. John’s Wort 14 98
Placebo 5 102

Note 1: The St. John’s Wort problem is precisely a the time to remind any reader of Mike’s Biostatistics Book: under no circumstances is medical advice implied from my presentation. From the National Center for Complementary and Integrative Medicine: “It has been clearly shown that St. John’s wort can interact in dangerous, sometimes life-threatening ways with a variety of medicines.”

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 \chi^2 test.

Note 2: The preferred analysis is to use a logistic regression approach, Chapter 18.3, because additional covariates can be included in the model.

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.

Treatment Variable and Levels Outcome Variable(s) Reference
Lead exposure

Levels: Low, Medium, High

 Intelligence and development Bellinger et al 1987
Wood preservatives

Levels: Borates vs. Chromated Copper Arsenate

 Effectiveness as fungicide Hrastnik et al 2013
Antidepressants

Levels: St. John’s Wort, conventional antidepressants, placebo

 Depression relief Linde et al 2008
Coral reefs

Levels: Protected vs Unprotected

 Fish community structure Guidetti 2006
Aspirin therapy

Levels: low dose vs none

 Cancer incidence in women Cook et al 2013
Aspirin therapy

Levels: low dose vs none

 Cancer mortality in women Cook et al 2013

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

  Outcome
  
Exposure or Treatment group Yes No Marginal total
Exposure (Treatment) a b Row1 = a + b
Nonexposure (Control) c d Row2 = c + d
Marginal total Column1
= a + c		 Column2
= a + c		 N

Contingency tables are all of the form (Table 5)

Table 5.  Basic format of a 2X2 table

Outcome 1 Outcome 2
Treatment 1 Yes No
Treatment 2 Yes No

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)

Lived Died Row totals
HG+ 103 41 144
HG- 95 123 218
Column totals 198 164 362

Note 3: HG virus is no longer called a hepatitis virus, but instead is referred to as GB virus C, which, like hepatitis viruses, is in the Flaviviridae virus family. For more about the GB virus C, see review article by Bhattarai and Stapleton (2012).

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.

Lived Died Row totals
HG+ a b a + b
HG- c d c + d
Column totals a + c b + d N

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

HO: 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

Lived Died Row totals
HG+ 103 41 144
HG- 95 123 218
Column totals 198 164 362

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)

Lived Died Row totals
HG+ 78.768 65.232 144
HG- 119.25 98.75 218
Column totals 198 164 362

Now we are ready to calculate the Chi-Square Value

Recall the formula for the chi-square test

    \begin{align*} \chi ^2 = \sum_{i = 1}^{k}\frac{\left (O_{i} - E_{i} \right )^2}{E_{i}} \end{align*}

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

Cell \left (O_{i} - E_{i} \right )^2 \chi^2
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
\chi^2 = 27.3452

Adding all the parts we have the chi-square test statistic \chi^2 = 27.3452. To proceed with the inference, we test using the chi-square distribution.

Determine the Critical Value of \chi^{2} 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. 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 \chi^2 test statistic is greater than the Critical Value. We Accept the Null Hypothesis when the calculated \chi^2 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.

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

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

    \begin{align*} \chi^2 = \frac{\left (ad-bc \right )^{2}\cdot N}{r_{1} \cdot r_{2} \cdot c_{1} \cdot c_{2}} \end{align*}

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 \chi^2 distribution to calculate the p-value (one tailed test).

Note 4: This formulation was provided as equation 1 in Yates 1984 (referred to in Serra 2018), but is likely found in the earlier papers on \chi^2 dating back to Pearson.

Table 11. Example spreadsheet with formulas for odds ratio (OR), Pearson’s \chi^2, and p-value from \chi^2 distribution.

A B C D E F G
1
2 a 103 OR =(B2*B5)/(B3*B4)
3 b 41
4 c 95
5 d 123 chisq, p =(((B2*B5-B3*B4)^2)*B6)/
(SUM(B2:B3)*SUM(B4:B5)*
SUM(B2,B4)*SUM(B3,B5))		 =CHIDIST(F5,1)
6 N 632

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

\chi^{2} 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 \chi^2 that’s φ, phi for the φ (phi) coefficient.

    \begin{align*} \phi = \sqrt{\frac{\chi^{2}}{N}} \end{align*}

where N is the total. Phi for our example is

    \begin{align*} \phi = \sqrt{\frac{27.3452}{362}} \end{align*}

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.

Effect size Interpretation
< 0.2 Small, weak effect
0.5 Moderate effect
> 0.8 Large effect

For our example, \phi 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

    \begin{align*} \kappa = \frac{p_{o}-p_{c}}{1 - p_{c}} \end{align*}

where po is the proportion in agreement and pc 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

    \begin{align*} se_{\kappa}= \sqrt{\frac{p_o\left ( 1-p_o \right )}{N\left ( 1 - p_c \right )}} \end{align*}

and 95% confidence intervals are then \pm 1.96 se{\k}.

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.

Note 5: Base R includes the function kappa(), but that returns an estimate for the condition number of the matrix, or how close to singular the matrix solution may be, ie, determinant is zero and its inverse is unavailable.

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.

tails heads
flips1 10 14
flips2 14 10

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.

Cohen’s kappa Agreement?
< 0 none
0.01 — 0.20 slight
0.21 — 0.40 fair
0.41 — 0.60 moderate
0.61 — 0.80 substantial
0.81 — 1 nearly perfect

 

Note 6: Wikipedia page for Cohen’s kappa suggests that this interpretation began with the Landis and Koch 1997 paper in the journal Biometrics. I haven’t confirmed this — regardless, this paper is also highly influential, with more than 89K citations (Fall 2024).

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).

Screenshot Rcmdr menu, enter contingency table information and select statistics

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 \chi^2 test statistic is done when you select the “Chi-square test of independence” check-box (Fig 2).

2x2 HIV HG coinfection data set

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.

Note 7: R Commander often cleans up after itself by removing objects like .Test and .Table. This is not necessary for the code to work, but does make it easier for you to go back and modify the code without worrying about confusing objects.

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.

Value of the test statistic 27.339
Degrees of freedom 1
p-value 0.0000001708

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).

Note 7: Simple enough to get R to report numbers as you need. For example, R code

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

returns

[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 \chi^2 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

Chapter 9 contents

7.6 – Confidence intervals

Quantifying uncertainty.

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

There are three components of statistical analysis

  1. Estimation
  2. Inference
  3. Modeling

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

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

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

Confidence interval for proportions

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

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

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

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

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

Confidence interval for ARR

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

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

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

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

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

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

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

Confidence interval for NNT

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

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

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

where SE is the standard error for NNT

The CI is approximately then NNT \pm 2 X SE

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

Odds ratio Standard error and 95% confidence interval

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

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

R code

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

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

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

where exp is the exponential function 

    \[e^{x}\]

and ln is natural logarithm.

R code

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

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

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

Our example the upper limit was 2.19.

R code

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

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

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

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

log(x, base = exp(1))

P-values from confidence intervals

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

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

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

We need

The OR estimate, myEST, which was 1.89

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

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

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

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

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

Questions

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

Quiz Chapter 7.6

Confidence intervals

Chapter 7 contents

7.5 – Odds ratio

What are the odds.

We introduced the concept of odds 7.1 — Epidemiology definitions. As a reminder, odds are a way to communicate the chance (likelihood) that a particular event will take place. Odds are calculated as the number of individuals with the event divided by the number of individuals without the event.

Odds ratio definition: is a measure of effect size for the association between two binary (yes/no) variables. It is the ratio of the odds of an event occurring in one group to the odds of the same event happening in another group. The odds ratio (OR) is a way to quantify the strength of association between one condition and another.

Note 1: Effect size — the size of the difference between groups — is discussed further in Chapter 9.2 and Chapter 11.4.

And note, it’s association, not correlation. In statistics, correlation is a specific kind of relationship (linear) among the variables, whereas “association” means there is some relationship between the variables without specifying the shape of the relationship.

How are odds ratios calculated? The probabilities are conditional; recall that conditional probability of some event A, given the occurrence of some other event B.

Let p_{y-y} equal probability of the event occurring (y = Yes) in A, p_{y-n} equal probability of the event not occurring (n = No) in A, p_{n-y} equal probability of the event occurring in B, and p_{n-n} equal probability of the event not occurring in B.

A
Yes No
B Yes p_{y-y} p_{y-n}
No p_{n-y} p_{n-n}

These sum to one: p_{y-y}+p_{y-n}+p_{n-y}+p_{n-n} = 1

The conditional probabilities are

A
Yes No
B Yes \frac{p_{y-y}}{\left ( p_{y-y}+p_{y-n} \right )} \frac{p_{y-n}}{\left ( p_{y-y}+p_{y-n} \right )}
No \frac{p_{n-y}}{\left ( p_{n-y}+p_{n-n} \right )} \frac{p_{n-n}}{\left ( p_{n-y}+p_{n-n} \right )}

and finally then, the odds ratio (OR) is

    \begin{align*} OR = \frac{p_{y-y} \cdot p_{n-n}}{p_{y-n} \cdot p_{n-y}} \end{align*}

If you have the raw numbers you can calculate the odds ratio directly, too.

A
Yes No
B Yes a b
No c d

and the odds ratio is then

    \begin{align*} OR = \frac{a \div b}{c \div d} \end{align*}

or, equivalently

    \begin{align*} OR = \frac{a \cdot d}{b \cdot c} \end{align*}

Example.

Comparing proportions is a frequent need in court. Gray (2002) provided an example from Title IX of the Education Act of 1972 case Cohen v. Brown University. Under the Act, discrimination based on gender is prohibited. The case concerned participation in collegiate athletics by women. The case data were that of the 5722 undergraduate students, 51% were women, but of the 987 athletes, only 38% were women. A mosaic plot shows graphically these proportions (Fig 1, males in red bars, females in yellow bars).

Mosaic plot of athletes to non-athletes in college. Males red, females yellow, data from Gray 2002.

Figure 1. Mosaic plot of athletes to non-athletes in college. Males red, females yellow, data from Gray 2002.

Alternatively, use a Venn diagram to describe the distribution (Fig 2). Circles that overlap show regions of commonality.

Venn diagram

Figure 2. Venn Diagram of athletes to non-athletes in college. Female athletes (n = 375), male athletes (n = 612), data from Gray 2002.

where the orange region

    \begin{align*} Students \cap Female Athletes \end{align*}

R code for the Venn diagram was

library(VennDiagram)
area1 = 5722
area2 = 987 
cross.area = 375 
draw.pairwise.venn(area1,area2,cross.area,category=c("Students","Athletes"),
euler.d = TRUE, scaled = TRUE, inverted = FALSE, print.mode = "percent",
fill=c("Red","Yellow"),cex = 1.5, lty="blank", cat.fontfamily = rep("sans", 2),
cat.cex = 1.7, cat.pos = c(0, 180), ext.pos=0)

The question raised before the court was whether these proportions meet the demand of “substantially proportionate.”  What exactly the law means by “substantially proportionate” was left to the courts and the lawyers to work out (Gray 2002). Title IX suggests that “substantially proportionate” is a statistical problem and the two sides of the argument must address the question from that perspective.

What is the chance that an undergraduate student was an athlete and female? 38% And the chance that an undergraduate student was an athlete and male? 62% Clearly 38% is not 62%; did the plaintiffs have a case?

Graphs like Figure 1 and Figure 2 help communicate but can’t provide a sense of whether the differences are important. Let’s start by looking at the numbers. Working with the proportions we have the following break down for numbers of students  (Table 1) or as proportions (Table 2).

Table 1. Gray’s raw data displayed in a 2 x 2 format.

Athletes
Yes No
Undergraduates Male 612 2192
Female 375 2543

Together, the numbers total 5,722.

The Odds Ratio (OR) would be

    \begin{align*} OR = \frac{612 \cdot 2543}{2192 \cdot 375} = 1.89 \end{align*}

Or from the proportions (Table 2)

Table 2. Data from Table 1 as proportions.

Athletes
Yes No
Undergraduates Male 0.107 0.383
Female 0.066 0.444

adding all of these frequencies together equal 1. Carry out the calculation of odds (Table 3), the conditional probabilities (in bold).

Table 3. Odds calculated from Table 2 inputs.

Athletes
Yes No
Undergraduates Male 0.218

= \frac{0.107}{\left ( 0.107+0.383\right )}

 0.782

= \frac{0.383}{\left ( 0.107+0.383\right )}
Female 0.129

= \frac{0.066}{\left ( 0.066+0.444\right )}

 0.871

= \frac{0.444}{\left ( 0.066+0.444\right )}

Calculate the odds ratio

    \begin{align*} OR = \frac{0.2182 \cdot 0.871}{0.129 \cdot 0.782} = 1.89 \end{align*}

Thankfully, whether we use the raw number format or the proportion format, we got the same results!

Odds ratio interpretation.

Because the Odds Ratio (OR) was greater than 1, males students were more likely to be athletes than female students. If there was no difference in proportion of male and female athletes, the odds ratio would be close to one. That is a test of statistical inference (eg, a contingency table), but for now, if one is included in the confidence interval, then this would be evidence that there was no difference between the proportions.

And in R? Simple enough, just create a matrix then apply the Fisher test. which we will discuss further in Chapter 9.5.

title9 <- matrix(c(612, 2192, 375, 2543), nrow=2))
fisher.test(title9)

and results

p-value < 2.2e-16
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
1.641245 2.185576
sample estimates:
odds ratio 
1.893143

Thankfully, they agree. But note, we now have confidence intervals and a p-value, which we use to conduct inference: were the odds “significantly different from 1?” We would conclude, yes!  Between the lower limit (1.64) and the upper limit (2.19), the value “1” was excluded. Moreover, the p-value at 2.2e-16 was much less than the standard type I error cut-off of 5% (see Chapter 8).

Before we leave the interpretation, sometimes a calculated odds-ratio is less than one. If our calculated odds ratio for the Title IX case described in Table 1 was less than 1, (say, the numbers were flipped, Table 4) we  then the interpretation would be females were more likely to be athletes on college campus.

Table 4. Table 1 data, but order of entry changed.

Athletes
Yes No
Undergraduates Female 375 2543
Male 612 2192

Now, calculating the odds via Fisher exact test, the odds ratio is less than one (0.53):

Fisher's Exact Test for Count Data
p-value < 2.2e-16
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
0.4575452 0.6092936
sample estimates:
odds ratio 
0.5282222

Note this result is for the same data (Table 1 vs Table 4), just the order by which the groups are specified changed. Of course they are related to each other mathematically, and in a simple way. Note that taking the inverse (reciprocal) odds ratio.

    \begin{align*} OR = \frac{1}{0.5282222} = 1.89 \end{align*}

will return the comparison we wanted in the first place — the odds a student athlete was male. As long as you keep track of the comparison, of the groups, it may be easier to communicate results when the reported odds ratio is greater than one.

Relative risk v. odds ratio.

We introduced another way to quantify this association as the Relative Risk (RR) and Absolute Risk Reductions in the previous section. Both can be used to describe the risk of the treatment (exposed) group relative to the control (nonexposed) group. RR is the ratio the treated to control group. OR is the ratio between odds of treated (exposed) and control (nonexposed). What’s the difference? OR is more general — it can be used in situations in which the researcher chooses the number of affected individuals in the groups and, therefore, the base rate or prevalence of the condition in the population is not known or is not representative of the population, whereas RR is appropriate when prevalence is known (this is a general point, but see Schechtman 2002 for a nice discussion).

The odds ratio is related to relative risk, but not over the entire range of possible risk. Odds of an event is simply the number of individuals with the event divided by the number without the event. Odds of an event therefore can range from zero (event cannot occur) to infinity (event must occur). For example, odds of eight (1.89:1) means that nearly two male students were student athletes at Brown University for every one female student.

In contrast, the risk of an event occurring is the number of individuals with the event divided by the total number of people at risk of having that event. Risk is expressed as a percentage (Davies et al 1998).  Thus, for our example, odds of 1.89:1 correspond to a risk of 1.89 divided by (1 + 1.89) equals 65%.

To get the relative risk we can use

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

or 1.7% for our example.

In this example we could use either odds or relative risk; the key distinction is that we knew how many events happened in both groups. If this information is missing for one group (eg, control group of the case-control design), then only the odds ratio would be appropriate.

From cumulative wisdom in the literature (eg, Tamhane et al 2107), if prevalence is less than ten percent, OR ≈ RR. We can relate RR and OR as

    \begin{align*} RR=OR\cdot \frac{1+\frac{n_{21}}{n_{22}}}{1+\frac{n_{11}}{n_{12}}} \end{align*}

where n11 and n21 are the frequency with the condition for group 1 and group 2, respectively, and n12 and n22 are the frequency without the condition for group 1 and group 2, respectively. For the examples on this page group 1 is the treatment group and group 2 is the control group.

Hazard ratio

The hazard ratio is the ratio of hazard rates. Hazard rates are like the relative risk rates, but are specific to a period of time. Hazard rates come from a technique called Survival Analysis (introduced in Chapter 20.9). Survival analysis can be thought of as following a group of subjects over time until something (the event) happens. By following two groups, perhaps one group exposed to a suspected carcinogen vs. another group matched in other respects except the exposure, at the end of the trial, we’ll have two hazard rates: the rate for the exposed group and the rate for the control group. If there is no difference, then the hazard ratio will be one.

Hazard ratios are more appropriate for clinical trials; relative risk is more appropriate for observational studies.

For a hazard ratio, it is often easier to think of it as a probability (between 0 to 1). To translate a hazard ratio to a probability use the following equation

    \begin{align*} p =\frac{hazard \ ratio}{1+hazard \ ratio} \end{align*}

 

Questions

  1. Distinguish between odds ratio, relative risk, and hazard ratio.
  2. Refer to problem 1 introduced in 7.4 – Epidemiology: Relative risk and absolute risk, explained.

Quiz Chapter 7.5

Odds ratio

Chapter 7 contents