g01aff
g01aff
© Numerical Algorithms Group, 2002.
Purpose
2
G01AFF Two-way contingency table analysis, with chi /Fisher's exact test
Synopsis
[nobs,pred,chis,p,npos,ndf,m1,n1,ifail] = g01aff(nobs<,num,ifail>)
Description
The data consist of the frequencies for the two-way
classification, denoted by n , for i=1,2,...,m; j=1,2,...,n with
ij
m,n>1.
A check is made to see whether any row or column of the matrix of
frequencies consists entirely of zeros, and if so, the matrix of
frequencies is reduced by omitting that row or column. Suppose
the final size of the matrix is m by n (m ,n >1), and let
1 1 1 1
n
1
--
R = > n , the total frequency for the ith row,
i -- ij
j=1
i=1,2,...,m ,
1
m
1
--
C = > n , the total frequency for the jth column,
j -- ij
i=1
j=1,2,...,n ,
1
m n
1 1
-- --
and T= > R = > C , the total frequency.
-- i -- j
i=1 j=1
There are two situations:
(a) If m >2 and/or n >2, or m =n =2 and T>40, then the matrix
1 1 1 1
of expected frequencies, denoted by r , for i=1,2,...,m ;
ij 1
2
j=1,2,...,n , and the test statistic, X , are computed,
1
where
r =R C /T , i=1,2,...,m ; j=1,2,...,n
ij i j 1 1
and
m n
1 1
2 -- -- 2
X = > > [|r -n |-Y] /r ,
-- -- ij ij ij
i=1 j=1
where
{ 1
{ - if m =n =2
Y = { 2 1 1
{
{ 0 otherwise
is Yates' correction for continuity.
Under the assumption that there is no association between
2
the two classifications, X will have approximately a chi-
square distribution with (m -1)*(n -1) degrees of freedom.
1 1
An option exists which allows for further 'shrinkage' of
the matrix of frequencies in the case where r <1 for the (
ij
i,j)th cell. If this is the case, then row i or column j
will be combined with the adjacent row or column with
smaller total. Row i is selected for combination if
R *m <=C *n . This 'shrinking' process is continued until
i 1 j 1
r >=1 for all cells (i,j).
ij
(b) If m =n =2 and T<=40, the probabilities to enable Fisher's
1 1
exact test to be made are computed.
The matrix of frequencies may be rearranged so that R is
1
the smallest marginal (i.e., column and row) total, and
C >=C . Under the assumption of no association between the
2 1
classifications, the probability of obtaining r entries in
cell (1,1) is computed where
R !R !C !C !
1 2 1 2
P = ------------------------------ , r=0,1,...,R .
r+1 T!r!(R -r)!(C -r)!(T-C -R +r)! 1
1 1 1 1
The probability of obtaining the table of given frequencies
is returned. A test of the assumption against some
alternative may then be made by summing the relevant values
of P .
r
Parameters
g01aff
Required Input Arguments:
nobs (:,:) integer
Optional Input Arguments: <Default>
num integer 0
ifail integer -1
Output Arguments:
nobs (:,:) integer
pred (:,:) real
chis real
p (21) real
npos integer
ndf integer
m1 integer
n1 integer
ifail integer