Statistical Methods in Medical Research 4ed

510 Empirical methods for categorical data
Table 15.2 Numbers of mothers by state of grief and degree of support (data of Tudehope et al.,
1986).
Grief state
Row
score
xi
Support
Good Adequate Poor
Total
Ri
Sum of
column
scores Yi
I 1 17 9 8 3459
II 2 6 5 1 12 19
III 3 3 5 412 25
IV 4 1 2 5 8 20
Total, Cj 27 21 18 66 123
Col. score, yj 1 2 3
Sum of row scores, Xj 4 2 4 2 4 2 126
Mean score, xj 1 56 2 00 2 33
3 Calculate the sum of squares of row scores for all the mothers,
P 2
x ˆ…34 1†‡…12 4†‡…12 9†‡…8 16† ˆ318:
Correct this sum of squares for the mean.
Sxx ˆ 318 126 2 =66 ˆ 77 455:
4 A test of the equality of the mean scores xj may now be carried out. The test statistic
is
X 2 ˆ…N 1†‰ P …X 2
j =Cj† … P x† 2 =NŠ=Sxx
ˆ 65
77 455
ˆ 5 70:
422 422 422
‡ ‡
27 21 18
126 2
66
…15:11†
This may be regarded as a x2 …c 1† , i.e. a x2 …2† statistic. This is not quite significant at the
5% level (P ˆ 0 058), but is sufficiently close to allow the possibility that a test for the
apparent trend in the column means, xj, may be significant. We now, therefore, make
use of the column scores, yj.
5 Repeat steps 2 and 3, working across rows instead of down columns; it is not necessary
to calculate the mean scores.
P P 2
y ˆ 123, y ˆ 273, Syy ˆ 273 123 2 =66 ˆ 43 773:
6 Calculate the sum of products of the sum of row scores, Xj, and the corresponding
column scores, yj.
P Xjyj ˆ…42 1†‡…42 2†‡…42 3† ˆ252:
This total is P xy over all mothers and correcting for the means