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# *G*-test for Count Data
`g.test()` performs chi-squared contingency table tests and
goodness-of-fit tests, just like
[`chisq.test()`](https://rdrr.io/r/stats/chisq.test.html) but is more
reliable (1). A *G*-test can be used to see whether the number of
observations in each category fits a theoretical expectation (called a
***G*-test of goodness-of-fit**), or to see whether the proportions of
one variable are different for different values of the other variable
(called a ***G*-test of independence**).
## Usage
``` r
g.test(x, y = NULL, p = rep(1/length(x), length(x)), rescale.p = FALSE)
```
## Source
The code for this function is identical to that of
[`chisq.test()`](https://rdrr.io/r/stats/chisq.test.html), except that:
- The calculation of the statistic was changed to \\2 \* sum(x \* log(x
/ E))\\
- Yates' continuity correction was removed as it does not apply to a
*G*-test
- The possibility to simulate p values with `simulate.p.value` was
removed
## Arguments
- x:
a numeric vector or matrix. `x` and `y` can also both be factors.
- y:
a numeric vector; ignored if `x` is a matrix. If `x` is a factor, `y`
should be a factor of the same length.
- p:
a vector of probabilities of the same length as `x`. An error is given
if any entry of `p` is negative.
- rescale.p:
a logical scalar; if TRUE then `p` is rescaled (if necessary) to sum
to 1. If `rescale.p` is FALSE, and `p` does not sum to 1, an error is
given.
## Value
A list with class `"htest"` containing the following components:
- statistic:
the value the chi-squared test statistic.
- parameter:
the degrees of freedom of the approximate chi-squared distribution of
the test statistic, `NA` if the p-value is computed by Monte Carlo
simulation.
- p.value:
the p-value for the test.
- method:
a character string indicating the type of test performed, and whether
Monte Carlo simulation or continuity correction was used.
- data.name:
a character string giving the name(s) of the data.
- observed:
the observed counts.
- expected:
the expected counts under the null hypothesis.
- residuals:
the Pearson residuals, `(observed - expected) / sqrt(expected)`.
- stdres:
standardized residuals, `(observed - expected) / sqrt(V)`, where `V`
is the residual cell variance (Agresti, 2007, section 2.4.5 for the
case where `x` is a matrix, `n * p * (1 - p)` otherwise).
## Details
If `x` is a [matrix](https://rdrr.io/r/base/matrix.html) with one row or
column, or if `x` is a vector and `y` is not given, then a
*goodness-of-fit test* is performed (`x` is treated as a one-dimensional
contingency table). The entries of `x` must be non-negative integers. In
this case, the hypothesis tested is whether the population probabilities
equal those in `p`, or are all equal if `p` is not given.
If `x` is a [matrix](https://rdrr.io/r/base/matrix.html) with at least
two rows and columns, it is taken as a two-dimensional contingency
table: the entries of `x` must be non-negative integers. Otherwise, `x`
and `y` must be vectors or factors of the same length; cases with
missing values are removed, the objects are coerced to factors, and the
contingency table is computed from these. Then Pearson's chi-squared
test is performed of the null hypothesis that the joint distribution of
the cell counts in a 2-dimensional contingency table is the product of
the row and column marginals.
The p-value is computed from the asymptotic chi-squared distribution of
the test statistic.
In the contingency table case simulation is done by random sampling from
the set of all contingency tables with given marginals, and works only
if the marginals are strictly positive. Note that this is not the usual
sampling situation assumed for a chi-squared test (such as the *G*-test)
but rather that for Fisher's exact test.
In the goodness-of-fit case simulation is done by random sampling from
the discrete distribution specified by `p`, each sample being of size
`n = sum(x)`. This simulation is done in R and may be slow.
### *G*-test Of Goodness-of-Fit (Likelihood Ratio Test)
Use the *G*-test of goodness-of-fit when you have one nominal variable
with two or more values (such as male and female, or red, pink and white
flowers). You compare the observed counts of numbers of observations in
each category with the expected counts, which you calculate using some
kind of theoretical expectation (such as a 1:1 sex ratio or a 1:2:1
ratio in a genetic cross).
If the expected number of observations in any category is too small, the
*G*-test may give inaccurate results, and you should use an exact test
instead ([`fisher.test()`](https://rdrr.io/r/stats/fisher.test.html)).
The *G*-test of goodness-of-fit is an alternative to the chi-square test
of goodness-of-fit
([`chisq.test()`](https://rdrr.io/r/stats/chisq.test.html)); each of
these tests has some advantages and some disadvantages, and the results
of the two tests are usually very similar.
### *G*-test of Independence
Use the *G*-test of independence when you have two nominal variables,
each with two or more possible values. You want to know whether the
proportions for one variable are different among values of the other
variable.
It is also possible to do a *G*-test of independence with more than two
nominal variables. For example, Jackson et al. (2013) also had data for
children under 3, so you could do an analysis of old vs. young, thigh
vs. arm, and reaction vs. no reaction, all analyzed together.
Fisher's exact test
([`fisher.test()`](https://rdrr.io/r/stats/fisher.test.html)) is an
**exact** test, where the *G*-test is still only an **approximation**.
For any 2x2 table, Fisher's Exact test may be slower but will still run
in seconds, even if the sum of your observations is multiple millions.
The *G*-test of independence is an alternative to the chi-square test of
independence
([`chisq.test()`](https://rdrr.io/r/stats/chisq.test.html)), and they
will give approximately the same results.
### How the Test Works
Unlike the exact test of goodness-of-fit
([`fisher.test()`](https://rdrr.io/r/stats/fisher.test.html)), the
*G*-test does not directly calculate the probability of obtaining the
observed results or something more extreme. Instead, like almost all
statistical tests, the *G*-test has an intermediate step; it uses the
data to calculate a test statistic that measures how far the observed
data are from the null expectation. You then use a mathematical
relationship, in this case the chi-square distribution, to estimate the
probability of obtaining that value of the test statistic.
The *G*-test uses the log of the ratio of two likelihoods as the test
statistic, which is why it is also called a likelihood ratio test or
log-likelihood ratio test. The formula to calculate a *G*-statistic is:
\\G = 2 \* sum(x \* log(x / E))\\
where `E` are the expected values. Since this is chi-square distributed,
the p value can be calculated in R with:
p <- stats::pchisq(G, df, lower.tail = FALSE)
where `df` are the degrees of freedom.
If there are more than two categories and you want to find out which
ones are significantly different from their null expectation, you can
use the same method of testing each category vs. the sum of all
categories, with the Bonferroni correction. You use *G*-tests for each
category, of course.
## References
1. McDonald, J.H. 2014. **Handbook of Biological Statistics (3rd
ed.)**. Sparky House Publishing, Baltimore, Maryland.
## See also
[`chisq.test()`](https://rdrr.io/r/stats/chisq.test.html)
## Examples
``` r
# = EXAMPLE 1 =
# Shivrain et al. (2006) crossed clearfield rice (which are resistant
# to the herbicide imazethapyr) with red rice (which are susceptible to
# imazethapyr). They then crossed the hybrid offspring and examined the
# F2 generation, where they found 772 resistant plants, 1611 moderately
# resistant plants, and 737 susceptible plants. If resistance is controlled
# by a single gene with two co-dominant alleles, you would expect a 1:2:1
# ratio.
x <- c(772, 1611, 737)
g.test(x, p = c(1, 2, 1) / 4)
#>
#> G-test of goodness-of-fit (likelihood ratio test)
#>
#> data: x
#> X-squared = 4.1471, p-value = 0.1257
#>
# There is no significant difference from a 1:2:1 ratio.
# Meaning: resistance controlled by a single gene with two co-dominant
# alleles, is plausible.
# = EXAMPLE 2 =
# Red crossbills (Loxia curvirostra) have the tip of the upper bill either
# right or left of the lower bill, which helps them extract seeds from pine
# cones. Some have hypothesized that frequency-dependent selection would
# keep the number of right and left-billed birds at a 1:1 ratio. Groth (1992)
# observed 1752 right-billed and 1895 left-billed crossbills.
x <- c(1752, 1895)
g.test(x)
#>
#> G-test of goodness-of-fit (likelihood ratio test)
#>
#> data: x
#> X-squared = 5.6085, p-value = 0.01787
#>
# There is a significant difference from a 1:1 ratio.
# Meaning: there are significantly more left-billed birds.
```