Chest pain

Description

From Example 4.6

A 2×2 contingency table cross-classifying chest-pain status (pain / no pain) against gender (male / female) for 1073 patients. Used to illustrate the chi-squared test for independence and odds-ratio computation.

	pain	no pain
male	46	474
female	37	516

Total: 1073 patients. Marginal proportions: $\hat{P}(X=\text{male})=0.485$, $\hat{P}(Y=\text{pain})=0.077$.

Expected counts

From Example 4.6

Under $H_0$ (independence), the joint probability factors:

$$\hat{P}(X=\text{male},Y=\text{pain})=0.485\times 0.077\approx 0.04$$

Expected count for the male-pain cell: $0.04\times 1073=42.92$.

General formula:

$$\text{Expected count}=\frac{\text{Row total}\times \text{Column total}}{\text{Total sample size}}$$

R code

chisq_output <- chisq.test(chest_tab)
chisq_output$expected

Python code

chisq_output.expected_freq

All expected counts exceed 5, so the chi-squared test is valid.

Chi-squared test

From Example 4.6

Formula (with continuity correction):

$${\chi }^2=\sum \frac{(|\text{observed}-\text{expected}|-0.5{)}^2}{\text{expected}}$$

R code

x <- matrix(c(46, 37, 474, 516), nrow=2)
dimnames(x) <- list(c("male", "female"), c("pain", "no pain"))
chest_tab <- as.table(x)
 
chisq_output <- chisq.test(chest_tab)
chisq_output

Python code

from scipy import stats
import numpy as np
 
chest_array = np.array([[46, 474], [37, 516]])
chisq_output = stats.chi2_contingency(chest_array)
 
print(f"The p-value is {chisq_output.pvalue:.3f}.")
## The p-value is 0.228.
print(f"The test-statistic value is {chisq_output.statistic:.3f}.")
## The test-statistic value is 1.456.

Conclusion: p-value = 0.228, do not reject $H_0$ at the 5% level. Not enough evidence to conclude gender and chest pain are associated.

Odds ratio

From Example 4.10 (Chest Pain and Gender Odds Ratio)

Sample odds ratio:

$$\widehat{OR}=\frac{n_{11}\cdot n_{22}}{n_{12}\cdot n_{21}}=\frac{46\times 516}{474\times 37}$$

95% CI via log-OR + Normal approx:

$$\log \widehat{OR}\pm z_{0.025}\times \text{ASE}(\log \widehat{OR})$$

where $\text{ASE}=\sqrt{1/n_{11}+1/n_{12}+1/n_{21}+1/n_{22}}$, then exponentiate the endpoints.

R code

library(DescTools)
OddsRatio(chest_tab, conf.level = .95)

Python code

import statsmodels.api as sm
chest_tab2 = sm.stats.Table2x2(chest_array)
print(chest_tab2.summary())

The CI for the OR includes 1 — consistent with the chi-squared test, no evidence of association.

See Odds Ratio & Relative Risk for further interpretation notes, and Chi-Square & Fisher for the independence test in general.

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See also: L4 Exploring Categorical Data · Chi-Square & Fisher · Odds Ratio & Relative Risk