
Re: Our first control group result is precisely how we want AML's program to behave with control group data ....
Posted:
Apr 30, 2012 6:54 PM


Thank you so much, Ray. I think I understand now, and also understand why I was confused.
But please let me play it back to you in my own words so you can see if I have it right.
1. Procedure 1: using the "likelihoods" to get a chisquare to take into the table.
If I follow the procedure based on the "likelihoods":
"Fit the model to the pooled data for the two groups, and take the Converged value of 2 Log Likelihood. Subtract the sum of the Converged values from fitting the same model to each group separately. The result is a chisquare with df = the number of parameters in the model (in this case, 4: the three predictors, plus the intercept). If the chisquare is significant then the two groups differ. If it's not significant then you can't say the groups are the same; all you can say is that there isn't sufficient evidence to say that the groups differ. "
then WITHOUT using any 2x2 "abcd" table, I will get a chisquare that I can take directly into the table (with df=4). This is the chi square that results simply from:
a) pooling the study and control group data, running the model on these pooled data, and getting a "pooled" converged value b) summing the converged values from the separate study group and control group runs; c) substracting this sum from the "pooled" converged value.
Is what I just wrote correct? If so, then the only reason why I was confused is that I incorrectly thought that I had to use an "abcd" table somewhere in the procedure.
2. Procedure 2: getting chisquares from the individual n0's and n1's for each corresponding row of study and control group. (This is the proceduree I always understood, except I didn't understand the relationship between chisquare and z (chisquare = square of z.)
In particular, this is the procedure in which I constrruct an "abcd" table like the one you gave:
n0 n1 n prop Study a = 1521 b = 95 1616 95/1616 = .0588 Control c = 28 d = 6 34 6/34 = .1765
from each pair of corresponding rows in the study and the control group.
After constructing each abcd table, I can either compute z manually as you indicated, or get the chisquare using the tool.
If I compute the z's manually, I sum them and take the resulting chi square into the table with df = "number of cells" (now typically 32, except for the case of the a3 fold where it was reduce to 12.)
If I let the tool compute chisquare for each abcd table, I sum these and take the resulting chisquarre into the table with df = "number of cells".
Is what I just wrote correct?
If so, one other question: if I wanted to find a table online that I can take z's into, instead of an online table that I can take chi aquares into, how would you advise that I Google to find one? In other words, what is the technical name for this "ztable"?
3. Finally, if I understand your previous posts correctly, then of the two procedures which I've recapitulated above, the first is best for our immediate purposes, because it "abstracts" away from the particulars of the situation (like whether the regressions are "real" logistic regressions, etc.
Correct?
If all the above is correct, then I thank you very very much for your patience and time you took to clarify.

