9.3 Partial confounding

When the design is replicated, the same effect need not be confounded in every replicate. Confounding a different effect in each replicate, called partial confounding, keeps some information on every effect. The effect confounded in a replicate is estimated from the other replicates only. (When the same effect is confounded in every replicate, the design is completely confounded, and that effect cannot be estimated at all.)

Example 9.3.1. An experiment studied three factors \(A\), \(B\), \(C\) at two levels each. Homogeneous blocks of eight runs were not available, so two replicates were run, each in two blocks of four:

replicate I
replicate II
block \(1\) block \(2\) block \(3\) block \(4\)
\((1) = 5\) \(a = 21\) \((1) = 6\) \(b = 13\)
\(ab = 18\) \(b = 13\) \(a = 21\) \(ab = 17\)
\(c = 8\) \(ac = 13\) \(bc = 6\) \(c = 8\)
\(abc = 18\) \(bc = 7\) \(abc = 18\) \(ac = 17\)

The total is \(209\) and \(SST = 482.9375\).

Which effects are confounded? In block \(1\) every run has an even number of the letters \(a\), \(b\) (none or both), so the \(AB\) column is \(+\) throughout, and in block \(2\) it is \(-\) throughout: \(AB\) is confounded in replicate I. In replicate II the same holds for \(B\) and \(C\): block \(3\) has \(BC = +\) throughout and block \(4\) has \(BC = -\). So \(AB\) is confounded in replicate I and \(BC\) in replicate II.

Estimates. \(A\), \(B\), \(C\), \(AC\) and \(ABC\) are estimated from all \(16\) runs, as contrasts divided by \(8\). \(AB\) is estimated from replicate II alone and \(BC\) from replicate I alone, each as a contrast over eight runs divided by \(4\): \begin {gather*} A = 77/8 = 9.625 , \qquad B = 11/8 = 1.375 . \end {gather*} In replicate II the runs \((1)\), \(a\), \(bc\), \(abc\), \(b\), \(ab\), \(c\), \(ac\) have \(AB\) signs \(+\), \(-\), \(-\), \(+\), \(-\), \(+\), \(+\), \(-\). In replicate I the runs \((1)\), \(ab\), \(c\), \(abc\), \(a\), \(b\), \(ac\), \(bc\) have \(BC\) signs \(+\), \(-\), \(-\), \(+\), \(+\), \(-\), \(-\), \(+\). So \begin {align*} \widehat {AB} &= \tfrac 14(6 - 21 - 6 + 18 - 13 + 17 + 8 - 17) = -8/4 = -2.00 , \\ \widehat {BC} &= \tfrac 14(5 - 18 - 8 + 18 + 21 - 13 - 13 + 7) = -1/4 = -0.25 . \end {align*}

Analysis of variance. The block sum of squares, from the four block totals \(49\), \(54\), \(51\), \(55\), is \(5.6875\) on \(3\) degrees of freedom; it splits into replicates (\(0.5625\), \(1\) df) and blocks within replicates (\(5.125\), \(2\) df). Each unconfounded effect has \(SS = \text {contrast}^2/16\), and each partially confounded effect has \(SS = \text {contrast}^2/8\) from the replicate in which it is not confounded:

source SS df \(F_0\) \(p\)
blocks \(5.6875\) \(3\)
replicates \(0.5625\) \(1\)
blocks within replicates \(5.1250\) \(2\)
\(A\) \(370.5625\) \(1\) \(237.2\) \(< 0.0001\)
\(B\) \(7.5625\) \(1\) \(4.84\) \(0.079\)
\(C\) \(22.5625\) \(1\) \(14.44\) \(0.013\)
\(AB\) (replicate II only) \(8.0000\) \(1\) \(5.12\) \(0.073\)
\(AC\) \(0.5625\) \(1\) \(0.36\) \(0.57\)
\(BC\) (replicate I only) \(0.1250\) \(1\) \(0.08\) \(0.79\)
\(ABC\) \(60.0625\) \(1\) \(38.44\) \(0.002\)
error \(7.8125\) \(5\)
total \(482.9375\) \(15\)

The error is found by subtraction, with \(15 - 3 - 7 = 5\) degrees of freedom, and \(MSE = 1.5625\). \(A\), \(C\) and \(ABC\) are significant at \(5\%\) (\(F_{0.05,1,5} = 6.61\)). The partially confounded effects \(AB\) and \(BC\) are each estimated with half the information of the others: their sums of squares are based on eight runs, and their estimates have twice the variance, \(\sigma ^2/2\) against \(\sigma ^2/4\). The relative information on a partially confounded effect is the fraction of replicates in which it is not confounded, here \(\frac 12\).

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