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In signal detection theory using a 2-alternative forced choice task, when d' = 1, threshold is found at the stimulus intensity which is perceived 76% of the time (https://www.oxfordreference.com/view/10.1093/oi/authority.20110803110358463).

Why 76%?

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d' is equal to |meanA - meanB|/SD when you have two gaussian distributions of equal variance and different means.

Imagine SD=1, meanA = 0, meanB = 1, with B>A as the "correct" answer (perhaps B is the signal, and A is the noise). This will give you a d' of 1 because the difference between the means is 1, divided by a standard deviation of 1.

The % correct would be the fraction of the time that a random draw from the distribution with larger mean is larger than a draw from the distribution with smaller mean. That's what an ideal observer would do given the task to choose the larger distribution: pick the larger number on that trial. This is equivalent to detecting a signal if you think of the other distribution as representing the "noise".

For this example (or any example that satisfies the condition that d' = 1), that rate will be about 76%. I don't believe there is a simple closed form equation to calculate this, and my calculus class days for rearranging integrals are a bit dated, but you can simulate it easily by taking (pseudocode):

a = rand(mu=0,sigma=1,n=100000)
b = rand(mu=1,sigma=1,n=100000)
sum(b>a)/100000

Note that there isn't any special reason d' = 1 is chosen to make this 76% number be true, it's just a consequence of the assumption of normal distributions and probabilities. If d' = 0, the % "correct" will be 50%.

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  • $\begingroup$ Thanks @Bryan, I understand this, but my question is Why 76%? I've never seen a satisfactory derivation of this figure. I am specifically interested in where the 76% figure comes from. $\endgroup$ Dec 22 '21 at 11:41
  • $\begingroup$ @CaptainProg Uh, I've explained it here, no? Draw from two normal distributions; if d' for those distributions = 1, then the one that is larger on average will be larger in your draw 76% of the time. I don't believe there is a simple closed form equation to calculate this, and my calculus class days for rearranging integrals are a bit dated, but you can simulate it easily by taking (pseudocode) a = rand(mu=0,sigma=1), b= rand(mu=1,sigma=1) a bunch of times and checking how often b>a $\endgroup$
    – Bryan Krause
    Dec 22 '21 at 15:12
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    $\begingroup$ Thanks for editing – yes, this now makes sense. I've run this in MATLAB and finally happy that I can see how 76% was derived (with 100,000 random samples generated, I got 75.99%) $\endgroup$ Dec 22 '21 at 15:36

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