# s = (1/2)(((Log(H/1-H))-log(F/1-F)) Compute s for the pair (H, F) for thepairs (H, F) = (0.80, 0.20), (H, F) = (0.8, 0.6) and (H, F) = (0.80, 0.90). In a psychophysical experiment designed to measure performance in a recognitiontask, a subject is presented with a set of pictures of people’s faces. Later, thesubject is presented with a second set of pictures which contains the previouslyshown pictures and some new ones. The subject then is asked to answer “yes” or“no” to the question “Do you recognize this face?” We would like to determinea measure of the observer’s ability to discriminate between the previously shownpictures and the new ones.If a subject correctly recognizes a face as being one of the previously shownones, it is called a “hit.” If a subject incorrectly states that they recognize a face,when the face is actually a new one, it is called a “false alarm.” The proportionof responses to previously shown faces which are hits is denoted by H, while theproportion of responses to new faces which are false alarms is denoted by F.A measure of the ability of the subject to discriminate between previously shownfaces and new ones is given by all are in log base 10 s = (1/2)(((Log(H/1-H))-log(F/1-F)) Compute h and F obtained in Problem 1 above.

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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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