Chpt10 - Chapter 10 Experimental Design Section 10.1 1....

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Chapter 10 Experimental Design Section 10.1 1. Replication allows you to obtain an estimate of the experimental error , which is sometimes thought of as the "noise". That is, we expect a certain amount of natural variation between experimental results, even when all factors are held fixed, and we call this variation the experimental error. Knowing the magnitude of the experimental error allows you to know when a factor's effect is important or not; important/significant factors are those whose effect on the response variable causes changes/variation that is larger in magnitude than the experimental error. 2. (a) An estimate of the effect of changing factor A from level 1 to level 2 is: () 25 . 2 3 . 6 4 . 7 2 1 0 . 4 2 . 5 2 1 = + + (b) An estimate of the effect of changing factor B from level 1 to level 2 is: 15 . 1 3 . 6 0 . 4 2 1 4 . 7 2 . 5 2 1 = + + 3. (a) and (b): The following surface plot of the function f(x,y) was created in MathCAD. The surface is a dome whose maximum point sits over the point x = 2, y = 5 in the x-y plane :
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Chapter 10 Experimental Design 2 (c) For a contour level of c, we set f(x,y) = c. Taking natural logarithms of both sides of the equation [ ] 2 2 2 1 ) 5 ( ) 2 ( + y x e = c, we find - 2 1 [(x-2) 2 +(y-5) 2 ] = ln(c). Note that c must be positive (because the exponential function is always positive) and less than 1 (since the expression in the exponent is always negative and the exponential function is less than 1 for negative arguments). Thus, ln(c) must be negative. Multiplying through by -2 then yields the familiar equation (x-2) 2 +(y-5) 2 = -2ln(c), where - 2ln(c) is a positive number. This is the equation of a circle in the plane with radius equal to the square root of -2ln(c). That is, all contours for f(x,y) are circles in the plane. (d) Sketching the contours should show that the maximum is achieved when the expression in the exponent of f(x,y) is 0. Because (x-2) 2 +(y-5) 2 is nonnegative, it can only equal 0 when both x = 2 and y = 5. Another method of obtaining the maximum point would be to take partial derivatives of f(x,y), set them equal to 0, and solve. For example, the equations x [ ] 2 2 2 1 ) 5 ( ) 2 ( + y x e = -(x-2) [ ] 2 2 2 1 ) 5 ( ) 2 ( + y x e = 0 and y [ ] 2 2 2 1 ) 5 ( ) 2 ( + y x e = -(y-5) [ ] 2 2 2 1 ) 5 ( ) 2 ( + y x e = 0 have the unique solution x = 2, y =5. 4. (a) The following surface plot of the function f(x,y) was created in MathCAD. The surface is an inverted trough that falls off symmetrically on either side of the line y = x in the x-y plane. Note from the graph that there are infinitely many maxim, which all occur on the line y = x. (b) Pick any value of c between 0 and 1; i.e., 0 < c < 1. To find the contour level for c , set f(x,y) = c, take natural logarithms of both sides, and solve for y: -(x-y) 2 = ln(c), or , (x-y) 2 = -ln(c) = k ,
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Chapter 10 Experimental Design 3 So, x-y = ± k 1/2 and therefore y = x ± k 1/2 .
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Chpt10 - Chapter 10 Experimental Design Section 10.1 1....

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