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statstudent
Questions (4)
How do I generate the bezier curve control points to represent a normal (Gaussian) distribution?
0 answers
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I have a simple set of 10 data points
My ten Data Points 2 3 3 4 5 8 9 11 11 13 (mean = 6.9) My prediction nodel predicts the
1 answer
906 views
I'm trying to work through the proof for
SST = SSM + SSE MEAN = ∑(X)/N SST = ∑((x - MEAN)^2) = ∑(x^2 - 2 * x1 * MEAN +
4 answers
785 views
Where can I find a proof for:
SST = SSM + SSE
4 answers
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Answers (5)
I don't follow this at all: = + + 2 Trying to follow your logic, for the left side: = = - ^2 = SST For SSM + SSE: ^2> + = + = - 2 + + - 2 + ^2> = 2 + 2 - 2 - 2 And I'm stuck...
I don't follow this: = Of course, if k is constant and x is variable: = k = k = k^2 but... != ^2
OK, SST makes sense, but I can't see how to derive your SSM or SSE formulas: I get this: SSM = ∑((MODEL - MEAN)^2) = ∑(MODEL^2 - 2 * MODEL * MEAN + MEAN^2) = ∑(MODEL^2) - 2 * MEAN * SIGMA(MODEL) + N * MEAN^2 SSE = ∑((X - MODEL)^2) = ∑(X^2 - 2 * X
Got it... SST = ∑((x - MEAN)^2) = ∑(x^2 - 2 * x1 * MEAN + MEAN^2) = ∑(x^2) - 2 * MEAN * ∑(x) + N * MEAN^2 MEAN = ∑(x)/N = ∑(x^2) - 2 * ∑(x)^2/N + ∑(x)^2/N = ∑(x^2) - ∑(x)^2/N Awesome! Thanks!
Wow. That's right, but how do you get from SST = ∑((x - MEAN)^2) to this SST = ∑(x^2) - (∑(x))^2/N ? The former is the definition I'm used to. The latter is what you used in your simple proof. I tested it out and they are equal, but can you prove
