Dear Professor Ricco RAKOTOMALALA:

the Reference is:
<An Empirical Comparison of Selection Measures for Decision-Tree Induction>

The following datasets from above reference
try to find what feature is relevant to Cancer Recurrence.

Datasets(totally 10 items):

Radiation	menopause	Class
No	＜60	Recur
No	≥60	Recur
No	＜60	Recur
No	NOT	Recur
Yes	≥60	Not Recur
Yes	＜60	Not Recur
Yes	≥60	Not Recur
No	NOT	Not Recur
No	＜60	Not Recur
No	＜60	Recur

for datasets as follows:

we have Chi-square Test formula:
$\chi^2=\sum_{i=1}^r \sum_{j=1}^{s} \frac{(N_{ij}-E_{ij})^2}{E_{ij}}=\sum_{i=1}^r\sum_{j=1}^s \frac{(N_{ij}-\frac{N_{i·}N_{·j}}{n})^2 }{\frac{N_{i·}N_{·j}}{n}}$

where $E_{ij}=\frac{N_{i·}N_{·j}}{N}$
-------------------All the following computation are identical with the above article(skip please)-----------

For Radiation:

Radiation\Class	Recur	Not Recur	Total
Yes	0	3	3
No	5	2	7
Total	5	5	10

$\chi^2= \frac{(N_{11}-E_{11})^2}{E_{11}}+ \frac{(N_{12}-E_{12})^2}{E_{12}}+ \frac{(N_{22}-E_{22})^2}{E_{22}}+ \frac{(N_{21}-E_{21})^2}{E_{21}}$
where
$N_{11}=0$
$N_{12}=3$
$N_{21}=5$
$N_{22}=2$

$E_{11}=\frac{N_{1·}N_{·1}}{N}=\frac{3·5}{10}=1.5$
$E_{12}=\frac{N_{1·}N_{·2}}{N}=\frac{3·5}{10}=1.5$
$E_{21}=\frac{N_{2·}N_{·1}}{N}=\frac{7·5}{10}=3.5$
$E_{22}=\frac{N_{2·}N_{·2}}{N}=\frac{7·5}{10}=3.5$

∴$\chi^2= \frac{(N_{11}-E_{11})^2}{E_{11}}+ \frac{(N_{12}-E_{12})^2}{E_{12}}+ \frac{(N_{22}-E_{22})^2}{E_{22}}+ \frac{(N_{21}-E_{21})^2}{E_{21}} \\=\frac{(0-1.5)^2}{1.5}+\frac{(3-1.5)^2}{1.5}+\frac{(5-3.5)^2}{3.5}+\frac{(2-3.5)^2}{3.5}=4.285713$

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