On Tue, 30 Jan 2001, Kathleen Bloom wrote:
> If you have unequal n's, and want to determine linear parameters, you can
> develop new coefficients by taking the normal unweighted coefficients
> (e.g., -1, 0, +1, for three group design) and the formula:
> n1(X1) + n2(X2) + n3(X3)/ n1+n2+n3 where the X's are 1, 2, and 3
> because you have 3 groups. This gives you a new mean of the Xs... (i.e., no
> longer 1+2+3/3 = 2), and from there you calculate the new coefficients
> (e.g., 1 - ?, 2 - ?, 3 - ?, gives you the new linear coefficients) for the
> 3-group design with unequal n's.
You get the same results if you use X's corresponding to the unweighted
coefficients (-1, 0, +1). I should suppose that for quadratic estimates
you'd play the same game with the quadratic unweighted coefficients
(+1, -2, +1). However, I've never played around much with weighted
trend analyses, so my supposition may possibly be incorrect. It may be
better to retain the grand mean calculated from (-1,0,+1), equal to your
"?" minus 2 (let's call that "&"), and generate coefficients from the
unweighted quadratic coefficients as 1-&, -2-&, 1-&.
I note in passing that your decision to pursue weighting-by-sample-size
implies that you have decided to assign equal weight to individual cases
and NOT to assign equal weight to each subgroup. (Had you chosen equal
weight for each subgroup, you'd use the "unweighted" (they're not really
UNweighted, they're _equally_ weighted) coefficients directly.) I've not
encountered situations where it seemed necessary to give equal importance
to each individual case, enough to make it worth the extra effort to
weight the coefficients -- and think about what it means, that the "grand
mean" for the data depends on which trend you're currently pursuing, and
that the several trends (linear, quadratic, cubic, ...) are explicitly
NOT orthogonal.
> From there you can do things like determine
> the the weighted linear estimated parameters. They are given in the spss
> oneway printout... as I understand it... i.e., the weighted (for sample
> size) beta for the linear contrast.
Notice that all this does is change the distance between each group mean
and the grand mean; it does not change the relative distances between
groups, which are still equally spaced. It has never been very clear to
me what advantage one gets from the weighted parameters, especially as
those estimates depend on the accident of how many observations you were
able to find for each group. For this reason (among others) I am
inclined to favor equal weighting in general. If it turns out that the
choice of weighting influences the conclusion(s) to be drawn, one has
compelling arguments for repeating the experiment, this time with a
proper (equal-numbers-of-cases) design and carefully random selection of
cases.
< snip >
> ... My means are 2.05, 6.38, and 12.08 for the three groups
> respectively. In other words.. what does one calculate and how?
You might reasonably try using the regression module (rather than one-way
anova) to compare output: predict Y from X1 (X1 = 1,2,3 for the three
groups, and X1 = -1,0,+1 if you want to confirm that the results are the
same for this coding); and for an alternate (quadratic) model, predict
Y from X1 and X2 (X1 = -1,0,+1; X2 = +1,-2,+1).
You have not, by the way, said what you're doing this analysis FOR, so
it's a bit difficult to know whether one is offering useful advice. Or
not.
-- Don.
----------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED]
MSC #29, Plymouth, NH 03264 (603) 535-2597
Department of Mathematics, Boston University [EMAIL PROTECTED]
111 Cummington Street, room 261, Boston, MA 02215 (617) 353-5288
184 Nashua Road, Bedford, NH 03110 (603) 471-7128
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