<p>1</p><p>Determining the Effect of a Factor</p><p>Basic 2 level, 2 variable Design (22 Design)</p><p>22 Experimental Design with a Measured Response Experiment (Yates Factor A Factor B Response Order) 1 (1) -1 -1 6 2 (a) 1 -1 3 3 (b) -1 1 7 4 (ab) 1 1 10</p><p>Computing the Average Effect of Factor A</p><p>The Effect of a factor is defined as the change in the response produced by a change in the level of that factor. </p><p>The AVERAGE effect of factor A at its LOW level is (-6 -7)/2 = - 6.5</p><p>The AVERAGE effect of factor A at its HIGH level is (+3 + 10)/2 = + 6.5</p><p>The Average effect of A FOR THE ENTIRE EXPERIMENTAL EFFORT is the sum of the average of the low and the average high level settings </p><p>= [(3 + 10)2] – [(6 + 7)/2] = ((3 + 10) – (6 + 7))/2 = 0</p><p>Thus the average effect of factor A = 6.5 – 6.5 = 0</p><p>Computing the Average Effect of Factor B </p><p>In a similar fashion – assigning the signs of the column associated with Factor B to the responses we have the following</p><p>The AVERAGE effect of factor B at its LOW level is (-6 -3)/2 = - 4.5</p><p>The AVERAGE effect of factor B at its HIGH level is (+7 + 10)/2 = + 8.5</p><p>The Average effect of B FOR THE ENTIRE EXPERIMENTAL EFFORT is the sum of the average of the low and the average high level settings </p><p>Thus the average effect of factor B = 8.5 – 4.5 = 4 2</p><p>The expression of the calculation in Terms of Yates Order</p><p>Effect of Factor A = ((a + ab) – (1 + b))/2 which gives ((3 + 10) – (6 + 7))/2 = 0</p><p>Effect of Factor B = ((b + ab) – (1 + a))/2 which gives ((7 + 10) – (6 + 3))/2 = 4</p><p>Item of Interest – note well – Yates order is just a short hand way of describing the makeup of a particular experiment. If the letter is present then that corresponds to the factor setting at a high level (+1) and if it is absent then that corresponds to the factor setting at a low level (-1). Thus for the design above, the first experiment has both factor A and factor B at a low level corresponding to the Yates designation of (1). The second experiment has factor A at a high level and factor B at a low level thus the Yates designation of (a) etc.</p><p>How Can the Average Effect of A be Distinguished from the Effect of B using the same set of Experiments?</p><p>The Average effort of A = Average of the sum of the values of the response when A is at the high level minus the average of the sum of the values of the response when A is at the low level = [ (a +ab)/2] – [(1 + b)/2] = ((a +ab) – (1 + b))/2. </p><p>Experiment (a) has factor B at its low level Experiment (1) has factor B at its low level Experiment (b) has factor B at its high level Experiment (ab) has factor B at its high level</p><p>THEREFORE the AVERAGE EFFECT of FACTOR B when computing THE AVERAGE EFFECT OF A IS:</p><p>((Blow + Bhigh) -(Blow + Bhigh))/2 = 0</p><p>The Extrapolation</p><p>If we have a three variable, 2 level design (23) the factors will be A, B, and C.</p><p>The 8 experiments in the design will be (remember the Yates designation): (1), a, b, ab, c, ac, bc, and abc.</p><p>Therefore the average effect of Factor A is ((a + ab + ac+ abc) – (1 + b+ c +bc))/4</p><p>Note the division by 4 since we are now taking averages of 4 experiments with the factor settings at either the high or the low level. 3</p>