# a=10s Show that a=sqrt((H(1-F)/(F(1-H)) 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))   Another measure of sensitivity is given by a=10s. Show that a=sqrt((H(1-F)/(F(1-H))

You need to prove that `a = 10 s` , if `s = (1/2) (log(H/(1-H)) - log(F/(1 - F)))` , hence, using logarithmic identities yields:

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

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

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

Converting the power into a square root yields:

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

You need to equate 10 s and a such that:

`a = 10 s => a = 10 log (sqrt((H(1 - F))/(F(1 - H)))) => a = log (((H(1 - F))/(F(1 - H)))^(10/2)) `

`a = log (((H(1 - F))/(F(1 - H)))^5) `

Hence, evaluating the value of a, under the given conditions, yields that `a = log (((H(1 - F))/(F(1 - H)))^5)`  and the indicated result `a = (sqrt((H(1 - F))/(F(1 - H)))) ` does not hold.

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