After several unsuccessful days with failed attempts I got an algorithm able to achieve the goal to find a Confidence Interval for the difference of two sample variances using the general intra-permutation no-parametric method where the exhaustive no-replacement draws of the two source sample items are performed in order to construct the pseudo-samples. __1__The Procedure Let X be a real-world sample, size nX, mean mX, whose items we randomly draw in order to form a pseudo-sample X´. After to to get it we have the current variance j given by: sdX(j)= (1<=i<=nX) Sum (X´(g) - mX)^2*Wx(i)/(nX-1) where Wx(i) = i/(1+2+?+n), index showing the order the item is drawn: g=INT(nX*RND)+1. Once an item is chosen it is made unable to be redrawn in the current sample. Same operations for the sample Y, leading to d(j) as the generic member d(j) = sdX(j) - sdY(j).
__2__Illustrative example: the spider´s data (size 30) presented at my Feb 24, 2013 11:11 AM post.(see program below)
A 95% Confidence interval 20´000 iterations, provided [0.528, 1.165] for the difference of variances. So, one can conclude that at least the difference is 0.528, with 5% significance. Repeating with 1 million it was got an interval [0.525, 1.164]. As expected these intervals are centered at the source difference variances 0.846 approx. Luis A. Afonso REM "varSPID" REM CLS PRINT : PRINT "______VARSPID______"; PRINT " Spiderïs intra-Permutation___variances" pi = 4 * ATN(1) DIM X(30), Y(30), xx(30), YY(30) DIM W(9000), wx(30), wy(30) INPUT " many= "; many REM male DATA 4.70,4.70,4.80,5.20,5.20,5.40,5.50,5.65,5.65,5.70 DATA 5.75,5.75,5.75,5.80,5.85,5.85,5.90,5.95,5.95,6.10 DATA 6.20,6.20,6.35,6.35,6.45,6.55,6.80,6.95,7.00,7.50 REM female DATA 5.90,6.10,6.30,6.60,7.00,7.05,7.05,7.50,7.55,7.55 DATA 7.80,7.95,8.00,8.00,8.10,8.25,8.30,8.30,8.35,8.45 DATA 8.70,8.75,9.00,9.10,9.30,9.50,9.60,9.80,9.95,10.00 REM n = 30 FOR i = 1 TO n: READ Y(i): YY(i) = Y(i) msY = msY + Y(i) / n: YY = YY + Y(i) * Y(i): NEXT i FOR i = 1 TO n: READ X(i): xx(i) = X(i) msX = msX + X(i) / n: xx = xx + X(i) * X(i): NEXT i varX = (xx - n * msX * msX) / (n - 1) varY = (YY - n * msY * msY) / (n - 1) PRINT "*** var female - var male ---> "; PRINT USING "#.### "; varX - varY REM FOR t = 1 TO 30: si = si + t: NEXT t FOR t = 1 TO 30: W(t) = t / si: NEXT t REM REM PRINT : COLOR 14 FOR tur = 1 TO many RANDOMIZE TIMER LOCATE 5, 55 PRINT USING "#########"; many - tur FOR i = 1 TO n: xx(i) = X(i): NEXT i FOR i = 1 TO n: YY(i) = Y(i): NEXT i REM gxx = 0 FOR t = 1 TO n 1 g = INT(n * RND) + 1 IF xx(g) = 123456 THEN GOTO 1 gxx = gxx + W(t) * (xx(g) - msX) * (xx(g) - msX) xx(g) = 123456 NEXT t gyy = 0 FOR t = 1 TO n 2 g = INT(n * RND) + 1 IF YY(g) = 123456 THEN GOTO 2 gyy = gyy + W(t) * (YY(g) - msY) * (YY(g) - msY) YY(g) = 123456 NEXT t varX = n * gxx / (n - 1): REM gxx is a mean value varY = n * gyy / (n - 1) d = varX - varY d1 = INT(1000 * d + .5) REM REM REM W(d1) = W(d1) + 1 NEXT tur REM COLOR 14 PRINT " Confidence Interval 95% " u(1) = .025: u(2) = 1 - u(1) FOR uu = 1 TO 2 sum = 0 FOR gi = 0 TO 8000 sum = sum + W(gi) / many IF sum > u(uu) THEN GOTO 45 NEXT gi 45 PRINT USING "##.### .### "; gi / 1000; sum NEXT uu END