# s = (1/2)(((Log(H/1-H))-log(F/1-F)) Compute H and FIn a psychophysical experiment designed to measure performance in a recognitiontask, a subject is presented with a set of pictures of people’s...

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

Compute H and F

In a psychophysical experiment designed to measure performance in a recognition

task, a subject is presented with a set of pictures of people’s faces. Later, the

subject is presented with a second set of pictures which contains the previously

shown 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 determine

a measure of the observer’s ability to discriminate between the previously shown

pictures and the new ones.

If a subject correctly recognizes a face as being one of the previously shown

ones, 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 proportion

of responses to previously shown faces which are hits is denoted by H, while the

proportion 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 shown

faces and new ones is given by

all are in log base 10

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

1. Suppose that the subject responds “yes” to 20 and “no” to 5 of the previously

shown pictures, while the subject gives 10 “yes” and 15 “no” responses to the new

pictures.

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The quantity s is a log odds ratio. It tells us about the subject's ability to discriminate between faces they know and faces they don't.

`H` is the proportion of hits out of the group of faces they have seen. They have seen 20 + 5 faces = 25 so` `

`H = 20/25 = 4/5`

`F` is the proportion of faces they think they know out of of the group they haven't seen; the proportion of false positives. They have seen 10 + 15 = 25 faces they don't know so

`F = 10/25 = 2/5`

We can compute `s` from this and test whether `s` is significantly different from 1/2 (there is no difference between their recognition of faces they should know and faces they haven't seen before) by computing the standard error of `s` and checking that the lower limit of the confidence interval is greater than 1/2.

**We have H = 4/5 and F=2/5**