Part2: Analysis Prepared by: Paul Funkenbusch, Department of Mechanical Engineering, University of Rochester
Review ANOM ANOVA ◦ Error estimate Replication vs. Pooling ◦ ANOVA table ◦ Judging statistical significance DOE mini-course, part 2, Paul Funkenbusch, 2015 2
Example Terms (measure the volume of a balloon as a function of temperature and pressure) Temperature, Factors variables whose influence you want to study. Pressure Levels specific values given 50 C, 100 C to a factor during experiments 1Pa, 2Pa (initially limit ourselves to 2- levels) Set T = 50 C, P = 1Pa Treatment condition one running of the experiment and measure volume Response result measured measured volume for a treatment condition DOE mini-course, part 2, Paul Funkenbusch, 2015 3
Leve vel Use X1, X2, etc. to designate factors. Factor -1 +1 Use -1, +1 to designate levels 50 100 X1. Temperature ( C) X2. Pressure (Pa) 1 2 X1 at level -1 means T = 50 C y response y = volume of balloon TC TC X1 X1 X2 X2 y Use a table to show factor levels and response (a) for each 1 -1 -1 y 1 treatment condition. 2 +1 -1 y 2 For example, during TC2, 3 -1 +1 y 3 set T = 100 C and P = 1Pa, 4 +1 +1 y4 measure the balloon volume = y 2 etc. DOE mini-course, part 2, Paul Funkenbusch, 2015 4
Test all combinations Leve vel Factor -1 +1 Responses 4 DOF 50 100 X1. Temp ( C) ◦ 4 measured (y) values X2. Pressure (Pa) 1 2 Effects 4 DOF ◦ 4 values calculated TC TC X1 X1 X2 X2 X1*X2 *X2 y ◦ m* = (y 1 +y 2 +y 3 +y 4 )/4 1 -1 -1 +1 y 1 ◦ D X1 = (y 3 +y 4 )/2 - (y 1 +y 2 )/2 ◦ D X2 = (y 2 +y 4 )/2 - (y 1 +y 3 )/2 2 -1 +1 -1 y 2 ◦ D 12 = (y 1 +y 4 )/2 - (y 2 +y 3 )/2 3 +1 -1 -1 y 3 4 +1 +1 +1 y 4 Model 4 DOF ◦ 4 constants in model ◦ y pred = a o +a 1 X 1 +a 2 X 2 +a 12 X 1 X 2 DOE mini-course, part 2, Paul Funkenbusch, 2015 5
Most basic level of analysis Which effects are largest (which factors/interactions most important)? Which levels produce the best (highest or lowest) responses? Just based on the D values ◦ you’ve already done this D average response at level 1 - average response at level 1 m - m 1 1 DOE mini-course, part 2, Paul Funkenbusch, 2015 6
Graphically Sign and Magnitude of D Positive D ◦ +1 level should increase the response m* ◦ -1 level should decrease the response Negative D reversed A -1 A +1 B -1 B +1 Magnitude of D Choose A -1 and B +1 to increase response ◦ Indicates relative importance A is more important DOE mini-course, part 2, Paul Funkenbusch, 2015 7
Can treat interactions terms the same way Larger D more important interaction If interactions are large “best settings” to increase (or decrease) the response will depend on the combination of factor levels Use model to test different combinations ◦ y pred = a o +a 1 X 1 +a 2 X 2 +a 12 X 1 X 2 DOE mini-course, part 2, Paul Funkenbusch, 2015 8
From part I. Leve vel Factor -1 +1 Removal rate of X1. applied load (kg) 2 3 osteotomy drills vs. applied load and X2. previous cuts 0 20 number of cuts. (new) TC TC X1 X1 X2 X2 X1*X2 *X2 Remov oval al rate Which effects are (mm3/s) /s) most important? 1 -1 -1 +1 3 How can you 2 -1 +1 -1 2 increase the 3 +1 -1 -1 5 removal rate? 4 +1 +1 +1 2 DOE mini-course, part 2, Paul Funkenbusch, 2015 9
Factor Leve vel Effects -1 +1 ◦ D X1 = +1 X1. applied load (kg) 2 3 ◦ D x2 = -2 X2. previous cuts 0 20 ◦ D 12 = -1 (new) Number of previous cuts is most important new drill (level -1) will increase removal rate the most Applied load and the interaction are comparable higher load (level +1) will increase removal rate but need to test combinations (since interaction is important too). DOE mini-course, part 2, Paul Funkenbusch, 2015 10
y pred = 3.0 + (0.5)X 1 - (1.0)X 2 - (0.5)X 1 X 2 Set X2 = -1 (Much larger than X1 and interaction) What is best level for X1? ◦ For X1 = -1, X2 = -1 y pred = 3.0 ◦ For X1 = +1, X2 = -1 y pred = 5.0 Best settings X1 = +1 (3kg load), X2 = -1 (new drill) Note; This is a synergistic interaction, ◦ Best level for interaction (-1) corresponds to best factor levels [X1X2 = (+1)(-1) = -1] ◦ Interaction enhances effects of “best” factor level choices For an anti-synergistic interaction, ◦ Conflict between best factor settings and best interaction level ◦ Best overall settings then depend on relative strength of the interaction vs. factor DOE mini-course, part 2, Paul Funkenbusch, 2015 11
Second level of analysis Which of the observed effects are statistically significant? ◦ Based on comparing observed effects against an estimate of error. ◦ Compares D 2 for factors and interactions with D 2 for error. ◦ Actually compare “mean square” or “MS” proportional to D 2 How much does each factor/interaction contribute to the total variance in system? DOE mini-course, part 2, Paul Funkenbusch, 2015 12
Pooling of higher-order Replication interactions Repeat (replicate) each of the Assume that higher-order treatment conditions interactions are unimportant/zero Independent experimental runs Must choose these interactions (not multiple measurements upfront (before examining results) from the same TC) these form a “pool” for error Differences in the responses Effects measured for pooled measured for identical TCs run interactions are used to estimate at different times provide error error “Pure error” not dependent on “Error” includes modeling error modeling assumptions (i.e. assumptions about interactions) Best way to estimate error, but Requires less experimental effort, greatly increases effort but error estimate is not as good DOE mini-course, part 2, Paul Funkenbusch, 2015 13
TC TC X1 X1 X2 X2 y TC TC X1 X1 X2 X2 y 1a -1 -1 y1a 1 -1 -1 y1 2a -1 +1 y2a 2 -1 +1 y2 3a +1 -1 y3a 2X 3 +1 -1 y3 4a +1 +1 y4a 4 +1 +1 y4 1b -1 -1 y1b 2b -1 +1 y2b Original design two factors at 2- 3b +1 -1 y3b levels 4 DOF m*, D X1 , D x2 , D 12 4b +1 +1 y4b Replicated design (2x) + 4 DOF for error 8 DOF total Contrast responses measured 4 DOF m*, D X1 , D x2 , D 12 under nominally identical TC DOE mini-course, part 2, Paul Funkenbusch, 2015 14
Test more factors. Increase size and DOF. TC TC X1 X1 X2 X2 X3 X3 y 1 -1 -1 -1 y1 Example: three 2-level factors instead of two. 2 -1 -1 +1 y2 8 DOF total 3 -1 +1 -1 y3 1 DOF for m* 4 -1 +1 +1 y4 3 DOF for factors D X1 , D X2 , D X3 3 DOF for 2-factor interactions D 12 , D 13 , D 23 5 +1 -1 -1 y5 1 DOF for 3-factor interaction D 123 6 +1 -1 +1 y6 Pool all interactions 7 +1 +1 -1 y7 decide before examining results 8 +1 +1 +1 y8 assess m*, D X1 , D X2 , D X3 (4 DOF) use D 12 , D 13 , D 23 , D 123 for error (4 DOF) Note: alternative choice (pool only the highest order , 3- factor,interaction) is possible, but only leaves 1 DOF for the error estimate not desirable. For larger experiments this is not as much of a constraint (more higher-order interactions that can be pooled). DOE mini-course, part 2, Paul Funkenbusch, 2015 15
Four 2-level factors Five 2-level factors 2 5 = 32TC = 32 DOF 2 4 = 16TC = 16 DOF ◦ m* 1 DOF ◦ m* 1 DOF ◦ factors 5 DOF ◦ factors 4 DOF ◦ 2-factor inter. 10 DOF ◦ 2-factor inter. 6 DOF ◦ 3-factor inter. 10 DOF ◦ 3-factor inter. 4 DOF ◦ 4-factor inter. 5 DOF ◦ 4-factor inter. 1 DOF ◦ 5-factor inter. 1 DOF Pool 4 and 5 factor int. Pool 3 and 4 factor int. ◦ Assess factors and 2-factor ◦ Assess factors and 2-factor and 3-factor interactions interactions ◦ 6 DOF for error estimate ◦ 5 DOF for error estimate DOE mini-course, part 2, Paul Funkenbusch, 2015 16
“Pure error” Assess more (no modeling factors for same assumptions effort (or same needed) number of factors for less effort) Best for: Best for: ◦ large numbers of factors ◦ small numbers of factors ◦ systems with strong ◦ systems with large time/cost constraints on uncertainties experimental size DOE mini-course, part 2, Paul Funkenbusch, 2015 17
Source ce SS SS DOF MS MS F p % SS A 100 1 100 20 0.011 56 B 50 1 50 10 0.034 28 AxB 10 1 10 2 0.230 6 error 20 4 5 -- -- 11 Total 180 7 -- -- -- 100 Typical way of presenting ANOVA results. Explain each column so you can interpret these results. Most software packages will output their analysis in some variant of this format. Will also show how you can do the calculations. DOE mini-course, part 2, Paul Funkenbusch, 2015 18
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