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
I may need help figuring out the best way to code the recursion at the very bottom of the spec at the end of this email (regarding the "Holm" implementation of the "Bonferoni correction") ... but I will try to get thru it on my own first ... that's why you're cc'd on this post to Ray Koopman at Google group sci.stat.math . . . ]
Ray-
I am extremely grateful for the detailed specification of your significance comparison algorithm. Now that I've automated the error- prone manual process-points, there are no longer any "rate-limiting" clerical factors, and I can very rapidly generate all the required regressions for all the folds, plus error-free summary and detail files for you in case you're interested in investigating anything particular upon your return.
So as soon as I've gone thru all the folds obtaining all the required regressions, I will turn my attention to coding up the spec as you provided it. (Or, I may attack it in pieces whenever I find my eyes glazing over from too much point and click in Excel ...)
Thank you so much again.
> Yes, they're unequal-variance (aka "heteroscedastic") t-tests, > but not the standard variety. You'll have to code them yourself. > Here's how. The basic data are n pairs of triples, each one of the > form {{x,vx,nx},{y,vy,ny}}. > > x & y are intercepts (or slopes), > vx & vy are the squares of their estimated standard errors, > nx & ny are their sample sizes. > > From each pair of triples you create a new triple {d,vd,fd}, where > > d = x-y, vd = vx+vy, fd = (vx+vy)^2 / (vx^2/(nx-2) + vy^2/(ny-2)). > > Each vd is the estimated variance of the corresponding d, and each > fd is the degrees of freedom of that estimate. As a point of interest > and a possible check on the computations, the degrees of freedom of > vx & vy are fx = nx-2 & fy = ny-2, and min(fx,fy) <= fd <= fx + fy. > > Then t = (sum d)/sqrt(sum vd), with degrees of freedom = > > ft = (sum vd)^2 / sum(vd^2/fd), analogous to each fd. > > As with each fd, min(vd) <= ft <= sum(vd). Note that the number of > {x,y} pairs (i.e., the number of d's) plays no explicit role in the > analysis. Also, it is assumed that all 2n triples are mutually > independent. > > I used the stepwise (aka "Holm") Bonferroni adjustment > to the threshold for declaring a p "significant". > > 0. Let alpha = the desired maximum familywise Type I error rate. > Let n = the number of tests. > Let k = 0. > > 1. Let alpha' = alpha/(n-k). Then check all the p's, > declaring a p "significant" if it's < alpha'. > > 2, Let k = the total number of significant p's. > If the new k > the old k then go to 1. > Otherwise quit. > > For the results I reported, alpha = .05 and n = 48 = > 3 pairs of datasets * > 2 types of comparand (intercept, slope) * > 2 types of comparison (level, pattern) * > 2 DV's (e, u) * > 2 types of c (s, a).- Hide quoted text - > > - Show quoted text -
> Yes, they're unequal-variance (aka "heteroscedastic") t-tests, > but not the standard variety. You'll have to code them yourself. > Here's how. The basic data are n pairs of triples, each one of the > form {{x,vx,nx},{y,vy,ny}}. > > x & y are intercepts (or slopes), > vx & vy are the squares of their estimated standard errors, > nx & ny are their sample sizes. > > From each pair of triples you create a new triple {d,vd,fd}, where > > d = x-y, vd = vx+vy, fd = (vx+vy)^2 / (vx^2/(nx-2) + vy^2/(ny-2)). > > Each vd is the estimated variance of the corresponding d, and each > fd is the degrees of freedom of that estimate. As a point of interest > and a possible check on the computations, the degrees of freedom of > vx & vy are fx = nx-2 & fy = ny-2, and min(fx,fy) <= fd <= fx + fy. > > Then t = (sum d)/sqrt(sum vd), with degrees of freedom = > > ft = (sum vd)^2 / sum(vd^2/fd), analogous to each fd. > > As with each fd, min(vd) <= ft <= sum(vd). Note that the number of > {x,y} pairs (i.e., the number of d's) plays no explicit role in the > analysis. Also, it is assumed that all 2n triples are mutually > independent. > > I used the stepwise (aka "Holm") Bonferroni adjustment > to the threshold for declaring a p "significant". > > 0. Let alpha = the desired maximum familywise Type I error rate. > Let n = the number of tests. > Let k = 0. > > 1. Let alpha' = alpha/(n-k). Then check all the p's, > declaring a p "significant" if it's < alpha'. > > 2, Let k = the total number of significant p's. > If the new k > the old k then go to 1. > Otherwise quit. > > For the results I reported, alpha = .05 and n = 48 = > 3 pairs of datasets * > 2 types of comparand (intercept, slope) * > 2 types of comparison (level, pattern) * > 2 DV's (e, u) * > 2 types of c (s, a).- Hide quoted text - > > - Show quoted text -
