On Oct 25, 2010, at 8:25 AM, Wayne W Lukens Jr wrote:
A more useful way to look at this is that the probabilities that A, B and C are present are 99.99999999%, 93%, and 77%, respectively.
An excellent post, Wayne, but I don't think that last statement is quite right. If the F-test gives a probability of 0.23 for material C, I believe it's saying that there is a 23% that, given the noise level in the data, the fit would indicate that C was present when it was not. That is not the same thing as saying there is a 77% chance of C being present. To see this, imagine very, very noisy data. Including C in the fit might very well improve the fit in the sense of an R-factor--maybe, in fact, there's a 45% chance of a modest improvement with a given set of very noisy data, even if there's no C present. That does not mean that a result like that should lead to the conclusion that C is more likely than not present (55%). --Scott Calvin Sarah Lawrence College
Another point is that even if the F-test calls out a .00001% probability that A appears just due to noise, that's no guarantee that what looks like A isn't something else which is "A-like" enough to turn up in the LCF. It's probably something similar, but need not be exactly A unless you have some other reason for believing that A is the only species present that can look like that. Yes, this is an obvious point, but one which should be kept in mind. mam On 10/25/2010 9:07 AM, Scott Calvin wrote:
On Oct 25, 2010, at 8:25 AM, Wayne W Lukens Jr wrote:
A more useful way to look at this is that the probabilities that A, B and C are present are 99.99999999%, 93%, and 77%, respectively.
An excellent post, Wayne, but I don't think that last statement is quite right. If the F-test gives a probability of 0.23 for material C, I believe it's saying that there is a 23% that, given the noise level in the data, the fit would indicate that C was present when it was not. That is /not/ the same thing as saying there is a 77% chance of C being present.
To see this, imagine very, very noisy data. Including C in the fit might very well improve the fit in the sense of an R-factor--maybe, in fact, there's a 45% chance of a modest improvement with a given set of very noisy data, even if there's no C present. That does not mean that a result like that should lead to the conclusion that C is more likely than not present (55%).
--Scott Calvin Sarah Lawrence College
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Hi Scott,
You are correct. That is a much better way to explain it.
Sincerely,
Wayne
On Mon, Oct 25, 2010 at 9:07 AM, Scott Calvin
On Oct 25, 2010, at 8:25 AM, Wayne W Lukens Jr wrote:
A more useful way to look at this is that the probabilities that A, B and C are present are 99.99999999%, 93%, and 77%, respectively.
An excellent post, Wayne, but I don't think that last statement is quite right. If the F-test gives a probability of 0.23 for material C, I believe it's saying that there is a 23% that, given the noise level in the data, the fit would indicate that C was present when it was not. That is *not* the same thing as saying there is a 77% chance of C being present.
To see this, imagine very, very noisy data. Including C in the fit might very well improve the fit in the sense of an R-factor--maybe, in fact, there's a 45% chance of a modest improvement with a given set of very noisy data, even if there's no C present. That does not mean that a result like that should lead to the conclusion that C is more likely than not present (55%).
--Scott Calvin Sarah Lawrence College
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participants (3)
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Matthew Marcus -
Scott Calvin -
Wayne W Lukens Jr