 # Alternative Tests for Contingency Tables

## David C. Howell  This page was designed to go with an entry that I wrote for the International Encyclopedia of Statistical Sciences, Lovric (2010). In that paper I discussed four different designs under which one could derive a contingency table, and pointed out that there were randomization tests that could replace Pearson's chi-square for at least three of them. (I haven't figured out a sampling scheme for the fourth.) The main purpose of this page is to present those randomization tests as functions (or programs) in R, which is a freely available and very popular software package. R can be download from the R Project at http://www.r-project.org/.

Pearson's chi-square was originally developed by Pearson in papers published in 1900 and 1904. It has had somewhat of a controversial past, beginning with the fact that Pearson got the degrees of freedom wrong and was most unhappy when Fisher pointed that out. (He must have been especially annoyed because Fisher used data from Pearson's own son (Egon)). Essentially, Pearson, with Fisher's correction, showed that the statistic
$χ 2 =Σ (O−E) 2 E$
was distributed approximately as the chi-square distribution. For an r × c table it will be distributed on (r-1)(c-1) df. And when the sample sizes are reasonably large with at an expected value of 5 in each cell, the approximation is quite good. Certainly good enough for our purposes.

The problem arises when the expected frequency in one or more cells is too small. When that happens, the chi-square statistic can only take on a limited number of different values, and hence its distribution cannot be well approximated by the chi-square test. For example, when we have a 2 × 2 table with marginal frequencies of 4 in each row and column, there are only 3 possible values of chi-square (0, 2, and 8). Certainly we cannot expect a continuous distribution to be a good fit under such extreme conditions.

In an attempt to adjust for this problem, Yates(1934) proposed a correction that simply involved reducing the absolute value of each numerator by 0.5 before squaring. This correction actually worked reasonably well, but it was still an approximation. Yates' correction has been falling out of favor for many years, but most software still produces it, sometimes without notice. In its place Fisher's Exact Test Fisher (1935), sometimes known as the Fisher-Irwin Test, has taken its place in many situations. This test is not dependent on the chi-square distribution and is, under certain conditions, exact.

## Alternative Data Collection Designs

### All Marginals Fixed

Fisher's tea-tasting experiment is a classic example of a data collection procedure in which the marginal totals are fixed. Muriel Bristol, a scientist the the Rothamstead Experiment Station, claimed that she could tell the difference between a cup of tea in which the milk had been poured before the tea and one in which the tea had been poured before the milk. Fisher devised an experiment in which he made 8 cups of tea. Four of them had the milk poured first and 4 had the tea poured first. This fixes the row marginals at 4 because every replication of this experiment would have row totals of 4. Ms Bristol was asked to taste each cup and assign 4 cups to each order of pouring. This also fixes the column totals because they will always be 4 and 4.

As an aside, let me say something about Muriel Bristol, because others have suggested that this story was just made up. According to Fisher’ s daughter, Joan Box, it is a true story. By the way, Muriel was no slouch. She was a Ph.D. scientist, back in the days when women were not Ph.D. scientists and she established the Rothamstead Experiment Station in 1919. This was the place that Fisher was later to make famous. I think you should know what Muriel looked like, so
here she is - taken without permission from the Web. If you apply the standard chi-square test (one -sided) to these data you will have:

```### Traditional chi-square test
Bristol <- matrix(c(3,1,1,3), nrow = 2,
dimnames = list(Row = c("1","2"), Col = c("1","2")))
cat("Pearson's (two-sided) chi-square test without Yates' correction","\n")
chisq <- chisq.test(Bristol, correct = FALSE)
print(chisq)
#_________________________________________________________________________
Pearson's Chi-squared test

data:  Bristol
X-squared = 2, df = 1, p-value = 0.1573

```

# =============================================================

Instead of applying Pearson's chi-square test to these data, Fisher employed the hypergeometric distribution. Essentially the test computes the probability of 0 to 4 cups in cell11 when all of the marginals are 4. This is an exact probability given fixed marginals, and Fisher proposed it as the optimal solution to the problem. The probabilities of each possible outcome are shown in the table below

Observed
# Cell 11
Probability
0 .014
1 .229
2 .514
3 .229
4 .014

Ideally, Muriel should have placed 4 cups in cell11 and 4 cups in cell22. The probability that she would do that well by chance in only .014, and we would likely conclude that she had a strange ability with respect to tea tasting. But, unfortunately, she had cell totals of 3/1/1/3, which would have a cumulative probability of .013 + .229 = .243, which would not lead to rejection of the null hypothesis. (Others have claimed that she guessed correctly on all trials.) In this particular example, and in all of the other examples that my limited imagination can think of, we will only get excited if the judge does very well. We won't particularly care if she gets it all wrong. So what we really have here is a one-tailed test. We sum the probabilities of 4 and 3 correct responses, but not 1 and 2 correct. A two-tailed test does not make sense here.

I should point out that although this is a real example, which is partly why I used it, it is not a very good example from the point of understanding tea. With only 4 trials, only one outcome could lead to a statistically significant result. Fisher should have made up mamy more cups of tea to give Muriel a fair shot.

A function already exists for R that will calculate Fisher's exact probability. It is named "fisher.test" and is in the "stats" package, which is part of the base package of R and does not need to be installed separately. The code to compute Fisher's Exact Test follows.

```Bristol <- matrix(c(3,1,1,3), nrow = 2,
dimnames = list(Row = c("1","2"), Col = c("1","2")))
cat("Fisher's exact test--one-sided","\n")
ft <- fisher.test(Bristol, alternative = "greater", conf.int = TRUE)
#For a two-tailed test change to alternative = "two.sided"
#two.sided and conf.int apply only to 2x2
print(ft)
#  ________________________________________________________________

Fisher's Exact Test for Count Data

data:  Bristol
p-value = 0.2429
alternative hypothesis: true odds ratio is greater than 1
95 percent confidence interval:
0.3135693       Inf
sample estimates:
odds ratio
6.408309

# =================================================================
```

In this case we are only concerned whether Ms Bristol scored better than chance, so I have specified a one-tailed by adding 'alternative = "greater" .' The results follow the program,and, again, they are not statisticlly significant.

## One set of marginals fixed

Fisher's Exact Test is an exact test only if both sets of marginals are fixed. In that case the reference distribution consists of probability values associated with all possible arrangements of data preserving those marginal totals. But what if only the row (or column) marginals are fixed. An example of such a situation is taken from my entry referred to above.

In 2000 the Vermont legislature approved a bill authorizing civil unions. The results of that vote, broken down by the gender of the legislator, is shown below.

Vote
For Against Total
Women 35 9 44
Men 60 41 101
Total 95 50 145

This was an important vote, and all legislators were there. So if the vote were to be repeated over and over again, there would always be 44 women and 101 men. In other words the row totals are fixed. But people can change their mind, so the column totals would not be known in advance, so they are random. If we apply the standard Pearson chi-square test to these data to test the hypothesis that there is a gender difference in the voting pattern, we have χ2 = 5.50 on 1 df, with an associated probability of .019.

From a slightly different perspective, this test reduces to the binomial distribution. In this case the test is simply the test that the proportion of cases in each column is the same. You can solve this problem in R by using the prop.test() function.You will obtain the same result from prop.test() and chisq.test() so long as you do, or do not, call for Yates' correction with each test.

If we wish to create a randomization test for this design, the appropriate reference distribution would be the probabilities associated with all possible outcomes having those specific row totals. This is no longer the hypergeometric distribution because the column totals are not longer fixed. The R code for such a sampling design is shown below. The first thing that we do is to calculate the obtained chi-square for the data (here, 5.50) To create the random samples, we first calculate the marginal column probabilities. For this example they are .655 and .345. We then draw 44 uniform random numbers for the first row and assign cases to cell11 with probability = .655. To do this we draw 44 uniformly distributed random numbers between 0 and 1. If a number is less than 0.655, that case is assigned to the "For" category. Otherwise it is assigned to the "Against" category. This process is repeated for 101 cases in row 2. We then calculate a chi-square statistic, though several other measures are possible, such as the frequency of cell11, but chi-square is a good choice because it does not depend on the dimensionality of the table. (It is important to point out that although we are calculating a chi-square statistic, we are calculating it as a measure of deviation from expected, NOT as a standard chi-square test that would be compared to the chi-square distribution.) This process is repeated a large number of times (here 10,000) and the chi-square values for each random table are recorded. These are the chi-square values expected under a true null hypothesis. Finally, the "exact" probability is computed as the proportion of the 10,000 tables whose χ2 values exceed the obtained χ2 for the original data table.

The probability in this case, and in the next, is not truly "exact" because we generate a random set of tables rather than the full set of all possible tables with those row totals. However, with 10,000 samples the probability will be very close to exact. And because we have large cell frequencies in our example, it should be close to the probability given by Pearson's chi-square. The code also produces a likelihood ratio chi-square and its probability.

The R-code for this analysis is shown below. Notice that I have several different examples from which you can draw, most of which can be found in my book.

```
# Chi-square with only one set of marginal totals fixed.

# Alternative data sets follow--You must set the default directory and specify which file equals "data."
# These data sets vary in terms of which variable you can assume to be fixed. Sometimes none seem reasonable.
### You must remove "###" before two lines for one set of data.

### Visintainer <- read.table("Visintainer.dat", nrows = 2, header = TRUE)  #2 x 3 example
### data <- Visintainer
### Jankowski <- matrix(c(512, 54, 227, 37, 59, 15, 18, 12), nrow = 4, byrow = TRUE)
#### data <- Jankowski
### UnahBoger <- matrix(c(33, 251, 33, 508), nrow = 2, byrow = TRUE)
### data <- UnahBoger
### Hout <- matrix(c(7, 7, 2, 3, 2, 8, 3, 7, 1, 5, 4, 9, 2, 8, 9, 14), byrow = TRUE, nrow = 4)
### data <-Hout

civilUnion <- matrix(c(35,9,60,41), nrow = 2, byrow = TRUE)   #Civil Union example from Ex. 6.11
data <- civilUnion
result <- chisq.test(data, correct = F)
obtchisq <- result\$statistic
dfs <- result\$parameter
p <- result\$p.value
N <- sum(data)
cat("The original data","\n")
print(data)
cat("\n")
print(result)

rowmarg <- rowSums(data)
nrow <- length(rowmarg)
colmarg <- colSums(data)
ncol <- length(colmarg)
colprob <- colmarg/N

chisquare <- numeric(nreps)
results <- numeric(nreps)
extreme <- 0
# Get cumulative proportions over columns
# We might find that 60% of the obs are in col1, 25% in col2, and 15% in col3
# Then draw as many numbers as the row marginal, and assign to col based on
# those probabilities. That does not mean that we will always have 60, 25, and
# 15%, because assignment depends on whether the random number exceeds some value.
cum <- numeric()
cum <- colprob
for (j in 2:ncol) {cum[j] <- cum[j-1] + colprob[j]}

# Now resample nreps times
# nreps <- 10000
count <- c()
countx <- c()
for (i in 1:nreps) {
randomtable <- c()
for (j in 1:nrow) {
randnum <- runif(rowmarg[j],0,1)
for (j in 1:ncol) {count[j] <- length(randnum[randnum <= cum[j]])}
countx <- count
for (k in 2:ncol) {countx[k] <- count[k] - count[k-1]}
randomtable <- rbind(randomtable, countx)
}
chisquare[i] <- chisq.test(randomtable, correct = FALSE)\$statistic

}
print( "Ignore warnings if present!")
signif <- length(chisquare[chisquare >= obtchisq]) /nreps
cat("The corresponding p value is ", signif)

### Highlight the above code, run it repeatedly, and compare to standard chi-square
###
par(mfrow=c(2,2))
hist(chisquare, breaks = 50, main = "Distribution of chi-square under null")
library(car)
qqPlot(chisquare, dist = "chisq", df = dfs, id = FALSE, grid = TRUE)
cat("Pearson's chi-square statistic is ",obtchisq,"\n")
cat("That chi-square has a p value of ",p, "\n")
cat("\n")
cat("The proportion of chi-square values greater than the obtained chi-square is  ", signif, "\n")
### This is the desired probability.
```

### No Marginals Fixed

Suppose that instead of asking the Vermont Legislators to take a vote, we went out and drew a random sample of 145 Vermont residents and asked their opinion. In this case neither the row nor the column marginals would be fixed because we do not know in advance how many men and women will be in our sample, nor how many will vote "for and "against" civil unions. This type of design is quite common.

I have struggled with the solution to this example for a very long time, but I am not at all satisfied with my answer. Perhaps it makes sense, but when I come back tomorrow I will probably think about what an idiot I am. I think that I have just replaced "fixed marginal totals" with "fixed marginal proportions." That leaves some slack in the resulting marginal totals after resampling, but not a lot.) But let me continue with my original solution, in hopes that someone can write to me and prove that I am correct--or suggest an alternative approach.

The appropriate reference distribution of random tables is different in this case. What I did was to compute the proportions in each row and the proportions in each column. If Vote is independent of Gender, the probability of an observation falling in each cell is the product of its row and column proportions. I then sampled N uniformly distributed random numbers between 0 and 1, and made cell assignments based on cell probabilities. For example, if in a 2 × 3 table 40% of observations fell in row 1 and 30% of observations fell in column 3, then I would expect 40% × 30% = 12% to fall in cell13 if rows and columns are independent. Assume that the corresponding percentages in row 1 were 20% and 5% for cell11 and cell12 Then a random number between 0 and .20 would be assigned to cell11, a number between .20001 and .25 would be assigned to cell12, a number between .2500001 and .37 would be assigned to cell13, and so on. The cell frequencies that result would be a random sample of cell frequencies having a total sample size of N. The rest of the process is the same as above. Out of 10,000 tables, and their associated chi-squares, the probability under the null would be the proportion of them that equaled or exceeded the chi-square for the obtained data. The R code to carry out this analysis is shown below.

####################### (No Fixed Marginals) ######## ######################################################## # Analysis of contingency tables conditional only on the total sample size # I have changed this to make the cell freq under H0 equal to (on average) the # expected frequencies with those marginal totals. # It is the only way that I can think of to set up a sampling scheme. BUT I am # not restricting the sampling to those marginal totals. # The qq plot fails if you change nreps away from 10000 because of the # way I create length of variables.I should fix that some day. #read in the data civilUnion <- matrix(c(35, 9, 60, 41), nrow = 2, byrow = TRUE) #data <- matrix(c(3,1,1,3), nrow = 2 ) #data <- read.table("DeathModified.dat", header = TRUE, nrows = 2) #data <- read.table("Visintainer.dat", header = TRUE) #data <- read.table("Darley.dat", header = TRUE) #data <- matrix(c(12,2,1,3,2,4,3,4,187,17,11,9,16,7,8,6), nrow = 2, byrow = TRUE) #data <- matrix(c(13,36,14,30), nrow = 2, byrow = TRUE) #rownames(oata) <- c("Drug","Placebo") #colnames(data) <- c("Relapse","Success") #data <- matrix(c(12,2,1,3,2,4,3,4,187,17,11,9,16,7,8,6), nrow = 2, byrow = TRUE) # Establish sample size and number of rows and columns (rs and cs) data <- civilUnion N <- sum(data) nr <- nrow(data) nc <- ncol(data) ncells <- nr*nc # Establish row and column marginal totals rowtots <- rowSums(data) coltots <- colSums(data) # Calculate expected cell probabilities based on original marginal totals # and convert to vector expected <- matrix(data = NA, nrow = nr, ncol = nc, byrow = TRUE) for (i in 1:nr) { for (j in 1:nc) { expected[i,j] <- rowtots[i] * coltots[j] /N^2 } } # Convert to vector expectedvector <- as.vector(t(expected)) # This can now be sampled # Calculate chi-square on original data chisq.test(data, correct = F) obtchisq <- chisq.test(data, correct = F)\$statistic nreps <- 10000 xx <- numeric(N) results <- numeric(nreps) greater <- 0 cells <- numeric(ncells) cells <- c(1:ncells) # Now resample nreps times for (i in 1:nreps) { xx <- sample(cells,N,replace = TRUE, prob = expectedvector) vv <- matrix(0,1,ncells) for (j in 1:N) { b <- xx[j] vv[b] = vv[b] + 1 } x <- matrix(vv,nrow = nr, byrow = TRUE) # x is now the contingency table for ith random sample results[i]<- chisq.test(x, correct = FALSE)\$statistic greater <- ifelse(results[i] >= obtchisq, greater + 1, greater + 0) } probGreater <- greater/nreps cat("With no fixed marginals p(chi-square) = ",probGreater) dfs <- (nr - 1)*(nc - 1) cat("For a chi-square distribution the mean should be = ", dfs, ". It is actually ",mean(results, na.rm = TRUE),"\n") cat("\n") cat("For a chi-square distribution the var. should be = ", 2*dfs,". It is actually ",var(results, na.rm = "TRUE"), "\n") cat("\n") cat(" Look at the graphic for the distribution and a qq plot", "\n") cat("\n") par(mfrow=c(2,2)) hist(results, xlab = "Chi-Square", ylab = "Frequency", breaks = 50, main = "Chi-square distribution") # # # qqPlot(results, dist = "chisq", df = dfs) #Using Monte-Carlo simulation cat("The following is the result of letting R do a simulation for chisq.test","\n") cat("\n") print(chisq.test(data, simulate.p.value = TRUE, B = 10000)) # cat("The following is the result of Fisher's exact test ","\n") cat("\n") print(fisher.test(data, conf.level = 0.95)) cat("\n") cat("Do not be alarmed by warnings if you have some small marginals. ", "\n") cat("They appear because I used the chi-square statistic as my indicator of disparity.", "\n")

### Nothing Fixed

The logical next step would be to consider a study where we went into classrooms and asked students to vote. In this case we not only do not know how many men and women there will be, nor the number of For and Against votes, but we don't even know the total sample size.

It may be possible to generate random samples with this design, but I have not figured out how to do it. I would first have to draw a random sample size, then repeated the calculations in the previous section, then draw another random sample size, and so on. I seriously doubt that it is worth the computational time needed to carry out this process. I suppose that it could be done, but I'm not going to do it.

## Conclusion

In many situations the above procedures might be considered overkill. When we have large expected frequencies in all of the cells, the Pearson chi-square test is quite appropriate. The difference in probability values between that statistic and the ones given here will be small. However when we have one or more small expected frequencies things start to fall apart. Campbell (2007) in an extensive study of 2 × 2 tables, concluded that Fisher's Exact Test works well enough when we have small expected frequencies. He also concluded that a modified Pearson's chi-square [χ2(N/(N-1))] is satisfactory when the smallest expected frequency is greater than 1 for 2 × 2 tables. (I believe that this adjustment can be attributed to Karl Pearson's son Egon, who was also a noted statistician.) However there is no reason why we need to be satisfied with "well enough" when we have a more appropriate solution.

×

Campbell, I. (2007). Chi-squared and Fisher-Irwin tests of two-by-two tables with small sample recommendations. Statistical in Medicine, 26, 3661-3675.

Fisher, R. A. (1935). The Design of Experiments. Edinburgh, Oliver and Boyd.

Howell, D. C. ,(2010). Chi-square test: Analysis of contingency tables. In Lovric, M. (2011). International Encyclopedia of Statistical Science, Springer-Verlag, Berlin.

Yates, F. (1934). Contingency tables involving small numbers and the χ2 test. Journal of the Royal Statistical Society Supplement 1:217-235.  Send mail to: David.Howell@uvm.edu

Last revised 6/25/2015