# An unbiased observer is one for which H=1-F. Give the simplest possible formula for computing the measure of sensitivity s for an unbiased subject.In a psychophysical experiment designed to...

An unbiased observer is one for which H=1-F. Give the simplest possible formula for computing the measure of sensitivity s for an unbiased subject.

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))

An unbiased observer is one for which H=1-F. Give the simplest possible formula for computing the measure of sensitivity s for an unbiased subject.

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to substitute `1-F`  for `H`  in equation of `s = (1/2)(log(H/(1-H))-log(F/(1-F)))`  such that:

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

Using the equation `H = 1- F => F = 1 - H,`  yields:

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

Using logarithmic identity `log(a/b) = log a - log b`  yields:

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

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

Factoring out 2 yields:

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

Since `1-F = H` , yields:

`s = logH - log(1-H)`

Converting the difference of logarithms into a quotient yields:

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

Hence, evaluating the simplest formula for computing the measure of sensitivity s yields `s = log (H/(1-H)).`