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 ?
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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.`
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