Lecture 13: Proportions and Categorical data

BIOS 600 - Spring 2026

Announcements

  • HW 5 due Sunday
  • HW 6 will be posted today
  • Lab 6 due tomorrow

Overview

  • Chi-square test and example

  • Fisher exact test and example

Very optional reading

  • P&G Chapters 14, 15

  • OI: Sections 5.2, 5.3, 6.1, 6.2, 6.4

Pneumococcal disease

The World Health Organization estimates that in developing countries 814,000 children under the age of five die annually from invasive pneumococcal disease (IPD), with an estimated 1.6 million deaths affecting all ages globally.

Several recent studies have identified associations between pneumococcal serotypes (species variations) and patient outcomes from IPD. We consider data from a study of pneumococcal serotypes and mortality (Inverarity et al. (2011)).

Pneumococcal disease

Serotype Alive Dead Total
Serotype 10 37 7 44
Serotype 31 24 10 34
Total 61 17 78

Is there an association between serotype and mortality?

Chi-square tests

\(H_0\): There is no association between serotype and mortality

\(H_1\): There is an association between serotype and mortality

Under \(H_0\), we have independence between serotype and mortality, implying that joint probabilities should be equal to products of marginal distributions:

Serotype Alive Dead Total
Serotype 10 ? ? 44
Serotype 31 ? ? 34
Total 61 17 78

What counts might we expect under \(H_0\)?

Intuition: Joint and Marginal Probabilities

  • Suppose we look at two variables: Serotype and Mortality (Alive or Dead).

  • The joint probability describes the probability of being in both categories (e.g., “Serotype 10 and Alive”).

  • The marginal probabilities describe totals across one variable:

    • P(Serotype 10) = proportion of all patients with Serotype 10
    • P(Alive) = proportion of all patients who survived

Marginal totals

Serotype Total
10 44
31 34
Total 78

Marginal totals

Outcome Total
Alive 61
Dead 17
Total 78

What would happen if these two variables were independent?

  • Under the null hypothesis of independence, the chance of being both “Serotype 10” and “Alive” should just be:

\[ P(\text{Serotype 10 and Alive}) = P(\text{Serotype 10}) \times P(\text{Alive}) \]

  • We can multiply these probabilities by the total number of observations to find the expected counts.

For example:

\[ E_{10,Alive} = P(\text{Serotype 10}) \times P(\text{Alive}) \times N \]

  • Using our totals:

\[ P(\text{Serotype 10}) = \frac{44}{78}, \quad P(\text{Alive}) = \frac{61}{78}, \quad N = 78 \]

Compute expected count under independence

# Compute expected count under independence
expected_10_alive <- (44/78) * (61/78) * 78
expected_10_alive
[1] 34.41026

  • We can repeat this for each cell to find what we’d expect if serotype and mortality were truly unrelated.

  • The chi-square test then measures how far our observed counts deviate from these expected counts.

\[\chi^2 = \sum\frac{(O-E)^2}{E}\]

Intuition

  • If observed counts are close to what we’d expect under independence, \(\chi^2\) will be small, then we’d fail to reject \(H_0\).

  • If they’re far from expected, \(\chi^2\) will be large, which might lead to evidence of an association.

Visualizing Observed vs Expected Counts

Let’s visualize how the observed counts compare to the expected counts under independence.

Chi-square tests

The chi-square test compares the observed frequencies in each cell to the expected count under \(H_0\); if the total differences are large enough, we reject \(H_0\):

\[\chi^2 = \sum_{i=1}^{r \times c}\frac{(O_i - E_i)^2}{E_i}\]

which has a \(\chi^2_{(r - 1) \times (c-1)}\) distribution, under \(H_0\).

Chi-square tests

data <- matrix(c(37, 7, 24, 10), 
               byrow = T, nrow = 2)
data
     [,1] [,2]
[1,]   37    7
[2,]   24   10
chisq.test(data)

    Pearson's Chi-squared test with Yates' continuity correction

data:  data
X-squared = 1.3359, df = 1, p-value = 0.2478

What might we conclude at \(\alpha = .05\)?

Conclusion

  • Conclusion: Since the p-value is greater than 0.05, we fail to reject the null hypothesis. There is not enough evidence to conclude that survival differs between Serotype 10 and Serotype 31.

Fisher’s exact test

  • The chi-square test relies on asymptotic results (i.e. large sample size), which might not be appropriate if we have small cell counts (common rules of thumb are >5 or >10 observed in each cell).

  • Fisher’s exact test calculates an exact p-value under a specific distributional assumption.

Serotype Alive Dead Total
Serotype 10 37 7 44
Serotype 31 24 10 34
Total 61 17 78

Fisher’s exact test

Fisher’s exact test relies on the hypergeometric distribution (don’t worry about this, seriously).

Serotype Alive Dead Total
Serotype 10 a c a + c
Serotype 31 b d b + d
Total a + b c + d a + b + c + d

Conditionally on the margins (assumed fixed), then each of the individual cells is distributed according to the hypergeometric distribution. We can thus calculate the exact probabilities of obtaining specific tables.

Fisher’s exact test

Observed data:

Serotype Alive Dead Total
Serotype 10 37 7 44
Serotype 31 24 10 34
Total 61 17 78

We could use the hypergeometric distribution to calculate the probability of seeing this table, assuming 61 total alive patients, 17 dead, and 44 and 34 infections of serotypes 10 and 31, respectively (lots of factorials, so you could imagine it wasn’t fun before modern computers).

Fisher’s exact test

Another “possible” table:

Serotype Alive Dead Total
Serotype 10 36 8 44
Serotype 31 25 9 34
Total 61 17 78

  • Here we’ve displayed another table with the same margins, but with some slightly different potential cell counts.

  • Again we could use the hypergeometric distribution to calculate the probability of observing the cell counts conditionally on fixed margins.

Fisher’s exact test

Fisher’s exact test relies on constructing all possible contingency tables with the same margins, and then summing up probabilities of all tables as extreme or more extreme than the observed data.

Question

What might “more extreme” mean in the context of contingency tables?

Fisher’s exact test

\(H_0\): The proportion of deaths is the same for Serotype 10 and Serotype 31 (no association between serotype and outcome).

\(H_A\): The proportion of deaths differs between serotypes (there is an association)

data <- matrix(c(37, 7, 24, 10), byrow = T, nrow = 2)
data
     [,1] [,2]
[1,]   37    7
[2,]   24   10

Conducting the Fisher’s Exact test

fisher.test(data)

    Fisher's Exact Test for Count Data

data:  data
p-value = 0.1758
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.6465347 7.7636744
sample estimates:
odds ratio 
  2.179552
  • What might we conclude at \(\alpha = 0.05\)?

Conclusion

  • The Fisher’s Exact Test yielded a p-value of 0.17, which is greater than the conventional significance level of 0.05. Therefore, we fail to reject the null hypothesis.

  • There is no statistically significant evidence of an association between serotype and survival status in this sample.

Odds ratios

Serotype Alive Dead Total
Serotype 10 37 7 44
Serotype 31 24 10 34
Total 61 17 78

  • Odds of death among serotype 10 patients = \(\frac{7/44}{37/44} = 7/37\)

  • Odds of death among serotype 31 patients = \(\frac{10/34}{24/34} = 10/24\)

  • Odds ratio of death comparing 31 to 10 is thus \(\frac{10/24}{7/37} \approx 2.2\)

r by c tables

Serotype Alive Dead Total
Serotype 10 37 7 44
Serotype 15 60 12 72
Serotype 20 97 9 106
Serotype 31 24 10 34
Total 218 38 256

We can also extend the chi-square test and Fisher’s exact test to more general contingency tables (although the notion of an odds ratio would need to be adjusted).

Winter Olympics training

Suppose we were interested in whether a particular training regimen was associated with “success” as measured by whether an athlete qualified for being on the US Winter Olympic Team for giant slalom.

Winter Olympics training

Qualified Did Not Qualify
New Regimen 129 97
Old Regimen 112 114 

    Pearson's Chi-squared test with Yates' continuity correction

data:  data
X-squared = 2.2755, df = 1, p-value = 0.1314

Practice

Write out the code to 1) construct the data matrix and 2) to conduct a chi square test in R. What might we conclude using alpha = .05?

Overview

  • Chi-square test and example

  • Fisher exact test and example (use if sample sizes aren’t large enough)

Next Class

  • Tuesday’s class: Non-Parametric Methods