On Sep 17, 3:54 pm, djh <halitsk...@att.net> wrote: > I've got the CI analysis "by thirds" for the other five folds (as in > my post of 9/15@10:08 for the a1 fold) , but don't want to bother > posting the data if the approach is entirely wrong - i.e. if one can't > draw conclusions the way I'm doing it. > > So whenever you have a chance, please let me know if: > > a) the "thirds" approach is technically legit ? > > b) if so, can one draw conclusions from the "1:3's" and "4:0"'s? > > c) if not, how do you want me to look at the data ? > > Thanks as always ...
The "thirds" approach is misleading, and changing "thirds" to any other fixed set would be no better. Always look at the plot, then decide where to split.
M L H -0.211 -0.297 -0.124 2 R2 C uH -0.087 -0.131 -0.043 3 R3 C uH -0.046 -0.104 0.012 3 R3 S uL -0.043 -0.074 -0.011 3 R3 C uL -0.029 -0.077 0.020 2 R2 C uL -0.004 -0.080 0.072 2 R2 S uL 0.016 -0.034 0.067 1 R1 C uH 0.051 0.008 0.093 3 R3 S uH 0.065 0.013 0.103 1 R1 S uH 0.065 0.030 0.100 1 R1 C uL 0.077 0.018 0.136 1 R1 S uL 0.081 0.033 0.130 2 R2 S uH
Let LL be the lowest L, and HH the highest H. Pick the range, A...B, of print positions you want the plot to be in. Define a position function: x -> P(x) = A + (x-LL)*(B-A)/(HH-LL), rounded to the nearest integer. Then for each M,L,H triple: put a "-" in positions P(L)...P(H), and an "m" in position P(M).
Here is a plot of the above data with A,B = 1,69. View it in a fixed-width font such as Courier.
The plot makes it obvious that the only possible split for these intervals is the first vs the rest. If you're worried about the overlap of the first two intervals, remember that 95% is just a convenient decimal value. If our number system were hexadecimal then the "standard" confidence level might well be 15/16, which is .9375 in decimal. Think of the intervals as having fuzzy endpoints. Don't interpret them as sharp legalistic cutoffs.
Also, what multiplicity adjustment did you use? Eventually, it will have to 72.