# The Association Between Exposure and Disease Outcome

Consider a standard epidemiological table that cross-classifies members of a study population by their disease and exposure status. The columns of the two-way table are the disease state (“disease”, “no disease”) and the rows are the exposure state (“exposed”,“not exposed”).

library(hwriter)
epitab <- matrix(c("a", "b",
"c", "d"), nr=2, nc=2, byrow=TRUE)
colnames(epitab) <- c("Disease","No Disease")
rownames(epitab) <- c("Exposed", "Not Exposed")
epitab
##             Disease No Disease
## Exposed     "a"     "b"
## Not Exposed "c"     "d"
cat(hwrite(epitab, border = 1, center=TRUE, width='300px', br=TRUE, table.style=list('padding: 50px'),  col.style=list('text-align:center'), row.names=TRUE))
 Disease No Disease Exposed a b Not Exposed c d

The values in the cells of this table are counts. For example, a is the count of disease cases who experienced exposure. The row and column sums are known collectively as the marginal totals of the table: a+b and c+d are the row sums and a+c and b+d are the column sums. The grand sum is n=a+b+c+d.

There are two broad ways that we could construct an epidemiological table like this. First, we could select a population and follow it through time, noting exposures and the occurance of disease. This is known as a cohort design. Cohort designs are difficult, expensive, and impractical for many infectious diseases, particularly emerging infectious diseases. A common design begins with a group of cases of the disease and then seeks a comparable sample of non-diseased individuals as controls, ascertaining for both groups their exposure status. This is known as a case-control design.

## Prevalence, Absolute Risk, and Risk Difference

The prevalence is simply the number of cases divided by the total population: $$(a+c)/(a+b+c+d)$$. The absolute risk or attack rate is $$a/(a+b)$$. The Risk difference is the difference between the absolute risk of exposed and unexposed individuals:

$RD = \frac{a}{a+b} - \frac{c}{c+d}$ In a cohort design, the risk difference is typically known as the attributable risk.

Something not easily captured in our simple $$2 \times 2$$ table is incidence, which is the number of new cases in some time interval divided by the person-time units. For example, if there are 3 new cases in a week in a population with 100 people at risk, then the incidence is $$3/100$$. Suppose that each case is fatal and that the average number of person-weeks lived by those who contract the disease is 0.5 weeks. In this case, the incidence would be slightly higher at $$3/98.5$$ (i.e., 97 person-weeks for those not afflicted and a total of $$3 \times 0.5 = 1.5$$ person-weeks for those dying).

## Risk Ratios and Odds Ratios

The risk ratio or relative risk is the ratio of the risk to those exposed relative to the risk of those not exposed:

$\widehat{RR} = \frac{a/(a+b)}{c/(c+d)}$ If the entries of our table are large enough a normal approximation for the binomial distribution applies and we can use normal theory to calculate standard errors and place confidence bounds around $$\widehat{RR}$$. It turns out that $$\log \widehat{RR}$$ has a sampling distribution better approximated by a normal, so we work with that. The standard error of $$\log \widehat{RR}$$ is

$\mathrm{se}[\log \widehat{RR}] \approx \sqrt{\frac{b}{an_1} + \frac{d}{cn_2}},$

where $$n_1=a+b$$ and $$n_2 = c+d$$ are the row sums.

Using the standard approach for calculating confidence intervals from the normal approximation, the 95% confidence interval on the relative risk are calculated from:

$c_1 = \log \widehat{RR} - 1.96 \sqrt{\frac{b}{an_1} + \frac{d}{cn_2}}$

$c_2 = \log \widehat{RR} + 1.96 \sqrt{\frac{b}{an_1} + \frac{d}{cn_2}}$

And the CI is thus, $$[e^{c_1}, e^{c_2}]$$. Just as a reminder, we multiply by 1.96 when calculating a 95% confidence interval using a normal approximation because 95% of the probability mass of a normal distribution lies within 1.96 standard deviations of the mean.

round(c(pnorm(1.96), pnorm(-1.96)),3)
## [1] 0.975 0.025
round(c(qnorm(0.975), qnorm(0.025)),3)
## [1]  1.96 -1.96

The relative risk is a pretty intuitive idea, but a problem with it is that it is constrained by the denominator. If $$c/(c+d) = 0.5$$, then the biggest the relative risk could be is $$\widehat{RR}=2$$. This is because the biggest $$a/(a+b)$$ can be is unity since this would happen when $$b=0$$, therefore $$\widehat{RR} = 2$$. In addition, we often (usually?) don’t have a cohort design for our data collection. Without the prospective design of a cohort study, relative risks don’t make much sense since we shouldn’t believe that $$a/(a+b)$$ is a good estimator of the risk associated with the exposure of interest. We can get around this by using an odds ratio.

The odds ratio is the ratio of odds of exposure among cases to those among controls:

$OR = \frac{a/c}{b/d} = \frac{ad}{bc}$ It is the cross-product of the cells of our epidemiological table. For rare diseases, the OR is a very good approximation of the RR. This is because, if the disease is rare, $$a \ll b$$ and $$c \ll d$$. This means:

$\frac{a/(a+b)}{c/(c+d)} \approx \frac{a/b}{c/d} = \frac{ad}{bc}.$ We can calculate confidence intervals on an odds ratio using the normal approximation (assuming the cells are large enough). Again, we work with the logarithm of the measure of association. The standard error of the logarithm of the odds ratio is,

$\mathrm{se}[\log OR] = \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}},$

and the 95% confidence intervals for the log-odds ratio are thus:

$c_1 = \log OR - 1.96 \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}$

$c_2 = \log OR + 1.96 \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}}$

Back-transforming, we get the confidence interval on the unit scale of $$[e^{c_1}, e^{c_2}]$$.

## SARS

As China increased its cooperation with WHO and other health organizations, the details of the early epidemic in Guangdong Province became known. The first case in Guangdong to have signs of the disease consistent with the WHO case definition occurred in Foshan city. Unfortunately, the patient died before samples could be collected for virological investigation. The second case presented itself on 17 December 2002. A chef from Heyuan presented with severe atypical pneumonia. It seems that this chef worked in a restaurant that specialized in exotic game and had regular contact with several species of live caged animals.

Sampling effort thus turned to the exotic animal markets of Guangdong. A team from the University of Hong Kong and the Guangdong CDC sampled animals from a market in Shenzhen. A total of seven common game animals to the market were subjected to both nasal and fecal sampling. Virus was repeatedly detected in palm civets. These researchers also took nasal samples from the animal traders, in addition to hospital workers, Guandong CDC employees, and health visitors to a clinic as a controls. A total of 70 people in the market were SCoV+, 66 of which were animal traders. 579 were SCoV-, of which 442 were animal traders (Yu et al. 2003).

require(hwriter)
cc <- matrix(c(66L, 442L,
4L, 133L), nr=2, nc=2, byrow=TRUE)
colnames(cc) <- c("SCoV+","SCoV-")
rownames(cc) <- c("Animal Trader", "Hospital Worker")
cc
##                 SCoV+ SCoV-
## Hospital Worker     4   133
cat(hwrite(cc, border = 1, center=TRUE, width='300px', br=TRUE, table.style=list('padding: 50px'),  col.style=list('text-align:center'), row.names=TRUE))
 SCoV+ SCoV- Animal Trader 66 442 Hospital Worker 4 133

Calculate the odds ratio and confidence intervals.

cc <- matrix(c(66, 442,
4, 133), nr=2, nc=2, byrow=TRUE)
colnames(cc) <- c("SCoV+","SCoV-")
rownames(cc) <- c("Animal", "Hospital")
cc
##          SCoV+ SCoV-
## Animal      66   442
## Hospital     4   133
(row.totals <- apply(cc,1,sum))
##   Animal Hospital
##      508      137
(col.totals <- apply(cc,2,sum))
## SCoV+ SCoV-
##    70   575
## absolute risk
(ar <- cc[1,1]/row.totals[1])
##    Animal
## 0.1299213
# risk difference
(rd <- cc[1,1]/row.totals[1] - cc[2,1]/row.totals[2])
##    Animal
## 0.1007242
## rate ratio (watch the parentheses!)
(rr <- (cc[1,1]/row.totals[1])/(cc[2,1]/row.totals[2]))
##   Animal
## 4.449803
#odds ratio
(or <- (cc[1,1]*cc[2,2])/(cc[1,2]*cc[2,1]))
## [1] 4.964932
se <- sqrt(1/cc[1,1]+1/cc[2,2]+1/cc[1,2]+1/cc[2,1])
#lower
exp(log(or)-(1.96*se))
## [1] 1.776584
#upper
exp(log(or)+(1.96*se))
## [1] 13.87525

## Fisher Exact Test

When the cells are small and the normal approximation won’t work. Note that the odds ratio will be undefined if $$c=0$$ but you could imagine a situation where there is a really strong association between exposure and disease, and where the sample size is relatively small, where there are no cases in unexposed people (i.e., $$c=0$$). We can use Fisher’s exact test in situations like this. Fisher’s exact test calculates the probabilities of the table directly from a hypergeometric distribution.

Keele et al. (2009) examined the demographic effects of SIV infection among wild chimpanzees. They report that, in the nine years of their study, four of nine SIV-infected females give birth, whereas 22 of 30 uninfected females gave birth in that same period. We can set up an epidemiological table from this information and test for an association between infection status and fertility.

gombe.fert <- matrix(c(4,22,5,8),2,2,byrow=TRUE)
colnames(gombe.fert) <- c("SIV+","SIV-")
rownames(gombe.fert) <- c("Birth", "No Birth")
gombe.fert
##          SIV+ SIV-
## Birth       4   22
## No Birth    5    8
fisher.test(gombe.fert, alternative="less")
##
##  Fisher's Exact Test for Count Data
##
## data:  gombe.fert
## p-value = 0.1146
## alternative hypothesis: true odds ratio is less than 1
## 95 percent confidence interval:
##  0.000000 1.401445
## sample estimates:
## odds ratio
##  0.3014952

We can see that while the odds ratio is substantially less than one (meaning that SIV infection appears to reduce fertility), the confidence intervals on the odds ratio cross one, suggesting that this result can not be distinguished from a null association. Keele et al. (2009) actually used a different method – that accounted for chimp-years of exposure – to calculate the association. This example is just for illustration of the method.

## Logistic Regression

Logistic regression gives you the same answer as cross-multiplying a two-way table. We can analyze the SARS table using a logistic regression model (i.e., a binomial-family glm). In R, the left-hand side of a formula for a binomial-family glm can be specified in one of several ways. One of those is to pass a two-column matrix in which the first column is the number of “successes” and the second column is the number of “failures”. This happens to be exactly the form of our $$2\times 2$$ table, which is convenient. All we need is to construct a factor for exposure and call the model:

expose <- factor(c("yes","no"))
out <- glm(cc ~ expose, family=binomial)
summary(out)
##
## Call:
## glm(formula = cc ~ expose, family = binomial)
##
## Deviance Residuals:
## [1]  0  0
##
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  -3.5041     0.5075  -6.905 5.02e-12 ***
## exposeyes     1.6024     0.5243   3.056  0.00224 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance:  1.4450e+01  on 1  degrees of freedom
## Residual deviance: -3.1086e-14  on 0  degrees of freedom
## AIC: 13.127
##
## Number of Fisher Scoring iterations: 4
## odds
exp(out\$coef[2])
## exposeyes
##  4.964932
## now compare to cross-multiplication
(or <- (cc[1,1]*cc[2,2])/(cc[1,2]*cc[2,1]))
## [1] 4.964932

We can see that the two ways of calculating the odds-ratio give essentially identical answers.