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so for us for any certain task what we're going to have is a noise threshold so we're going to have a noise distribution and the noise distribution might kind of look something like this so this is just background noise so what this basically shows is if we were to take a bunch of individuals and we showed them a bunch of we had experimentally tested signal detection Theory you would get this graph just kind of indicating the noise and then we'd get a second graph which is kind of shifted over to the right a little bit and this is the signal distribution so this is the signal distribution and in the blue over here we have the noise distribution so the difference between the means of these two distributions is D Prime so if the signal distribution was shifted over here to the right then D prime would be really big it would be a really easy task it would be something kind of like this whether there's a green dot on the screen or not but on the other hand if the signal distribution was shifted over to the left then D prime would be super small and it'd be something more like this like a more difficult task so the x-axis here we have the intensity of the stimulus so that would be how easy the stimulus is to distinguish from the background okay so we've got the first variable which is D prime and the second variable is C and that is the strategy of the individual so strategy can actually be strategy can be expressed via the choice of the threshold so what does the individual deem as necessary what what threshold is necessary to surpass in order for them to say yes versus no so we're just going to label the different strategy so there's B there's D there C and there's beta so these are different strategies and they're just variables given to the different strategies so if we were to use this B strategy this would basically say okay I'm going to choose a certain threshold so let's say that I choose this threshold over here so let's say that I choose to so anything that is greater than 2 I will say yes to and anything that's less than two I will say no to so in that case the probability of a hit is this area over here and the probability of a false alarm is this area over here so that would be the B strategy the D strategy basically says okay so the position of my threshold is going to be relative to the signal distribution so basically what that comes out to be is D prime so the signal distribution if it's over to the right will have a big D prime so D prime minus B so we choose a threshold let's say - and let's say that D prime and in this example is 1 then we would have 2 minus 1 and we get we'd get 1 so if we were using the D strategy then anything above a 1 would get a yes anything below would get a no so the C strategy would be an ideal observer so this would be someone that would be minimizing the possibility of a Miss and of a false alarm so the C strategy is basically if we were to write an equation for it would be B minus D prime over 2 so in our example it would be 2 minus 1 divided by 2 so that is equal to 1.5 so if you are using a C strategy that of an ideal observer then we would say anything above 1.5 would get a yes anything below would get a no and so we said over here that the C this C variable is indicative of the strategy that a person uses so when C equals zero then the participant is an ideal observer if C is less than 1 we say that the participant is liberal and if C is greater than 1 we say the participant is conservative in their strategy and this would be ideal so when we say conservative that means that they respond no more often than an ideal observer and when we say liberal then we then that means that the participant says no less often than the ideal observer so the final variable that we have to talk about is beta so if we're using this beta approach we've set the value of the threshold equal to the ratio of the height of the signal distribution to the height of the noise distribution and so it's easier to actually look at beta in this way so if we were actually writing an equation we would say the natural log of beta is equal to D prime times C and so in this case it would equal one times one point five which equals one point five