# 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...

Suppose that (H,F) = (0.6, 0.2). If F is unchanged, what would H have to be

to double ? If H is unchanged, what would F have to be to double ?

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))

Suppose that (H,F) = (0.6, 0.2). If F is unchanged, what would H have to be

to double ? If H is unchanged, what would F have to be to double ?

*print*Print*list*Cite

### 1 Answer

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.` **