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Topic: Interpretation of coefficients in multiple regressions which model
linear dependence on an IV

Replies: 146   Last Post: Dec 15, 2012 6:44 PM

 Messages: [ Previous | Next ]
 Halitsky Posts: 600 Registered: 2/3/09
OK – I think I’m set, at least till we get to c
on (e, u, u*e).

Posted: Nov 27, 2012 9:19 AM

1.

You wrote:

?slope = dy/dx = a1 + 2*a2*x?

Oh! That slope! (Of the tangent to the quadratic at x = 0.) Duh!

I was construing ?slope? as just a alternative term for ?coefficient?
? don?t know why. Someday I?ll learn if that if I can?t make sense of
something you?ve written, then I?ve construed something wrongly.

2.

When I re-run the analysis, I will use (u/1+u) instead of u. So I
assume I would use ln(u/(1+u)), and (ln(u+(1/u)))^2, parallel to ln(u)
and (ln(u))^2? (Unless you don?t want me to take logs ? please
clarify here.)

3. You wrote:

?For instance, you might get the slope at each data point and then use
their literal average, a1 + 2*a2*mean_x. But what if cells with
different x-means give the same a1 and a2? Should their
"average slope" measures be the same or different? That's the kind of
question you have to ask yourself.?

Since I can?t think that far ahead in the abstract (as you can), I
will use ?a1 + 2*a2*mean_x? initially, and see if any peculiarities
arise of the sort you mention (or others.)

4.

Apart from correcting the general theoretical deficiency which you
perceive as arising from uL/uH dichotomization itself, another benefit
of doing the re-analysis with the three new regressions is presumably
that it will help us improve the two tables I presented at the end of

Table I:

Extent to which Significant 3-way Interactions
across Length Intervals
are Exhibited at Specific Length Intervals
(Data Not Shown for 2RS and 2RC)

3-way u Non-Random
Coeff Len Int p Lev 2S 2C % Chg
eS ALL .014 L 2.288 6.075 165.5%
H 7.993 8.309 4.0%

1 .035 L 1.578 6.545 314.6%
H 5.719 4.760 -16.8%

euSe ALL .005 L 2.402 6.154 156.2%
H 8.265 7.975 -3.5%

1 .025 L 2.076 6.640 219.8%
H 6.002 3.926 -34.6%

4 .011 L 1.250 7.881 530.3%
H 10.755 7.658 -28.8%

uS ALL .0003 L 0.277 -0.082 -129.6%
H 0.050 -0.848 -1796.0%

5 .040 L -0.287 -0.345 20.0%
H 0.289 -0.325 -212.3%

9 .0002 L 0.548 -0.078 -114.2%
H 0.436 -1.056 -342.4%

10 .011 L 0.563 -0.006 -101.0%
H 0.261 -1.103 -522.9%

11 .032 L 0.617 -0.142 -123.1%
H 0.354 -1.178 -432.9%

euSu ALL .036 L 0.240 -0.037 -115.4%
H -0.190 -0.760 300.0%

9 .008 L 0.482 -0.067 -113.8%
H 0.141 -0.829 -686.3%

11 .012 L 0.415 -0.083 -120.1%
H 0.185 -1.253 -777.3%

Table II:

Bonferronu Correction of p?s in Table I

TstVal=
.025/ p(j)-
j Category p(j) (14-j) TstVal

1 uS-9 .0002 .00192 -.00177
2 uS-ALL .0003 .00208 -.00178
3 euSe-ALL .005 .00227 .00264
4 euSu-9 .008 .00250 .00559
5 uS-10 .011 .00278 .00779
6 euSe-4 .011 .00313 .00784
7 euSu-11 .012 .00357 .00819
8 eS-ALL .014 .00417 .01032
9 euSe-1 .025 .00500 .02016
10 uS-11 .032 .00625 .02528
11 eS-1 .035 .00833 .02701
12 euSu-ALL .036 .01250 .02366
13 uS-5 .040 .02500 .01523

In particular:

(A) with respect to the equivalent of Table I that will be constructed
from the new 2-ways, I am hoping first that more of the new 2-ways at
specific length-intervals will show probabilities < .05, as opposed to
the paltry yield of ?good? interval-specific 3-ways in the current
table.

(B) with respect to the equivalent of Table II that will be
constructed from the p?s in the new Table I, I am hoping that more of
these p?s (both cross-interval and interval-specific) will withstand
plausible Bonferroni correction.

Is there any reason to assume A PRIORI that (A-B) will not be the
case, due to some subtle theoretical reason that I?m too ignorant to
see? And more generally, is there a better pair of tables that should
be generated for the new 2-ways, instead of tables like (I-II):

Also, just out of curiousity, I?m wondering if it would be legitimate
to ?allow? row 3 in Table II as well as rows 1 and 2? (You mentioned
once that some manual tinkering with Bonferroni results is allowable.)

Thanks as always for your considered consideration of these matters.
It will take me 2-3 days to add code and re-run for c on (u,u^2)with
computation of a first-cut ?average? slope, depending on day-job
demands.

In the meantime, I?m hoping you?ll continue your exegesis with an
explication of c on (e, u, u*e), the second of the three new
regressions.

Date Subject Author
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