Association test

• If independence holds, we should observe pij = pi.p.j

• We do not observe probabilities, but frequencies (counts)

Xn-1 A C G T

Xn

A YAA YCA YGA YTA

C YAC YCC YGC YTC

G YAG YCG YGG YGA

T YAT YCT YGT YTT

• We compare Yij against Eij = Yi. Y.j /Y..

• Under null hypothesis of independence, the index I has ?2 distribution with (4-1)(4-1)=9 df

• Usual testing theory applies (Section 3.4.3, Example 4)

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