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(The R code for Barnard’s exact test is at the end of the article, and you could also just download it from here) About half a year ago, I was studying various statistical methods to employ on contingency tables. I came across a promising method for 2×2 contingency tables called “Barnard’s exact test“. Barnard’s test is a non-parametric alternative to Fisher’s exact test which can be more powerful (for 2×2 tables) but is also more time-consuming to compute (References can be found in the Wikipedia article on the subject).

The test was first published by George Alfred Barnard (1945). Mehta and Senchaudhuri (2003) explain why Barnard’s test can be more powerful than Fisher’s under certain conditions:

When comparing Fisher’s and Barnard’s exact tests, the loss of power due to the greater discreteness of the Fisher statistic is somewhat offset by the requirement that Barnard’s exact test must maximize over all possible p-values, by choice of the nuisance parameter, π. For 2 × 2 tables the loss of power due to the discreteness dominates over the loss of power due to the maximization, resulting in greater power for Barnard’s exact test. But as the number of rows and columns of the observed table increase, the maximizing factor will tend to dominate, and Fisher’s exact test will achieve greater power than Barnard’s.

### About the R implementation of Barnard’s exact test

After finding about Barnard’s test I was sad to discover that (at the time) there had been no R implementation of it. But last week, I received a surprising e-mail with good news. The sender, Peter Calhoun, currently a graduate student at the University of Florida, had implemented the algorithm in R. Peter had  found my posting on the R mailing list (from almost half a year ago) and was so kind as to share with me (and the rest of the R community) his R code for computing Barnard’s exact test. Here is some of what Peter wrote to me about his code:

On a side note, I believe there are more efficient codes than this one.  For example, I’ve seen codes in Matlab that run faster and display nicer-looking graphs.  However, this code will still provide accurate results and a plot that gives the p-value based on the nuisance parameter.  I did not come up with the idea of this code, I simply translated Matlab code into R, occasionally using different methods to get the same result.  The code was translated from:

Trujillo-Ortiz, A., R. Hernandez-Walls, A. Castro-Perez, L. Rodriguez-Cardozo. Probability Test.  A MATLAB file. URL

My goal was to make this test accessible to everyone.  Although there are many ways to run this test through Matlab, I hadn’t seen any code to implement this test in R.  I hope it is useful for you, and if you have any questions or ways to improve this code, please contact me at [email protected]

p.s: I added some minor cosmetics to the code, like allowing the input to be a table/matrix and the output to be a list.

### The R function for Barnard’s exact test

```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 # published on: # http://www.r-statistics.com/2010/02/barnards-exact-test-a-powerful-alternative-for-fishers-exact-test-implemented-in-r/   Barnardextest<-function(Ta,Tb =NULL,Tc =NULL,Td =NULL, to.print = F, to.plot = T) { # The first argument (Ta) can be either a table or a matrix of 2X2. # Or instead, the values of the table can be entered one by one to the function   #Barnard's test calculates the probabilities for contingency tables. It has been shown that for 2x2 tables, Barnard's test #has a higher power than Fisher's Exact test. Barnard's test is a non-parametric test that relies upon a computer to generate #the distribution of the Wald statistic. Using a computer program, one could find the nuisance parameter that maximizes the #probability of the observations displayed from a table. #Despite giving lower p-values for 2x2 tables, Barnard's test hasn't been used as often as Fisher's test because of its #computational difficulty. This code gives the Wald statistic, the nuisance parameter, and the p-value for any 2x2 table. #The table can be written as: # Var.1 # --------------- # a b r1=a+b # Var.2 # c d r2=c+d # --------------- # c1=a+c c2=b+d n=c1+c2   #One example would be # Physics # Pass Fail # --------------- # Crane 8 14 # Collage # Egret 1 3 # --------------- # #After implementing the function, simply call it by the command: #Barnardextest(8,14,1,3) #This will display the results:   #"The contingency table is:" # [,1] [,2] #[1,] 8 14 #[2,] 1 3 #"Wald Statistic:" #0.43944 #"Nuisance parameter:" #0.9001 #"The 1-tailed p-value:" #0.4159073   #On a side note, I believe there are more efficient codes than this one. For example, I've seen codes in Matlab that run #faster and display nicer-looking graphs. However, this code will still provide accurate results and a plot that gives the #p-value based on the nuisance parameter. I did not come up with the idea of this code, I simply translated Matlab code #into R, occasionally using different methods to get the same result. The code was translated from: # #Trujillo-Ortiz, A., R. Hernandez-Walls, A. Castro-Perez, L. Rodriguez-Cardozo. Probability Test. A MATLAB file. URL #http://www.mathworks.com/matlabcentral/fileexchange/loadFile.do?objectId=6198 # #My goal was to make this test accessible to everyone. Although there are many ways to run this test through Matlab, I hadn't #seen any code to implement this test in R. I hope it is useful for you, and if you have any questions or ways to improve #this code, please contact me at [email protected].     # Tal edit: choosing if to work with a 2X2 table or with 4 numbers: if(is.null(Tb) | is.null(Tc) | is.null(Td) ) { # If one of them is null, then Ta should have an entire table, and we can take it's values if(is.table(Ta) | is.matrix(Ta)) { if(sum(dim(Ta) == c(2,2))==2) { Tb <- Ta[1,2] Tc <- Ta[2,1] Td <- Ta[2,2] Ta <- Ta[1,1] } else {stop("The table is not 2X2, please check it again...") } } else {stop("We are missing value in the table") } }     c1<-Ta+Tc c2<-Tb+Td n<-c1+c2 pao<-Ta/c1 pbo<-Tb/c2 pxo<-(Ta+Tb)/n TXO<-abs(pao-pbo)/sqrt(pxo*(1-pxo)*(1/c1+1/c2)) n1<-prod(1:c1) n2<-prod(1:c2)   P<-{} for( p in (0:99+.01)/100) { TX<-{} S<-{} for( i in c(0:c1)) { for( j in c(0:c2)) { if(prod(1:i)==0){fac1<-1} else {fac1<-prod(1:i)} if(prod(1:j)==0){fac2<-1} else {fac2<-prod(1:j)} if(prod(1:(c1-i))==0){fac3<-1} else {fac3<-prod(1:(c1-i))} if(prod(1:(c2-j))==0){fac4<-1} else {fac4<-prod(1:(c2-j))}   small.s<-(n1*n2*(p^(i+j))*(1-p)^(n-(i+j))/(fac1*fac2*fac3*fac4)) S<-cbind(S,small.s) pa<- i/c1 pb<-j/c2 px <- (i+j)/n if(is.nan((pa-pb)/sqrt(px*(1-px)*((1/c1)+(1/c2))))) { tx<-0 } else { tx <- (pa-pb)/sqrt(px*(1-px)*((1/c1)+(1/c2))) } TX<-cbind(TX,tx) } }   P<-cbind(P,sum(S[which(TX>=TXO)])) }   np<-which(P==max(P)) p <- (0:99+.01)/100 nuisance<-p[np] pv<-P[np]     if(to.print) { print("The contingency table is:") print(matrix(c(Ta,Tc,Tb,Td),2,2)) print("Wald Statistic:") print(TXO) print("Nuisance parameter:") print(nuisance) print("The 1-tailed p-value:") print(pv) } if(to.plot) { plot(p,P,type="l",main="Barnard's exact P-value", xlab="Nuisance parameter", ylab="P-value") points(nuisance,pv,col=2) }   return( list( contingency.table = as.table(matrix(c(Ta,Tc,Tb,Td),2,2)), Wald.Statistic = TXO, Nuisance.parameter = nuisance, p.value.one.tailed = pv ) ) }     Barnardextest(matrix(c(8,1,14,3),2,2)) fisher.test(matrix(c(8,1,14,3),2,2))     Convictions <- matrix(c(2, 10, 15, 3), nrow = 2, dimnames = list(c("Dizygotic", "Monozygotic"), c("Convicted", "Not convicted"))) Convictions fisher.test(Convictions, alternative = "less") Barnardextest(Convictions)```

Fisher’s Exact Test for Count Data

data: Convictions
p-value = 0.0004652
alternative hypothesis: true odds ratio is less than 1
95 percent confidence interval:
0.0000000 0.2849601
sample estimates:
odds ratio
0.04693661

\$contingency.table
A B
A 2 15
B 10 3

\$Wald.Statistic
 3.609941

\$Nuisance.parameter
 0.4401

\$p.value.one.tailed
 0.0001528846

Final note: I would like to thank Peter Calhoun again for sharing his code with the rest of us – Thanks Peter!