Planejamento e Otimização de Experimentos

Transcription

1 Planejamento e Otimização de Experimentos The Analysis of Variance Prof. Dr. Anselmo E de Oliveira anselmo.quimica.ufg.br anselmo.disciplinas@gmail.com

2 An Example Integrated circuits Wafers Plasma etching process is employed to remove unwanted material Energy is supplied by a radiofrequency (RF) generator causing plasma to be generated in the gap between the electrodes RF power setting etch rate C 2 F 6 and 0.80 cm (gap) RF power: 160, 180, 200, 220 W (four levels) Five wafers at each RF level

3 Single-factor experiment a = 4 levels of the factor n = 5 replicates 20 runs Random order How to generate the run order?» Ex: random numbers using a random function in a spreadsheet; sort by that number Prevent the effects of unknown nuisance variables» Nonrandomized order ( 5x160; 5x180...) Ex: the etching tool exhibits a warm-up effect: the longer it is on, the lower the observed etch rating readings will be It will destroy the validity of the experiment

4 Power (W) Totals Averages > load etch.oc > whos > etch > boxplot(etch) > tics ("x", 1:4, {"160"; "180"; "200";"220"}); > xlabel ("Power (W)"); ylabel("etch rate (A/min)");

5 Etch rate increases as the power setting increases. There is no strong evidence to suggest that the variability in each rate around the average depends on the power setting. Suppose that we wish to test for differences between the mean etch rates: t-test for all six possible pairs of means. It takes a lot of effort It inflates the type I error (rejecting the null hypothesis when it is in fact true) Analysis of variance

6 The Analysis of Variance a treatments or different levels of a single factor Treatment (Level) Observations Totals Averages 1 y 11 y y 1n y 1. y 1. 2 y 21 y y 2n y 2. y a y a1 y a2... y an y a. y a. y.. y..

7 (Linear Statistical) Models for the Data Means model i = 1,2,, a y ij = μ i + ε ij j = 1,2,, n y ij is the ijth observation μ i is the mean of the ith factor level or treatment ε ij is the random error It is convenient to think of the errors as having mean zero, so that E y ij = μ i Effects model μ i = μ + τ i, i = 1,2,, a y ij = μ + τ i + ε ij μ is the overall mean τ i is the ith treatment effect i = 1,2,, a j = 1,2,, n one-way or single-factor analysis of variance (ANOVA)

8 The experimental design is a completely randomized design The experiments must be performed in random order The environment in which the treatments are applied is as uniform as possible (experimental units) Hypotesis test y ij ~N μ + τ i, σ 2 Observations mutually independent

9 The Analysis of Fixed Effects Model a treatments Test hypotheses about the treatment means, and the conclusions will apply only to the factor levels considered in the analysis y i. = y.. = n j=1 a i=1 y ij n j=1 y ij y i. = y i. n y.. = y.. N i = 1,2,, a N = an the total number of observations

10 Testing the equality of the a treatment means E y ij = μ + τ i = μ i, i = 1,2,, a H 0 : μ 1 = μ 2 = = μ a H 1 : μ i μ j for at least on pair i, j μ = a i=1 a μ i a τ i = 0 i=1 H 0 : τ 1 = τ 2 = = τ a = 0 H 1 : τ i 0 for at least one i the treatment or factor effects can be thought of as deviations from the overall mean Thus, we speak of testing the equality of treatment means or testing that the treatment effects (the τ i ) are zero

11 Decomposition of the Total Sum of Squares The name Analysis of variance is derived from a partitioning of total variability into its component parts a n SS T = y ij y.. 2 i=1 j=1 a = n y i. y.. 2 i=1 sum of squares of the differences between the treatment averages and the grand mean SS T = SS Treatments + SS E a n + y ij y i. 2 i=1 j=1 sum of squares of the differences of observations within treatments from the treatment average

12 Degrees of freedom SS T : N 1 SS Treatments : a 1 SS E : a n 1 = an a = N a Mean squares MS Treatments = SS Treatments a 1 MS E = SS E N a Expected values E MS E = σ 2 E MS Treatments = σ 2 + n 2 i=1 τ i a 1 a if there are no differences in treatment means (τ i = 0), MS Treatments also estimates σ 2

13 Statistical Analysis If the null hypothesis of no difference in treatment means is true, the ratio F 0 = SS Treatments a 1 SS E N a = MS Treatments MS E is distributed as F with a 1 and N a degrees of freedom. Reject H 0 if F 0 > F α,a 1,N a

14 ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Between treatments SS Treatments = 1 n a i=1 y i. 2 y.. 2 N Mean Square a 1 MS Treatments F 0 Error (within treatments) SS E = SS T SS Treatments N a MS E F 0 = MS Treatments MS E Total a n SS T = y ij y.. 2 i=1 j=1 N 1

15 Power (W) Totals y i. Averages y i a n SS T = y ij y.. 2 i=1 j=1 4 5 = y ij 2 y.. 2 i=1 j=1 = ,3552 = 72, SS Treatments = 1 n 4 i=1 y i. 2 y.. 2 = ,3552 = 66, SS E = SS T SS Treatments = N N y.. =12,355 y.. =

16 ANOVA Table Source of Variation Sum of Squares Degrees of Freedom Mean Square RF Power (between-treatment mean square) Error (within-treatment) it is unlikely that the treatment means are equal F 0.05,3,16 = 3.24 (F table) Because F 0 = > 3.24, we reject H 0 and conclude that the treatment means differ F 0 p-value RF Power 66, , <0.01 Error ,70 Total 72, The RF power setting significantly affects the mean etch rate

17 > anova (etch)

18 Estimation of the Model Parameters y ij = μ + τ i + ε ij μ = y.. τ = y i. y.. y ij = y i. A 100(1 a) percent confindence interval on the ith treatment mean μ i : y i. t α 2,N a MS E n μ MS E i y i. + t α 2,N a n A 100(1 a) percent confindence interval on the difference in any two treatment means: y i. y j. t α 2,N a 2MS E n μ i μ j y i. y j. + t α 2,N a 2MS E n

19 Model Adequacy Checking The Normality Assumption Normal probability plot of the residual cd octavedocs load etch.oc etch m = mean(etch) one = ones(5,4) M = one.* m e = M etch normplot(e(:)) y i. y ij = y i. e ij = y i. y ij

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21 Standardized residuals Outliers e ij d ij = MS E If the erros are N 0, σ 2, the standardized residuals should be approximately normal with mean zero and unit variance ~68% of d ij should fall within the limits 1 ~95%... 2 ~100%... 3 Using the largest standardized residual: sort(e(:)) d = 25.6/sqrt(333.7) d = should cause no concern

22 Plot of Residuals in Time Sequence Plotting the residuals in time order of data collection is hepful in detecting strong correlation between the residuals A tendency to have runs of positive and negative residuals indicates positive correlation: The independence assumption of the errors would have been violated

23 plot(etch_run(:),e(:),"*") xlabel ("Run order or time"); ylabel("residuals");

24 Plot of Residuals Versus Fitted Values If the model is correct and the assumptions are satisfied, the residuals should be structureless; in particular, they should be unrelated to any other variable including the predicted response Plasma etching experiment y ij = y i. e ij = y i. y ij

25 plot(m(:),e(:),"*") xlabel ("Predicted"); ylabel("residuals"); There is not inequality of variance: the variance of the observations does not increase as the magnitude of the observation increases

26 Statistical Test for Equality of Variance (in each treatment) H 0 : σ 2 1 = σ 2 2 = = σ 2 a = 0 H 1 : σ 2 i 0 for at least one i Residual plots Bartlett s test χ 2 distribution with a 1 degrees of freedom a random samples from independent normal populations χ 0 2 = q c q = N a logs p 2 n i 1 c = a 1 a i=1 a i=1 logs i 2 n i 1 1 N a 1 a 2 S 2 i=1 n i 1 S i p = N a and S 2 i is the sample variance of the ith population. 2 The quantity q is large when the sample variances S i differ greatly and is equal to zero when all S 2 i are equal. Reject H 0 χ > χ α,α 1

27 var_p=sum(4/(20-4)*var(etch)) q = (20-4)*log10(var_p)-4*sum(log10(var(etch))) c = 1+1/(3*(4-1))*(4*1/4-1/(20-4)) chi_0=2.3026*q/c The tests return a p-value that describes the outcome of the test. Assuming that the test hypothesis is true, the p-value is the probability of obtaining a worse result than the observed one. Large p-values corresponds to a successful test. Usually a test hypothesis is accepted if the p-value exceeds chi2cdf(chi_0,3) ans = we cannot reject the null hypothesis there is no evidence to counter the claim that all four variances are the same (plot residuals x fitted values)

28 Practical Interpretation of Results After conducting the experiment, performing the statistical analysis, and investigating the underlying assumptions, the experimenter is ready to draw practical conclusions about the problem he or she is studying

29 A Regression Model The factors involved in an experiment can be either quantitative or qualitative Empirical model of the process An interpolation equation for the response variable in the experiment Regression analysis Method of least squares

30 quadratic model linear model

31 Comparing Pairs of Treatment Means Tukey s Test Following an ANOVA in which the null hypothesis of equal treatment mean was reject Test all pairwise mean comparisons (all i j) H 0 : μ i = μ j H 1 : μ i μ j Equal or unequal (Tukey-Kramer procedure) sample sizes Distribution of the studentized range statistics, q table T α = q α a, f MS E n

32 Plasma etching experiment a = 4 α = 0.05 f = 16 degrees of freedom N a MS E = n = 5 q ,16 q table q ,16 = T 0.05 = = y 1. = y 2. = y 3. = y 4. = y 1. y 2. = 36.2 y 1. y 3. = 74.2 any pairs of treatment averages that differ in absolute value by more than 33.1 would imply that the corresponding pair of population means are significantly different the Tukey procedure indicates that all pairs of means differ

33 Nonparametric Method in the Analysis of Variance The Kruskal-Wallis Test In situations were the normality assumption is unjustified Reject H 0 2 H > χ α,α 1 [pval, chisq, df] = kruskal_wallis_test(etch(:,1),etch(:,2),etch(:,3),etch(:,4)) pval = e-004 we reject the null hypothesis chisq = df = 3 This is the same conclusion as given by the usual analysis of variance F test