(dataTable <- matrix(c(19, 122, 2473, 7768), nrow = 2, byrow = TRUE, dimnames = list(c("Cancer", "No cancer"), c("Coffee", "No coffee")))) Coffee No coffee
Cancer 19 122
No cancer 2473 7768Contingency Analysis
October 16, 2023
Meme-of-the-day
Class Project!
Class Announcements
- Reading Assignment #7 (WITH QUIZ) due Wednesday (Chapter 9)
- Homework #5 posted (due Monday, October 23 at 11:59 pm)
- Lab #4 posted (due Monday, October 23 at 11:59 pm)
Contingency analysis
Our data in this chapter consists of two categorical variables.
We are interested in:
- Estimating association in 2 x 2 tables (i.e. special case of two categoricals with only 2 levels each)
- Testing if there is an association (or dependence) between two categorical variables [\(\chi^2 \!\) contingency test]
Quick example
So far, we’ve said things like “Yeah, that looks like those variables are associated.” It’s time to quantify the evidence.
Left: Death of adult passengers following Titanic shipwreck
Right: Mosaic plot if death and sex were independent
Coffee consumption and healthy prostate
Practice Problem #1
Wilson et al. (2011) followed a set of male health professionals for 20 years. Of all the men in the study, 7890 drank no coffee and 2492 drank on average more than 6 cups per day. In the “no coffee” group, 122 developed advanced prostate cancer during the course of the study, and 19 in the “high coffee” group did.
Coffee consumption and healthy prostate
Coffee consumption and healthy prostate
Definition: The
odds of success are the probability of success divided by the probability of failure. <br > \[ O = \frac{p}{1-p} \]
Discuss: What is the estimated odds of developing cancer in the coffee and no coffee groups?
Coffee consumption and healthy prostate
Discuss: What is the estimated odds of developing cancer in the coffee and no coffee groups?
Answer: \[ \begin{eqnarray*} \hat{p}_{c} & = & \frac{19}{2492} = 0.0076244 \\ \hat{O}_{c} & = & \frac{0.0076244}{1 - 0.0076244} = 0.007683 \end{eqnarray*} \]
Coffee consumption and healthy prostate
Discuss: What is the estimated odds of developing cancer in the coffee and no coffee groups?
Answer: \[ \begin{eqnarray*} \hat{p}_{nc} & = & \frac{122}{7890} = 0.0154626 \\ \hat{O}_{nc} & = & \frac{0.0154626}{1 - 0.0154626} = 0.0157055 \end{eqnarray*} \]
Coffee consumption and healthy prostate
Coffee No coffee
Cancer 19 122
No cancer 2473 7768Definition: The
odds ratio is the odds of success in one group divided by the odds of success in a second group.
\[ \begin{eqnarray*} \hat{O}_{c} & = & 0.007683 \\ \hat{O}_{nc} & = & 0.0157055 \\ \hat{OR} & = & \frac{\hat{O}_{c}}{\hat{O}_{nc}} = 0.4891915 \end{eqnarray*} \]
Coffee consumption and healthy prostate
Treatment Control
Success "a" "b"
Failure "c" "d" Notes:
- The odds and odds ratio are population parameters.
- The odds of success is a ratio of conditional probabilities, \[\hat{O}_{t} = \frac{\frac{a}{a+c}}{1-\frac{a}{a+c}} = \frac{a}{c}\]
- The odds ratio can be calculated as follows: \[\hat{OR} = \frac{ad}{bc}\]
Coffee consumption and healthy prostate
Definition: The
standard error for the log-odds ratio is given by \[ \mathrm{SE}[\ln(\hat{OR})] = \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}} \]
You can use this with the “1.96 rule of thumb” to calculate a 95% confidence interval. We use log-odds ratios as the sampling distribution for the odds ratio is highly skewed (more on why this matters in Chapter 13).
\[ \begin{array}{c} \ln(\hat{OR}) - 1.96\times\mathrm{SE}[\ln(\hat{OR})] < \ln(OR) < \ln(\hat{OR}) + 1.96\times\mathrm{SE}[\ln(\hat{OR})] \\ c_{L} < \ln(OR) < c_{R} \\ e^{c_{L}} < OR < e^{c_{R}} \end{array} \]
Coffee consumption and healthy prostate
(dataTable <- matrix(c(19, 122, 2473, 7768), nrow = 2, byrow = TRUE, dimnames = list(c("Cancer", "No cancer"), c("Coffee", "No coffee")))) Coffee No coffee
Cancer 19 122
No cancer 2473 7768\[ \begin{align} \mathrm{SE}[\ln(\hat{OR})] & = \sqrt{\frac{1}{19} + \frac{1}{122} + \frac{1}{2473} + \frac{1}{7768}} = 0.248 \\ \ln(\hat{OR}) & = \ln\bigl(\frac{ad}{bc}\bigr) = \ln\bigl(\frac{19\cdot 7768}{122\cdot 2473}\bigr) = -0.715 \end{align} \]
Coffee consumption and healthy prostate
Coffee No coffee
Cancer 19 122
No cancer 2473 7768[1] -1.2005175 -0.2294851[1] 0.3010384 0.7949428Discuss: Conclusion?
Coffee consumption and healthy prostate
Using epitools package
Coffee No coffee
Cancer 19 122
No cancer 2473 7768Assumes table is in the form:
| treatment | control | |
|---|---|---|
| sick | a | b |
| healthy | c | d |
Coffee consumption and healthy prostate
Using epitools package
Coffee consumption and healthy prostate
$data
Coffee No coffee Total
Cancer 19 122 141
No cancer 2473 7768 10241
Total 2492 7890 10382
$measure
NA
odds ratio with 95% C.I. estimate lower upper
Cancer 1.0000000 NA NA
No cancer 0.4891915 0.3010411 0.7949357
$p.value
NA
two-sided midp.exact fisher.exact chi.square
Cancer NA NA NA
No cancer 0.001956364 0.002722787 0.003208084
$correction
[1] FALSE
attr(,"method")
[1] "Unconditional MLE & normal approximation (Wald) CI"Coffee consumption and healthy prostate
TMI!!
Coffee consumption and healthy prostate
Can also get hypothesis testing information for association…
Coffee consumption and healthy prostate
Before moving on to hypothesis testing, know that there is another measure of association: relative risk. More specific context…
Definition:
Relative risk is the probability of an undesired outcome in the treatment group divided by the probability of the same outcome in a control group.
| treatment | control | |
|---|---|---|
| undesired | a | b |
| desired | c | d |
\[ \hat{RR} = \frac{\mathrm{Pr[undesired \ | \ treatment]}}{\mathrm{Pr[undesired \ | \ control]}} = \frac{\frac{a}{a+c}}{\frac{b}{b+d}} \]
Coffee consumption and healthy prostate
Coffee No coffee
Cancer 19 122
No cancer 2473 7768
Sum 2492 7890Coffee consumption and healthy prostate
Now using epitools package. Strangely enough, we need the table to be both transposed and levels reversed for the riskratio function to work.
Coffee consumption and healthy prostate
$data
No cancer Cancer Total
No coffee 7768 122 7890
Coffee 2473 19 2492
Total 10241 141 10382
$measure
NA
risk ratio with 95% C.I. estimate lower upper
No coffee 1.0000000 NA NA
Coffee 0.4930861 0.3047198 0.7978932
$p.value
NA
two-sided midp.exact fisher.exact chi.square
No coffee NA NA NA
Coffee 0.001956364 0.002722787 0.003208084
$correction
[1] FALSE
attr(,"method")
[1] "Unconditional MLE & normal approximation (Wald) CI"Coffee consumption and healthy prostate
Remember, the odds ratio is 0.4891915. Hmmm, so close!!!
The relative risk and odds ratio will be very close when the undesired outcome is rare in the population.
Which one do we use to measure association?
Case-control studies
Definition: A
case-control study is a method of observational study in which a sample of individuals having a disease or other focal condition (“cases”) is compared to a second sample of individuals who do not have the condition (“controls”).
Total number of cases and controls are chosen by researcher and not sampled randomly from the population. This means we can’t estimate probability of undesired result in control and treatment. Thus, we cannot estimate relative risk.
Thus begins our venture into parasitology…
A few parasitology examples…
A few parasitology examples…
A few parasitology examples…
A few parasitology examples…
…starting with Toxoplasma gondii
Example of case-control study
Toxoplasma gondii is a protozoan parasite that can infect the brains of many birds and mammals, including humans… In humans, toxoplasmosis may be associated with some mental illnesses, and it may be associated with risky behavior. Yereli et al. (2006) compared the prevalence of Toxoplasma gondii in a sample of 185 drivers between 21 and 40 years old who had been involved in a driving accident (cases) with a sample of 185 drivers of similar age and sex who had not had accidents (control). The researchers were interested in whether Toxoplasma infection may cause a change in the probability of an accident.
Example of case-control study
Example of case-control study
Example of case-control study
infected uninfected
accidents 61 124
no accidents 16 169Question: Why can’t we estimate relative risk?
Answer: Because we can’t estimate the necessary probabilities!!!
\[ \mathrm{Pr[accidents \ | \ infected]} = \frac{\mathrm{Pr[accidents \ AND \ infected]}}{\mathrm{Pr[infected]}} \]
Example of case-control study
| infected | uninfected | |
|---|---|---|
| accidents | 61 | 124 |
| no accidents | 16x | 169x |
There exists a scaling factor \(x\) that scales the “no accidents” counts that will make the proportions representative (i.e. with no bias) of the population.
The odds ratio thus becomes
\[ OR = \frac{61\times 169x}{124\times 16x} = \frac{61\times 169}{124\times 16} = 5.1960685. \]
Ah hah! So, \(x\) cancels out, and the odds ratio can be computed without worrying about under- or over-representation of the levels of the “cases” in the population.
Example of case-control study
Discuss: Turn the result \(OR = 5.1960685\) into a statement.
Answer: The odds of having an accident when infected with toxoplasma is about 5 times larger than having an accident when not infected.
Example of case-control study
estimate lower upper
5.196069 2.859352 9.442394 Discuss: Given this 95% confidence interval, what do you conclude?
Conclusion: Since the 95% confidence interval does not contain 1, an odds ratio of 1 is not a plausible parameter, and thus there is evidence that there is increased odds of having an accident when infected with toxoplasma.
Flipside of the parasitic coin
Estimation of association for 2 x 2 contingency tables: odds ratios and relative risks
What about hypothesis testing?
- \(H_{0}\): There is no association between two categorical variables
- \(H_{A}\): There is an association between two categorical variables
Remember: Hypothesis tests will give you a yes/no answer, but will NOT give you the magnitude of the effect if there is one (hence, always use confidence intervals as well, if possible)
\(\chi^2\) contingency test
Example of contingency test
Example 9.4
Many parasites have more than one species of host, so the individual parasite must get from one host to another to complete its life cycle. Trematodes of the species Euhaplorchis californiensis use three hosts during their life cycle.
Example of contingency test
Example 9.4
Lafferty and Morris (1996) tested the hypothesis that infection influences risk of predation by birds. A large outdoor tank was stocked with three kinds of killfish: unparasitized, lightly infected, and heavily infected. This tank was left open to foraging by birds… The numbers of fish eaten according to their levels of parasitism is given as follows:
Uninfected Lightly Highly Sum
Eaten 1 10 37 48
Not eaten 49 35 9 93
Sum 50 45 46 141Example of contingency test
Uninfected Lightly Highly Sum
Eaten 1 10 37 48
Not eaten 49 35 9 93
Sum 50 45 46 141
Example of contingency test
If we call our two categorical variable “fate” and “infection status”, then what we want to know is are these two variables independent.
How do we compute the frequency table we would expect if the variables were independent? If two variables are independent, we have
\[\mathrm{Pr[A \ and \ B]} = \mathrm{Pr[A]}\times\mathrm{Pr[B]}\]
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | Pr[Eaten and Uninfected] | * | * | Pr[Eaten] |
| Not eaten | * | * | * | * |
| Sum | Pr[Uninfected] | * | * | * |
Example of contingency test
The observed table (in relative frequencies) is given by:
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | Pr[Eaten and Uninfected] | * | * | Pr[Eaten] |
| Not eaten | * | * | * | * |
| Sum | Pr[Uninfected] | * | * | * |
Assuming independence, the expected (relative frequencies) table should have:
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | Pr[Eaten]\(\cdot\) Pr[Uninfected] | * | * | Pr[Eaten] |
| Not eaten | * | * | * | * |
| Sum | Pr[Uninfected] | * | * | * |
Example of contingency test
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | \(141\cdot\frac{48}{141}\cdot\frac{50}{141}\) | * | * | 48 |
| Not eaten | * | * | * | 93 |
| Sum | 50 | 45 | 46 | 141 |
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | 17.0 | \(141\cdot\frac{48}{141}\cdot\frac{45}{141}\) | * | 48 |
| Not eaten | * | * | * | 93 |
| Sum | 50 | 45 | 46 | 141 |
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | 17.0 | 15.3 | \(141\cdot\frac{48}{141}\cdot\frac{46}{141}\) | 48 |
| Not eaten | * | * | * | 93 |
| Sum | 50 | 45 | 46 | 141 |
Example of contingency test
Expected frequency table
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | 17.0 | 15.3 | 15.7 | 48 |
| Not eaten | 33.0 | 29.7 | 30.3 | 93 |
| Sum | 50 | 45 | 46 | 141 |
Observed frequency table
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | 1 | 10 | 37 | 48 |
| Not eaten | 49 | 35 | 9 | 93 |
| Sum | 50 | 45 | 46 | 141 |
Example of contingency test
The \(\chi^2\) contingency test is just a special case of the \(\chi^2\) goodness-of-fit test, so the test statistic is the same.
\[ \chi^2 = \sum_{i=1}^{r}\sum_{j=1}^{c} \frac{(Observed_{ij}-Expected_{ij})^2}{Expected_{ij}}, \]
where \(r\) and \(c\) are number of rows and columns, respectively. The number of degrees of freedom is given by
\[ \begin{align} df & = rc - 1 - (r-1) - (c - 1) \\ & = rc - r - c + 1 \\ & = (r-1)\times(c-1) \end{align} \]
Example of contingency test
Assumptions of the \(\chi^2\) contingency test are the same as the \(\chi^2\) goodness-of-fit test.
What do you do if the assumptions are violated?
- If table larger than 2 x 2, you can combine row and/or columns.
- If table is 2 x 2, you can use Fisher’s exact test.
- You can use a permutation test (see Chapter 13).
Example of contingency test
Expected frequency table
| Uninfected | Lightly | Highly | Sum | |
|---|---|---|---|---|
| Eaten | 17.0 | 15.3 | 15.7 | 48 |
| Not eaten | 33.0 | 29.7 | 30.3 | 93 |
| Sum | 50 | 45 | 46 | 141 |
Assumptions seem to be met, so let’s use R.
Example of contingency test
First, let’s see how the table was created in R:
Example of contingency test
Note: correct = FALSE means no Yates correction (see p. 251)
Example of contingency test
Pearson's Chi-squared test
data: parTable
X-squared = 69.756, df = 2, p-value = 7.124e-16Conclusion?
Conclusion: Since \(p\)-value is less than 0.05 (actually less than 0.001), then we can reject the null hypothesis of independence.
Other type of contingency tests
Fisher’s exact test: (2 x 2 tables only) Examines the independence of two categorical variables, even with small expected values
\(G\)-test: (any table) Derived from principles of likelihood.
Pro : Great with complicated experimental designs with multiple explanatory variables.
Con : Can be less accurate for small sample sizes.
Magnitudes of association
Situation: You’ve found a statistically significant association between two categorical variables using the \(\chi^2\) contingency test.
Question: Where is the association? In other words, in which levels of the categories is the association present and how large is the association?
We need to estimate the magnitude of the association, which the \(P\)-value does not give us!
We can estimate odds ratios or relative risks for 2 \(\times\) 2 sub-tables within the contingency table by either subsetting or collapsing.
Magnitudes of association
Uninfected Lightly Highly
Eaten 1 10 37
Not eaten 49 35 9Magnitudes of association
Uninfected Highly
Eaten 1 37
Not eaten 49 9
Magnitudes of association
$data
Uninfected Highly Total
Eaten 1 37 38
Not eaten 49 9 58
Total 50 46 96
$measure
NA
odds ratio with 95% C.I. estimate lower upper
Eaten 1.000000000 NA NA
Not eaten 0.004964148 0.0006020703 0.04093004
$p.value
NA
two-sided midp.exact fisher.exact chi.square
Eaten NA NA NA
Not eaten 1.110223e-16 6.861412e-17 4.140762e-15
$correction
[1] FALSE
attr(,"method")
[1] "Unconditional MLE & normal approximation (Wald) CI"Introduction to Biostatistics, Fall 2023