0.1. Statistical Design and Analysis of Experiments Part One. Lecture notes Fall semester Henrik Spliid 0.2. Foreword

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1 0.1 Statistical Design and Analysis of Eperiments Part One Lecture notes Fall semester 007 Henrik Spliid Informatics and Mathematical Modelling Technical University of Denmark Foreword The present collection af lecture notes is intended for use in the courses given by the author about the design and analysis of eperiments. Please respect that the material is copyright protected. The material relates to the tetbook: D.C. Montgomery, Statistical Design and Analysis, 6th ed., Wiley. The notes have been prepared as a supplement to the tetbook and they are primarily intended to present the material in both a much shorter and more precise and detailed form. Therefore long eplanations and the like are generally left out. For the same reason the notes are not suited as stand alone tets, but should be used in parallel with the tetbook. The notes were initially worked out with the purpose of being used as slides in lectures in a design of eperiments course based on Montgomery s book, and most of them are still in a format suited to be used as such. Some important concepts that are not treated in the tetbook (especially orthogonal polynomials, Duncan s and Newman-Keuls multiple range tests and Yates algorithm) have been added and a number of useful tables are given, most noteworthy, perhaps, the epected mean square tables for all analysis of variance models including up to 3 fied and/or random factors August 007 : A strict mathematical presentation is not intended, but by indicating some of the eact results and showing eamples and numerical calculations it is hoped that a little deeper understanding of the different ideas and methods can be achieved. In all circumstances, I hope these notes can inspire and assist the student in studying and learning a number of the most fundamental principles in the wonderful art of designing and analyzing scientific eperiments. A major revision was carried out. No new material, but (hopefully) better organized. In part 11 a new and very easy way of computing epected mean squares (EMS) is introduced. Henrik Spliid August 007 The present version is a revision of the previous (003) notes. Some of the material is reorganized and some additions have been made (sample size calculations for analysis of variance models and a simpler calculation of epectations of mean squares (005)). July 004 A moderate revision has been made in January 006 in which, primarily, the page references have been changed to the 6th edition of Montgomery s tetbook. January 006 A larger revision was undertaken in August 006. The format is now landscape. A number of slides I considered less important have been taken out. I hope this has clarified the subjects concerned. c Henrik Spliid, IMM, DTU

2 List of contents 1.1: Introduction, ANOVA 1.6: A weighing problem (eample of DE) 1.10: Some elementary statistics (repetition) 1.16: The paired comparison design 1.19: Analysis of variance eample 1.5: Designs without or with structure (after ANOVA analyses) 1.31: Patterns based on factorial designs 1.33: Polynomial effects in ANOVA List of contents, cont..1: Multiple range and related methods..3: LSD - Least significant differences.5: Newman-Keul s (multiple range) test.7: Duncan s multiple range test.9 Comparison: Newman-Keuls versus Duncan s test.11: Dunnett s test.13: The fied effect ANOVA model.14: The random effect ANOVA model, introduction.17: Eample of random effects ANOVA.3: Choice of sample size fied effect ANOVA.4: Choice of sample size random effect ANOVA List of contents, cont..5: Sample size for fied effect model - eample.9: Sample size for random effect model - eample 3.1: Block designs - one factor and one blocking criterion 3.: Confounding (unwanted!) 3.3: Randomization 3.5: Eamples of factors and blocks List of contents, cont. 3.7: Confounding again : Model and ANOVA problems 3.9: Randomization again : Model and ANOVA problems 3.11: The balanced (complete) block design 3.13: The Latin square design 3.15: The Graeco-Latin square design 3.17: Overview of interpretation of slides : The (important) two-period cross-over design 3.1: A little more about Latin squares 3.3: Replication of Latin squares (many possibilities) 3.5: To block or not to block 3.3: Two alternative designs - which one is best? 7 8

3 List of contents, cont. 3.8: Test of additivity (factor and block) 3.30: Choice of sample size in balanced block design 4.1: Incomplete block designs 4.: Small blocks, why? 4.3: Eample of balanced incomplete block design 4.5: A heuristic design (an inadequate design eample) 4.7: Incomplete balanced block designs and some definitions 4.9: Data and eample of balanced incomplete block design (BIBD) 4.10: Computations in balanced incomplete block design 4.11: Epectations and variances and estimation List of contents, cont. 4.1: After ANOVA based on Q-values (Newman-Keuls, contrasts) in BIBD 4.13: Analysis of data eample of BIBD 4.17: Youden square design (incomplete Latin square) 4.1: Contrast test in BIBD and Youden square (eample) : Tables over BIBDs and Youden squares 5.1: An eample with many issues 5.1: Construction of the normal probability plot 5.17: The eample as an incomplete block design Supplement I I.1: System of orthogonal polynomials I.4: Weights for higher order polynomials I.8: Numerical eample from slide Supplement II Determination of sample size - general II.8: Sample size determination in general - fied effects II.11: Sample size determination in general - random effects Supplement III III.1: Repetition of Latin squares and ANOVA Design of Eperiments (DoE) What is DoE? E: Hardening of a metallic item Variables that may be of importance: Factors 1: Medium (oil, water, air or other) : Heating temperature 3: Other factors? Dependent variables: Response 1: Surface hardness : Depth of hardening 3: Others? 11 1

4 Sources of variation (uncertainty) 1: Uneven usage of time for heating : All items not completely identical 3: Differences in handling by operators Mathematical model Y = f(a, B,...)+E Factors (A, B,... etc) How do we study the function f(.). The 5% rule. Input item Hardening process Y 3 Y Responses Y 3 Sources of noise Design of Eperiments Model of process temperatures, heating time, etc. determines Factors in general based on a priori knowledge) Laboratory Number of measurements resources Practical eecution decide Handling and staff Conclusions How are data to be analyzed wanted Which factors are important Which sources of uncertainty are important Estimation of effects and uncertainties Demands: You must have a reasonable model idea and you must have some idea about the sources of uncertainty. Aims: 1) To identify a good model, ) estimate its parameters, 3) assess the uncertainties of the eperiment in general, and 4) assess the uncertainty of the estimates of the model in particular

5 A weighing problem Three items Standard weighing eperiment: Measurement (1) a b c Meaning No with with with item A B C Model for (1) = µ + E 1 responses a = µ + A + E b = µ+b +E 3 c = µ+c +E 4 A B C µ = offset (zero reading) of weighing device A = weight of item A B = weight of item B C = weight of item C E 1,E,E 3 and E 4 are the 4 measurement errors The natural estimates of A, B and C are A = a (1) and the corresponding for B and C An alternative eperiment: (1) ac bc ab No with with with item A and C B and C A and B The alternative weighing design Model of (1) = µ + E 5 responses ac = µ + A + C + E 6 bc = µ + B + C + E 7 ab = µ + A + B + E 8 Which design is preferable and why? A (1) + ac bc + ab = = A + 4 errors Var{Â} =σ E Var{Â } = 4σ E = σ E Conclusion The alternative design is preferable because 1) The two designs both use 4 measurements but ) The second design is (much) more precise than the first design. The reason for this is that In the first design not all measurements are used to estimate all parameters, which is the case in the second design. This is a basic property of (most) good designs. 19 0

6 Some repetition of elementary statistics Factor: Types of mortar with levels Response: Strength of cement Table -1: Portland cement strength Observation Modified Unmodified number Mortar mortar : : : The eperiment represents a comparative (not absolute) study (it assesses differences between types of mortar). Two treatments: the t-test can be applied Two distributions to compare z z z z zzz z zzz z zzz y 1 y z z zz =y 1 z=y Model: Y ij = µ i + E ij = µ + τ i + E ij ; i = {1, } with τ 1 + τ =0 Test of H 0 : µ 1 = µ τ 1 = τ =0 t= (Y 1 Y ) (µ 1 µ ) s 1/n 1 +1/n s = s pooled = (n 1 1)s 1 +(n 1)s n 1 1+n The t-test and the conclusion ± t(18) 0.05 t(18) distribution Analysis of variance for cement data Two levels Two treatments. The test (of the hypothesis of no difference between treatments) can be formulated as an analysis of variance (one-way model): Eample p. 36: Y 1 =16.76, Y =17.9, s =0.84 µ 1 = µ t = /10 + 1/10 = 9.13 Source of SSQ df s F variation value Between treatments Within treatments Total Variation The difference is strongly significant 3 4

7 The reference distribution is an F-distribution: Conclusions to formulated: F(1,18) distribution 95% fractile The t-test and the one-way analysis of variance with two treatments give the same results. The F-value in the analysis of variance is the t-value squared: t (f) F (1,f) Point estimates µ 1 and µ for µ 1 µ σe Confidence intervals µ 1 and µ for µ 1 µ σ E and a suitable verbal formulation of the obtained result Two alternative eperimental designs An alternative design using blocks (items) test item Treatm. AorB Design I : 0 items used Method A Method B Y 1,A Y 1,B Y,A Y,B : : Y 10,A Y 10,B Allocation of treatments to items by randomization Treatm. B part 1 part Treatm. A Design II : 10 items used Item Method A Method B 1 Y 1,A Y 1,B Y,A Y,B : : : 10 Y 10,A Y 10,B Allocation of treatments to the two parts by randomization The method of analysis? Answer: One-way analysis of variance (or t-test) The proper mathematical model is a two-way analysis of variance model. Formulate the two models for designs I and II. Which design is preferred? Why? 7 8

8 Detailed mathematical models Design I : Y i,a = µ A + E i,a + U i,a Y i,b = µ B + E i,b + U i,b Design II : Var{Y A Y B } = σ E +σ U n Y i,a = µ A + E i + U i,a Y i,b = µ B + E i + U i,b Y i,a Y i,b = D i = µ A µ B + U i,a U i,b Var{Y A Y B } = σ U n Conclusion: Design II eliminates the variation between items. Design II is preferable. The analysis is a paired t-test or a two-way analysis of variance with treatments and 10 blocks. Analysis of variance eample What are the problems with this design? Sequence of measurements Factor is % cotton 15% 0% 5% 30% 35% The table displays a systematic sequence of measurements An alternative design: Randomized sequence Factor is % cotton 15% 0% 5% 30% 35% 7 (15) 1 (8) 14 (5) 19 (11) 7 (4) 7 (1) 17 (9) 18 () 5 () 0 (10) 15 (4) 1 (3) 18 (18) (13) 16 (0) 11 (1) 18 (1) 19 (14) 19 (7) 15 (17) 9 (19) 18 (16) 19 (3) 3 (5) 11 (6) The table displays both the data and the random sequence of measurements in (.) What is achieved by randomizing the sequence? Mathematical model for randomized design Y ij = µ + τ j + E ij Factor is % cotton 15% 0% 5% 30% 35% sum Sum Complete randomization assumed 31 3

9 SSQ tot = = SSQ treatm = = ANOVA table for cotton eperiment Source SSQ f s EMS F-value Cotton σ E +5φ τ Residual σ E Total SSQ resid = SSQ tot SSQ treatm = F(4,0) distribution f tot = N 1=5 1=4 f treatm = a 1=5 1=4 f resid = a(n 1) = 5(5 1) = % fractile Conclusion: Since >>.87 the percentage of cotton is of importance for the strength measured Model identified: Y ij = µ + τ j + E ij Parameter µ τ j σe Estimate Y.. Y.j Y.. s E Value = from data Design without or with structure - how to analyse after ANOVA Design without structure A B C D E Y 11 Y 1 Y 13 Y 14 Y 15 Y 1 Y Y 3 Y 4 Y 5 : : : : : Y n1 Y n Y n3 Y n4 Y n5 Scaled t or range distribution E A D B C Design with structure A=control B 1 B Y 11 Y 1 Y 13 Y 1 Y Y 3 : : : Y n1 Y n Y n3 Natural comparisons? Use orthogonal contrasts (two!) How can they be constructed? 35 36

10 Important eample of orthogonal contrasts Design with structure A=Control Tablet=B 1 Inject=B Present Two alternative method methods ANOVA table for drug eperiment Source SSQ f s F-value Treatm Residual Total F(,9) distribution 95% fractile F (, 9) 0.05 = 4.6, such that the variation between treatments is (just) significant at the 5% significance level. What now? We can suggest reasonable contrasts: C A B = T A (T B1 +T B )=7.1 CA B SSQ A B = 4 ( +( 1) +( 1) = 16.60, f =1 ) C B1 B =0 T A +T B1 T B = 5.1 SSQ B1 B = CB 1 B 4 (0 +1 +( 1) =3.5, f =1 ) Splitting up the variance between treatments in two parts: Detailed ANOVA table for drug eperiment Source SSQ f s F-value Between A and B: A B Between the two B s: B 1 B Residual Total F (1, 9) 0.05 =5.1, such that A B is significant, but B 1 B is far from. The variation between all three treatments has been split up in variation between A and the B s and variation between the two B s. The B s are probably not (very) different while A has significantly higher response than the B s. Some patterns leading to orthogonal contrasts Design I A B 1 B Contrasts T A T B1 T B T B1 T B Design II A 1 A B 1 B Contrasts T A1 +T A T B1 T B T A1 T A T B1 T B Design III A B 1 B B 3 Contrast 3T A1 T B1 T B T B3 (artificial) T B1 T B T B3 (artificial) T B T B

11 In the design III eample the SSQ s from the two artificial contrasts [T B1 T B T B3 ] and [T B T B3 ] add up to the variation between the three B s. An ANOVA table could in principal look like Source SSQ f s F-value A B SSQ A B 1 Between B s SSQ B Residual SSQ res N-1-3 Total SSQ tot N-1 Patterns in two-way factorial designs Factor Factor B A B1 B A1 T 11 T 1 A T 1 T Totals T 11 T 1 T 1 T Effect Coeffi Amain cients Bmain AB interaction A3 design Factor Factor B A B1 B Control (C) T 01 T 0 A1 T 11 T 1 A T 1 T Totals T 01 T 0 T 11 T 1 T 1 T Effect Main A-C effects A B Inter (A-C) B actions A B The two last contrasts correspond to interactions. They are easily constructed by multiplication of the coefficients of the corresponding main effects. All 5 contrasts are orthogonal. Polynomial effects in ANOVA Model : Y ij = µ + τ j + E ij Concentration 5% 7% 9% 11% Sum ANOVA of response Source SSQ d.f. s F Concentration Residual (sign) Total

12 Plot of data and approimating 3. order polynomium: Polynomial estimation in ANOVA Possible empirical function as a polynomial: Y ij = β 0 + β 1 j + β j + β 3 3 j + E ij With 4 -points a polynomial of degree (4 1)=3 can be estimated using standard (polynomial) regression analysis. Alternative (reduced) models: Y ij = β 0 + β 1 j + β j + E ij Y ij = β 0 + β 1 j + E ij Y ij = β 0 + E ij (ultimately) By the general regression test method these models can be tested successively in order to identify the proper order of the polynomial. An alternative method to identify the necessary (statistically significant) order of the polynomial is based on orthogonal polynomials. The technique uses the concept of ortogonal regression and it is much similar to the orthogonal contrast technique. The technique is shown in the supplementary section I. Eercise 3-1 Tensile strength A B C D ANOVA for miing eperiment Source SSQ df s F Methods Residual Total

13 ..3 How can we try to group the treatments? 56.63=s mean LSD: Least Significant Difference For eample A versus B: D 666 scaled t(1) CA B Y A Y B s res 1/nA +1/n B t(f res ) s mean = s residual / n mean = 186/ 4= Which averages are possibly significantly different? Y A Y B <s res 1/nA +1/n B t(f res ) 0.05 Here n A = n B =4,s res = 113.5, f res =1 Y A Y B > /4+1/4.179 = 174.5? A B = = 185 significant A C = = 38 not significant A D = = 305 significant B C = = 3 significant B D = = 490 significant C D = = 3 significant Newman - Keuls Range Test Sort averages increasing: Y (1), Y (), Y (3), Y (4) Range = Y (4) Y (1) Table VII (gives q α ) : Criterion Y (4) Y (1) >s mean q α (4,f res )? 666 D CA B s mean = s res / n mean = 113.5/ 4=56.63 q 0.05 (4, 1) = 4.0 Conclusion? All pairs multiple testing - any problems? 51 5

14 .6.7 Range including 4: LSR 4 = = 37.8 Range including 3: LSR 3 = = 13.5 Range including : LSR = = B - D: = 490 > 37.8 (LSR 4 ) sign. B - C: = 3 > 13.5 (LSR 3 ) sign. B - A: = 185 > (LSR 3 ) sign. A - D: = 305 > 13.5 (LSR 3 ) sign. A - C: = 38 < (LSR ) not s. Duncans Multiple Range Test Sort averages increasing: Y (1), Y (), Y (3), Y (4) Range = Y (4) Y (1) Criterion (from special table find r α ): Conclusion: Y (4) Y (1) >s mean r α (4,f res )? s mean = s res / n mean = 113.5/ 4=56.63 r 0.05 (4, 1) = 3.33 D CA B Range including 4: LSR 4 = = Range including 3: LSR 3 = = 18.9 Range including : LSR = = B - D: = 490 > (LSR 4 ) sign. B - C: = 3 > 18.9 (LSR 3 ) sign. B - A: = 185 > (LSR 3 ) sign. A - D: = 305 > 18.9 (LSR 3 ) sign. A - C: = 38 < (LSR ) not s. Conclusion is the same as for Newman -Keuls here: Newman - Keuls & Duncans test Works alike, but use different types of range distributions. For eample: Duncan Newman - Keuls r(6, 1) 0.05 =3.40 q(6, 1) 0.05 =4.75 r(5, 1) 0.05 =3.36 q(5, 1) 0.05 =4.51 r(4, 1) 0.05 =3.33 q(4, 1) 0.05 =4.0 r(3, 1) 0.05 =3.3 q(3, 1) 0.05 = D CA B r(, 1) 0.05 =3.08 q(, 1) 0.05 =3.08 More significances More conservative 55 56

15 A grouping of averages that is significant according to Newman - Keuls test is more reliable No structure on treatments = Use Newman Keuls or Duncans test (LSD method not recommendable) Structure on treatments = Use contrast method or f Dunnetts test (below) Dunnetts test Alternative Control Treatments A B C D Parameters µ A µ B µ C µ D H 0 : µ A = µ B = µ C = µ D H 1 : One or more of (µ B, µ C,µ D ) different from µ A Eample: Eercise 3-1 with A as control (f) Two sided criterion: Y A Y B >s res 1/nA +1/n B d(4 1, 1) 0.05 d(3, 1) 0.05 (two sided) =.68 = critical difference = 186 1/4+1/4.68 = 14.7 One sided criterion: Y A Y B >s res 1/nA +1/n B d(4 1, 1) 0.05 d(3, 1) 0.05 (one sided) =.9 = critical difference = 186 1/4+1/4.9 = The fied (deterministic) effect ANOVA model 4 treatments Filter Clean Heat Nothing Assumptions: j τ j =0 and E ij N(0,σ E) Model for response: Y ij = µ + τ j + E ij The 4 treatment effects are deterministic (µ and τ j are constants) More reliable (and correct) than LSD if relevant 59 60

16 The random effect ANOVA model (see chapter 13 in 6th ed. of book) Eample: choose 4 batches among a large number of possible batches and measure some response (purity for eample) on these batches: 4 batches B-101 B-309 B-84 B-11 Model for response: Y ij = µ + B j + E ij The 4 batch effects are random variables (B j are random variables) Fied effect model: Y ij = µ + τ j + E ij ANOVA for fied effect model Source SSQ df s EMS=E{s } F Methods SSQ τ f τ s τ σe + n φ τ s τ/s E Residual SSQ E f E s E σe Total SSQ tot f tot φ τ = j τ j /(a 1), and τ j =Y.j Y.. Fied (deterministic) effects: temperature, concentration, treatment, etc. Assumptions: B j N(0,σB) and E ij N(0,σE) σe and σb are called variance components: They are the variances within and between (randomly chosen) batches, respectively Random effect model: Y ij = µ + B j + E ij Eample 13-1, p 487, typical eample of random effect model σ B =V{B},and σ B=(s B s E)/n ANOVA for random effect model Source SSQ df s EMS=E{s } F Batches SSQ B f B s B σe + n σb s B/s E Residual SSQ E f E s E σe Total SSQ tot f tot Looms Model for tensile strength: Y ij = µ + L j + E ij The 4 looms are randomly chosen with effects L j (being random variables) Random effects: batches, days, persons, eperimental rounds, litters of animals, etc. Assumptions: L j N(0,σ L) and E ij N(0,σ E) 63 64

17 One-way ANOVA for loom eample ANOVA for variation between looms Source SSQ df s E{s } F Looms σ E +4 σ L Residual σ E Total F (3, 1) 0.05 =3.49 << = significance! σ E =1.90=1.38 σ L = ( )/4 =6.96 =.64 Newman-Keuls or Duncans test on looms First: s Y = 1.90 = 1.38 = s Y = 1.90/4 =0.69 Eample: Newman - Keuls test: Find least significant ranges (q(.,.)) from studentized range table and multiply with standard deviation of group means: LSR q 0.05 (4, 1) = 4.0 s Y = q 0.05 (3, 1) = 3.77 s Y =.60 q 0.05 (, 1) = 3.08 s Y =.13 How do we further analyze this result? Compare group means: The smallest and the largest first and continue if difference is significant. Then net largest versus smallest, etc.: = 6.00 >.90 : significant = 1.75 <.60 : not significant = 5.50 >.60 : significant = 4.5 >.13 : significant Confidence interval for σl Interval for σl/σ E can be constructed Lower< σl/σ E < Upper Lower = s L s E F (a 1,N a) α/ n Upper = s L s E F (N a, a 1) α/ 1 1 n Conclusion: loom no is significantly different from the other looms 67 68

18 ..3 Looms: Lower = [15.65/4.47 1]/4 = 0.65 Upper = [ ]/4 =55.85 An alternative: Lower 1+Lower < σ L σ L +σ E < Upper 1+Upper Choice of sample size i A B C 1 y 11 y 1 y 13 y 1 y y 3 : : : : n y n1 y n y n3 Problem : Choose sample size n with k treatment/groups Fied effect model : Y ij = µ + τ j + E ij, i τ i =0 Requirements: 1) Know or assume σ E ) Which τ s are of interest to detect 3) How certain do we want to be to detect Random effect model : Y ij = µ + B j + E ij,v(b)=σb Requirements: 1) Know or assume σe ) Which σb is of interest to detect 3) How certain do we want to be to detect Eample fied effect model Assume (based on previous knowledge) : σe 1.5 Interesting values for τ (f) : {.00, 0.00, +.00} Criterion: P {detection} 0.80 (for eample) Try n =5(to start with) The tetbook has graphs for both cases pp Below, after the eamples based on the tetbook, some mere general results are presented. Compute Φ =(n jτj)/(a σ E) =5 ( +0 + )/(3 1.5 )=5.9 Compute Φ= 5.9 =

19 .6.7 Read off graph page 613: 0.0 ν 1 = a 1=3 1= ν =a(n 1) = 3(5 1)=1 Acceptance probability with α = 0.05 ν 1 = ν =1 Will 4 be enough? Compute Φ =(n jτj)/(a σ E) =4 ( +0 + )/(3 1.5 )=4.74 Compute Φ= 4.74 =.18 ca 0.10 Read off graph page 613: ν 1 = a 1=3 1= ν =a(n 1) = 3(4 1) = 9.43 Φ The graph shows, that n =5is enough ca 0.18 Acceptance probability with α = 0.05 ν = 1 ν =9 Eample random effect model Assume (based on previous knowledge) : σ E 1.5 Interesting values (for eample) for σ B :.0.18 ν =1 Φ Criterion: P {detection} 0.90 (for eample). Try n =5(to start with) The graph shows, that with n =4and testing with level of significance α =0.05 the probability of acceptance is about 18%. The probability of rejection (detection of significant τ s) is about 8%. n =4is thus enough. Compute λ = σ E +n σ B σe = =

20 Read off graph page 617 : ν 1 = a 1=3 1= ν =a(n 1) = 3(5 1)=1 Will 10 be enough? λ = σ E +n σ B σe = =4.33 Note: The degrees of freedom labeling is wrong - for the α =0.05 curves. It should be as shown for the α =0.01 curves and for all graphs with ν 1 4. Read off graph page 617: ν 1 = a 1=3 1= ν =a(n 1) = 3(10 1) = 7 Acceptance probability with α = 0.05 Note: Remember the degrees of freedom labeling again! 0.40 ca ν 1 = ν =1 ca 0. Acceptance probability with α = 0.05 ν = 1 ν = The graph shows, that n =5is not enough λ 4.13 ν =7 λ The graph shows, that with n =10and testing with level of significance α =0.05 the probability of acceptance is still about 0. (it should be ma. 0.10). n = 10 is thus not enough. The graph p. 617 shows, that for λ = 5. the acceptance probability It will require about n =15for σe =1.5 and σb =. In the supplementary part III the eact determination of sample size is described for bth deterministic and random effects models. Block designs - one factor and one blocking criterion Sources of uncertainty (noise) Day-to-day variation Batches of raw material Litters of animals Persons (doing the lab work) Test sites or alternative systems Treatment A B C Batch B-X B-V B-II Data Y 11 Y 1 Y 13 Y 1 Y Y 3 : : : Y n1 Y n Y n

21 One factor and one block, but they vary in the same way! Alternative to confounded design Mathematical model : Y ij = µ + τ j + B j + E ij Is the model correct? How can we analyze it? What can and what cannot be concluded? Is there a problem? Confounding? The inde for the factor and the block is the same: 100% confounding. Treatment A B C Data Y 11 (B-II) Y 1 (B-XI) Y 13 (B-IV) Y 1 (B-IX) Y (B-I) Y 3 (B-VI) : : : Y n1 (B-III) Y n (B-XX) Y n3 (B-IIX) In the design the batches used for the individual measurements are shown in parentheses The batches are selected randomly Eamples of factors Mathematical model : Y ij = µ + τ j + B ij + E ij How can this model be analyzed? What does the randomization do with respect to the mean and variance of Y ij? Compared to the above design: any problems solved? Have any new problems been introduced? Concentration of active compound in eperiment: (%,4%,6%,8%) Electrical voltage in test circuit (10 volt, 1 volt, 14 volt) Load in test of strength: (10 kp/m,15kp/m,0kp/m ) Alternative catalysts: (A, B, C, D) Alternative cleaning methods: (centrifuge treatm., filtration, electrostatic removal) Gender of test animal: (, ) Can the second design be improved even more (how)? 83 84

22 Eamples of blocks Batches of raw material: (I, II, III, IV) (can be of limited size) Collections of eperiments conducted simultaneously (dates f): (/-1990, 9/3-1990, 4/1-1990) Groups of participants in an indoor climate eperiment: (Test team 1, Test team, Test team 3) Litters of test animals: (Litter 1, Litter, Litter 3, Litter 4) Position in test equipment: (position 1, position, position 3) Design with inadequate confounding - schematic: Thermometers (= eperimental condition = block) are I, II and III. Y ij = µ + α j + T j + E ij Treatments A B C Data 5 (II) 16 (I) 19 (III) 4 (II) 15 (I) 0 (III) 4 (II) 17 (I) 0 (III) Total SSQ treat = = SSQ tot = ( ) = SSQ resid = SSQ tot SSQ treat Analysis of variance table for 100% confounded design One way ANOVA Source of var. SSQ d.f. s EMS = E{s } F-test Between treatm σ +3φ α +3σT Uncertainty (3 1) σ Total Design with thermometers randomized Thermometers are randomized (I, II,, X) Treatments A B C Data 6 (X) 0 (IV) 5 (II) 0 (II) 15 (V) 0 (VI) (I) 19 (III) (V) Total What can (or cannot) be concluded? Y ij = µ + α j +(T ij + E ij ) SSQ treat = =40.67 SSQ tot = ( ) =86.00 SSQ resid = SSQ tot SSQ treat =

23 Analysis of variance table for completely randomized design One way ANOVA Source of var. SSQ d.f. s EMS F-test Between treatm σ + σt +3φ α.69 Uncertainty (3 1) 7.56 σ + σt Total Balanced block design Thermometers are balanced (complete) blocks Treatments Thermometers A B C Total I II III Total What can (or cannot) be concluded? Y ij = µ + α j + T i + E ij SSQ treat = = SSQ therm = =13.56 SSQ tot = ( ) =64.89 SSQ resid = SSQ tot SSQ treat SSQ therm = Analysis of variance table for completely balanced block design Two way ANOVA Source of var. SSQ d.f. s EMS F-test Between treatm σ +3φ α Between therm σ +3σT (8.71) Uncertainty σ Total What can now be concluded? Latin square design Laboratory technicians (labor : (1) () (3)) are balanced against both treatments and thermometers Labor totals : (1)=59, ()=67, (3)=64 Treatments Thermometers A B C Total I 7 () 0 (3) 1 (1) 68 II 1 (1) 18 () 0 (3) 59 III 4 (3) 17 (1) () 63 Total Y ijk = µ + α j + T i + L k + E ijk 91 9

24 Analysis of variance table for Latin square design Two blocking criteria completely balanced block design : SSQ treat =48., SSQ therm =13.56, SSQ tot =7.89 SSQ labor = = SSQ resid =0. Triple balanced design = Graeco-Latin square Treatments Thermometers A B C Total I 8 ()(z) 1 (3)(y) 3 (1)() 7 II 3 (1)(y) 0 ()() 0 (3)(z) 63 III 5 (3)() 18 (1)(z) ()(y) 65 Total Latin square ANOVA Source of var. SSQ d.f. s EMS F-test Between treatm σ +3φ α Between therm σ +3σT (61.68) Between labor σ +3σL (49.54 ) Uncertainty σ Total Labor totals : (1)=64, ()=70, (3)=66 Batch totals : ()=68, (y)=66, (z)=66 Y ijkr = µ + α j + T i + L k + B r + E ijkr Analysis of variance table for Graeco-Latin square design SSQ batch = = SSQ treat =49.56, SSQ therm =14.89, SSQ labor =6. SSQ tot =71.56, SSQ resid =0.00 Graeco-Latin square ANOVA Source of var. SSQ d.f. s EMS F-test Between treatm σ +3φ α? Between therm σ +3σT? Between labor σ +3σL? Between batches σ +3σB? Uncertainty 0 0 (σ ) Total Comments to slides 3.3 to 3.15 All the eamples are created by numerical simulation using the corresponding models. 3.3: The variation of treatments is very significant, but it cannot be determined whether it is treatments or thermometers that cause it. If the eperiment is repeated at a later occasion we will presumably again find a significant, but probably different treatment effect (since thermometers would be 3 other thermometers). The eperiment is not reproducible and may lead to false conclusions. 3.4: The confounding treatments/thermometers is broken. However the variation between thermometers is causing a large uncertainty variance. The treatments are estimated with correct mean, but with a large variance. The treatments are not significant. The eample shows the principle, but of course, since there is no residual variance no tests can be carried out. An eternal variance estimate could be used if available 95 96

25 : The thermometers are now balanced out of the ANOVA, and the estimate of the treatment effect has correct mean plus a small variance. The treatment effect is significant. Note, that essentially the SSQ for treatments is as in the randomized design, but the SSQ for the residual is now free of the variation between thermometers and, thus, much smaller. 3.6: In the Latin square the same principle as used for thermometers is now used for the laboratory technicians. Variation between technicians is eliminated from the residual variance, causing improved precision (however again loosing degrees of freedom for the residual variance). 3.7: The same design principle (balance) is used to eliminate variation between batches from the residual variation. The (important) two-period cross-over design (page 14) Period Patient 1 1 A B B A 3 B A 4 A B : : : : : : n-1 A B n B A Y ijk = µ + τ i + per j + P k + E ijk ANOVA for the two period cross over design: Two period crossover ANOVA, eample with n=0 patients Source of var. SSQ d.f. s EMS F-test Treatments (τ i ) σ +0φ τ 4.54 Periods (per j ) σ +0φ per 0.40 Patients (P k ) (σ +σp ) (7.77) Uncertainty σ Total The design consists of R = n Latin squares repeated with different persons in all squares and identical periods (1 or ). The analysis of this design can take other forms if residual effects are suspected (effect from A on B different from the effect from B on A). A little more about Latin squares Table 4-8 page 136 Batches of Operators raw material A B C D E B C D E A 3 C D E A B 4 D E A B C 5 E A B C D Treatments A, B, C, D, E A standard Latin square Y ijk = µ + τ i + B j + O k + E ijk

26 ANOVA of Latin square eample ANOVA Source of var. SSQ d.f. s EMS F-test Treatments σ +5φ τ 7.73 Batches (σ +5σB) (1.59) Operators (σ +5σO ) (3.51) Uncertainty σ Total Interpretation of result of ANOVA What has been achieved by using this design? Replication of Latin squares O 1 O O 3 O 1 O O 3 O 1 O O 3 B 1 A B C B 1 B C A B 1 C A B B B C A B A B C B B C A B 3 C A B B 3 C A B B 3 A B C R 1 R R 3 3 squares with identical operators and batches O 1 O O 3 O 4 O 5 O 6 O 7 O 8 O 9 B 1 A B C B 1 B C A B 1 C A B B B C A B A B C B B C A B 3 C A B B 3 C A B B 3 A B C R 1 R R 3 3 squares with 9 operators and 3 batches O 1 O O 3 O 4 O 5 O 6 O 7 O 8 O 9 B 1 A B C B 4 B C A B 7 C A B B B C A B 5 A B C B 8 B C A B 3 C A B B 6 C A B B 9 A B C R 1 R R 3 3 squares with 9 operators and 9 batches Which design is probably the most precise (with smallest residual variance)? Answer: The third design, but why? How are the three different designs analyzed? In the supplementary part 6 the detailed ANOVA tables are indicated for each of the three cases. To block or not to block? Eample 4.1 Type of Test item (block) tip A B C D Y ij = µ + t i + B j + E ij ANOVA for block design (data scaled 10:1) Source SSQ df s EMS F Type of tip (t) σ +4φ t Test item (B) σ +4σ B (30.94) Residual σ Total

27 Two alternative designs - which one is best? Type of Test item (block) tip The second design is preferable. It is more precise, because the blocks are smaller (variance within blocks is smaller). A B C D Randomization is easier to do correct in small blocks and eperimental circumstances are easier to keep constant. 4 blocks of size 8. Double measurements for each treatment within the blocks. Round 1 Test item (block) A B C D 8 blocks of size 4. One measurements for each treatment within the blocks ANOVA test of additivity a treat- bblocks ments 1 : b A (1) B () : D (a) yy yy : yy yy yy : yy Basic model for block design with n measurements pr combination (the general case): Y ijk = µ + t i + B j + E ijk : : : : yy yy : yy If n>1start with Y ijk = µ + t i + B j + TB ij + E ijk and test the TB ij term (two way ANOVA with interaction term) against E ijk term If accepted, reduce model to ideal model and analyze as usual (two way ANOVA without interaction term) If rejected, use TB ij term to test the factor where i = {1,a}, j ={1,b} and k = {1,n}

28 Choice of sample size If : a treat- bblocks ments 1 : b A (1) B () : D (a) Y ijk = µ + t i + B j + E ijk yy yy : yy yy yy : yy If : : : : : yy yy : yy 3.30 Y ijk = µ + t i + B j + TB ij + E ijk Incomplete block designs Testing of 4 alternative etraction methods. Etraction of one production lasts 3 hours = only 3 methods can be tested on 1 day: Design Fine Normal % additive % additive material material Fine mat. Norm. mat. Day 1 X X X Day X X X Day 3 X X X Day 4 X X X Day 5 X X X Day 6 X X X 4.1 F treat = s treat/s E F (ν 1,ν ) Φ t =(bn i t i )/(a σ E ) ν 1 = a 1andν =abn a b +1 F treat = s treat /s TB F (ν 1,ν ) Φ t =(bn i t i )/(a (σe + nσtb)) Φ t (b i t i )/(a σtb) ν 1 = a 1andν =(a 1)(b 1) Y ij = µ + τ j + D i + E ij Is the design adequate. How could we improve the design. Which requirements should be made for the design. 1 day is an incomplete block: block size = Small blocks, why? The smallest block size = Eample: Test item that can be treated on two sides with surface hardening upside of test item downside of test item The intra-block variation is small for small blocks: That is a physical fact that the eperimenter can utilize: use small blocks! A balanced incomplete block design Four treatments, A, B, C and D. Two treatments per block. 4 treatments with block size A B C D Item 1 X X Item X X Item 3 X X Item 4 X X Item 5 X X Item 6 X X Problem: If systematic difference between upside and downside treatment results. Can that be handled? How?

29 Other classical eamples of incomplete blocks World Championship in football: 16 teams participate in 4 groups of 4 teams. In one group of 4 only teams can be on the field at the same time (1 match = 1 block of size ). 6 matches per group needed. World Championship in speedway with 1 participants: Groups of 4 drivers compete at the same time. Bridge tournament with 10 teams. In one round 5 tables are used each with teams. How many rounds are needed so that all 10 teams meet each other once. Football tournament with 10 teams. In one round 5 matches are played each with teams. How many rounds are needed so that all 10 teams meet each other once. How is the advantage of home matches handled in practice. A heuristic design (an inadequate design) Treatments Day A B C D I X X X II X X X III X X X IV X X X Y ij = µ + α j + D i + E ij α j is the fied factor effect (deterministic quantity) D i is the block effect. A random variable Estimate f : α B α A =(Y 1 + Y )/ (Y 11 + Y 1 )/ or : α B α A = Y B Y A Which one is best? Depends on σ E and σ D. Estimate f : α B α A =(Y 1 + Y )/ (Y 11 + Y 1 )/ and : α D α A =(Y 34 + Y 44 )/ (Y 11 + Y 1 )/ Which one is the most precise? Always α B α A Can the design be balanced, so that all comparisons are equally precise and independent of the actual blocks used? (Yes) Incomplete balanced block designs and some definitions Treatments Day A B C D I X X X II X X X III X X X IV X X X Y ij = µ + α j + D i + E ij k = 3 = block size a = 4 = number of treatments (some times called t ) b = 4 = number of blocks r = 3 = number of times each treatment is tried λ = = number of times any two treatments are in the same block = r (k-1)/(a-1) N = 1 = Total number of measurements = k b = a r

30 Eercise: Data from incomplete balanced block design Design Fine Normal % additive % additive material material Fine mat. Norm. mat. Day 1 X X Day X X Day 3 X X Day 4 X X Day 5 X X Day 6 X X Find k, a, b, r, λ and N for this design Blocks Treatments (days) A B C D T i N = t r = b k I t = a = 4 (treatments) II b = 4 (blocks) III k = 3 (block size) IV r = 3 (repeat. treat.) T j λ = (pairs in one block) Y ij = µ + α j + B i + E ij t j=1 α j =0, b i=1 B i =0, Var{E ij } = σ Computations for balanced incomplete block design Q j = T j 1 b k i=1 n ij T i SSQ α = k Q 1 +Q + +Q t λ t SSQ blocks = T 1 +T + +T b T k N n ij = { 0 if cell (i, j) is empty 1 if cell (i, j) is not empty Epectations and variances of computed quantities - estimation E{Q j } = λt k α j the estimate ˆα j = Q j λt k Var{Q j } = λ(t 1) k σ Var{ ˆα j } = t 1 t k λ Treatment difference estimate is ˆα i ˆα j and Var{Q i Q j } = λt k σ Var{ˆα i ˆα j } = k λt σ SSQ resid = SSQ tot SSQ α SSQ blocks ˆµ = Y, Var{Y } = σ /N Treatment mean estimate: [ˆµ +ˆα j ]=Y + Q j λt k Variance of treatment mean estimate: Var [ˆµ +ˆα j ]=σ k λt

31 After-ANOVA tests based on the Q-values Range-test (f Newman-Keuls test and table VII): ˆα (ma) ˆα (min) ˆσ k/λt Contrast-test procedure: = Q (ma) Q (min) s resid λt/k q( number,f resid ) contrast =[C]=Q 1 c 1 +Q c + +Q t c t SSQ contrast = k [C] λt(c 1 +c + +c t ) Analysis of data - eample λ =,a=t=4,b=4,r=3,k=3 Q 1 = ( ) = Q = ( ) = Q 3 = ( ) = 9.67 Q 4 = ( ) = Sum = SSQ tot = ij yij T.. /N = SSQ treat = 3 ( ) 4 = SSQ blocks = = Newman-Keuls test for treatments using Q s SSQ resid = SSQ tot SSQ treat SSQ blocks = f tot = 1 1 = 11 f treat = 4 1 = 3 f blocks = 4 1 = 3 s Q = s E λt k = = = q(p, 5) 0.05 = LSR = s Q q(p, 5) 0.05 = f resid = =

32 ( 3.00) = > sign ( 8.67) = > 47.0 sign = 1.33 < 37.5 not sign ( 3.00) = > 47.0 sign ( 8.67) = > 37.5 sign (3.00) = 3.33 < 37.5 not sign The Youden square (incomplete Latin square) Construction of a Youden square design Blocks Treatments (days) A1 A B1 B T i I??? II??? III??? IV??? T j Data from Youden square eperiment Youden square design Blocks Treatments (Days) A1 A B1 B T i I α β γ II β γ α III β γ α IV γ α β T j Blocks Treatments (days) A1 A B1 B T i I 5 (α) 75 (β) 57 (γ) 184 II 87 (β) 86 (γ) 53 (α) 6 III 54 (β) 68 (γ) 69 (α) 191 IV 50 (γ) 78 (α) 61 (β) 189 T j Analysis of Youden square The data are the same as on slide 4.9 and the eample primarily illustrates how the computations go. T α = = 5 T β = = 77 T γ = = 61 SSQ pos = ( )/4 790 /1 = The other SSQ s : See slide 4.9 (the same data) SSQ treat = SSQ blocks = SSQ tot = ANOVA for Youden Square Source SSQ df s E{s } F Treatm σ + c φ treat Blocks Posit σ +4 φ pos (1.03) Residual σ Total

33 Eample of computation for contrasts Consider the Youden square slide k=3,λ=,a=t=4,b=4,r=3. Q A1 = 3.00 Q A =31.00 Q B1 =9.67 Q B = 8.67 SSQ A B = (.00) k λt( ) = 0.38 SSQ A1 A = ( 63.00) 3 4(1 +1 ) = SSQ B1 B = (1 +1 ) = Sum = C A B = +Q A1 +Q A Q B1 Q B =.00 C A1 A = +Q A1 Q A = C B1 B = +Q B1 Q B = The sums of squares for the 3 orthogonal contrasts add up to the total sum of squares between treatments. Each of these sums of squares have 1 degree of freedom and can be tested against the residual sum of squares Tables of balanced incomplete block designs A, B, C,... = Treatments a = Number of treatments (often called t ) b = Number of blocks r = Number replications of each treatment k = Block size N = ar = bk = total number of measurements λ = number of times any two treatments occur in the same block =r(k 1)/(a 1) α, β, γ,... = positions within block for incomplete Latin squares (Youden squares) Balanced designs for a treatments with block size k = consist of all possible combinations of two treatments giving a(a-1)/ blocks of Block size Treatments Blocks Replications Pairings Total design k= a=3 b=3 r= λ=1 N=6 Symmetrical. A Youden square Treat- Blocks ments 1 3 A α β B β α C β α Blocksize Treatments Blocks Replications Pairings Total design k=3 a=4 b=4 r=3 λ= N=1 Symmetrical. A Youden square Treat- Blocks ments A α β γ B β γ α C γ α β D α β γ

34 Block size Treatments Blocks Replications Pairings Total design k=3 a=5 b=10 r=6 λ=3 N=30 All combinations of 3 treatments among 5 Block size Treatments Blocks Replications Pairings Total design k=3 a=6 b=10 r=5 λ= N=30 10 out of 0 possible combinations of 3 treatments among 6 (reduced) Treat- Blocks ments A B C D E Treat- Blocks ments A B C D E F Block size Treatments Blocks Replications Pairings Total design k=3 a=7 b=7 r=3 λ=1 N=1 7 out of 35 possible combinations of 3 treatments among 7 (reduced) Symmetrical. Youden square. Treat- Blocks ments A α β γ B β α γ C β γ α D α β γ E γ α β F α β γ G γ β α Block size Treatments Blocks Replications Pairings Total design k=3 a=9 b=1 r=3 λ=1 N=36 1 out of 84 possible combinations of 3 treatments among 9 (reduced) Treat- Blocks ments A B C D E F G H I

35 Block size Treatments Blocks Replications Pairings Total design k=4 a=5 b=5 r=4 λ=3 N=0 Symmetrical. Youden square Treat- Blocks ments A α β γ δ B β γ δ α C γ δ α β D δ α β γ E α β γ dy Block size Treatments Blocks Replications Pairings Total design k=4 a=7 b=7 r=4 λ= N=8 7 out of 35 possible combinations of 4 treatments among 7 (reduced) Symmetrical. Youden square. Treat- Blocks ments A α β γ δ B α β δ γ C β γ δ α D δ α γ β E β δ α γ F γ δ α β G δ γ α β Block size Treatments Blocks Replications Pairings Total design k=4 a=8 b=14 r=7 λ=3 N=56 14 out of 70 possible combinations of 4 treatments among 8 (reduced) Block size Treatments Blocks Replications Pairings Total design k=5 a=6 b=6 r=5 λ=4 N=30 Symmetrical. Youden square. Treat- Blocks ments A B C D E F G H Treat- Blocks ments A α β γ δ ɛ B β γ δ ɛ α C γ δ ɛ α β D δ ɛ α β γ E ɛ α β γ δ F α β γ δ ɛ

36 Block size Treatments Blocks Replications Pairings Total design k=5 a=11 b=11 r=5 λ= N=30 11 out of 46 possible combinations of 5 treatments among 11 (reduced) Symmetrical. Youden square. 4.3 Block size Treatments Blocks Replications Pairings Total design k=6 a=7 b=7 r=6 λ=5 N=4 Symmetrical. Youden square Treat- Blocks ments A α ɛ δ γ β B α ɛ δ γ β C α ɛ δ γ β D α ɛ δ γ β E β α ɛ δ γ F γ β α ɛ δ G δ γ β α ɛ H δ γ β α ɛ I ɛ δ γ β α J ɛ δ γ β α K ɛ δ γ β α Treat- Blocks ments A α φ ɛ δ γ β B β α φ ɛ δ γ C γ β α φ ɛ δ D δ γ β α φ ɛ E ɛ δ γ β α φ F φ ɛ δ γ β α G φ ɛ δ γ β α Block size Treatments Blocks Replications Pairings Total design k=6 a=9 b=1 r=8 λ=5 N=7 1 out of 84 possible combinations of 6 treatments among 9 (reduced) 4.34 Block size Treatments Blocks Replications Pairings Total design k=6 a=11 b=11 r=6 λ=3 N=66 11 out of 46 possible combinations of 6 treatments among 11 (reduced) Symmetrical. Youden square 4.35 Treat- Blocks ments A B C D E F G H I Treat- Blocks ments A φ ɛ δ γ β α B α φ ɛ δ γ β C β α φ ɛ δ γ D γ β α φ ɛ δ E γ β α φ ɛ δ F γ β α φ ɛ δ G γ β α φ ɛ δ H δ γ β α φ ɛ I δ γ β α φ ɛ J ɛ δ γ β α φ K φ ɛ δ γ β α

37 An eample with many issues Practical problems Load One test item can only be used once The testing of one item takes time (is epensive) supports Test tablet How do we find a representative collection of independent tablets (randomization) Other problems about sampling and preparation in lab Response : Strength of tablet (load to breakage) 1 factor = Humidity in powder for tablets Other possible factors: Load during production Time of pressing in production Size distribution in powder etc Design problems Approimately, how large is the load to be measured (select a suited apparatus to do the tests) Which humidity percentages are relevant (f 5% 50%) What is an interesting difference in load to detect (f =±1 g) The anticipated measurement uncertainty variance (f σ (15 g) ) (measurement + tablet variation) These questions must be answered before the eperiment is started Sources of uncertainty Temperature in laboratory Different handling by different operators Day-to-day variation of measurement devices Variation in the eperimental setup (geometry) The order of magnitude of these variations must be known or assessed before the eperiment is started

38 Design must be based on knowledge Guess (or study) how an eperiment may turn out is a possible (good) way: How will the ANOVA look when I have collected the data? Design Humidity 5% 0% 35% 50% Day 1 Day Day b Plot how you think (or hope) the data will turn out. Do you believe, that you will find what you are looking for: The optimal humidity for obtaining a high strength, f. Source of var. SSQ d.f. for design with b blocks Humidity SSQ treat ν 1 = a 1=3 Days (blocks) SSQ blocks ν 3 = b 1=b 1 Residual SSQ resid ν =(a 1)(b 1) = 3(b 1) Total SSQ tot ν tot = ab 1=4b Determine the necessary number of days (blocks) Y ij = µ + τ j + D i + E ij Is there a reasonable probability to detect the prescribed differences? Look up probability of acceptance page 648: Possible sizes of τ j to detect τ j = j = Compute Φ = b a σ j τj = b 4 15 (576) = 0.64 b Φ=0.80 b 0.0 ca 0.08 Acceptance probability with α = 0.05 ANOVA test ν =3 1 ν =1 Try f b =8 Φ=.6, ν 1 =3,ν =3 7=1 Looks reasonable. The chance of overlooking the above τ s is less than 10% (lucky punch). b =8could be worthwhile trying..6 Φ

39 Analysis of the data from the eperiment Day 5% 0% 35% 50% Sums Data from 31 4 eperiment SSQ resid = Sums Source of var. SSQ d.f. s EMS F Humidity σ +8 φ τ 9.3 Days (blocks) (σ +4 σd ) (.6) Residual σ Total Mean response 40 0 regression estimate Estimates: y j = τ j = µ = σ = 34 = 15.3 ( σ D = (609 34)/4 =93.8=9.7 ) Regression function estimate (5 j 50): Y ij = j j 155 Further analysis of days (blocks) Draw normal probability plot for daily averages, f: Standard normal fractiles Dayly averages 156 (i 0.5)/n