### K. M. Jenkins and A. J. Boulton. 2003. Connectivity in a dryland river: short-term aquatic microinvertebrate recruitment following floodplain inundation. Ecology 84:2708–2723.

Appendix A. The statistical design for ANOVA showing factors, degrees of freedom, multipliers, and formulas to determine estimated mean squares, variance components, and F-ratio tests. A pdf version is also available.

Factor R has i =…a levels; factor L(R) has j =…b levels; factor S(L( R)) has k =…c levels; factor FT has l =…d levels and FD has m =…e levels.  All combinations of between sample effects are replicated n times (r =…n).  All combinations of within sample effects are replicated N times (s =…N).  There are three orthogonal factors and 2 nested factors.  Factors Reach (R), Flood Type (FT) and Flood Day (FD) are fixed and Lakes within Reaches (L(R)) and Sites within (L(R)) are random.  Multipliers and formula’s to determine estimated mean squares (E(MS)) and variance components, denominators for F-ratio tests (den.) and degrees of freedom (df) are shown.

Xijkl= m + Ri + L(R)j(i) + S(L(R))k(j(i))  + FTl + Ri x FTl + FTl x L(R)j(i) + FTl x S(L(R))k(j(i))  + E r[j(i)kl]

Xijklm= m + FDm + FDm x  Ri + FDm x  L(R)j(i) + FDm x  S(L(R))k(j(i))  + FDm x  FTl + FDm x  Ri x FTl + FDm x  FTl x L(R)j(i) + FDm x  FTl x S(L(R))k(j(i))  + E s[mj(i)kl]

 Source of variation Multipliers E(MS) Den. df Variance component Between samples i j k l r - - 1. Ri 1 b c d n - - 2e+ dn2S(L(R))+ cdn2(L(R))+bcdn2R 2 1,2 2R = [MSR- MSL(R)  ] / bcdn 2. L(R)j(i) 1 1 c d n - - 2e + dn2S(L(R))+ cdn2(L(R))+ 3 2,8 2L(R) = [MSL(FH)- MSS(L(R)) ] / cdn 3. S(L(R))k(j(i)) 1 1 1 d n - - 2e + dn2S(L(R)) 8 8,96 2S(L(R)) = [MSS(L(R))- MSe ] / dn 4. FTl a b c 0 n - - 2e + n2FTS(L(R))+ cns2FTL(R)+ bcn2FTR+ abcns2FT 5 1,1 2FT = [MSFT- MSFTR ] / abcn 5. FT x Rli 1 b c 0 n - - 2e + n2FTS(L(R))+ cns2FTL(R)+ bcn2FTR 6 1,2 2FTR = [MSFTR  - MSFTL(FH)] / bcn 6. FT x L(R) lj(i) 1 1 c 0 n - - 2e + n2FTS(L(R))+ cns2FTL(R) 7 2,8 2FTL(R) = [MSFTL(R) - MSFTS(L(R))  ] / cn 7. FT x S(L(R)) lk(j(i)) 1 1 1 0 n - - 2e + n2FTS(L(R)) 8 8,96 2FTS(L(R)) = [MSFTS(L(R)) – MSe] / n 8. e r[ijkl] 1 1 1 1 1 - - 2e 2e =  MSe Within samples i j k l - m 9. FDm a b c d - 1 N 2E  + dN2FDS(L(R))+ cdNs2FDL(R)+bcdN2FDR+abcdN2FD 10 3,3 2FD = [MSFD - MS FDR ] / abcdN 10. FDxRi 0 b c d - 1 N 2E  + dN2FDS(L(R))+ cdNs2FDL(R)+bcdN2FDR 11 3,6 2FDR = [MSFDR - MS FDL(R) ] / bcdN 11. FDxL(R) mj(i) 1 1 c d - 1 N 2E  + dN2FDS(L(R))+ cdNs2FDL(R) 12 6,16 2FDL(R) = [MS FDL(R) - MS FDS(L(R)) ] / cdN 12. FDxS(L(R)) mk(j(i)) a b 0 d - 1 N 2E  + dN2FDS(L(R)) 17 24,288 2FDS(L(R)) = [MSFD S(L(R)) - MS E ] / dN 13. FDxFTml a b c 0 - 1 N 2E  + N2FDFTS(L(R))+bN2FDFTL(R)+bcN2FDFTR+abcN2FDFT 14 3,3 2FDFT = [MSFDFT - MS FDFTR ] / abcN 14. FDxFTxRmli 0 b 0 d - 1 N 2E  + N2FDFTS(L(R))+ bN2FDFTL(R)+ bcN2FDFTR 15 3,6 2FDFTR = [MSFDFTR- MS FDFTL(R)] / bcN 15. FDxFTxL(R) m lj(i) 0 b c 0 - 1 N 2E  + N2FDFTS(L(R))+ bN2FDFTL(R) 16 6,24 2FDFTL(R) = [MSFDFTL(R) -MSFDFTS(L(R))] / bN 16. FDxFTxS(L(R)) m lk(j(i)) a b 0 0 - 1 N 2E  + N2FDFTS(L(R)) 17 24,288 2FDFTS(L(R)) = [MSFDFTS(L(R)) - MS E ] / N 17. E s[mijkl] 1 1 1 1 - 1 1 2E 2FD = MSE

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