Now we have the formula `s = (1/2) (log(H/(1-H)) - log(F/(1-F)))`

Using the result `loga - logb = log(a/b)`

we have that ` ` `s = (1/2) log((H "/" 1-H)/(F "/" 1-F))`

` ` Pair 1 : (H, F) = (0.8,0.2)

H/1-H = 0.8/0.2 = 4

F/1-F = 0.2/0.8 = 1/4

So (H/1-H)/(F/1-F) = 4/(1/4) = 16

s = 1/2 * log_10(16) = 1/2*1.20412 = 0.602 **answer**

Pair 2: (H, F) = (0.8,0.6)

H/1-H = 0.8/0.2 = 4 same as before

F/1-F = 0.6/0.4 = 3/2

So (H/1-H)/(F/1-F) = 4/(3/2) = 8/3

s = 1/2*log_10(8/3) = 1/2* 0.42597 = 0.213 **answer**

Pair 3: (H,F) = (0.8,0.9)

H/1-H = 0.8/0.2 = 4 same as before

F/1-F = 0.9/0.1 = 9

So (H/1-H)/(F/1-F) = 4/9

s = 1/2*log_10(4/9) = 1/2*(-0.35218) = -0.176 **answer**

The hit rate stays the same in all 3 pairs, but the false alarm rate increases leading to increasing s. When the false alarm rate is higher than the hit rate, s is negative.

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