Topic:
The same four proportional weighting factors work for each 00/01/10/11 when 0.25 is subtracted from each !!!
Replies:
506
Last Post:
Nov 20, 2012 9:21 PM
On Jul 23, 4:59 pm, djh <halitsk...@att.net> wrote: > Thanks for the prompt response. If it's not allowed to t-test > variances, then how about this approach. > > Consider the following two sets of variances of residuals, obtained > for the a1 and a3 folds by running ln(c/u) on ln(c/L) on each of our > 12 length intervals, using dicodon set C1058 and u restricted to uL. > > a1 fold a3 fold > Variance Set A Variance Set B > > Li N Vars N Vars > > 1 346 0.014528036 324 0.041647529 > 2 404 0.009185699 381 0.031573403 > 3 378 0.008822013 366 0.027733517 > 4 358 0.010882797 357 0.022499068 > 5 355 0.007926638 395 0.015636086 > 6 305 0.006239904 319 0.014493505 > 7 310 0.007451552 327 0.014627494 > 8 338 0.006225212 299 0.011507795 > 9 293 0.005164239 283 0.010064917 > 10 240 0.005487285 294 0.009341207 > 11 274 0.004696066 294 0.009341207 > 12 298 0.004245657 218 0.009237589 > > The linear regressions of Variance Sets A and B on the Li's 1-12 give > the following confidence interval for slopes and intercepts: > > Slopes: > > Lower Upper > > A: -9.96171E-04 -4.67755E-04 > B: -3.51061E-03 -1.80244E-03 > > Intercepts: > > Lower Upper > > A: 1.03845E-02 1.42735E-02 > B: 2.91235E-02 4.16953E-02 > > So according to what you've taught me, the two regressions can be said > to differ significantly (pre-Bonferroni) because neither the CI's of > their slopes not the CI's of their intercepts overlap at all. > > Is this the correct way to show that the pattern of variances in > Variance Set A differs from the pattern of variances in variance set > B? > > If not, how then to show this? > > Thanks as always for whatever time you can afford to spend considering > this question.
If the df are large and the usual linear regression assumptions are not too false then the sample standard error of prediction -- i.e., the adjusted SD of the sample residuals -- will be approximately normal, with variance approximately equal to the true standard error divided by the square root of 2df. For your Ns & SDs, it should be OK to regress the adjusted SDs of the residuals on the length indices and then compare the slopes & intercepts of those regressions.
