This package brings together a number of routines for the two sample problem for univariate data. There are two data sets x and y and we want to test whether they were generated by the same probability distribution.
The highlights of this package are:
We generate two data sets of size 100 and 120 respectively from standard normal distributions and run the test:
x1 = rnorm(100)
y1 = rnorm(120)
twosample_test(x1, y1, B=500)
#> maxProcessor set to 1 because parallel overhead would dominate this short computation
#> $statistics
#> KS Kuiper CvM AD ZA ZK ZC Wassp1
#> 0.1383 0.1666 0.3363 1.7056 -3.2530 2.4087 -3.1940 0.2224
#> EppsS HHG ES large ES small EP large EP small
#> 5.2636 1.8584 10.0100 5.2000 9.7100 8.8000
#>
#> $p.values
#> KS Kuiper CvM AD ZA ZK ZC Wassp1
#> 0.2280 0.4320 0.0920 0.1280 0.3640 0.4300 0.3420 0.1360
#> EppsS HHG ES large ES small EP large EP small
#> 0.2760 0.2240 0.6147 0.5183 0.7171 0.4559In this case the the null hypothesis is true, both data sets were generated by a standard normal distribution. And indeed, all 14 tests have p values larger than (say) 0.05 and therefore correctly fail to reject the null hypothesis.
Alternatively one might wish to carry out a select number of tests, but then find a correct p value for the combined test. This can be done as follows:
twosample_test_adjusted_pvalue(x1, y1)
#> p values of individual tests:
#> ES small : 0.5183
#> ZA : 0.3742
#> ZK : 0.4252
#> Wassp1 : 0.1424
#> Kuiper : 0.4236
#> adjusted p value of combined tests: 0.328By default for continuous data this runs five tests (chi square with a small number of bins, ZA, ZK, Wassp1 and Kuiper). It then finds the smallest p value (here 0.142, for Wassp1) and adjusts it for the multiple testing problem, resulting in an overall p value of 0.328.
Say we wish to use two new tests for continuous data. One is based on the difference in standardized means and the other is based on the difference in standard deviations:
\[ \begin{aligned} &TS_1 = sd(x) - sd(y) \\ &TS_2 = \bar{x}/sd(x) - \bar{y}/sd(y) \\ \end{aligned} \]
myTS1 = function(x, y) {
out = c(0, 0)
out[1] = abs(sd(x) - sd(y))
out[2] = abs(mean(x)/sd(x) - mean(y)/sd(y))
names(out) = c("std", "std t test")
out
}Then we can simply run
twosample_test(x1, y1, TS=myTS1, B=500)
#> maxProcessor set to 1 because parallel overhead would dominate this short computation
#> $statistics
#> std std t test
#> 0.06396 0.18060
#>
#> $p.values
#> std std t test
#> 0.538 0.190The arguments have to be x and y for the two data sets and (optionally) a list called TSextra for any additional input needed to find test statistic.
Note that the output vector of the routine has to be a named vector. If the routine is written in Rcpp parallel programming is not available.
Here we generate two data sets of size 1000 and 1200 respectively from a binomial distribution with 5 trials and success probabilities of 0.5 and 0.55, respectively.
x2 = table(c(0:5,rbinom(1000, 5, 0.5)))-1
y2 = table(c(0:5,rbinom(1200, 5, 0.55)))-1
rbind(x2, y2)
#> 0 1 2 3 4 5
#> x2 27 157 301 346 133 36
#> y2 15 121 335 418 253 58
twosample_test(x2, y2, vals=0:5, B=500)$p.values
#> maxProcessor set to 1 because parallel overhead would dominate this short computation
#> KS Kuiper CvM AD ZA Wassp1 EppsS HHG large small
#> 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.008 0.000 0.000In this case the the null hypothesis is false, all ten tests for discrete data have p values much smaller than (say) 0.05 and therefore correctly reject the null hypothesis.
Notice, it is the presence of the vals argument that tells the twosample_test command that the data is discrete. The vectors x and y are the counts. Note that the lengths of the three vectors have to be the same and no value of vals is allowed that has both x and y equal to 0.
Again we might want to run a different test, say based again on the difference in standardized means. This time we will implement the new test using Rcpp:
(For reasons to do with submission to CRAN this routine is already part of R2sample)
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
NumericVector myTS2(IntegerVector x,
IntegerVector y,
NumericVector vals) {
Rcpp::CharacterVector methods=CharacterVector::create("std t test");
int const nummethods=methods.size();
int k=x.size(), n=sum(x), i;
double meanx=0.0, meany=0.0, sdx=0.0, sdy=0.0;
NumericVector TS(nummethods);
TS.names() = methods;
for(i=0;i<k;++i) {
meanx = meanx + x[i]*vals[i]/n;
meany = meany + y[i]*vals[i]/n;
}
for(i=0;i<k;++i) {
sdx = sdx + x[i]*(vals[i] - meanx)*(vals[i] - meanx);
sdy = sdy + y[i]*(vals[i] - meany)*(vals[i] - meany);
}
sdx = sqrt(sdx/(n-1));
sdy = sqrt(sdy/(n-1));
TS(0) = std::abs(meanx/sdx - meany/sdy);
return TS;
}Now
twosample_test(x2, y2, vals=0:5, TS=R2sample::myTS2, B=500)
#> Parallel Programming is not possible if custom TS is written in C++. Switching to single processor
#> $statistics
#> std t test
#> 0.2405
#>
#> $p.values
#> std t test
#> 0.006The arguments have to be x and y for the two data sets (the counts) and vals for the set of values where \(P(X=vals)>0\). Also (optionally) a list called TSextra for any additional input needed to find test statistic.
Note that the output vector of the routine has to be a named vector.
If the routine is written in Rcpp parallel programming is not available.
For all the tests except the chi square variants the p values are found via the permutation method. The idea of a permutation test is simple. Say we have data sets \(x_1,..,x_n\) and \(y_1,..,y_m\). They are combined into one large data set, permuted and split again in sets of sizes n and m. Under the null hypothesis these new data sets come from the same distribution as the actual data. Therefore calculating the tests statistics for them and repeating many times one can build up the distributions of the test statistics and find p values from them.
In the discrete case the situation is somewhat more complicated. Say we have data sets with values \(v_1,..,v_k\) and counts \(x_1,..,x_n\), \(y_1,..,y_k\). One can then mimic the description above by expanding these to yield a large data set with \(x_1+y_1\) \(v_1\)’s, \(x_2+y_2\) \(v_2\)’s and so on. Then this vector is permuted and split as described above. This method is used with the argument samplingmethod=“independence”. The drawback of this sampling method is that its calculation speed increases with the sample size.
Alternatively R2sample provides the following option. Let \(n=\sum x_i\) and \(m=\sum y_i\). Given the class totals \(x_i+y_i\), the counts \(X^*\) of the permuted x data set have a multivariate hypergeometric distribution:
\[P(X^*=a|x,y)=\frac{\prod_{j=1}^k{{x_j+y_j}\choose{a_j}}}{{n+m}\choose{n}}\] for any \(a\) such that \(0\le a_i \le x_i+y_i; i=1,..,k\) and \(\sum a_i=n\). One can sample from this distribution sequentially: \(X^*_1\sim Hyper(x_1+y_1, n+m-(x_1+y_1), n)\), that is the number of white balls when drawing \(n\) balls from an urn with \(x_1+y_1\) white and \(n+m-(x_1+y_1)\) black balls. Then \(X^*_2\) is drawn the same way from the remaining \(n+m-(x_1+y_1)\) balls, with \(n-X^*_1\) balls drawn, and so on. This gives exactly the same distribution as the permutation method, but requires only \(k\) random numbers, so its speed does not depend on the sample sizes. This method is used with the argument samplingmethod=“Binomial” (the name is kept for backward compatibility).
The default is samplingmethod=“Binomial” with a simulation size of 5000 for testing and 1000 for power estimation.
The user can also provide a routine rnull, which generates new data under the null hypothesis. This is needed in the following situation: we wish to perform a goodness-of-fit test \(H_0:X\sim F(p)\), where \(F\) is some distribution function that depends on unknown parameters \(p\). These parameters can be estimated from the data by \(\hat{p}\). Unfortunately \(F\) is complicated and can not be evaluated directly. We can, however, sample from \(F(\hat{p})\). So we generate a second data set \(y\) from \(F(\hat{p})\). In this setup the permutation method fails, and the p values have to be found via the parametric bootstrap.
Say we have a data set \(x\) and wish to test whether it comes from a normal distribution with unknown mean and standard deviation. Let’s assume we do not know how to evalute the normal cdf, so instead we sample from \(N(\bar{x}, sd(x))\):
rnull=function(dta) {
list(x=dta$x,
y=rnorm(length(dta$y), mean(dta$x), sd(dta$x)))
}
twosample_test(x1, y1, rnull=rnull, B=500)
#> maxProcessor set to 1 because parallel overhead would dominate this short computation
#> $statistics
#> KS Kuiper CvM AD ZA ZK ZC Wassp1
#> 0.1383 0.1666 0.3363 1.7056 -3.2530 2.4087 -3.1940 0.2224
#> EppsS HHG ES large ES small EP large EP small
#> 5.2636 1.8584 10.0100 5.2000 9.7100 8.8000
#>
#> $p.values
#> KS Kuiper CvM AD ZA ZK ZC Wassp1
#> 0.0680 0.1840 0.0080 0.0080 0.0980 0.1420 0.0780 0.0080
#> EppsS HHG ES large ES small EP large EP small
#> 0.0620 0.0220 0.6147 0.5183 0.7171 0.4559The routine rnull has to have as its argument a list with the data sets and return an object of the same type.
In the continuous case we have a sample x of size of n and a sample y of size m. In the discussion below both are assumed to be ordered. We denote by z the combined sample. In the discrete case we have a random variable with values v, and x and y are the counts. We denote by k the number of observed values.
We denote the edf’s of the two data sets by \(\widehat{F}\) and \(\widehat{G}\), respectively. Moreover we denote the empirical distribution function of the combined data set by \(\widehat{H}\).
In the case of continuous data, four chi square tests are run: the data is binned into either nbins[1] (default 50, ES large and EP large) or nbins[2] (default 10, ES small and EP small) bins. ES bins have equal size on the range of the combined data, EP bins have equal probability, that is they are found from the quantiles of the combined data. In the case of discrete data (large and small) the classes come from vals, and if there are more than nbins[1] or nbins[2] classes neighboring classes are combined until there are that many.
In all cases neighboring bins are also combined until the smallest expected count is at least minexpcount (default 5). Then the standard chi square test for homogeneity is run, and the p value is found using the chi square approximation with (number of bins - 1) degrees of freedom. Many previous studies have shown that a chi square test with a large number of classes often has poor power. For a literature review of chi square tests see (W. Rolke 2021).
For all other tests the p values are found using a permutation test. They are
This test is based on the quantity \(\max\{|\widehat{F}(x)-\widehat{G}(x)|:x\in \mathbf{R}\}\). In the continuous case (without ties) the function \(\widehat{F}(x)-\widehat{G}(x)\) either takes a jump of size \(1/n\) up at a point in the x data set, a jump of size \(1/m\) down if x is a point in the y data set, or is flat. Therefore the test statistic is
\[\max\{\vert \sum_{i=1}^{j} \left(\frac1n I\{z_i \in x\}-\frac1m I\{z_i \in y\}\right) \vert:j=1,..,n+m\}\]
In the discrete case the jumps have sizes \(x_i/n\) and \(y_j/m\), respectively and the test statistic is
\[\max\{ \vert \sum_{i=1}^{j} \left(\frac{x_i}n -\frac{y_i}m\right) \vert:j=1,..,k\}\]
This test was first proposed in (Kolmogorov 1933), (Smirnov 1939) and is one of the most widely used tests today.
There is a close connection between the edf’s and the ranks of the x and y in the combined data set. Using this one can also calculate the KS statistic, and many statistics to follow, based on these ranks. This is used in many software routines, for example the ks.test routine in R. However, when applied to discrete data these formulas can fail badly because of the many ties, and so we will not use them in our routine.
This test is closely related to Kolmogorov-Smirnov, but it uses the sum of the largest positive and negative differences between the edf’s as a test statistic:
\[\max_z\left\{\widehat{F}(z)-\widehat{G}(z)\right\}+\max_z\left\{\widehat{G}(z)-\widehat{F}(z)\right\}\] The discrete case follows directly. This test was first proposed in (Kuiper 1960).
This test is based on the integrated squared difference between the edf’s:
\[\frac{nm}{(n+m)^2}\sum_{i=1}^{n+m} \left( \hat{F}(z_i)-\hat{G}(z_i) \right)^2 \] the extension to the two-sample problem of the Cramer-vonMises criterion is discussed in (Anderson 1962).
This test is based on
\[nm\sum_{i=1}^{n+m-1} \frac{\left( \hat{F}(z_i)-\hat{G}(z_i) \right)^2}{i(n+m-i)} \]
This is the two-sample version of the test proposed in (Anderson et al. 1952), see (Pettitt 1976). In the discrete case the sum runs over all the n+m positions of the combined data set, with the edf’s evaluated at the value of each position.
These tests are based on the paper (Zhang 2006). They use the edf’s adjusted at the observations, \(\widehat{F}(z_i)-\frac{1}{2n}\) if \(z_i\) is an x and \(\widehat{G}(z_i)-\frac{1}{2m}\) if \(z_i\) is a y. In case of ties (and for discrete data) the “midpoint” edf’s \(\frac1n\left(\#\{x_j<z_i\}+\frac12\#\{x_j=z_i\}\right)\) are used, which reduce to Zhang’s definition if there are no ties. For ZA and ZC small values lead to rejection of the null hypothesis, so in order for all tests to reject for large values the routine returns \(-Z_A\) and \(-Z_C\). Only ZA is available for discrete data.
A test based on the Wasserstein p1 metric. It compares the quantiles of the x and y in the combined data set. If n=m the test statistic in the continuous case is very simple:
\[ \frac1n\sum_{i=1}^n |x_i-y_i|\] If \(n\ne m\) it is first necessary to find the respective quantiles of the x’s and y’s in the combined data set. In the discrete case the test statistic is
\[\sum_{i=1}^{k-1} \vert \sum_{j=1}^i\frac{x_j}{n}-\sum_{j=1}^i\frac{y_j}{m} \vert(v_{i+1}-v_i)\]
For a discussion of the Wasserstein distance see (Vaserstein 1969).
This test, proposed in (Epps and Singleton 1986), compares the empirical characteristic functions of the two data sets, \(\hat{\phi}_x(t)=\frac1n\sum_{j=1}^n e^{itx_j}\) and \(\hat{\phi}_y(t)=\frac1m\sum_{j=1}^m e^{ity_j}\), at the two points \(t_1=0.4/\hat{\sigma}\) and \(t_2=0.8/\hat{\sigma}\), where \(\hat{\sigma}\) is the semi-interquartile range of the combined data set. Let \(g(z)=(\cos t_1z, \cos t_2z, \sin t_1z, \sin t_2z)\), let \(\bar{g}_x\) and \(\bar{g}_y\) be the means of \(g\) over the x and the y data, and let \(\hat{\Omega}=\frac{n+m}{n}\hat{\Sigma}_x+\frac{n+m}{m}\hat{\Sigma}_y\), where \(\hat{\Sigma}_x\) and \(\hat{\Sigma}_y\) are the sample covariance matrices of \(g\) over the x and the y data. The test statistic is
\[W=(n+m)(\bar{g}_x-\bar{g}_y)'\hat{\Omega}^{+}(\bar{g}_x-\bar{g}_y)\]
where \(\hat{\Omega}^{+}\) is the (pseudo) inverse of \(\hat{\Omega}\). If \(\max(n,m)<25\) the small sample correction of (Epps and Singleton 1986) is applied. In the discrete case the means and covariances are weighted by the counts, and if the interquartile range is 0, as can happen for discrete data with very few values, the standard deviation of the combined data set is used instead of \(\hat{\sigma}\). The implementation agrees with scipy.stats.epps_singleton_2samp; see also (Goerg and Kaiser 2009). Under the null hypothesis \(W\) has asymptotically a chi square distribution with degrees of freedom equal to the rank of \(\hat{\Omega}\) (usually 4), which is used for large data sets.
This test is based on (Heller et al. 2016). In its equipartition version, which is used here, the combined data set is divided into atoms: for continuous data 40 groups of consecutive order statistics of (nearly) equal size, for discrete data the distinct values. For each number of cells \(m=2,..,10\) the statistic \(S_m\) is the average of the likelihood ratio statistic of the \(2\times m\) table of counts over all partitions of the atoms into \(m\) contiguous cells. This average can be found exactly without going through all the partitions. The \(S_m\) are then combined with the MinP rule: each \(S_m\) is turned into a p value using its distribution under random relabeling of the combined data set (found once by simulation, with 2000 runs and a fixed seed, so it does not change the random numbers of anything else), and the test statistic is \(-\log(\min_m p_m)\).
Because it considers partitions with many cells the test has good power against alternatives where the densities differ in several places, for example multimodal ones. For continuous data the test is distribution-free, so its null distribution depends only on the sample sizes, and for large data sets the p value is found from that simulated null distribution without a permutation test. The HHG test is not available for weighted data or together with a user supplied routine TS.
The routine has the following arguments:
x and y are numeric vectors. In the continuous data case they are simply the data, in the discrete case the counts.
vals a numeric vector of values of the discrete random variable. If missing continuous data is assumed. In the discrete case x, y and vals have to be vectors of equal length.
TS a routine to calculate a test statistic. If missing built-in tests are used.
TSextra a list to be passed to TS.
wx,wy weights for data set (optional)
B=5000 number of simulation runs for permutation test. If B=0 only test statistics are found.
nbins=c(50, 10) number of bins to use for the chi-square tests, see above.
minexpcount=5 minimum expected count required in each bin for the chi square tests
maxProcessor if >1 parallel computing is used.
UseLargeSample for large continuous data sets (both sample sizes at least 10000, the default) p values are found without a permutation test: from large sample theory for KS, Kuiper, CvM, AD and EppsS, and from the simulated null distribution for HHG. Only these six methods are run in that case.
samplingmethod “independence” or “Binomial” (exact hypergeometric sampling, the default) for discrete data
rnull routine to generate data for parametric bootstrap.
SuppressMessages=FALSE suppress informative messages
doMethods a character vector giving the names of the methods to include.
Say for example we want to use only the KS and AD tests:
A list with vectors statistics (the test statistics) and p values.
The routine twosample_power allows the estimation of the power of the various tests, and plot_power draws the corresponding power graph with the methods sorted by their mean power.
Note that due to the use of permutation tests the power calculations are fairly slow. Because of the time constraints imposed by CRAN the following routines are not run. A full power study with (say) 25 alternatives and fairly large data sets might take several hours to run.
New tests can be used in the same way as above.
Say we wish to find the powers of the tests when one data set comes from a standard normal distribution with sample size 100 and the other from a t distribution with 1-10 degrees of freedom and sample size 200.
The arguments TS, TSextra, nbins, minexpcount , UseLargeSample, samplingmethod, rnull, SuppressMessages, maxProcessor and B are the same as in twosample_test. In addtion we have
f A function that generates data sets x and y, either continuous or the counts of discrete variables. In the discrete case care needs to be taken that the resulting vectors have the same length as vals. The output has to be a list with elements x and y. In the case of discrete data the list also has to include a vector vals, the values of the discrete variable.
… arguments passed to f. The most common case would be a single vector, but two vectors or none is also possible.
alpha=0.05 type I error probability to use in power calculation
The arguments of plot_power are a matrix of powers created by twosample_power and (optional) Smooth=FALSE if no smoothing is desired and a name of the variable on the x axis.
We wish to find the powers of the tests when the data set x are 100 observations comes from a binomial n=10, p=0.5 and y 120 observations from from a binomial n=10 and a number of different success probabilities p. Note that in either case not all the possible values 0-10 will actually appear in every data set, so we need to take some care in assuring that x, y and vals match every time:
plot_power(twosample_power(
function(p) {
vals=0:10
x=table(c(vals, rbinom(100, 10, 0.5))) - 1 #Make sure each vector has length 11
y=table(c(vals, rbinom(120, 10, p))) - 1 #and names 1-10
vals=vals[x+y>0] # only vals with at least one count
list(x=x[as.character(vals)],y=y[as.character(vals)], vals=vals)
},
p=seq(0.5, 0.6, length=5)
), "p")We want to assess the effect of the sample size when one data set comes from a standard normal and the other from a normal with mean 1 and standard deviation 1:
As no single test can be relied upon to consistently have good power, it is reasonable to employ several of them. We would then reject the null hypothesis if any of the tests does so, that is, if the smallest p-value is less than the desired type I error probability \(\alpha\).
This procedure clearly suffers from the problem of simultaneous inference, and the true type I error probability will be much larger than \(\alpha\). It is however possible to adjust the p value so it does achieve the desired \(\alpha\). This can be done as follows:
We generate a number of data sets under the null hypothesis. Generally about 1000 will be sufficient. Then for each simulated data set we apply the tests we wish to include, and record the smallest p value. Here is an example. .
pvals=matrix(0,1000,14)
for(i in 1:1000)
pvals[i, ]=R2sample::twosample_test(runif(100), runif(100), B=1000)$p.valuesNext we find the smallest p value in each run for two selections of four methods. One is the selection found to be best above, namely Zhang’s ZC and ZK methods, the methods by Kuiper and Wasserstein as well as a chi square test with a small number of bins. As a second selection we use the methods by Kolmogorov-Smirnov, Kuiper, Anderson-Darling and Cramer-vonMises, which for this null data turn out to be highly correlated.
colnames(pvals)=names(R2sample::twosample_test(runif(100), runif(100), B=1000)$p.values)
p1=apply(pvals[, c("ZK", "ZC", "Wassp1", "Kuiper", "ES small" )], 1, min)
p2=apply(pvals[, c("KS", "Kuiper", "AD", "CvM")], 1, min)Next we find the empirical distribution function for the two sets of p values and draw their graphs. We also add the curves for the cases of four identical tests and the case of four independent tests, which of course is the Bonferroni correction. The data for the cdfs is in the inst/extdata directory of the package.
tmp=R2sample::pvaluecdf
Tests=factor(c(rep("Identical Tests", nrow(tmp)),
rep("Correlated Selection", nrow(tmp)),
rep("Best Selection", nrow(tmp)),
rep("Independent Tests", nrow(tmp))),
levels=c("Identical Tests", "Correlated Selection",
"Best Selection", "Independent Tests"),
ordered = TRUE)
dta=data.frame(x=c(tmp[,1],tmp[,1],tmp[,1],tmp[,1]),
y=c(tmp[,1],tmp[,3],tmp[,2],1-(1-tmp[,1])^4),
Tests=Tests)
ggplot2::ggplot(data=dta, ggplot2::aes(x=x,y=y,col=Tests))+
ggplot2::geom_line(linewidth=1.2)+
ggplot2::labs(x="p value", y="CDF")+
ggplot2::scale_color_manual(values=c("blue","red", "Orange", "green"))The package includes the routine run.studies, which provides 100 case studies each for the continuous and the discrete case. This allows the user to easily compare the power of the methods included in the package to a different one of their choice.
The 100 studies fall into 5 types. In the first four, with 20 cases each, we have data sets that differ in exactly one population parameter: either mean, variance, skewness or kurtosis. In type 5 (=“mixed”) any of the four parameters can be different.
As an example consider a type 1 (=“mean”) study. Study #1 has one data set from a uniform distribution on \([0,1]\). Therefore we have mean=0.5, variance=1/12, skewness=0 and kurtosis=1.8. Next we find a distribution with the same variance, skewness and kurtosis, but with a mean of 0.53. Their densities look as follows:
f=R2sample::case_studies("uniform", "mean", n=1e4)
x=0:100/100
dta=data.frame(x=c(x,x),
y=c(f$dnull(x), R2sample::funs_list[["cont"]][["uniform"]][["mean"]]$density(x)),
Distribution=rep(c("Uniform",
"Alternative"), each=101))
ggplot(dta, aes(x, y, col=Distribution)) +
geom_line() +
scale_y_continuous(limits = c(0, 2)) +
labs(y="Density")These second distributions are designed so that at sample sizes of 500 and a true type I error of \(\alpha=0.05\) the best method as a power of between \(70\%\) and \(90\%\).
The package includes two example “new” tests, both simple chi-square tests, for either continuous (myTS_cont) or discrete data (myTS_disc). Either can return the value of the test statistic, or a p value, depending on what is specified in TSextra.
These cases can be run easily with the routine run.studies:
TSextra=list(nbins=5, statistic=FALSE)
R2sample::run.studies(1:2, 1:2,
TS=R2sample::myTS_cont, TSextra=TSextra,
With.p.value = TRUE)
#> Average rank of a method:
#> ZA ZC ZK EppsS HHG AD EP small EP large
#> 98.3 95.0 85.0 81.7 73.3 55.0 51.7 50.0
#> ES small ES large Chi Kuiper Wassp1 CvM KS
#> 41.7 40.0 31.7 30.0 30.0 23.3 13.3
#> Chi KS Kuiper CvM AD ZA ZK ZC Wassp1 EppsS
#> uniform/mean 0.418 0.299 0.111 0.366 0.525 0.655 0.570 0.659 0.405 0.617
#> linear/mean 0.304 0.215 0.093 0.273 0.482 0.915 0.828 0.912 0.294 0.669
#> uniform/variance 0.076 0.071 0.157 0.078 0.152 0.812 0.655 0.799 0.085 0.394
#> linear/variance 0.075 0.051 0.148 0.064 0.134 0.749 0.564 0.736 0.066 0.404
#> HHG ES large ES small EP large EP small
#> uniform/mean 0.564 0.208 0.419 0.206 0.444
#> linear/mean 0.569 0.264 0.319 0.326 0.422
#> uniform/variance 0.278 0.168 0.122 0.220 0.144
#> linear/variance 0.267 0.154 0.131 0.188 0.136Arguments of run.studies:
Arguments of routine to calculate new test:
Continuous data:
x, y the two data sets
TSextra (optional): a list passed to the routine with any object needed
to do the calculation.
Discrete data:
x, y the two data sets (their counts) vals: the possible values of the
discrete random variable
TSextra (optional): a list passed to the routine with any object needed
to do the calculation.
The output of the routine has to be a named vector with either the test statistic(s) or the p value(s).
The case studies are as follows. As discussed above, there are 20 studies with one distribution chosen from the usual list, and the second so it matches the three of the four parameters mean, variance, skewness and kurtosis but differs in the fourth. The distributions are:
For the exact definitions run case.studies:
R2sample::case_studies("noncentral")[2:4]
#> $rnull
#> function (n)
#> rchisq(n, 5, 0.875)
#> <bytecode: 0x000001bfce9746c8>
#> <environment: 0x000001bfdd27c620>
#>
#> $dnull
#> function (x)
#> dchisq(x, 5, 0.875)
#> <bytecode: 0x000001bfce9748c0>
#> <environment: 0x000001bfdd27c620>
#>
#> $pnull
#> function (x)
#> pchisq(x, 5, 0.875)
#> <bytecode: 0x000001bfce947200>
#> <environment: 0x000001bfdd27c620>In addition there are another 20 case studies where most of the parameters are different:
uniform.linear\(\hspace{3cm}\) U[0,1] vs a linear model on [0,1] with slope s.
uniform.quadratic\(\hspace{2.5cm}\) U[0,1] vs a quadratic model with vertex at 0.5 and some curvature a.
uniform.bump\(\hspace{3.1cm}\) U[0,1] vs U[0,1]+N(0.5,0.05).
uniform.sine\(\hspace{3.3cm}\) U[0,1] vs U[0,1]+Sine wave
beta22.betaaa\(\hspace{3cm}\) Beta(2,2) vs Beta(a,a)
beta22.beta2a\(\hspace{3cm}\) Beta(2,2) vs Beta(2,a)
uniform.uniformmixture\(\hspace{1.9cm}\) U[0,1] vs. \(\alpha\)U[0,1/2]+(1-\(\alpha\))U[1/2,1]
uniform.betamixture\(\hspace{2.5cm}\) U[0,1] vs. \(\alpha\)U[0,1/2]+(1-\(\alpha\))Beta(2,2)
uniform.triangular\(\hspace{2.9cm}\) U[0,1] vs. triangular
double.exponential\(\hspace{3.5cm}\) with different rates.
normal.onestretch\(\hspace{3.1cm}\) N(0,1) vs 1/2N(0,1)+1/2N(0, \(\sigma\))
normal.t\(\hspace{4.2cm}\) N(0,1) vs t(df)
normal.outlier1\(\hspace{3.1cm}\) N(0,1) vs N(0,1)+N(5, 1)
normal.outlier2\(\hspace{3.1cm}\) N(0,1) vs N(0,1)+U[-5,-3]+U[3,5]
normal.normalmixture\(\hspace{3.1cm}\) N(0,1) vs N(-a,1)+N(a,1)
exponential.gamma\(\hspace{2.3cm}\) Exp(1) vs Gamma(1,b)
exponential.weibull\(\hspace{2.5cm}\) Exp(1) vs Weibull(1,b)
exponential.bump\(\hspace{2.7cm}\) Exp(1) vs Exp(1)+N(0.5,0.05)
chisquare.noncentral\(\hspace{2.3cm}\) \(\chi^2(5)\) vs. \(\chi^2(5, \tau)\)
gamma.gammamixture\(\hspace{2.3cm}\) \(\Gamma(2, 3)\) vs \((1-\alpha)\Gamma(2, 3)+ \alpha\Gamma(3,3)\)
Generally a user will wish to apply their test to all the case studies. This can be done easily with:
run.studies can also be used to run the studies for the included methods but for different values of the sample size n and/or alpha.