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Statistical Methods in Medical Research 4ed

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D 2 …zA zB†<br />

ˆ<br />

2<br />

…13:5†<br />

variance of z with<strong>in</strong> groups<br />

is as large as possible. The estimated variance of z will, <strong>in</strong> general, be different <strong>in</strong><br />

the two groups, but a pooled estimate could be calculated as <strong>in</strong> the two-sample t<br />

test.<br />

The solution of this problem is as follows:<br />

1 Calculate the pooled sums of squares and products of the xs with<strong>in</strong> groups,<br />

divide by nA ‡ nB 2 to obta<strong>in</strong> the matrix of with<strong>in</strong>-group variances and<br />

covariances, and obta<strong>in</strong> the <strong>in</strong>verse matrix with general term cij. Here, nA and<br />

nB are the numbers of <strong>in</strong>dividuals <strong>in</strong> groups A and B, respectively.<br />

2 Calculate the bs as follows:<br />

b1 ˆ c11…xA1 xB1†‡c12…xA2 xB2†‡...‡ c1p…xAp xBp†<br />

b2 ˆ c21…xA1 xB1†‡c22…xA2 xB2†‡...‡ c2p…xAp xBp†<br />

bp ˆ cp1…xA1 xB1†‡cp2…xA2 xB2†‡...‡ cpp…xAp xBp†:<br />

…13:6†<br />

Here, xAi is the mean of xi <strong>in</strong> group A and so on.<br />

To use (13.4) for allocat<strong>in</strong>g future <strong>in</strong>dividuals to one of the two groups we<br />

need an end-po<strong>in</strong>t, or cut-po<strong>in</strong>t, to discrim<strong>in</strong>ate between A and B. A completely<br />

symmetrical rule would be to use the mean, z0, ofzA and zB. From (13.4),<br />

and<br />

whence<br />

z0 ˆ b1<br />

xA1 ‡ xB1<br />

2<br />

zA ˆ b1xA1 ‡ b2xA2 ‡ ...‡ bpxAp<br />

zB ˆ b1xB1 ‡ b2xB2 ‡ ...‡ bpxBp,<br />

‡ b2<br />

xA2 ‡ xB2<br />

2<br />

13.3 Discrim<strong>in</strong>ant analysis 465<br />

‡ ...‡ bp<br />

xAp ‡ xBp<br />

2<br />

…13:7†<br />

: …13:8†<br />

If zA > zB the allocation rule is: allocate an <strong>in</strong>dividual to A if z > z0 and to B if<br />

z < z0. Many computer programs calculate the value of z for each <strong>in</strong>dividual <strong>in</strong><br />

the two orig<strong>in</strong>al samples. It is thus possible to count how many <strong>in</strong>dividuals <strong>in</strong> the<br />

two groups would have been wrongly classified by the allocation rule. This<br />

unfortunately gives an overoptimistic picture, because the allocation rule has<br />

been determ<strong>in</strong>ed to be the best (<strong>in</strong> a certa<strong>in</strong> sense) for these two particular<br />

samples, and it is likely to perform rather less well on the average with subsequent<br />

observations from the two groups (see Hills, 1966).<br />

Various methods have been proposed to overcome this problem (Krzanowski<br />

& Marriott, 1995, §§9.32±9.44). One method is to split the sample randomly <strong>in</strong>to<br />

two subsamples. The first subsample, the tra<strong>in</strong><strong>in</strong>g sample, is used to estimate the

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