10.0 REPLICATED FULL FACTORIAL DESIGN

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1 10.0 REPLICATED FULL FACTORIAL DESIGN (Updated Spring, 001) Pilot Plant Example ( 3 ), resp - Chemical Yield% Lo(-1) Hi(+1) Temperature 160 o 180 o C Concentration 10% 40% Catalyst A B Test# Temp Conc Catalyst Yield x 1 x x 3 y i1 y i y X Concentration 46, 44 79, ,58 69, 67 X 3 Catalyst 50, 54 81, 85-59, 61 74, X 1 Temperature Standard Order 1st variable column (-1,+1,-1...) Alternates every 0 value nd variable column (-1,-1,+,+,-,-,+,+) Alternates every 1 values 3rd variable column (-,-,-,-,+,+,+,+,-,-,-,-,+,+,+,+,...) Alternates every values 59

2 k th variable - alternates every k-1 values Underlying Model y= + b 0 + b 1 x 1 + b x + b 3 x 3 b 1 x 1 x + b 13 x 1 x 3 + b 3 x x 3 + b 13 x 1 x x 3 + ε Avg. of all responses at high level of x = = Avg. of all responses at low level for x = = E 1 = =3.0 Geometrical interpret E 1 = Number of + Signs = 8 E 1 = E 1 = = y s (We will work with these) Calculation Matrix Test I Y i

3 Avg = E 1 = E 1 = = = = E 13 = = Summary Avg = 64.5 E 1 =1.5 E 1 = 3.0 E 13 =10.0 E = -5.0 E 3 = 0.0 E 3 = 1.5 E 13 =0.5 Which of these are important? Which of the effects are distinguishable from the noise in the experimental environment? Lets say we run experiment numerous times. Sample effects, E s, are distributed normally. E = Linear Combination of y s We will look at three ways to figure out the important effects Replication of experiment Normal Prob. Plot Make an assumption about higher - order effects being negligible. Replication -> Important Effects Our experiment is replicated, calculate a sample variance for each test. 61

4 Test y i1 y i y S y Let s assume σ y is constant for all the tests. Calculate pooled sample variance estimate of σ y Σs P S P = = = 8 Pilot Plant Example: Repl. 3 Factorial Test s p x 1 x x 3 y i1 y i y

5 Calculated effects: E i is an estimate of Avg = 64.5 E 1 =1.5 E 1 = 3.0 E 13 =10.0 E = -5.0 E 3 = 0.0 E 3 = 1.5 E 13 =0.5 µ E S P = m i = 1 s i m where m= k, n is the number of trials for each test condition, N is total number of tests, and N=mn Under H o : ( µ E1 = µ E = = 0) The E s have arisen from a normal distribution centered at 0. We may, based on each calculated effect, develop a confidence interval for the true mean effect E i + t s eff -> How do we estimate σ eff First let s develop a confidence interval for the true average Avg in our case m n i = 1 j = 1 y ij = = 64.5 mn Var (Avg) = Var y + y + + y N = Var N [ y + y + + y] Nσ y = = > as expected N N σ y 63

6 8 = = = 0.5 N 16 s S p avg ---- s avg = ( 1 α )% Confidence interval for average (95% C.I.) is Avg ± t αs avg = 64.5 ± t 8, s avg = 64.5 ± (.306) ( 0.707) ν 1, --- Since 0 is not on C.I., reject H o that = 0 µ avg Confidence Intervals for Other Effects We calculate effects, generally speaking by: E = y (+) - y (-) Each based on N/ observations (8 for our example) Var[E] = Var [y (+)] + Var [y (-)] y+ y + + y Var[y (+)] = Var N 4 = Var [y+y+...+y] = N N ---σ y = ---σ N y N Var [E] = ---σ N y + ---σ N y For our example, 4 s eff = 4 σ eff = ---σ N y 4 ---s N P = = 16 64

7 Total number of trials = 16, s eff = Sometimes called the Std. error of an effect 100 ( 1 α) % confidence interval Avg ± t α ν 1 95% confidence interval is E i + (.306) (1.414) E i % confidence interval for each effect E 1 = * E 1 = E = * E 13 = * E 3 = E 3 = E 13 = , s avg does not lie on C.I. for E 1, E, & E 13. The calculated (sample) effects E 1, E, & E 13 have arisen from a normal distribution not centered at 0. Interpreting Significant Effects E 1 = 3.0 Collapse or average the design in X and X 3 directions 65

8 . y Average y for all (+) levels of X E 1 = X 1 E = Collapse or average the design in X 1 and X 3 directions y Average y for all (-) levels of X E = X E 13 = 10.0 Collapse or average the design in X direction. 66

9 X Since X 1 is involved in interaction, much care must be used in interpreting the main effect, E X 1 y +X X X y X 3 +X X Mathematical Model For 3 factorial design pilot plant example we tacitly assume that response can be characterized as: y = b 0 + b 1 x 1 + b x + b 3 x 3 + b 1 x 1 x + b 13 x 1 x 3 + b 3 x x 3 + b 13 x 1 x x 3 + ε We ran tests and fit this equation to the data: ŷ = bˆ 0 + bˆ 1x 1 + bˆ x + bˆ 3x 3 + bˆ 1 x 1 x + bˆ 13 x 1 x 3 + bˆ 3 x x 3 + bˆ 13 x 1 x x 3 67

10 We now know that only bˆ 0, bˆ 1, bˆ & bˆ 13 are important. The fitted model becomes: ŷ = bˆ 0 + bˆ 1x 1 + bˆ x + bˆ 13 x 1 x 3 ŷ = E 1 Avg E x E 13 x x1 x 3 ŷ = x x +5x 1 x 3 For each unique combination of x 1, x, & x 3 a predicted response may be calculated. ŷ = (-1) -.5 (-1) +5(-1) (-1) Test# x 1 x x 3 y ij y i ŷ i e ij = y ij - ŷ We must examine the model residuals to search for model inadequacies. Plots: e ij vs. run order e vs. y e ij vs. x 1, x,... (y ij - y i ) vs. y i Using the Model for Process Improvement ŷ = x x +5x 1 x 3 What values of x 1, x, & x 3 maximize predicted yield? How does the response behave as a function of x 1, x, & x 3? What is the shape of the response surface? 68

11 x 1 = 1 Temp -- ( Temp HI + Temp LO ) ( Temp HI Temp LO ) Temp MidTemp x 1 = = UnitChangeInTemp Temp When Temp = x 1 = = When x 1 = -0. Temp = > Temp = 170 +(-.) (10) = Only can use this transformation for quantitative variables. Recall that: Low (-1) High(+1) x 1 Temp x Conc x 3 Catalyst A B Based on other criteria, catalyst set to A, i.e., x 3 = -1 So, ŷ = x x +5x 1 (-1) or ŷ = x x The presence of interactions warp the plane. Remember, 3 ways to figure out the significant effects: 1. Use replication to get (estimate of ) (estimate of ) s p. Assume that higher order interactions are negligible (estimate of σ eff ) 3. Normal probability plot. σ y s eff s eff σ eff 69

12 Higher Order Interactions Negligible Study surface finish (Ra value) produced by a machining operation. Variable Low (-1) High(+1) 1-Speed (ft/min) Feed (in./rev) Depth of Cut (in.) Nose Radius (in.) 1/64 3/ SCEA 15 o 30 o Calculated Effects - Effect Estimates Avg = E 13 =.658 E 1 = E 14 = E = E 15 = E 3 = E 134 = E 4 = E 135 = E 5 = 0.04 E 145 = E 1 = E 34 = E 13 = E 35 = E 14 = E 45 = E 15 = -1.4 E 345 = E 3 = E 134 = E 4 = E 135 = E 5 = E 145 = E 34 = E 1345 = E 35 =.041 E 345 = -0.9 E 45 = E 1345 =

13 Test Table Test R a

14 Comments As we go from generally the proportion of sig. effects decreases Response surfaces generally fairly smooth - absence of higher order interactions Rarely find 3 - factor interactions significant, even rarer to find 4 - factor interactions significant even rarer to find 5 - factor interactions significant and so on. Let s assume that higher order interactions are negligible. Specifically, let s assume that 3 - factor and higher order interactions are negligible. The calculated effect estimates are then non-zero only because of experimental error. Normal Dist N (0, σ eff ) Calculated Effects Value (True effect value=0 for high order interactions) Under our assumption, all calculated 3-factor interactions come from a normal dist n with a standard deviation of centered at 0. Use the σ eff calculated effects to estimate. σ eff ( ) Number of 3rd and higher order E i s eff = All Higher Order Effects In above equation, the number of higher order effects equals the number of degrees of freedom. This is because that we have already assumed zero mean for the normal distribution. If we take 3rd and higher order interactions, (.658) + (.134 ) + + ( 4.1) = = s eff 7

15 s eff =.536 is an estimate of. σ eff Construct 100 ( 1 α) % Confidence Interval for each effect estimate E i ± t α ν 1, s eff --- where ν is the number of 3 - factor & higher order interactions used to calculate s eff. For our example, ν = 16, α = 0.05, The C.I. is E i ± (.1) (.536) = E i ± We can conclude that the average, E, E 4, & E 4 are significant. One final way to figure out which effects are significant is the normal probability plot. Normal Probability Plot Draw 100 values from a normal dist n, and plot the CDF for it. Draw another 100 values from an unknown dist n and plot the CDF for it too. Is the unknown dist n normal? Does the S shaped appearance in the data match the S shapes of the normal CDF? Let s change the vertical scale on the CDF plot so that the normal CDF becomes a straight line Linear Scale Scale streched out about Mean of the data x 73

16 Normal Not Normal If the sampled data from the unknown dist n appears to be well described by a line on such a plot, then the unknown dist n may be reasonably approximated as normal. How can we use a normal probability plot to help us decide which effects are important? Calculate the effects Rank the effects from smallest to largest (don t include the average) Calculate the cumulative prob. associated with each effect i -- P i = N Plot the points (E i, P i ) on Normal probability paper (Home-made or purchased). 74