FunChisq (version 2.5.4)

cp.fun.chisq.test: Comparative Chi-Squared Test for Model-Free Functional Heterogeneity

Description

Comparative functional chi-squared tests on two or more contingency tables.

Usage

cp.fun.chisq.test(
  x, method = c("fchisq", "nfchisq", "default", "normalized"),
  log.p = FALSE
)

Value

A list with class "htest" containing the following components:

statistic

functional heterogeneity statistic if method = "fchisq" (equivalent to "default"), or normalized statistic if method = "nfchisq" (equivalent to "normalized").

parameter

degrees of freedom.

p.value

p-value of the comparative functional chi-squared test. By default, it is computed by the chi-squared distribution. If method = "normalized", it is the p-value of the normalized statistic computed by the standard normal distribution.

Arguments

x

a list of at least two matrices representing contingency tables of the same dimensionality.

method

a character string to specify the method to compute the functional chi-squared statistic and its p-value. The default is "fchisq" (equivalent to "default"). See Details.

Note: "default" and "normalized" are deprecated.

log.p

logical; if TRUE, the p-value is given as log(p). Taking the log improves the accuracy when p-value is close to zero. The default is FALSE.

Author

Yang Zhang and Joe Song

Details

The comparative functional chi-squared test determines whether the patterns underlying the contingency tables are heterogeneous in a functional way zhang2014nonparametricFunChisq. Specifically, it evaluates whether the column variable is a changed function of the row variable across the contingency tables.

Two methods are provided to compute the functional chi-squared statistic and its p-value. When method = "fchisq" (or "default"), the p-value is computed using the chi-squared distribution; when method = "nfchisq" (or "normalized") a normalized statistic is obtained by shifting and scaling the original statistic and a p-value is computed using the standard normal distribution Box2005FunChisq (Box et al., 2005). The normalized test is more conservative on the degrees of freedom.

References

See Also

For comparative chi-squared test that does not consider functional dependencies, cp.chisq.test.

Examples

# \donttest{
x <- matrix(c(4,0,4,0,4,0,1,0,1), 3)
y <- t(x)
z <- matrix(c(1,0,1,4,0,4,0,4,0), 3)
data <- list(x,y,z)
cp.fun.chisq.test(data)
cp.fun.chisq.test(data, method="nfchisq")
# }