Please take a run through my online experiment! (as in NOW!)

Ah, a comment lament Thanks for pointing it out. I was assuming some context that some people seemed to pick up on (judging from the results), but that absolutely should have been made explicit. This may have made my result less strong than I wanted it to be... but maybe not. We'll never know Regardless, the result from this part of the experiment was highly significant It's OK. I really appreciate your (and everybody else's) willingness to take a run through it.I'll try and explain the experiment in my next post. Feel free to nitpick away on anything you read there; having critical analysis will help me improve.Thanks!
Ben

This experiment was run to investigate "human reason" via the "Wason Card Selection Task." There were 6 versions of this task in my experiment. The task itself is, given a conditional experession, choose the cards that MUST be turned over to show the conditional to be true or false.The experimental design is always the same. A conditional is given:IF [p] THEN [q]and 4 cards (always logically equivalent):[p] [~p]
[q] [~q]However, it has been noted many, many times, that performance on the task varies as you negate each components of the conditional. Thus, all negation cases must be tested:IF p then q
IF p then ~q
IF ~p then q
IF ~p then ~q

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For my experiment, I was interested in two things:

People tend to focus on doing "verification" in some of the conditions. In other words, they try to prove things to be TRUE. It's been shown that, in some circumstances, people are actually good at falsification (when given a falsification task)--yet they seem not to do it. I wanted to see if I could trigger people to give correct answers on the WCST if I told them to focus on falsification.

Results vary in the 4 negation cases. They vary in some really interesting ways... I came up with an argument of why negation of the antecedent ("p") is DIFFERENT than negation of the consequent ("q"). I didn't actually try to test that. But what I DID try to test is one of the premises within that argument.

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One of the premises is that negation is "cognitively complex." This means that it more than a single "concept". "Not X" is not unitary in the same way that "X" or "Y" is. So for example, "not the letter A" is not reducible to a concept that any of us have. However, in some cases, "Not X" DOES correspond to some unitary "concept." For example, in the context of numbers, "not even" means "odd". In the context of brightness, very often "not bright" means "dark." (note: even though LOGICALLY speaking this isn't the case, cognitively it is often the case).

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In these cases where a reduction is POSSIBLE, I wanted to see if the reduction could be done. It's almost "obvious" from "symmetry" that it would be done--after all, there's no reason I would have chosen to negate "even" or "odd" in the first place. "Not odd" is "even" and "not even" is "odd", but each of the concepts stands on it's own, and if I tested you on either one without negation, I should get exactly the same result.

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Anyway, this was a step towards showing that negation is "cognitively complex". Once negation is considered to be cognitively "complex", then I can proceed with the argument in my paper (which I won't bother with here.

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Ugly and boring. Questions / comments are encouraged. And if anybody wants "answers", wants to point out ambiguities, or wants to talk about their reasoning process in trying to solve the problem... feel free. I'll do my best to respond... I'm still swamped. This paper was due on Friday ; even though I finished (finally), I'm still way behind.........

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Thanks again all!
Ben

Are we allowed to discuss this yet?

Yup. Test's over. Sorry, I had edited the OP but didn't post that clearly.Discuss away!Thanks!
Ben

Here's an easy graph to see how people did on each question:In this view, each question (along the bottom) had the answer "Logical P" and "Logical ~Q". "Logical P" has a complex relationship to "P"; same with "Logical ~Q" and "Q"; this view "normalizes" for those factors. So just look at the "Logical P" and "Logical ~Q" bars. Those are the #s for right answers. The other bars are rates for wrong answers.Notice that performance varied by question. This is totally typical.Actually EvC-ers were

higher in questions #2 and #6 for the rate of "~Q" selection (which is the "hard" answer) compared to college undergrads.

marginally better at avoiding choosing "Q" (the fallacy of "affirming the consequent") compared to college undergrads

I can give more details as to what the "actual" answers to specific questions were, listing of questions, etc... if requested. I'm putting this up at nwr's request for now.Thanks!
BenThis message has been edited by Ben, Monday, 2005/10/10 12:54 PMThis message has been edited by Ben, Monday, 2005/10/10 12:57 PM

OK, I see you did modify the OP to let us know you'd explain it later.I basically just want to knjow what the "right" answers are anyway, but I'll report my response to the test anyway. Since you've removed the URL I'm not sure I remember the test all that clearly but here goes.I wasn't patient enough to spend much time on it but I did reason it through. The ambiguity of "Did you know that 'not a consonant' means a vowel" I just decided to overlook, and treat it as a definition. It was a very odd way of putting it but I can't see any reason for the oddness. Have you explained that or are you going to? However, I was aware that the numbers are also not consonants and took that into consideration in my answers.I didn't see any logical difference between thinking it out in terms of what would show what was true versus what was false. It seemed to me that in all cases merely one choice would suffice to determine whether the statement was either true OR false. If the statement was about what was on the back of a consonant or "not a consonant" I figured whatever was on the back of the named item would either verify OR falsify the statement. if there was more than one of the named item (a vowel and two numbers for "not a consonant" for instance) I figured that since the definition purported to be inclusive that only one choice would determine whether it was true OR false. Could choose either what was said to be on the front or on the back and it would work as well. I ended up with one choice for all questions, all determined by what was identified in the statement.So am I right or wrong?

Please do give the specific answers to the questions. The graph doesn't tell me much (it's also not very legible on my monitor).This message has been edited by Faith, 10-10-2005 04:01 PM

The instructions are kind of classic, so I won't take much flak for the wording but... each card is independent. The rule may hold for one card, but not for another. So ANY card that might show you whether the rule is true or false should be selected. That's right. Basically that's called "modus ponens". If p then q.... check the p card; if there's ~q on the other side, you're screwed (rule's wrong). Otherwise you're OK.The other one you need to do is "modus tollens" If p then q... check the ~q card. If on the other side is p, you're screwed (rule's wrong). Otherwise, you're OK.The other cards... none of them can show the rule is wrong. They can only be consistent. The key to this task is falsification. Most people try to prove the rule is "true" though (I think).BenThis message has been edited by Ben, Monday, 2005/10/10 01:18 PM

1. Rule: If there is a B on one side of the card, then there will be a 3 on the other side.
Cards: B, 3, U, 6
Answers: B, 62. Rule: If there is a W on one side of the card, then there will not be a 3 on the other side.
Cards: 2, W, 3, I
Answers: W, 33. If there is not a consonant on one side of the card, then there will be an even number on the other side.
Cards: 7, 4, A, C
Answers: A, 74. If there is not a consonant on one side of the card, then there will not be an even number on the other side.
Cards: P, 2, E, 5
Answers: E, 25. If there is a A on one side of the card, then there will be a 2 on the other side.
Cards: A, N, 2, 7
Answer: A, 76. If there is a U on one side of the card, then there will not be a 6 on the other side.
Cards: 6, U, 9, C
Answer: U, 6

How can you be screwed if the task is to determine whether the statement is either true OR false? Any one with the figure the statement identified should do it.So the idea is that the rule could be true for one card but not another? What kind of rule is that? Well, I guess I did it wrong. I don't get why two cards had to be identified. If one has on the back of it what the rule said it would, then the rule is true, and if it doesn't the rule is false, and I don't see why I need to check the other card with the identified figure on it to be certain of that.But if the rule can be true for one card and false for the other, the whole thing hits me as too irrational to bother about. But I suppose I'm not getting something here.This message has been edited by Faith, 10-10-2005 04:21 PM

I think I got all those right, except for the first one. Why 6? Do all the cards have a number on one side and a letter on the other? If so, I get it, but screwed up reading the instructions

It's an experimental task. It's not so interesting. Calling it "irrational" is fairly amusing though. I can understand why you'd say that, and in the sense that you mean it, I agree. It's funny because the task is a test of logic; it's in another sense completely rational.But to answer... the point is to check if the rule holds for ALL cards. So yes, you have to consider all of them. The rule might hold for one and fail for another.

"If B is on one side then 3 is on the other"You have to check 6 because if there's a B on the other side, you're screwed.I think you're right--maybe I missed something from the original instructions. Looks like you have to assume that there's a letter on one side and a number on the other. I don't remember seeing that in the written instructions at all.But that goes back to ringo316's objection; there was supposed to be a context there (numbers on one side, letters on another), but it just wasn't there. That's a problem!But that's the way the cookie crumbles!Ben

Check B only. If 3 is there, the rule is true; if it is not there the rule is false.Why on earth would I need to check ANYTHING else? If there's a 3 on the back of B, but a B on the back of 6 then the rule is true for the first and false for the second. This makes no sense to me at all.{Edit: OK I get it I guess. SOMETIMES the rule may be true, but it's not a rule unless it applies in all cases??? So then wouldn't you have to check all possible cases to find out? That is, besides checking the 6 you'd have to check the 3 also since it should have a B on its front. If you don't check ALL cases then how would you know if the rule is true or false?This message has been edited by Faith, 10-10-2005 04:45 PM

Faith,If I give you four apples, and I say "none of the apples have worms in them." If you eat three of the apples and they're nice & tasty, and for the 4th one you bite into a worm, ...It's a similar situation. You have to check all cards that could possibly show your rule to be wrong. Like in the above example... if I gave you an apple, an orange, a box that said "fruit with a worm in it" and a box that said "fruit with no worm in it", in order to know if none of the apples have worms in them, you better check the apple and the box that's labeled "fruit with a worm in it." If that apple has a worm in it, then the rule is wrong. If the fruit with a worm in it is an apple... then the rule's wrong.Everything else, you don't have to bother checking. There's no way it can make your rule wrong. The others would be consistent with your rule no matter what (either they would be an apple without a worm, or they wouldn't be an apple and so wouldn't matter). You don't need to check if you know it's either going to be true or not be relevant at all.AbE: Faith, check the last paragraph to read why you don't have to check all the cards. To reiterate, some cards will be consistent with the rule no matter what; either it will have a B and a 3, OR it won't have a B... and so doesn't apply at all. For those cards, you don't have to bother checking.This message has been edited by Ben, Monday, 2005/10/10 01:48 PM