Chapter 4 Chi-squared Distribution and Tests

4.1 chisq.test

  • performs chi-squared contingency table tests and goodness-of-fit tests.
chisq.test(x, y = NULL, correct = TRUE,
           p = rep(1/length(x), length(x)), rescale.p = FALSE,
           simulate.p.value = FALSE, B = 2000)

  • x, y = NULL: the input can be two numeric vectors as x and y (can both be factors), or a matrix as x.

  • correct: a logical indicating whether to apply continuity correction.

4.1.1 output components

  • Dataset
# Example in A01
birthwt <- data.frame("Low" = c(21054, 27126), "Normal" = c(14442, 3804294), row.names = c("Dead at Year 1", "Alive at Year 1"))
birthwt
##                   Low  Normal
## Dead at Year 1  21054   14442
## Alive at Year 1 27126 3804294
chi <- chisq.test(birthwt)
chi
## 
##  Pearson's Chi-squared test with Yates' continuity correction
## 
## data:  birthwt
## X-squared = 981695, df = 1, p-value < 2.2e-16
# use dollar sign($) to specify output
chi$statistic # the value the chi-squared test statistic
## X-squared 
##  981695.2
chi$parameter # the degrees of freedom
## df 
##  1
chi$p.value
## [1] 0
chi$method
## [1] "Pearson's Chi-squared test with Yates' continuity correction"
chi$data.name
## [1] "birthwt"
chi$observed
##                   Low  Normal
## Dead at Year 1  21054   14442
## Alive at Year 1 27126 3804294
chi$expected  # the expected counts under the null hypothesis
##                        Low     Normal
## Dead at Year 1    442.2639   35053.74
## Alive at Year 1 47737.7361 3783682.26
chi$residuals # the Pearson residuals, (observed - expected) / sqrt(expected).
##                       Low     Normal
## Dead at Year 1  980.10778 -110.08988
## Alive at Year 1 -94.33735   10.59637
chi$stdres    # standardized residuals, (observed - expected) / sqrt(V)
##                       Low    Normal
## Dead at Year 1   990.8294 -990.8294
## Alive at Year 1 -990.8294  990.8294

4.2 pchisq

  • Density, distribution function, quantile function and random generation for the chi-squared (\(\chi^2\)) distribution with df degrees of freedom and optional non-centrality parameter ncp.
pchisq(q, df, ncp = 0, lower.tail = TRUE, log.p = FALSE)

  • q: (vector of) quantile(s).

  • df: degrees of freedom (non-negative, but can be non-integer).

  • lower.tail: logical; if TRUE (default), probabilities are \(P[X \le x]\), otherwise, \(P[X > x]\).

4.2.1 Example

To find the p-value that corresponds with a \(\chi^2\) test statistic of 7 from a test with one degree of freedom:

pchisq(7, df=1, lower.tail = FALSE)
## [1] 0.008150972

  • The P-value is the probability of observing a sample statistic as extreme as the test statistic – the area to the left of the red line – “upper” tail

  • We always set lower.tail = FALSE when calculating P-value of \(\chi^2\) test.

  • Another method to get the p-value

# default: lower.tail = TURE
1 - pchisq(7, df=1)
## [1] 0.008150972

4.3 mantelhaen.test

Performs a Cochran-Mantel-Haenszel chi-squared test of the null that two nominal variables are conditionally independent in each stratum, assuming that there is no three-way interaction.

  • To test the null hypothesis that the exposure is independent of the disease when adjusting for confounding, we can use
mantelhaen.test(x, y = NULL, z = NULL,
                alternative = c("two.sided", "less", "greater"),
                correct = TRUE, exact = FALSE, conf.level = 0.95)

  • Input data:

    • A 3-dimensional contingency table in array form as x
    • Or three factor objects with at least 2 levels as x, y, and z.

  • correct = TRUE: Whether to apply continuity correction when computing the test statistic.

  • exact = FALSE: Whether the Mantel-Haenszel test or the exact conditional test (given the strata margins) should be computed.

# example
dat <- foreign::read.dta("./_data/diet.dta")
tab <- with(dat, table(diet, mort, by = act))
# see general code for more information about the table
mantelhaen.test(tab)
## 
##  Mantel-Haenszel chi-squared test with continuity correction
## 
## data:  tab
## Mantel-Haenszel X-squared = 1.1249, df = 1, p-value = 0.2889
## alternative hypothesis: true common odds ratio is not equal to 1
## 95 percent confidence interval:
##  0.5228808 1.1789039
## sample estimates:
## common odds ratio 
##         0.7851281