Earlier you asked this question regarding "degrees of freedom":Diatom
Sorry, but I find all the purely speculative suggestions about how Bem could have cheated a bit boring, unless they can be supported by some evidence that he did actually cheat.
Of course, I realise some people are approaching this from the point of view that he must have cheated, and working from there. But in that case, what's the point of all these elaborate scenarios? If he wanted to cheat, he could just edit the files containing the results, and save himself all the trouble. It would have been a lot simpler and a lot safer. And quite frankly, if he had been cheating, I think he'd have been a damn sight less sloppy about what he published!
Since you are now asking for evidence that this actually took place, I assume that you agree that it is a possible and sufficient explanation. Am I right?Most importantly, is it plausible that such factors are sufficient to explain the results he found?
Sorry, but I find all the purely speculative suggestions about how Bem could have cheated a bit boring, unless they can be supported by some evidence that he did actually cheat.
Since you are now asking for evidence that this actually took place, I assume that you agree that it is a possible and sufficient explanation. Am I right?
Do you see now the point I was trying to get at earlier?
Please clarify your position. Suddenly you want to change the subject. That's fine. But I don't see where we are now.Anything's possible. For example, it's possible that those p values could have arisen purely through chance, without any need for psychical phenomena on the one hand or multiple hypotheses, file drawers, or any thing of that sort on the other. But that's not a very useful statement.
I don't see any point in discussing evidence for your accusation of cheating before we haven't cleared up where you are coming from. I am calling it your accusation because you are the only one talking about cheating.My question was whether the explanation was plausible. That's where some evidence would come in handy.
I sometimes wish that just once in my life someone would flounce off and tell me they're never going to speak to me again because I'm such a jerk ... and actually keep their word. :(
I don't see any point in discussing evidence for your accusation of cheating before we haven't cleared up where you are coming from. I am calling it your accusation because you are the only one talking about cheating.
You started the talk about Bem cheating. That makes it your accusation. Own it.My "accusation of cheating"? You're crazy.
I get 54. But I'm not trying to be an ass.
Linda
I didn't, but I agree that that is also a possibility (given that's what he has done in some cases (e.g. the 300 series doesn't seem to have made an appearance, although it's possible that it got rolled into experiment 7 in some way)).Are you counting not using any of the studies (i.e., the empty set), as a possibility? I didn't count that, but I think it is reasonable to do so.
OK, so, given your more rigorous wording, here's my amended working:
[SNIP!]
4-set-combos, complete (one-and-one-and-one-and-one):
4C1*3C1*2C1*1C1
Now combining into equivalent sets of combinations, and then dividing by the number of possible permutations of each set of combinations:
4C4 + 4C3 + 4C2 + 4C1 + (4C3*1C1 + 4C1*3C1) / 2 + 4C2*2C2 + (4C2*2C1 + 4C1*3C2) / 2 + 4C1*3C1 + (4C2*2C1*1C1 + 4C1*3C2*1C1 + 4C1*3C1*2C2) / (3*2) + 4C1*3C1*2C1 + 4C1*3C1*2C1*1C1
= 1 + 4*3*2/(3*2*1) + 4*3/(2*1) + 4 + (4*3*2/(3*2*1)*1 + 4*3) / 2 + 4*3/(2*1)*1 + (4*3/(2)*1 + 4*3*2/(2*1)) / 2 + 4*3 + (4*3/(2*1)*2*1 + 4*(3*2)/(2*1)*1 + 4*3*1) / (3*2) + 4*3*2 + 4*3*2*1
I counted all 4 twice - once as one big study, and once as 4 separate studies. Is that the difference?
Diatom
Sorry, but I find all the purely speculative suggestions about how Bem could have cheated a bit boring, unless they can be supported by some evidence that he did actually cheat.
Of course, I realise some people are approaching this from the point of view that he must have cheated, and working from there. But in that case, what's the point of all these elaborate scenarios? If he wanted to cheat, he could just edit the files containing the results, and save himself all the trouble. It would have been a lot simpler and a lot safer. And quite frankly, if he had been cheating, I think he'd have been a damn sight less sloppy about what he published!
That's a bit hard to follow, but this jumped out at me:
1-set-combos, both complete and incomplete (individual sets of four, three, two and one):
4C1 + 4C2 + 4C3 + 4C4
2-set-combos, complete (individual sets of three-and-one, two-and-two):
4C3 + 4C2
2-set-combos, incomplete (individual sets of two-and-one, one-and-one):
4C2*2 + 4C2
3-set-combos, complete (individual sets of two-and-one-and-one):
4C2
3-set-combos, incomplete (one-and-one-and-one):
4C3
4-set-combos, complete (one-and-one-and-one-and-one):
4C4
Now, combining together:
= 4C1 + 4C2 + 4C3 + 4C4 + 4C3 + 4C2 + 4C2*2 + 4C2 + 4C2 + 4C3 + 4C4
= 2*4C4 + 3*4C3 + 6*4C2 + 4C1
= 2 + 3*4*3*2/(3*2*1) + 6*4*3/(2*1) + 4
= 54
Perhaps you're still misunderstanding my description of the problem.
Or, I can post my solution.
As usual, Chris, no one but you is talking about anybody lying or cheating.
I couldn't figure out a generic way to calculate each of the combinatorics
Here's my modified working, by which I get the same solution as Linda. I couldn't figure out a generic way to calculate each of the combinatorics, so I pieced together each using ad-hoc logic:
Please do that regardless. :)Or, I can post my solution.
Prototype Ways
------------------------------
Use all 4 experiments:
{{A, B, C, D}} 1
{{A, B, C}, {D}} 4
{{A, B}, {C, D}} 3*
{{A, B}, {C}, {D}} 6
{{A}, {B}, {C}, {D}} 1
Use exactly 3 experiments:
{{A, B, C}} 4
{{A, B}, {C}} 12
{{A}, {B}, {C}} 4
Use exactly 2 experiments:
{{A, B}} 6
{{A}, {B}} 6
Use only 1 experiment:
{{A}} 4
------------------------------
Total 51*
==============================