Chapter 9: Contingency Tables and Odds-Ratios

What is the range of values that the Odds and the Odds Ratios can assume?

Let p= the probability of success; 1−p = probability of failure. Then, as previously defined, the odds of success are calculated as O=p1−p and for two groups the odds ratio of success is O=O1O2. Table 1 (see below) maps out the minimum probability of success (=0) and the maximum probability of success (=1) for a single group. Note p=0 means success never occurs and p=1 means success always occurs, for this particular group. Expressing the probability as odds yields a corresponding range of values that are anchored below at 0 and above at ∞ … these are the limits of the odds.

The implication is that the odds follow an asymmetric, highly skewed distribution and hence require a transformation in order for us to come up with some estimate of the uncertainty surrounding the odds. The log-transformation achieves this by normalizing the distribution of the odds.

However, what would be the limits of odds ratios? The answer: It depends on what p is for any group. Let us see this with an example. Assume that the following fictional data represent passenger survival on the Titanic. Let us assume that there were a total of 100 passengers, 50 Male and 50 Female. Let us also assume that all Males and Females survived. What would the contingency table look like?

What then is the probability that a Male passenger survived? It is =5050=1 and hence the probability that a Male passenger died is 1−ˆpmale=1−1=0. What about the Female passengers? Well, the probability that a Female passenger survived is also =5050=1 and hence the probability that a female passenger died is 1−ˆpfemale=1−1=0. Now let us calculate the odds of survival for Male passengers =10=∞ and likewise the odds of survival for Female passengers =10=∞.

So this tells us that one extreme situation may be where for each group the odds of survival are ∞. When this happens the odds ratio will be ∞∞, and this is undefined so we would be unable to calculate an odds ratio!! The R code and output below shows you what happens.

M1 = matrix(c(50, 50, 0, 0), nrow=2)
rownames(M1) = c("Male", "Female")
colnames(M1) = c("Survived", "Died")
M1

##        Survived Died
## Male         50    0
## Female       50    0

library(mosaic)

## Warning: replacing previous import 'dplyr::vars' by 'ggplot2::vars' when
## loading 'mosaic'

oddsRatio(M1, verbose=TRUE)

## 
## Odds Ratio
## 
## Proportions
##     Prop. 1:  1 
##     Prop. 2:  1 
##   Rel. Risk:  1 
## 
## Odds
##      Odds 1:  Inf 
##      Odds 2:  Inf 
##  Odds Ratio:  NaN 
## 
## 95 percent confidence interval:
##   1 < RR < 1 
##   NaN < OR < NaN 
## NULL

## [1] NaN

Note that this result also obtains if no Male or Female passenger survived.

M2 = matrix(c(0, 0, 50, 50), nrow=2)
rownames(M2) = c("Male", "Female")
colnames(M2) = c("Survived", "Died")
M2

##        Survived Died
## Male          0   50
## Female        0   50

oddsRatio(M2, verbose=TRUE)

## 
## Odds Ratio
## 
## Proportions
##     Prop. 1:  0 
##     Prop. 2:  0 
##   Rel. Risk:  NaN 
## 
## Odds
##      Odds 1:  0 
##      Odds 2:  0 
##  Odds Ratio:  NaN 
## 
## 95 percent confidence interval:
##   NaN < RR < NaN 
##   NaN < OR < NaN 
## NULL

## [1] NaN

What if the distribution were different, where all Males died and all Females survived?

M3 = matrix(c(0, 50, 50, 0), nrow=2)
rownames(M3) = c("Male", "Female")
colnames(M3) = c("Survived", "Died")
M3

##        Survived Died
## Male          0   50
## Female       50    0

oddsRatio(M3, verbose=TRUE)

## 
## Odds Ratio
## 
## Proportions
##     Prop. 1:  0 
##     Prop. 2:  1 
##   Rel. Risk:  Inf 
## 
## Odds
##      Odds 1:  0 
##      Odds 2:  Inf 
##  Odds Ratio:  Inf 
## 
## 95 percent confidence interval:
##   NaN < RR < Inf 
##   NaN < OR < Inf 
## NULL

## [1] Inf

… and finally if things were reversed and all Males survived but no Females survived.

M4 = matrix(c(50, 0, 0, 50), nrow=2)
rownames(M4) = c("Male", "Female")
colnames(M4) = c("Survived", "Died")
M4

##        Survived Died
## Male         50    0
## Female        0   50

oddsRatio(M4, verbose=TRUE)

## 
## Odds Ratio
## 
## Proportions
##     Prop. 1:  1 
##     Prop. 2:  0 
##   Rel. Risk:  0 
## 
## Odds
##      Odds 1:  Inf 
##      Odds 2:  0 
##  Odds Ratio:  0 
## 
## 95 percent confidence interval:
##   0 < RR < NaN 
##   0 < OR < NaN 
## NULL

## [1] 0

Hence we see two limits when the odds ratio is defined: The confidence intervals are bounded below by 0 and bounded above by ∞.

As a rule, then, if in a 2×2 contingency table the count in any cell is zero the odds-ratio will be undefined. Consequently, when any cell is 0, in practice you will often see the odds-ratio calculated by adding a small constant (usually 0.5) to every cell frequency (see below).

Now the odds ratio and its associated confidence intervals can be calculated (see below):

M5 = matrix(c(50.5, 50.5, 0.5, 0.5), nrow=2)
rownames(M5) = c("Male", "Female")
colnames(M5) = c("Survived", "Died")
M5

##        Survived Died
## Male       50.5  0.5
## Female     50.5  0.5

oddsRatio(M5, verbose=TRUE)

## 
## Odds Ratio
## 
## Proportions
##     Prop. 1:  0.9902 
##     Prop. 2:  0.9902 
##   Rel. Risk:  1 
## 
## Odds
##      Odds 1:  101 
##      Odds 2:  101 
##  Odds Ratio:  1 
## 
## 95 percent confidence interval:
##   0.9621 < RR < 1.039 
##   0.01946 < OR < 51.38 
## NULL

## [1] 1

Fisher’s Exact Test

Fisher’s exact test is used if the assumptions of the χ2 are violated. In fact, as a rule just go with Fisher’s test if you are dealing with a 2×2 contingency table and have small samples. The example below runs the test with the data on vampire bats.

bats = read.csv(url("http://whitlockschluter.zoology.ubc.ca/wp-content/data/chapter09/chap09e4VampireBites.csv"))
head(bats); names(bats)

##   estrous bitten
## 1 estrous bitten
## 2 estrous bitten
## 3 estrous bitten
## 4 estrous bitten
## 5 estrous bitten
## 6 estrous bitten

## [1] "estrous" "bitten"

tab.b = table(bats$bitten, bats$estrous)
tab.b

##             
##              estrous no estrous
##   bitten          15          6
##   not bitten       7        322

H0: State of estrous and vampire bat attack are independent
HA: State of estrous and vampire bat attack are not independent
α=0.05; standard decision rule applies

fisher.test(tab.b, conf.int=TRUE, conf.level=0.95, alternative="two.sided")

## 
##  Fisher's Exact Test for Count Data
## 
## data:  tab.b
## p-value < 2.2e-16
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
##   29.94742 457.26860
## sample estimates:
## odds ratio 
##   108.3894

Given the p−value we can easily reject H0; the data suggest that state of estrous and vampire bat attack are not independent.

The G Test

The G−test is useful for large samples and when dealing with complex experimental research designs.

library(RVAideMemoire)
G.test(tab.b)

## 
##  G-test
## 
## data:  tab.b
## G = 71.451, df = 1, p-value < 2.2e-16

Problem 4

It has been hypothesized that the white rump of pigeons serves to distract predators like the peregrine falcons, and therefore it may be an adaptation to reduce predation. To test this, researchers followed the fate of 203 pigeons, 101 with white rumps and 102 with blue rumps. Nine of the white-rumped birds and 92 of the blue-rumped birds were killed by falcons.

1. Show the results in a frequency table (with the response as the rows and the explanatory variable as the columns). What association is suggested?

rumps = read.csv(url("http://whitlockschluter.zoology.ubc.ca/wp-content/data/chapter09/chap09q04PigeonRumps.csv"))
head(rumps); names(rumps)

##   rumpColor survivalWithFalcon
## 1      blue           survived
## 2      blue           survived
## 3      blue           survived
## 4      blue           survived
## 5      blue           survived
## 6      blue           survived

## [1] "rumpColor"          "survivalWithFalcon"

tab.p = table(rumps$survivalWithFalcon, rumps$rumpColor)
addmargins(tab.p)

##           
##            blue white Sum
##   killed     92     9 101
##   survived   10    92 102
##   Sum       102   101 203

tab.p2 = prop.table(tab.p, 1); tab.p2

##           
##                  blue      white
##   killed   0.91089109 0.08910891
##   survived 0.09803922 0.90196078

The table suggests that 91% of the blue-rumped pigeons were killed versus only 9% of the white-rumped pigeons.

Do the two kinds of pigeons differ in their rate of capture by falcons? Carry out an appropriate test.

H0: Predation by falcons is independent of rump-color
HA: Predation by falcons is not independent of rump-color
α=0.05

chisq.test(tab.p, correct=FALSE)

## 
##  Pearson's Chi-squared test
## 
## data:  tab.p
## X-squared = 134.13, df = 1, p-value < 2.2e-16

fisher.test(tab.p)

## 
##  Fisher's Exact Test for Count Data
## 
## data:  tab.p
## p-value < 2.2e-16
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
##   33.70502 270.81972
## sample estimates:
## odds ratio 
##   89.50586

Regardless of the test used, we can safely Reject H0 given that the p−value≈0. The data suggest that predation by falcons is not independent of rump-color

What is the estimated odds ratio for capture of the two groups of pigeons? What is the 95% confidence interval for the odds ratios?

Note: For the mosaic library, “successes” should be located in column 1, and the treatment of interest should be located in row 2. The odds ratio is calculated as (Odds row 2)/(Odds row 1). So we’ll have to recreate the table.

tab.p # The original table

##           
##            blue white
##   killed     92     9
##   survived   10    92

tab.pp = table(rumps$rumpColor, rumps$survivalWithFalcon)
tab.pp # The corrected table with "killed" as column 1 and "blue" as row 2

##        
##         killed survived
##   blue      92       10
##   white      9       92

library(mosaic)
oddsRatio(tab.pp, conf.level=0.95, verbose=TRUE)

## 
## Odds Ratio
## 
## Proportions
##     Prop. 1:  0.902 
##     Prop. 2:  0.08911 
##   Rel. Risk:  0.09879 
## 
## Odds
##      Odds 1:  9.2 
##      Odds 2:  0.09783 
##  Odds Ratio:  0.01063 
## 
## 95 percent confidence interval:
##   0.05279 < RR < 0.1849 
##   0.00413 < OR < 0.02738 
## NULL

## [1] 0.01063327

The estimated odds ratio is 0.0106 and the 95% confidence interval is [0.00413,0.02738]

Problem 7

Reed frogs, a species living in West Africa, have been observed hopping away from grass fires long before the heat of the fire reached the area they were in. This finding led to the hypothesis that the frogs might hear the fire and respond well before the fire reaches them. To test this hypothesis, researchers played three types of sound to samples of reed frogs and recorded their response. Twenty frogs were exposed to the sound of fire, 20 were exposed to the sound of the fire played backward (to control for the range of sound frequencies present in the real sound), and 20 were exposed to equally loud white noise. Of these 60 frogs, 18 hopped away from the sound of the fire, 6 hopped away from the sound of the fire played backward, and 0 hopped away from the sound of the white noise.

1. Illustrate these data with a frequency table. What association is suggested?

frogs = read.csv(url("http://whitlockschluter.zoology.ubc.ca/wp-content/data/chapter09/chap09q07FrogsFire.csv"))
head(frogs); names(frogs)

##   sound direction
## 1  fire      away
## 2  fire      away
## 3  fire      away
## 4  fire      away
## 5  fire      away
## 6  fire      away

## [1] "sound"     "direction"

tab.f = table(frogs$direction, frogs$sound)
tab.f

##           
##            backwards fire noise
##   away             6   18     0
##   not away        14    2    20

tab.ff = prop.table(tab.f, 2)
tab.ff

##           
##            backwards fire noise
##   away           0.3  0.9   0.0
##   not away       0.7  0.1   1.0

The table suggests an association between the type of sound and frogs hopping away.

Do the data provide evidence that reed frogs change their behavior in response to the sound of fire?

H0: Reed frogs do not change their behavior in response to the sound of fire
HA: Reed frogs do change their behavior in response to the sound of fire
α=0.05; the usual decision rule applies

fisher.test(tab.f)

## 
##  Fisher's Exact Test for Count Data
## 
## data:  tab.f
## p-value = 2.328e-09
## alternative hypothesis: two.sided

chisq.test(tab.f, correct=FALSE)

## 
##  Pearson's Chi-squared test
## 
## data:  tab.f
## X-squared = 35, df = 2, p-value = 2.511e-08

Regardless of the test used, the p−value is almost 0 and so we can easily reject H0. The data suggest that reed frogs do change their behavior in response to the sound of fire

Problem 12

Migraine with aura is a potentially debilitating condition, yet little is known about its causes. A case-control study compared 93 people who suffer from chronic migraine with aura to a sample of 93 healthy patients. The researchers used trans-esophageal echocardiography to look for cardiac shunts in all of these patients. (A cardiac shunt is a heart defect that causes blood to flow from the right to the left in the heart, causing poor oxygenation.) Forty-four of the migraine patients were found to have a cardiac shunt while only 16 of the people without migraine symptoms had this heart defect.

1. Is this an observational or experimental study?

It is an observational study

2. Show the association between migraines and cardiac shunts with a mosaic plot.

M = matrix(c(49, 44, 77, 16), nrow = 2,
       dimnames = list(Treatment = c("No Shunt", "Shunt"), Outcome = c("Migraines", "No Migraines"))
       )
M

##           Outcome
## Treatment  Migraines No Migraines
##   No Shunt        49           77
##   Shunt           44           16

mosaicplot(t(M), col=c("yellow", "firebrick"), main="Mosaic plot", cex.axis=0.75, ylab="Relative Frequency")

3. How strong is the association between migraines and cardiac shunts? Calculate the odds ratio for migraine, comparing the patients with and without cardiac shunts.

Remember: You need the successes in column 1 and the treatment in row 2. Our matrix “M” meets these criteria.

oddsRatio(M, verbose=TRUE)

## 
## Odds Ratio
## 
## Proportions
##     Prop. 1:  0.3889 
##     Prop. 2:  0.7333 
##   Rel. Risk:  1.886 
## 
## Odds
##      Odds 1:  0.6364 
##      Odds 2:  2.75 
##  Odds Ratio:  4.321 
## 
## 95 percent confidence interval:
##   1.444 < RR < 2.462 
##   2.2 < OR < 8.488 
## NULL

## [1] 4.321429

The odds ratio is 4.321429

4. What is the 95% confidence interval for this odds ratio?

The 95% confidence interval is [2.200,8.488]