# Compute s for the pairs (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)) 3. Compute s for the pair (H,F) obtained in this previous problem- s = (1/2)(((Log(H/1-H))-log(F/1-F)) Compute H and F as well as for thepairs (H,F) = (0.80, 0.20), (H,F) = (0.8, 0.6) and (H,F) = (0.80, 0.90).

You need to substitute 0.80 for H and 0.20 for F in equation of s to evaluate s, such that:

`s(0.80, 0.20) = (1/2)(log(H/(1-H))-log(F/(1-F))`

`s(0.80, 0.20) = (1/2)(log(0.80/(1-0.80))-log(0.20/(1-0.20))`

`s(0.80, 0.20) = (1/2)(log(0.80/0.20)-log(0.20/0.80))`

Using logarithmic identities, you may convert the logarithms of quotients in a difference of logarithms such that:

`s(0.80, 0.20) = (1/2)(log 0.80 - log 0.20 - log 0.20 + log. 0.80)`

Reducing like terms yields:

`s(0.80, 0.20) = (1/2)(2log 0.80 - 2log 0.20)`

Factoring out 2 yields:

`s(0.80, 0.20) = (2/2)(log 0.80 - log 0.20)`

Converting the difference of logarithms into the logarithm of quotient yields:

`s(0.80, 0.20) = log (0.80/0.20) => s(0.80, 0.20) = log 4`

Reasoning by analogy, you may evaluate `s(0.80, 0.60)`  and `s(0.80, 0.90)`  such that:

`s(0.80, 0.60) = log (0.80/0.60) => s(0.80, 0.60) = log 4/3`

`s(0.80, 0.90) = log (0.80/0.90) => s(0.80, 0.90) = log 8/9`

Hence, evaluating `s(0.80, 0.20), s(0.80, 0.60)`  and `s(0.80, 0.90)`  yields `s(0.80, 0.20) = log 4 , s(0.80, 0.60) = log 4/3 ,s(0.80, 0.90) = log 8/9. `

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