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quality engineering

  1. 1. M6141 QUALITY ENGINEERING CONTENTS 1. Introduction DESIGN OF EXPERIMENTS 2. Fundamentals of Probability and Statistics (cancelled) A/P Wu Zhang 3. Single Factor Experiments MAE, NTU 4. Randomized Blocks (cancelled) Office: N3.2-02-14 Tel: 6790 4445 5. Factorial Experiments Email: mzwu@ntu.edu.sg 6. 2k Experiments 7. Fractional ExperimentsTEXT 8. A Case Study Montgomery, D. C., (2009), Design and Analysis of Experiments, John Wiley & Sons. 1-1 1-21 INTRODUCTION Example 1.1: Two factors: Machine, Feed rate Response: Process capabilityA designed experiment is a test or series of tests, in which, theexperimenter chooses certain factors for study. He purposely varies Table 1.1those factors in a controlled fashion, and then observes the effect of Process capability Cpsuch factors on the response function. Feed rate (mm/sec.) Machine 0.2 0.4 0.6 USA 2.1, 2.2 1.7, 1.6 1.2, 1.3 Japan 0.9, 1.0 0.7, 0.6 0.4, 0.3 Conclusion: using USA machine at lowest feed rate will result in max. process capability. 1-3 1-4
  2. 2. Factor (input variable) (3) Run: A run is one set of levels for all factors in an experiment. In above experiment, there are three factors: (A) temperature,(1) Types (B) machine and (C) switch. The total number of runs is Quantitative: Temperature, time: They can be set at any value. R = 4 × 3 × 2 = 24 Qualitative: Machines, operators: They have different categories Switch: It has different status For example, [1] A = 10, B = Japan-made, C = on ..........(2) Levels [24] A = 40, B = UK-made, C = off Temperature (oC): 10, 20, 30, 40 [4 levels] Machine: made by Japan, by USA, by UK [3 levels] For single factor experiments, run is equivalent to level of the factor. The number R of runs is equal to the number a of levels. Switch: on, or off [2 levels] (4) Replicates (n): number of observations at each run. A larger value of n increases the accuracy of the experimental results, but makes the experiment more expensive. 1-5 1-6Run A B C Run A B C Response (or output, observation)No Temp. Machine Switch No Temp. Machine Switch 1 10 Japan On 13 30 Japan On Response is dependent on the values of the factors. 2 10 Japan Off 14 30 Japan Off 3 10 USA On 15 30 USA On Types of response: 4 10 USA Off 16 30 USA Off 5 10 UK On 17 30 UK On Larger the better: product yield, process capability, student’s 6 10 UK Off 18 30 UK Off mark 7 20 Japan On 19 40 Japan On Smaller the better: product cost, number of errors 8 20 Japan Off 20 40 Japan Off 9 20 USA On 21 40 USA On Nominal the best: diameter of a shaft 10 20 USA Off 22 40 USA Off 11 20 UK On 23 40 UK On By DOE, we can find out which factors have effects on the response, 12 20 UK Off 24 40 UK Off and the direction of the influence (i.e., whether the increase of a factor will increase or decrease the response ?) 1-7 1-8
  3. 3. Example 1.2: Single factor: A text book (a = 2, n = 3, R = 2) Example 1.3: Two factors: A text book, B studying hours Response: Students’ marks Response: Students’ marks Table 1.2 (a = 2, b = 3, n = 2, R = 6) Students’ marks Table 1.3 Text 1 76, 74, 78 Ave = 76 Students’ marks Text 2 48, 50, 52 Ave = 50 1 hr 2 hr 3 hr Text 1 45, 47 59, 58 76, 75 Text 2 32, 33 51, 54 70, 71 Text book (a qualitative factor) Text 1: book by Montgomery Text 2: book by Taguchi 1-9 1-10The objectives of an experiment may include What Makes Designed Experiment (DOE) Special?(1) Determining which factors xs are most influential on the response, (1) It conducts the experiments in a systematic and efficient way. y. Reduce design and development time, as well as the cost. (2) It presents experiment results in the simplest and clearest way.(2) Determining where to set the influential xs, so that y is near the desired requirement. (3) It extracts the maximum amount of information from a given set of experiments.(3) Determining where to set the influential xs, so that variability in y is small. (4) It draws the right conclusions, despite the variability presented in the data. Examples: Increase process capability Improve students’ study Reduce output variability. 1-11 1-12
  4. 4. Example 1.4 Reducing Defect Level in a Process In this application, DOEA manufacturing engineer is going to apply DOE to a process for (1) Determines which factors affect the occurrence of defectssoldering electronic components to printed circuit boards, in order toreduce defect levels even further. (2) Determines the direction of adjustment for the effective factors toFactors to be investigated: further reduce the number of defects per unit. Solder temperature Preheat temperature (2) Decides whether changing the factors together produces different Conveyor speed results than just adjusting individual factor separately? Flux type Flux specific gravity Solder wave depth Conveyor angle Thickness of the printed circuit board Types of components used on the board 1-13 1-14
  5. 5. 3 SINGLE FACTOR EXPERIMENTS Table 3.1 Tensile strength of paper ___________________________________________________________________________________________________________ Hardwood Observations3.1 Introduction _____________________________________________________________________________________________ Concen. (%) 1 2 3 4 5 6 Total Aver. _________________________________________________________________________________________________________________________________Single factor experiment can be used in the situation where the effect(influence) of a single factor on the response is dominant. The effects 5 7 8 15 11 9 10 60 10.00of all other factors are negligible. 10 12 17 13 18 19 15 94 15.67The procedures and formulae developed for the single factor 15 14 18 19 17 16 18 102 17.00experiments can be easily modified, and then used for the several 20 19 25 22 23 18 20 127 21.17factors experiments. __________________________________________________ 383 15.96 Study the effect (influence) of hardwood concentration on the tensile strength of paper. 3-1 3-2(1) One factor: hardwood concentration. Table 3.3 General data table for a single factor experiment(2) Four levels: 5%, 10%, 15%, 20%: a = 4. Level Observation Total Average 1 y11 y12 . . y1n y1. y1.(3) Response is the tensile strength of paper. 2 y21 y22 . . y2n y2. y2.(4) Each level has six observations or replicates: n = 6. . . . . . .(5) Total number of observations: N = an = 24. . . . . . . a ya1 ya2 . . yan ya. ya. y.. y.. yij the jth observation taken under the ith level of the factor. 3-3 3-4
  6. 6. yi. the total of the observations at the ith level 3.2 The Analysis of Variance (ANOVA) n ANOVA is the most important statistical tool used in DOE. Its task yi. = ∑ yij (3.1) j =1 is to decide whether and which factors or interactions have significant effect (influence) on the response.yi. the average of the observations at the ith level yi. = yi. / n (3.2) Model of ANOVA Yij = μ + τ i + ε ij = μi + ε ij (3.5)y.. the grand total of all observations a n a y.. = ∑∑ yij = ∑ yi. (3.3) μ the overall mean. μ ≈ y.. (mean ≈ average) i =1 j =1 i =1 μi the ith level mean. μ i ≈ yi.y.. the grand average of all observations. τi the ith level effect. τ i = μi − μ ≈ yi. − y.. y.. = y.. / N (3.4) ε ij the random error. ε ij = yij − μ i ≈ yij − yi.N = an is the total number of observations 3-5 3-6Machining Example (1) y1. = 2.1 + 2.0 + 1.9 = 6.0, y1. = y1. / 3 = 2.0Study the effect of machine on the process capability y 2. = 0.9 + 1.0 + 1.1 = 3.0, y 2. = y 2. / 3 = 1.0 Table 3.4 y.. = (2.1 + 2.0 + 1.9) + (0.9 + 1.0 + 1.1) = 9.0 Machine Process capability Cp y.. = 9.0 / 6 = 1.5 USA 2.1, 2.0, 1.9 Japan 0.9, 1.0, 1.1 μ1 ≈ y1. = 2.0, μ 2 ≈ y 2. = 1.0 μ ≈ y.. = 1.5(1) One factor: machine (qualitative). τ1 = μ1 − μ ≈ y1. − y.. = 2.0 − 1.5 = 0.5(2) Two levels: USA-made, Japan-made, a = 2. τ 2 = μ 2 − μ ≈ y2. − y.. = 1.0 − 1.5 = −0.5(3) Response is the process capability.(4) Each level has 3 observations or replicates: n = 3.(5) Total number of observations: N = an = 2 × 3 = 6. 3-7 3-8
  7. 7. Evaluation of Errors y ε11 = 0.1 For instance, if i = 1, j = 1, yij = y11 = 2.1 ε ij = yij − μi μ1 = 2 τ1 = 0.5 ε11 = y11 − μ1 ≈ y11 − y1. = 2.1 − 2.0 = 0.1 μ = 1.5 y11 = 2.1 τ2 = -0.5 Errors for the machining example (1) μ2 = 1 Table 3.5 Machine Errors USA 0.1, 0.0, -0.1 y11 = μ + τ1 + ε11 Japan -0.1, 0.0, 0.1 = 1.5 + 0.5 + 0.1 = 2.1 Figure 3.2 3-9 3-10 μ1 μ2 μ1 τ1 τ2 τ1μ1 μ2 μ1 μ2 μ3 μ μ τ1 τ2 τ3 μμ3 μ4 μ τ2 τ3 τ4 μ2 μ3 μ4 τ3 τ4 τ4 μ3 μ4 μ4 Figure 3.3 Figure 3.4 3-11 3-12
  8. 8. Hypothesis Test: to test if a factor has effect on the response. Null: Ho : the factor has no effect on the responseIf a factor has no effect on the response, the response is always the Alternative: H1: the factor has effect on the responsesame, regardless the change of the factor, Ho : τ1 = τ 2 =L = τ a = 0 (|τi| is small, no effect) μ1 = μ 2 = L = μ a = μ H1 : τ i = o for at lease one i / (|τi| is larger, has effect) (3.6) ( μ1 − μ ) = ( μ 2 − μ ) = L = ( μ a − μ ) = 0 τ 1 = τ 2 = Lτ a = 0 The larger the value of any |τi|, the more effective the factor will be. Here, the magnitude of τi, rather than its sign, makes sense.If a factor has effect, the value of the response will be different, alongwith the change of the factor. The more the response differs, the Type I error: the null hypothesis Ho is rejected when it is actuallygreater the effect of the factor is. true (The factor is concluded effective, when it is not) τ i = o for at least one i / Type II error: the null hypothesis is accepted when it is actually false (The factor is concluded ineffective, when it is effective). 3-13 3-14Machining Example (2) y Table 3.6 Machine Process capability Cp τ1 = τ2 = 0 USA 1.6, 1.5, 1.4 μ = μ1 = μ2 =1.5 Japan 1.4, 1.5, 1.6 Minor variation is caused by errors, rather than the machines μ1 ≈ 1.5, μ 2 ≈1.5 μ ≈ 1.5 τ 1 = μ1 − μ = 15 − 15 = 0 . . τ 2 = μ2 − μ = 15 − 15 = 0 . .Conclusion: in this example, the factor (machine) has no effect on the process capability. Because, when using different Figure 3.5 machines, the response is always the same, on average. 3-15 3-16
  9. 9. Analysis of Variance (ANOVA) (1) Calculate sum of squares (SS)Major Steps Three types of differences: [1] Difference between an individual observation and the grand(1) Calculate sum of squares (SS). average: yij - y..(2) Determine degrees of freedom (D.O.F.). [2] Difference between a level average and the grand average:(3) Calculate mean square (MS). yi. - y.. = τ i(4) Calculate the test ratio F0. [3] Difference between an individual observation and the corresponding level average:(5) Make conclusion. yij - yi. = ε ij Note: yij - y.. = ( yi. - y.. ) + ( yij - yi. ) 3-17 3-18 a n a n a n Sum of squares is the sum of squares of the above three differences.∑∑ ( yi =1 j =1 ij - y.. ) ∑∑ ( yi. - y.. ) i =1 j =1 ∑∑ ( y i =1 j =1 ij - yi. ) SST The total sum of squares, which is a measure of total a n a n a n variability in the data.∑∑ ( y - y.. ) ∑∑ ( yi. - y.. ) ∑∑ ( y - yi . ) 2 2 2 ij iji =1 j =1 i =1 j =1 i =1 j =1 a n SS T = ∑ ∑ ( yij - y.. ) 2 (3.7) i =1 j =1 SST = SSF + SSE SSF the sum of squares due to the factor, that is the sum of squares of differences between factor level averages and the grand average (difference between levels). a n a a SS F = ∑∑ ( yi. − y.. ) 2 = n∑ ( yi. − y.. ) 2 ≈ n∑ τ i2 (3.8) i =1 j =1 i =1 i =1 3-19 3-20
  10. 10. SSE the sum of squares due to error, that is the sum of squares Alternative Formulae of Calculating the Sum of Squares of differences between observations and their level averages (difference within levels). a n y . .2 a n a n SS T = ∑ ∑ yij 2 - i =1 j =1 an (3.11) SS E = ∑∑ ( yij − yi. ) 2 ≈∑∑ ε ij 2 (3.9) i =1 j =1 i =1 j =1 a yi. 2 y . .2 SS F = ∑ n an i =1 - (3.12) SS T = SS F + SS E (3.10) SS E = SST - SS F (3.13) Formulae (3.7) to (3.9) are used to describe the underlying ideas. Formulae (3.11) to (3.13) are used for actually computation. 3-21 3-22Machining Example (1) If SSF is large, it is due to differences among the means at the different factor levels. See equation (3.8), a large SSF means that τ2i Table 3.7 (or |τi|) is great. It is an indication that the factor has significant effect Machine Process capability Cp on the response. USA 2.1, 2.0, 1.9 Usually, SSF is standardized by taking SSF / SSE . By comparing SSF Japan 0.9, 1.0, 1.1 to SSE, we can see how much variability is due to changing factor levels and how much is due to error. a = 2, n = 3 y1⋅ = 6 y2⋅ = 3 y⋅⋅ = 9 Machining Examples y1⋅ = 2 y2⋅ = 1 y⋅⋅ = 15 . Table 3.8 Effective ? SSF/SSE SST = (2.12 + 2.0 2 + 1.9 2 + 0.9 2 + 1.0 2 + 1.12 ) − 9 2 / 6 = 1.54 Example 1 Yes 37.5 SS F = (6 2 + 32 ) / 3 − 9 2 / 6 = 1.50 Example 2 No 0 SS E = 1.54 − 1.50 = 0.04 3-23 3-24
  11. 11. (2) Determine degrees of freedom (D.O.F.). Machining Example (1) Table 3.9 total number of observations N: an Machine Process capability Cp degrees of freedom of SST: N - 1 = an - 1 (3.14) USA 2.1, 2.0, 1.9 degrees of freedom of SSF: a-1 (3.15) Japan 0.9, 1.0, 1.1 degrees of freedom of SSE: R (n - 1) = a (n – 1) (3.16) a = 2, n = 3 where, (n - 1) is the degrees of freedom of errors in a run. total number of observations N: 2×3=6 degrees of freedom of SST: N–1 =6–1=5Equation of DOF degrees of freedom of SSF: a–1 =2–1=1 an - 1 = (a - 1) + a (n - 1) (3.17) degrees of freedom of SSE: a (n-1) = 2 × (3 - 1) = 4 DOF of SST = DOF of SSF + DOF of SSE Check equation (3.17): 5 = 1 + 4 3-25 3-26(3) Calculate mean square (MS). (4) Calculate the test ratio F0. The ratio between a sum of squares and its DOF. MS F Fo = (3.20) MS E SS F MS F = (3.18) a − 1 If H0 is true, F0 follows a theoretical F distribution, which is completely determined by the two parameters ν1 and ν2. SS E MS E = (3.19) ν1 = a - 1 the D.O.F. of MSF (numerator) a ( n - 1) ν2 = a(n - 1) the D.O.F. of MSE (denominator) There is no need to calculate MS for the SST 3-27 3-28
  12. 12. f(x) (5) Make conclusion. ν1 = 1 ν2 = 4 (1) Decide the curve from SS F ν1 and ν2. MS F a (n − 1) SS F SS (2) Decide Fα,ν1,ν2 from α Fo = = a −1 = ⋅ = Q⋅ F MS E SS E a − 1 SS E SS E a (n − 1) a (n − 1) α Where, Q = is a constant. a −1 SS F Fα ,ν1 ,ν 2 F0 is equivalent to (the standardized SSF) except for a constant Q. SS E Figure 3.6 SS The density function curve of an example of the F distribution However, while F0 follows the theoretical F distribution, F does SS E not. 3-29 3-30 SS F Table 3.10 Control Limit Fα ,ν 1 ,ν 2 found from the F TableIf F0 is large ----> is large ---> SSF is large ----> τ2i is large ---> SS E |τi| is large ---> the factor has significant effect on the response. α = 0.05Control limit: Fα ,ν 1 ,ν 2 ν1 1 2 3 4 5 . ∞ ν2 ν1 = a - 1 and ν2 = a(n - 1) are the two parameters of an F 1 161.4 199.5 distribution 2 18.51 19.00 α is the probability of Type I error, which means concluding that the factor has significant effect on the response when it in fact has 3 10.13 9.55 Examples: no effect. Usually, set α = 1% or 5% 4 7.71 6.94 F0.05,2,1 = 199.5If Fo < Fα ,ν 1 ,ν 2 , conclude that the factor has no effect. . F0.05,1,4 = 7.71 ∞If Fo > Fα ,ν 1 ,ν 2 , conclude that the factor has significant effect. 3-31 3-32
  13. 13. Table 3.11 Table 3.12 3-33 3-34 Table 3.13 Machining Example (1) (a = 2, n = 3, N = 6), from Table 3.7 Analysis of Variance for a Single-Factor Experiment___________________________________________________ Table 3.14 ___________________________________________________Source of Sum of Degrees of Mean Fo Source of Sum of Degrees of Mean FoVariation Squares Freedom Square Variation Squares Freedom Square ______________________________________________________________________________________________________ Machine 1.50 1 (= ν1) 1.50 150Factor SSF a-1 MSF MS F Error 0.04 4 (= ν2) 0.01 MS E Total 1.54 5 ___________________________________________________Error SSE a(n-1) MSETotal SST an-1___________________________________________________ 3-35 3-36
  14. 14. Specify α = 0.05, Control limit Fα ,ν 1 ,ν 2 = F0.05,1, 4 = 7.71 Machining Example (2) (a = 2, n = 3, N = 6), Table 3.15Since F0 > Fα ,ν 1 ,ν 2 , machines have significant effect on the process Source of Sum of Degrees of Mean Fo capability. Variation Squares Freedom SquareSum of squares: 1.50 + 0.04 = 1.54 Machines 0 1 (= ν1) 0 0Degrees of freedom 1 + 4 = 5 Error 0.04 4 (= ν2) 0.01 Total 0.04 5 ___________________________________________________ Specify α = 0.05, Control limit Fα ,ν 1 ,ν 2 = F0.05,1, 4 = 7.71 Since F0 < Fα ,ν1 ,ν2 , machines don’t have significant effect on the process capability. 3-37 3-38 Table 3.16 Example: Tensile strength of paper 4 6 y.. 2___________________________________________________________________________________________________________ SS T = ∑ ∑ yij 2 - Hardwood Observations i =1 j =1 an _____________________________________________________________________________________________Concen. (%) 1 2 3 4 5 6 Total Aver. (383) 2_________________________________________________________________________________________________________________________________ = (7) 2 + (8) 2 + L + (20) 2 - = 512.96 24 5 7 8 15 11 9 10 60 10.00 4 yi.2 y..2 10 12 17 13 18 19 15 94 15.67 SS F = ∑ - i =1 n an 15 14 18 19 17 16 18 102 17.00 (60) 2 + (94) 2 + (102) 2 + (127) 2 (383) 2 = - = 382.79 20 19 25 22 23 18 20 127 21.17 6 24__________________________________________________ SS E = SS T - SS F 383 15.96 = 512.96 - 382.79 = 130.17a = 4, n = 6, N = an = 24. 3-39 3-40
  15. 15. Table 3.17 ANOVA for tensile strength of paper SUMMARY: ANOVA for Single Factor Experiment___________________________________________________________________________________________________________________________________Source of Sum of Degrees of Mean FoVariation Squares Freedom Square 1. Decide the factor to be investigated and the following two__________________________________________________________________________________________________________________________________ parameters:Hardwood 382.79 3 (= ν1) 127.60 19.61 1) number of levels a;concentration 2) replicate n.Error 130.17 20 (= ν2) 6.51 2. Carry out the experiments, obtain N (= an) observations yij by aTotal 512.96 23 random manner.______________________________________________________________________________________________________________________________________Specify α = 0.01, Control limit Fα ,ν 1 ,ν 2 = F0.01,3, 20 = 4.94 3. Check if the residuals (estimates of errors) satisfy the requirements.Since F0 > Fα ,ν 1 ,ν 2 , hardwood concentration has significant effect on the tensile strength of paper. 4. Calculate sum of squares SST, SSF and SSE (3.11, 3.12, 3.13). 3-41 3-425. Calculate the degrees of freedom for each of SST, SSF, SSE (3.14, 3.3 Test on Individual Level Means 3.15, 3.16). ANOVA results in a single parameter F0, which can tell whether a6. Calculate the mean squares MSF and MSE (3.18, 3.19). factor has effect on the response, or whether the mean response values will be different at different levels of the factor from an7. Calculate the ratio F0 between MSF and MSE (3.20). overall viewpoint.8. Specify a type I error α, and find the control value Fα ,ν 1 ,ν 2 from However, ANOVA cannot decide the direction of influence of the the F distribution Tables (ν1 = a - 1, ν2 = a (n – 1) ). factors, nor identify which factor level mean is different from another level mean.9. Conclude if the factor has significant effect on the response. 3-43 3-44
  16. 16. The plots indicate that changing the hardwood concentration has a strong effect on the paper strength; specifically, higher hardwood concentrations produce higher observed paper strength. Box plots show the variability of the observations within a factor level and the variability between factor levels. However, a less subjective approach is required to test the individual level means. Figure 3.7 Box plots of paper strength example 3-45 3-46Multiple Comparison Procedures One of the major problems with making comparisons among level means is that unrestricted use of these comparisons can lead to anIt is the techniques for making comparisons between two or more excessively high probability of a Type I error.level means subsequent to an analysis of variance.When we run an analysis of variance and obtain a significant F0 For example, if we have 10 levels in which the complete nullvalue, we have shown simply that the overall null hypothesis is false. hypothesis is true (with α = .05)We do not know which of a number of possible alternativehypotheses is true. Ho: µ1 = µ2 = µ3 = … = µ10 H1 : µ1 ≠ µ2 ≠ µ3 ≠ µ4 ≠ µ5 A series t tests between all pairs of level means will lead to making atOr H1: µ1 ≠ µ2 = µ3 = µ4 = µ5 least one Type I error 57.8% of the time. In other words, the experimenter who thinks he is working at the (α = .05) level ofMultiple comparison techniques allow us to investigate hypotheses significance is actually working at (α = 0.578).that involve means of individual levels. For example, we might beinterested in whether µ1 is different from µ2, or whether µ2 isdifferent from µ3. 3-47 3-48
  17. 17. The probability of making at least one Type I error increases as we Least Significant Difference (LSD) Test is a useful tool to test theincrease the number of independent t tests we make between pairs of differences between the individual level means.level means. While it is nice to find significant differences, it is notnice to find ones that are not really there. We need to find some way The basic requirement for a LSD test is that the F0 for the overallto make the comparisons we need but keep the probability of analysis of variance (ANOVA) must be significant. If the F0 was notincorrect rejections of Ho under control. significant, no comparisons between level means are allowed. You simply declare that there are no differences among the level means and stop right there. On the other hand, if the overall F0 is significant, you can proceed to make any or all pairwise comparisons between individual level means by the use of a t test. 3-49 3-50The t distributions To test between the ith level mean (µi) and jth level mean (µj), Ho: µi = µj H1: µi ≠ µj (3.21) we calculate yi⋅ − y j⋅ t= (3.22) 2 MS E / n The value of MSE has already been obtained during the overall ANOVA. The DOF of t is equal to the DOF of MSE, i.e., a(n-1) ort(30) is a t distribution with a DOF of 30. (N-a).t(∞) means a standard normal distribution with (µ = 0 and σ = 1). Then, using a two-tailed t test for a specified level of type I error α (e.g., α = 0.05), we can decide a pair of two-sided control limits ±tα(dof) from the t table. 3-51 3-52
  18. 18. If the t value determined by Eq (3.22) falls within the limits ±tα(dof), we cannot reject the null hypothesis in (3.21). We will therefore conclude that there is no difference between µi and µj. On the other hand, if the t value falls beyond the limits ±tα(dof), we will reject the null hypothesis and conclude that there is a difference between µi and µj. 3-53 3-54 The Example of Tensile Strength of Paper Problem: if hardwood concentrations 5% and 20% produce the Table 3.18 same paper strength ?___________________________________________________________________________________________________________ Hardwood Observations Ho : μ 1 = μ 4 _____________________________________________________________________________________________ Hypothesis:Concen. (%) 1 2 3 4 5 6 Total Aver. H1 : μ1 = μ 4 /_________________________________________________________________________________________________________________________________ 5 7 8 15 11 9 10 60 10.00 Five Steps: 10 12 17 13 18 19 15 94 15.67 (1) Use Eq (3.22) 15 14 18 19 17 16 18 102 17.00 20 19 25 22 23 18 20 127 21.17 yi ⋅ − y j ⋅ 10.00 − 21.17 t= = = −7.58__________________________________________________ 2 MS E / n 2 × 6.51 / 6a = 4, n = 6 (Note: MSE = 6.51 was calculated during ANOVA, which isy1⋅ = 10.00, y2⋅ = 15.67, y3⋅ = 17.00, y4⋅ = 21.17 always carried out before the test on individual level means) 3-55 3-56
  19. 19. (2) Determine the DOF, DOF = a (n – 1) = 4 × (6 – 1) = 20 General Steps to Test the Individual Level Means(3) Decide a type I error level. We use α = 0.01 as in the ANOVA for (1) Calculate t using Eq (3.22) and find the MSE value obtained this example. during ANOVA.(4) Find the limits from the t table: ±tα(dof) = ±t0.01(20) ≈ 3.00 (2) Determine the DOF, DOF = a (n – 1).(5) Make conclusion: since the calculated t value (= -7.58) falls (3) Specify a type I error α. beyond the limits ±3.00, we will reject the null hypothesis. (4) Find the control limits from the t table: ±tα(dof) based on α and Conclusion: There is a significant difference in paper strength DOF. between using hardwood concentrations 5% and 20%. Moreover, since t < 0, µ1 < µ4. (5) Make conclusion by comparing the calculated t value against the control limits. 3-57 3-58The Bonferroni Procedure To put this in a way that is slightly more useful to us, if you want the overall family-wise error rate to be no more than (α = 0.05), and youIn a Bonferroni procedure the family-wise error rate (Type I error) is want to run c tests, then each run of them should be at (α = 0.05/c).divided by the number of comparisons. To run these tests, you do exactly what you did in Fishers LSD test, though you omit any requirement about the significance of overall F.The basic idea behind this procedure is that if you run several tests(say c tests), each at a significance level (type I error) represented byα, the probability of at least one Type I error can never exceed cα.Thus, for example, if you ran 5 tests, each at α = 0.05, the family-wise error rate would be at most 5(0.05) = 0.25. But that is a too highType I error rate to make anyone happy.But suppose that you ran each of those 5 tests at the α = 0.01. Thenthe maximum family-wise error rate would be 5(0.01) = 0.05, whichis certainly acceptable. 3-59 3-60
  20. 20. a ni 2 y.. SST = ∑∑ yij −3.5 Unbalanced Experiments 2 DOF = N -1 i =1 j =1 NIn an unbalanced experiment, the number of observations taken undereach run is different. a yi2. y.. 2 SS F = ∑ − DOF = a -1ni is the number of observations taken under the ith run (for i =1 ni N single factor experiment, it is simply the ith level). a a SS E = SST − SS F DOF = ∑ (ni − 1) N = ∑ ni = n1 + n2 + L + na i =1 i =1 is the total number of observations.Note, in unbalanced experiments, N ≠ an. 3-61 3-62Machining Example SST = (2.12 + 2.0 2 + 1.9 2 ) + (0.9 2 + 1.0 2 ) - 7.9 2 / 5 = 1.348 DOET = N – 1 = 5 – 1 = 4 Machine Process capability Cp USA 2.1, 2.0, 1.9 SSF = (6.02/3 + 1.92/2) – 7.92/5 = 1.323 Japan 0.9, 1.0, DOEF = a – 1 = 2 – 1 = 1n1 = 3, n2 = 2, N = 3 + 2 = 5 SSE = 1.348 – 1.323 = 0.025y1. = 6.0, y2. = 1.9, y.. = 7.9 DOEE = (n1 – 1) + (n2 – 1) = (3 – 1) + (2 – 1) = 3 3-63 3-64
  21. 21. ANOVA for Machining Example (a = 2, n1 = 3, n2 = 2, N = 5) 3.7 Guidelines for Designing Experiments (1) Recognition and statement of the problem.Source of Sum of Degrees of Mean Fo Fully develop all ideas about the problem and the specificVariation Squares Freedom Square objectives of the experiment; Solicit input from all concerned parties -- engineering, quality,Machine 1.323 1 (= ν1) 1.323 158.8 marketing, customer, management and operators.Error 0.025 3 (= ν2) 0.0083 (2) Choice of factors and levels.Total 1.348 4 Choose the factors, their ranges and levels. Process knowledge is___________________________________________________ required.Specify α = 0.05, Control limit: Fα ,ν 1 ,ν 2 = F0.05,1,3 = 10.13 Investigate all factors that may be of importance and avoid being overly influenced by past experience.Since F0 > Fα ,ν 1 ,ν 2 , machines have significant effect on the process Keep the number of factor levels low (Most often two levels are capability. used.) in the early stage 3-65 3-66(3) Selection of the response (6) Data analysis Most often, the average or standard deviation (or both) of the Analyze the data so that results and conclusions are objective rather measured characteristic will be the response variable. than judgmental. Use software packages and simple graphical methods(4) Choice of experimental design. Carry out residual analysis. Decide the factorial fraction experiment, select the replicate (sample size) and run order for the experimental trials. (7) Conclusions and recommendations. Decide whether or not blocking or other randomization methods are involved. (8) Experimentation is an important part of the learning process. New hypotheses are formulated based on the investigation of the(5) Performing the experiment. tentative hypotheses. Don’t design a single, large comprehensive experiment at the start of a study. As a rule of thumb, the first Ensure that everything is being done accordingly to plan. experiment should spend no more than 25% of the total budget of the study. 3-67 3-68
  22. 22. 4 RANDOMIZED BLOCKS Response: coded roughness of a machined surface.4.1 Randomized Block Design Main factor (factor to be studied in the experiment)Nuisance factor probably has an effect on the response, but we are Cutter of a machine toolnot interested in that effect. Known nuisance factor (factor not to be studied in the experiment,For unknown and uncontrollable nuisance factors, use Complete but we are well aware of the existence of its effect on the response)Randomization technique to get rid of them. Different machine tools of the same typeFor known and controllable nuisance factors, use RandomizedBlocking technique instead. Unknown nuisance factor (unknown and uncontrollable factors, which may have some effect on the response)The variability of the nuisance factor will contribute to the variabilityobserved in the experimental data. As a result, the experimental error Temperature, humiditywill reflect both the random error of the unknown nuisance factorsand variability of the known nuisance factor. 4-1 4-2 Table 4.1 MS F MS F DOEE F0 = = = MS F (4.4) Machine tool (Block) MS E SS E SS E Cutter 1 2 3 4 DOEE 1 -2 -1 1 5 2 -1 -2 3 4 So, whether F0 will be increased by the addition of the known 3 -3 -1 0 2 nuisance factors depends on which of SSE or DOEE will increase 4 2 1 5 7 more. If SSE increases more by including the known nuisance factors, F0SST = SS F + SS known + SSunknown (4.1) becomes less sensitive. Therefore, Randomized Blocking technique is preferred.If the effects of the known nuisance factors are included in the error If DOEE increases more, F0 becomes more sensitive. Therefore, Complete Randomization technique is preferred.SS E = SS known + SSunknown (4.2) Note, the increase of DOEE also reduces the control limit Fα ,ν 1 ,ν 2 (ν2DOEE = DOEknown + DOEunknown (4.3) = DOEE), and therefore, makes F0 relatively more sensitive. 4-3 4-4
  23. 23. Randomized Block Design Block 1 Block 2 Block b It means blocking all the runs according to the different levels of a known nuisance factor. Within a block, the order in which the y11 y12 y1b runs of the main factors are tested is randomly determined. y21 y22 y2b The blocks form a more homogeneous experimental unit in terms of the known nuisance factor. ............. y31 y32 y3b Effectively, this design strategy improves the accuracy of the . . . comparisons of the responses by eliminating the variability of the . . . known nuisance factor. . . . ya1 ya2 yabExamples of blocks Equipment, machinery, batches of raw material, people, time slot Figure 4.1 4-5 4-6The statistical model Hypotheses of interest are ⎧ i = 1,2,L, a H0: μ1 = μ2 = = … μa yij = μ + τ i + β j + ε ij ⎨ (4.5) ⎩ j = 1,2,L, b H1: at least one μi ≠ μj (4.6)where, μ is an overall mean, τi is the effect of the ith run, βj is the Oreffect of the jth block, and εij is the usual NID(0, σ2) random error H0: τ1 = τ2 = … = τa = 0term. H1: at least one τi ≠ 0 (4.7) βj will not be studied, because it is the effect of the known nuisance factor. 4-7 4-8
  24. 24. yi. the total of all observations taken under run i (total of row) yi. the average of the observations taken under the ith runy.j the total of all observations in block j (total of column) y. j the average of the observations in block jy.. the grand total of all observations y.. the grand average of all observations.N = ab the total number of observations. b yi. = yi . / b y. j = y . j / a y.. = y.. / N (4.9) yi. = ∑ yij i = 1,2,L, a j =1 a y. j = ∑ yij j = 1,2,L, b (4.8) i =1 a b y.. = ∑∑ yij i =1 j =1 4-9 4-10Sum of square Degrees of Freedom SST = SS F + SS Block + SS E (4.10) SST N - 1 = ab - 1 a b y..2 SSF a-1 SST = ∑∑ yij 2 - (4.11) i =1 j =1 N SSBlocks b-1 a 2 1 y.. SS F = ∑ yi.2 - N b i=1 (4.12) SSE (ab − 1) − (a − 1) − (b − 1) = ab − a − b + 1 (4.15) 1 b 2 y..2 SS Block = ∑ y. j - (4.13) a j =1 N SS E = SST - SS F − SS Block (4.14) 4-11 4-12
  25. 25. Test Statistic Table 4.2 ANOVA Table _______________________________________________________ MS F Source of Sum of Degrees of Mean F0 F0 = MS E Variation Squares Freedom Square SS F _______________________________________________________ MS F = (4.16) a −1 SS F MS F Factor SS F a -1 Fo = SS E a −1 MS E MS E = ab − a − b + 1 SS Blocks Blocks SSBlocks b–1We would reject Ho in (4.7), if Fo > Fα,(a-1)(ab-a-b+1). b −1 SS E Error SS E ab – a – b + 1 ab − a − b + 1 Total SST N -1 _______________________________________________________ 4-13 4-14We may also examine the ratio of MSBlocks (= SSBlocks/(b-1)) to MSE. If Example 4.1this ratio is large, it implies that the blocking factor has a large effect Table 4.3and that the noise reduction obtained by blocking was probably Machine tool (Block) Cutter 1 2 3 4 yi.helpful in improving the effectiveness of the experiment. Otherwise, 1 -2 -1 1 5 3blocking may actually be unnecessary, i.e, the known nuisance factor 2 -1 -2 3 4 4should be handled as random error. 3 -3 -1 0 2 -2 4 2 1 5 7 16 y.j -4 -3 9 18 y.. = 20 4-15 4-16
  26. 26. The sums of squares are obtained as follows: Table 4.4 ANOVA Table _______________________________________________________ a b y..2 20 2 Source of Sum of Degrees of Mean F0 SST = ∑∑ yij 2 - = 154 − = 129 i =1 j =1 N 16 Variation Squares Freedom Square _______________________________________________________ = (3 + 4 + (−2) + 15 ) − 1 a 2 y..2 1 2 20 2 SS F = ∑ yi. - 2 2 2 = 38.5 Cutter 38.50 3 12.83 14.44 b i=1 N 4 16 Machine 82.50 3 27.50 ∑ y.2j - N = 4 ((−4) 2 + (−3) 2 + 92 + 182 ) − 16 = 82.5 b 2 2 (Block) 1 y.. 1 20 SS Block = a j =1 Error 8.00 9 0.89 SS E = SST - SS F − SS Block = 129 − 38.5 − 82.5 = 8 Total 129.00 15 _______________________________________________________ Specify α = 0.05, F0.05,3,9 = 3.86. Since 14.44 > 3.86, we conclude that the type of cutter affects the mean hardness reading. 4-17 4-18If the randomized block design is not used, F0.05,3,12 = 3.49, thehypothesis of equal mean values from the four cutters cannot berejected. Table 4.5 ANOVA TableSource of Sum of Degrees of Mean F0Variation Squares Freedom SquareCutter 38.50 3 12.83 1.70Error 90.50 12 7.54Total 129.00 15Here, the error is actually the sum of error and block in Table 4.4. 4-19
  27. 27. 5 FACTORIAL EXPERIMENTS Case one Two factors: A: Machine, B: Feed rate5.1 Introduction Response: Process capabilityFactorial experiments study the effects of several factors on theresponse. Table 5.1 Process capability CpIn factorial experiments, all possible runs (combinations of the levels Feed rate (mm/sec.)of the factors) are investigated. Machine 0.2 0.4 0.6For example, for two factors A: temperature (with a levels) and B: USA 2.1, 2.2 1.7, 1.6 1.2, 1.3humidity (with b levels), the factorial experiment has ab possible Japan 0.9, 1.0 0.7, 0.6 0.4, 0.3runs. a = 2, b = 3, n = 2, R =ab = 6If for each run, n observations are taken, then, the total number N of N = Rn = abn = 2 × 3 × 2 = 12the observations is equal to abn. 5-1 5-2Case two 5.2 ANOVA for Two-Factor Factorial Experiments Three factors: A, B and C. Mathematical model a = 2, b = 3, c = 4, n = 2, R = abc = 24 Yijk = μ + τ i + β j + (τβ )ij + ε ijk (5.1) N = Rn = abcn = 2 × 3 × 4 × 2 = 48 μ the overall mean τi the effect incurred at the ith level of factor ANumber of runs is equal to the product of the levels of all factors. βj the effect incurred at the jth level of factor B (τβ)ij the effect incurred at the interaction between the ith level of A and the jth level of B εijk the random error (normally and independently distributed, zero mean and constant variance) 5-3 5-4
  28. 28. Interaction indicates the influence of one factor on the effect of Table 5.2 Data Arrangement for a Two-Factor Factorial Design another factor, and vice versa. Factor B 1 2 … b Total AverageFactors A: temperature 1 y111, y112, y121, y122, … y1b1, y1b2, y1.. y1.. B: pressure Factor A …, y11n …, y12n …, y1bn 2 y211, y212, y221, y222, … Y2b1, y2b2, y2.. y 2..Response Process yield …, y21n …, y22n …, y2bn : … … … … … …If: (1) Increasing temperature alone cannot increase the yield; and a ya11, ya12, ya21, ya22, … yab1, yab2, ya.. y a.. (2) Increasing pressure alone cannot increase the yield; and …, ya1n …, ya2n …, yabn (3) Increasing temperature and pressure at the same time does Total y.1. y.2. … y.b. y… increase the yield Average y .1. y .2. … y .b. y ...Then: There is interaction between temperature (factor A) and yijk is the kth observation taken at the ith level of factor A and the jth pressure (factor B) level of factor B. The total number of observations: N = abn 5-5 5-6Total and average at the ith level of factor A: Hypotheses b n y yi .. = Σ Σ yijk y i .. = i .. i = 1,2,...,a Factor A: j =1 k =1 bn (5.2) H o : τ 1 = τ 2 = ... = τ a = 0 Factor A has no effect on yTotal and average at the jth level of factor B: a n y. j . H 1 : at least one τ i ≠ 0 Factor A has effect on y y. j . = Σ Σ yijk y. j. = j = 1,2,..., b i =1 k =1 an (5.3) Factor B:Total and average when A at the ith level and B at the jth level: H 0 : β1 = β2 =L = βb = 0 Factor B has no effect on y n y H 1 : at least one β j ≠ 0 yij . = Σ yijk y ij . = ij . Factor B has effect on y k =1 n (5.4) Interaction AB:Grand total and average a b n y... H o : (τβ ) ij = 0 for all i, j Interaction AB has no effect y y...= ΣΣΣ yijk y...= (5.5) i =1 j =1 k =1 abn H1 : at least one (τβ ) ij ≠ 0 Interaction AB has effect on y 5-7 5-8

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