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Re: Our first control group result is precisely how we want AML's
program to behave with control group data ....
Posted:
Apr 28, 2012 7:33 PM
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You wrote:
> I thought m1 was x1*x2 and had values 0,0,0,1. In any case,
> how do the four cells of the 2x2 design map into 0,1,2,3 ?
> Or is it something completely different?
Yes - you're correct m1 is 0,0,0,1 ... I was thinking of x1 = 0 0 1 1
when I typed that.
In any event, 0,1,2,3 are just a better way of "gradating" the pairs
of residuals we obtain from our two regressions:
lnc-lne on lnc-lnL (call this regression Re)
lnc-lnu on lnc-lnL (call this regression Ru)
For the new variable mv:
0 means residual above the median for both Re and Ru (like x1 = 0
x2 = 0)
1 means residual above the median for Re but below for Ru (like x1 = 0
x2 = 1)
2 means residual below the median for Re but above for Ru (like x1 = 1
x2 = 0)
3 means residual below the median for both Re and Ru (like x1 = 1
x2 = 1)
The idea here is that a residual below the median for Ru is better
than one above the median for Ru, but doesn't count "as much" as a
residual below the mean for Re.
You wrote:
"Yes, but (as I've said before) you still should test the difference
between the coefficients. Only if both coefficients are individually
significant and they have different signs can you skip testing the
difference.
Call the coefficients b1 & b2, and their standard errors s1 & s2.
Get z = (b1 - b2)/sqrt[s1^2 + s2^2], and refer it to the standard
normal distribution (as I described in my 10:48 am post earlier
today). "
I will do this for two reasons: i) because you've told me to; ii) I
want to learn how to do it.
But I want to wait to do this until after I've run the new model
{lnL,mv,lnLmv} against the remaining five folds. I want to see first
if the pattern holds up of the confidence intervals for mv and lnLmv
staying on either side of 1 for the study group, but crossing 1 for
the control group.
In this regard, I've run the new {lnL,mv,lnLmv} model against the a3
study and control groups and the pattern DOES hold up:
a3 Study Group:
14376 cases have Y=0; 1176 cases have Y=1.
Overall Model Fit...
Chi Square= 437.7746; df=3; p= 0.0000
Coefficients and Standard Errors...
Variable Coeff. StdErr p
1 4.2503 0.3094 0.0000
2 1.4813 0.6778 0.0289
3 -0.4283 0.1741 0.0139
Intercept -18.6238
Odds Ratios and 95% Confidence Intervals...
Variable O.R. Low -- High
1 70.1255 38.2413 128.5933
2 4.3988 1.1650 16.6082
3 0.6516 0.4632 0.9165
a3 Control Group:
466 cases have Y=0; 27 cases have Y=1.
Overall Model Fit...
Chi Square= 23.6075; df=3; p= 0.0000
Coefficients and Standard Errors...
Variable Coeff. StdErr p
1 1.9369 2.3111 0.4020
2 -6.9445 5.2765 0.1881
3 1.8649 1.3539 0.1684
Intercept -10.4753
Odds Ratios and 95% Confidence Intervals...
Variable O.R. Low -- High
1 6.9370 0.0748 643.2717
2 0.0010 0.0000 29.8908
3 6.4554 0.4544 91.6990
And I will post the results for b1 later tonight, but I'm pretty sure
these will maintain the pattern, because b1 worked so well even with
your original model (i.e. chi-squares relativized to sample size were
something like .29 and .16 for study and control group.)
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