# Suppose that (H,F) = (0.6, 0.2). If F is unchanged, what would H have to beto double ? If H is unchanged, what would F have to be to double ? 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)) Suppose that (H,F) = (0.6, 0.2). If F is unchanged, what would H have to beto double ? If H is unchanged, what would F have to be to double ? You need to evaluate s for `(0.6,0.2), ` using the equation of s such that:

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

Substituting 0.6 for H and 0.2 for F yields:

`s = (1/2)(log(0.6/(1- 0.6)) - log(0.2/(1- 0.2)))`

`s = (1/2)(log(0.6/0.4) - log(0.2/0.8))`

Using logarithmic identities yields:

`s = (1/2)(log0.6 - log 0.4- log 0.2 + log 0.8)`

`s = (1/2)(log(3*0.2) - log0.4 - log 0.2 + log (2*0.4))`

`s = (1/2)(log 3+ log 0.2 - log 0.4 - log 0.2 + log 2 + log 0.4)`

Reducing like terms yields:

`s = (1/2)(log 3 + log 2) => s = (1/2)(log(3*2)) => s = (1/2)log 6`

You need to double s for the same value of F to evaluate H, such that:

`2s = (1/2)(log(H/(1- H)) - log(0.2/0.8))`

`2*(1/2)log 6 = (1/2)(log(H/(1- H)) - log(0.2/0.8))`

`log 6 = (1/2)(log(H/(1- H)) - log(0.2/0.8))`

`2 log 6 = log(H/(1- H)) - log(0.2/0.8)`

`log 6^2 + log (0.2/0.8) = log (H/(1- H)) => log 36*(0.2/0.8) = log (H/(1- H)) => 36*(1/4) = (H/(1- H)) => 9 = (H/(1- H))`

`9(1-H) = H => 9 - 9H = H => 9 = 10H => H = 9/10 = 0.9`

Hence, evaluating the value of H under the given conditions yields `H = 0.9` .

You need to double s for the same value of H to evaluate F, such that:

`log 6 = (1/2)(log(0.6/0.4) - log(F/(1- F)))`

`log 36 = log(0.6/0.4) - log(F/(1- F)) => log(F/(1- F)) = log(0.6/0.4) - log 36`

`log(F/(1- F)) = log(0.6/(36*0.4)) => log(F/(1- F)) = log(0.1/(6*0.4)) => (F/(1- F)) = 1/24 => 24F = 1 - F => 25F = 1 => F = 1/25 =0 .04`

Hence, evaluating F under the given conditions yields `F = 0.04.`

Approved by eNotes Editorial Team

Posted on 