Georgia Southern University Digital Commons@Georgia Southern Electronic Theses & Dissertations Jack N. Averitt College of Graduate Studies COGS) Summer 204 Statistical Analysis of Unreplicated Factorial Designs Using Contrasts Meixi Yang Georgia Southern University Follow this and additional works at: http://digitalcommons.georgiasouthern.edu/etd Part of the Design of Experiments and Sample Surveys Commons Recommended Citation Yang, Meixi, "Statistical Analysis of Unreplicated Factorial Designs Using Contrasts" 204). Electronic Theses & Dissertations. 47. http://digitalcommons.georgiasouthern.edu/etd/47 This thesis open access) is brought to you for free and open access by the Jack N. Averitt College of Graduate Studies COGS) at Digital Commons@Georgia Southern. It has been accepted for inclusion in Electronic Theses & Dissertations by an authorized administrator of Digital Commons@Georgia Southern. For more information, please contact digitalcommons@georgiasouthern.edu.
STATISTICAL ANALYSIS OF UNREPLICATED FACTORIAL DESIGNS USING CONTRASTS by MEIXI YANG Under the Direction of Charles W. Champ) ABSTRACT Factorial designs can have a large number of treatments due to the number of factors and the number of levels of each factor. The number of experimental units required for a researcher to conduct a k factorial experiment is at least the number of treatments. For such an experiment, the total number of experimental units will also depend on the number of replicates for each treatment. The more experimental units used in a study the more the cost to the researcher. The minimum cost is associated with the case in which there is one experimental unit per treatment. That is, an unreplicated k factorial experiment would be the least costly. In an unreplicated experiment, the researcher cannot use analysis of variance to analyze the data. We propose a method that analyzes the data using normal probability plot of estimated contrast of the main effects and interactions. This method is applied to data and compared with Tukey s method that test for non-additivity. Our method is also discussed for use when the response is a multivariate set of measurements. Key Words: Contrast, normal probability plot, factorial design, Tukey 2009 Mathematics Subject Classification: 2K5, 2H99
STATISTICAL ANALYSIS OF UNREPLICATED FACTORIAL DESIGNS USING CONTRASTS by MEIXI YANG B.S., Shandong University of Science and Technology, China, 2009 M.S., Shandong University of Science and Technology, China, 202 A Thesis Submitted to the Graduate Faculty of Georgia Southern University in Partial Fulfillment of the Requirement for the Degree MASTER OF SCIENCE STATESBORO, GEORGIA 204
c 204 MEIXI YANG All Rights Reserved iii
STATISTICAL ANALYSIS OF UNREPLICATED FACTORIAL DESIGNS USING CONTRASTS by MEIXI YANG Major Professor: Charles W. Champ Committee: Broderick O. Oluyede Lili Yu Electronic Version Approved: July 24, 204 iv
ACKNOWLEDGMENTS I would like to thank my advisor Dr. Charles W. Champ for guidance of me in my research. I also want to thank my committee members Dr. Broderick O. Oluyede and Dr. Lili Yu for reviewing my thesis and their guidance in my academic endeavors. v
TABLE OF CONTENTS Page ACKNOWLEDGMENTS............................. v LIST OF TABLES................................. viii LIST OF FIGURES................................ ix CHAPTER Introduction.............................. 2 Two Factor Design Model with a Univariate Response....... 3 2. Introduction.......................... 3 2.2 The Two Factor Design with Replicates........... 3 2.3 One Observation per Treatment No Interaction Assumed. 7 2.4 Tukey s Method for One Observation per Treatment.... 9 2.5 Using a Normal Probability Plot............... 2. Some Examples........................ 3 2.7 Conclusion........................... 38 3 Three Factor Experiments...................... 39 3. Introduction.......................... 39 3.2 Tukey s Method for Three Factors.............. 40 3.3 Analyzing a Reduced Model................. 44 3.4 Analysis of Contrasts..................... 45 3.5 Unreplicated 3 k Factorial Designs.............. 45 vi
3. Example............................ 47 3.7 Conclusion........................... 8 4 Unreplicated Multivariate Factorial Designs............. 9 4. Introduction.......................... 9 4.2 Design and Data Models................... 9 4.3 Parameter Estimation and Contrasts............ 72 4.4 Example............................ 72 4.5 Conclusion........................... 80 5 CONCLUSION............................ 8 5. General Conclusions...................... 8 5.2 Areas for Further Research.................. 8 REFERENCES.................................. 83 vii
LIST OF TABLES Table Page 2. Montgomery s Example -2...................... 4 2.2 Kutner s Exercise 20.8......................... 2 3. Montgomery s Example -3...................... 47 4. Johnson s Example.......................... 72 viii
LIST OF FIGURES Figure Page 2. Montgomery Example, Probability Plot............... 9 2.2 Montgomery Example, Probability Plot and Fitted Line Points 20 2.3 Kutner Example, Probability Plot.................. 35 2.4 Kutner Example, Probability Plot and Fitted Line 9 Points.... 3 2.5 Kutner Example, Probability Plot and Fitted Line 8 Points.... 37 3. Montgomery Example, Probability Plot............... 3.2 Montgomery Example, Probability Plot and Fitted Line 8 Points. 7 4. Johnson Example, Probability Plot................ 79 4.2 Johnson Example, Probability Plot 2................ 80 ix
CHAPTER INTRODUCTION Analyzing data from a designed experiment using ANalysis Of VAriance ANOVA), generally requires at least two replicates for at least one treatment. There are, however, researchers who need to use unreplicated one observation per treatment) designs. Under the independent normal model with common variance, these designs do not provide enough data to independently estimate the overall mean, main effects, interactions, and the common variance. Montgomery 997) states concerning the analysis of data from an unreplicated two factor fixed effects design that there are no tests on main effects unless the interaction effect is zero. He also points out that even a moderate number of factors, the total number of treatment combinations in a 2 k factorial design is large. This is even a large number of total treatments for factorial designs in which the levels of one or more of the factors is greater than 2. Three methods are discussed in the literature for analyzing the response data in a two factor fixed effects model with one observation per treatment. The first of these is to assume there is no interaction between the factors. This is the additive model. The second method uses a regression model that elimates higher-order polynomials. The third method is a test developed by Tukey 949) for determining if there is an interaction. He states that the professional practitioner of the analysis of variance will have no difficulty in extending the process to more complex designs. These methods are discussed in Alin and Kurt 200) and Francka, Nielsenb, and Osbornec 203). We will examine an extension of Tukey s method to a three factor design. Various authors have examined method for evaluating the data from a 2 k factorial design with no replicates for a univariate response. One of these methods that is commonly recommended is the use of a normal probability plot of the estimates of the
2 main effects and interactions. We plan to study the use of normal probability plots in the analysis of unreplicated k factorial designs in which each of the factors has two or more levels with at least one factor having three or more levels. This will include cases in which there is a univariate response and there is a multivariate response. As we will demonstrate, the estimators of the main effects and interactions in an unreplicated fixed effects k factor design in which at least one factor has more than two levels are correlated. We propose a transformation of these estimators to a collection of independent random variables with common variance. Under the hypothesis of no main effects or interactions, these estimators under the independent normal model with common variance σ 2 Σ for a multivariate response) will be a random sample with common N 0, σ 2 ) N p 0, Σ) for a multivariate response) distributon. A normal probability plot of the transformed estimates of the main effects and interactions will be used to determine which linear combinations of the main effects and interactions are significantly different from zero. An examimation of the associated parameters will reveal which, if any, of the main effects and interactions are significantly different from zero.
CHAPTER 2 TWO FACTOR DESIGN MODEL WITH A UNIVARIATE RESPONSE 2. Introduction In a variety of studies, researchers are interested in studying the effect of two or more factors on a response variable. As has been shown by a several authors, factorial designs are the most efficient way to conduct such studies. A factor is a variable whose values are selected by the researcher. The possible values of a factor are called the levels of the factor. How the values of a factor are selected determines if the study is a fixed effect or random effect factorial design. If the levels of a factor are the only ones of interest to the researcher, then the study is a fixed effect factorial design with respect this factor. If random selection is used to select from a collection of possible values of a factor the levels of the factor to be studied, then the factorial design is a random effects factorial design with respect to this factor. The treatments in a factorial design are all the possible factor level combinatins. In our study, it will be convenient to discuss first two-factor design with replications before examining designs without replicates. 2.2 The Two Factor Design with Replicates We begin our study of factorial designs by examining two k = 2) factor designs. Under the additive model, the response variable Y ijl can be expressed as Y ijl = µ ij + ɛ ijl with µ ij = µ + τ ) i + τ 2 ) j + τ 2 ) ij
4 for i =,..., a, j =,..., b, and l =,..., n with n >. We have expressed the mean µ ij of the response variable Y ijl as the sum of an overall mean µ, the effect τ ) i due to setting the first factor at level i, the effect τ 2 ) j of setting the second factor at its jth level, and an effect τ 2 ) ij due to the interaction between the two factors when the first is set at its ith level and the second at its jth level. It is assumed that a a i= τ ) i = 0; b j= τ 2) j = 0; i= τ 2) ij = 0 for j =,..., b; and b j= τ 2) ij = 0 for i =,..., a. We also assume that the Y ijl s are independent and ɛ ijl N 0, σ 2 ij). We refer to these assumptions as the independent normal model. The model is further simplified by assuming a common variance, that is, σ 2 ijl = σ2 the common variance for i =,..., a, j =,..., b, and l =,..., n. The design is an unreplicated one if n =. Using matrix notation, we can write our additive model in the form Y = Xθ + ɛ, where Y is the abn vector of observations, X is the abn ab design matrix, θ is the ab vector of model parameters, and ɛ is the abn vector of error terms. An analysis of variance ANOVA) of the response data is based on the following partition of the sum of squares total SST ). SST = SSA + SSB + SSAB + SSE,
5 where SST = a SSA = a SSB = a SSAB = a SSE = a b n i= j= l= b n i= j= l= b n i= j= l= b n i= j= l= b n i= j= l= ɛ2 ijl. Yijl Y... ) 2 ; Y i.. Y... ) 2 ; Y.j. Y... ) 2 ; Y ij. Y i.. Y.j. + Y... ) 2 ; and It can be shown that SSA, SSB, SSAB, and SSE are stochastically independent under our independent normal model. The degrees of freedom of these sums of squares are df SST = abn ; df SSA = a ; df SSB = b ; df SSAB = a ) b ) ; and df SSE = n ) ab. The mean squares associated with SSA, SSB, SSAB, and SSE are, respectively, MSA = SSA df SSA, MSB = SSB df SSB, MSAB = SSAB df SSAB, and MSE = SSE df SSE. Note that if n =, then df SSE = 0 and the MSE is undefined. The null H 0 ) and alternative H ) hypotheses of interest can be written in terms of the following hypotheses. H A,0 : τ ) =... = τ ) a = 0 and H A, : H A,0 ; H B,0 : τ 2 ) =... = τ 2 ) b = 0 and H B, : H B,0 ; and H AB,0 : τ 2 ) =... = τ 2 ) ab = 0 and H AB, : H AB,0 The alternative hypothesis in this study is H : H AB, [H AB,0 H A, H B, )]
with the null hypothesis H 0 : H. The statistical test has decision rule that rejects the null hypotheis in favor of the alternative hypothesis if the observed value of MSAB MSE It can be shown that c AB [ MSAB MSE < c AB MSA MSE c A MSB )] MSE c B. SSA σ 2 χ 2 a,ξ, SSB A 2 σ 2 χ 2 b,ξ, SSAB B 2 σ 2 χ 2 a )b ),ξ, and SSE AB 2 σ 2 χ 2 n )ab, where ξa 2 = nb a i= τ ) 2 i, ξ 2 aσ 2 B = na b j= τ 2) 2 j, and ξ 2 bσ 2 AB = n a b i= j= τ 2) 2 ij σ 2 [a ) b ) + ]. If the null hypothesis is true, then we have ξ A = ξ B = ξ AB = 0. The size of the test α is given by [ MSAB MSAB MSA α = P MSE c AB MSE < c AB MSE c A MSB )]) MSE c B ) χ 2 a )b ) / [a ) b )] = P c χ 2 AB n )ab/ [n ) ab] ) χ 2 a )b ) / [a ) b )] χ 2 + P χ 2 n )ab / [n ) ab] < c AB, a / a ) χ 2 n )ab / [n ) ab] c A ) χ 2 a )b ) / [a ) b )] χ 2 b / b ) + P χ 2 n )ab / [n ) ab] < c AB, χ 2 n )ab / [n ) ab] c B = F Fa )b ),n )ab c AB ) + + 0 0 F χ 2 a )b ) n ) abxcab a ) b ) F χ 2 a )b ) n ) abxcab a ) b ) ) ) n ) abxca F χ 2 a a ) F χ 2 b n ) abxcb b f χ 2 n )ab x) dx ) f χ 2 n )ab x) dx,
7 where ) n ) abxca F χ 2 a a ) F χ 2 b n ) abxcb b ) n ) abxca = F χ 2 a a = F χ 2 b n ) abxcb b ). and The power of the test is determined by power = F Fa )b ),m ab,ξab c AB ) ) [ )] m ab) xcab m ab) xca + F χ 2 a )b ),ξab F 0 a ) b ) χ 2 a,ξa f a χ 2 m ab x) dx ) [ )] m ab) xcab m ab) xcb + F χ 2 a )b ),ξab F a ) b ) χ 2 b,ξb f b χ 2 m ab x) dx, 0 where at least one of the value ξ A, ξ B, and ξ AB is not equal to zero. 2.3 One Observation per Treatment No Interaction Assumed For the case of one observation per treatment n = ), there is only enough data to estimate independently the overall mean, the main effects, and the interactions but not the common variance in our model. One approach to analyzing the data for main effects is to assume there is no interaction between the two factors. In this case the model becomes µ ij = µ + τ ) i + τ 2 ) j. The null and alternative hypotheses are H 0 : τ ) =... = τ ) a = 0 and τ 2 ) =... = τ 2 ) b = 0; and H : H 0. The total sum of squares SST ) can be partitioned into the sum of square due to the first factor SSA), the sum of squares due to the second factor SSB), and the sum
8 of squares SSE) due to error. That is, SST = SSA + SSB + SSE. There respective degrees of freedom are df SST = ab, df SSA = a, df SSB = b, and df SSE = a ) b ). Under this model, it can be shown that SSA σ 2 χ 2 a,ξ, SSB A 2 σ 2 χ 2 b,ξ, and SSE B 2 σ 2 χ 2 n )ab, where ξa 2 = nb a i= τ ) 2 i, and ξ 2 aσ 2 B = na b j= τ 2) 2 j. bσ 2 The test based on the analysis of variances rejects H 0 in favor of H if the observed value of where MSA = b a ) 2 i= Y i. Y.. a MSE = MSA MSE c A MSB MSE c B, SST SSA SSB. a ) b ), MSB = a b ) 2 j= Y.j Y.. b, and The size of the test is ) ) ) ) MSA MSB MSA MSB α = P MSE c A + P MSE c B P MSE c A P MSE c B = P ) ) F a,a )b ) c A + P Fb,a )b ) c B P χ 2 a xc ) A P χ 2 b xc ) B f b a χ 2 a )b ) x) dx 0 = F Fa,a )b ) c A ) + F Fb,a )b ) c B ) ) ) xca xcb F χ 2 a F b χ 2 b f a χ 2 a )b ) x) dx, 0
9 where ) xca F χ 2 a b ) xca = F χ 2 a b ) ) xcb xcb and F χ 2 b = F a χ 2 b. a For example in the case in which a = 3 and b = 5, if the researcher selects the critical values c A and c B to be c A = F 3,3 )5 ),0.05 and c B = F 5,3 )5 ),0,05, then the actual size of the test is )) xfinv α; a, a ) b )) α = 2α ChiSquareDist ; a 0 b )) xfinv α; b, a ) b )) ChiSquareDist ; a a ChiSquareDen x; a ) b )) dx = 0.092225483. 2.4 Tukey s Method for One Observation per Treatment Tukey 949) developed a test for determining if there is an interaction between the two factors which assumes the interactions are of the form τ 2 ) ij = λ τ ) i τ 2 ) j. In this model, the a + b + 2 parameters including the common variance can be estimated. To determine these estimates using least squares we define the function Q = Q µ, τ ),..., τ ) a, τ 2 ),..., τ 2 ) b, λ) = a b ) 2 Y ij µ τ ) i τ 2 ) j λ τ ) i τ 2 ) j i= j= The least square estimates are the solutions to the following system of equations. Q µ = 0; Q = 0; τ ) i Q τ 2 ) j = 0; Q λ = 0.
The least squares estimates of the parameters µ, τ ),..., τ ) a, τ 2 ),...,τ 2 ) b, and λ are given, respectively, by 0 µ = Y.., τ ) i = Y i. Y.., τ 2 ) j = Y.j Y.., and a b ) ) i= j= Y i. Y.. Y.j Y.. Yij λ = a b ) 2 ) 2. i= j= Y i. Y.. Y.j Y.. We can now express Y ij as Y ij = Y.. + Y i. Y.. ) + Y.j Y.. ) + λ Y i. Y.. ) Y.j Y.. ) + ɛij. It follows that ɛ ij can be expressed as ɛ ij = Y ij Y.. Y i. Y.. ) Y.j Y.. ) λ Y i. Y.. ) Y.j Y.. ). The total sum of squares SST can now be partitioned into SST = SSA + SSB + SSAB + SSE, where SST, SSA, and SSB are defined in the previous section with SSAB = a SSE = a b i= j= b i= j= ɛ2 ij. λ 2 Y i. Y.. ) 2 Y.j Y.. ) 2 and The degrees of freedom of SSAB and SSE are, respectively, and ab a b. It can be shown that under H 0 : λ = 0, the random variables SSAB and SSE are stochastically independent with SSAB σ 2 χ 2 and SSE σ 2 χ 2 ab a b. The appropriate hypotheses in this case are H 0 : H with H : H λ, [H λ,0 H A, H B, )],
where H λ,0 : λ = 0 and H λ, : λ 0. The statistical test has decision rule that rejects the null hypotheis in favor of the alternative hypothesis if the observed value of where MSAB MSE The size of the test α is α = F F,ab a b c AB ) + xcab + F χ 2 ab a b 0 [ MSAB c AB < c MSE AB MSAB = SSAB MSA MSE c A MSB )] MSE c B, and MSE = xcab F χ 2 0 ab a b ) b ) xcb F χ 2 b ab a b ) SSE ab a b. ) a ) xca F χ 2 a f ab a b χ 2 ab a b x) dx ) f χ 2 ab a b x) dx. 2.5 Using a Normal Probability Plot The least squares estimate θ of the vector of parameters θ in the full model with no replicates is θ = X T X ) X T Y. It can be shown that if a > 2, then the estimators τ ),..., τ ) a of the parameters τ ),..., τ ) a are not independent under our independent normal model. Likewise, if b > 2, the estimators τ 2 ),..., τ 2 ) b of the parameters τ 2 ),..., τ 2 ) b are not stochastically independent as are the estimators τ 2 ) ij of the parameters τ 2 ) ij when a > 2 and/or b > 2. This can be seen by first noting that ) Σ θ = cov θ = X T X ) σ 2
and observing that X T X ) is not the identity matrix. However, it can be shown 2 that X T X ) has the form w 0 0 0 X T X ) 0 W a ) a ) A 0 0 = 0 0 W b ) b ) B 0 0 0 0 W a )b ) a )b ) AB, where w corresponds to the overall mean, W a ) a ) A effects due to factor A, W b ) b ) B B, and W a )b ) a )b ) AB and B. We can now see that, for example, whereas where is associated with the main is associated with the main effects due to factor is associated with the interactions between factors A cov τ )) = W a ) a ) A σ 2 cov τ, τ 2 )) = 0, τ = [ τ ),..., τ ) a ] T and τ2 = [ τ 2 ),..., τ 2 ) b ] T. Observe that X T X ) is a real symmetric matrix. In particular, observe that W A, W B, and W AB are real symmetric matrices. It follows that there exist matrices P A, P B, and P AB such that W A = P A P T A, W B = P B P T B, and W AB = P AB P T AB. We define P by P = w 0 0 0 0 P A 0 0 0 0 P B 0. 0 0 0 P AB
3 It follows that the vector of transformed estimators θ define by θ = P θ are stochastically independent since = cov P θ ) = P X T X ) ) Σ θ P T σ 2 = Iσ 2. The vector estimator θ is associated with the contrast given in W A, W B, and W AB of the vector of parameters θ. This suggest that under our null hypothesis of no main effects or interactions that the estimators θ 2,..., θ a )b ) are stochastically independent and identically distributed N 0, σ 2 ). A normal probability plot of the observed values of these estimators should reveal which if any of these linear combinations of the estimators of the main effects and interactions are different from zero. Exact plotting positions for a normal probability plot can be found in Harter 9) and Teichroew 95) for selected values of the sample size. Often the ith plotting position E Z i:n ) for a normal probability plot is usually approximated by ) i 0.375 E Z i:n ) = Φ, n + 0.25 where Z i:n is the ith order statistics of a random sample of size n from a standard normal distribution and Φ z) is the cumulative distribution function of a standard normal distribution. This approximation was originally proposed by Blom 958). A discussion on the selection of plotting positions are discussed in Champ and Vora 2005). 2. Some Examples Example :
4 Montgomery 997) gives an example of a two factor experiment in which n ij = for i =, 2, 3 and j =, 2, 3, 4, 5. He states in his Example -2 that the impurity present in a chemical product is affected by two factors presure and temperature. His data is presented in the following table. Table 2.: Montgomery s Example -2 Temperature 25 30 35 40 45 00 5 4 3 5 Pressure 25 3 4 2 3 50 3 2 Using Tukey s method, he conclude that there was no interaction effect but that the main effects due to both temperature and pressure are significant.
5 The design matrix for this experiment is 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 X = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.
It follows that w = 5, W A = W B = W AB = 2 5 5 5 2 5 4 5 5 5 5 5 5 5 4 5 5 5 5 5 5 4 5 5 4 5,, and 8 2 2 2 4 5 5 5 5 5 2 5 2 5 2 5 8 2 2 5 5 5 2 5 2 5 2 5 4 5 5 4 5 5 5 5 8 2 5 5 5 5 4 5 5 5 8 5 5 5 4 5 5 5 5 2 5 5 5 5 4 5 5 5 5 5 5 4 5 5 8 2 2 2 5 5 5 5 2 2 5 5 5 8 2 2 5 5 5 4 2 2 2 5 5 5 5 5 8 2 5 5 8 5.
7 It can be shown P A = P B = P AB = 0 0 30 30 30 0 30 0 5 30, 8 8 5 30 0 0 3 5 0 2 8 30 8 5 30 30 0 0 20 30 0 0 30 0 0 30 0 30 0 0 20 30 0 0 30 0 0 8 8 3, and 2 2 2 3 2 0 0 2 0 0 20 4 4 20 0 3 0 3 20 3 2 0 3 2 3 2 2 2 3 2 2 2 3 2 2 2 0 0 2 0 0 20 4 20 0 3 30 0 3 0 20 2 0 2 2 2 4 3 3 2 3 2 2. We observe that our least squares estimates are θ = X T X ) X T y [ = 44 5 5 3 3 4 5 5 7 4 5 5 3 3 0 2 3 3 2 3 0 3 ] T, where [ y = 5 4 3 5 3 4 2 3 3 2 ] T. It then follows that θ = P θ [ = 44 5 5 2 30 3 0 5 5 2 2 5 0 0 3 3 2 2 3 3 2 3 ] T.
8 Omitting the estimate for µ = 44 5/5, we find the observed order statistics for the remaining fourteen contrast estimates. From Teicherow 95), we obtain the plotting position for a normal probability plot for a sample size of fourteen. These ordered pairs are given in the following 4 2 matrix with the plotting position in the first column and the ordered data in the second..70338554 2.2079022754 3 0.9027039 5 5 0.737035 3 0.45550500 0 0.272970489 0 0.0885924 0.0885924 2 3 0.272970489 2 0.45550500 3 2 0.737035 2 5 0.9027039 3.2079022754 0 2.70338554 30 3
9 Figure 2.: Montgomery Example, Probability Plot This plots suggest that eleven of the points are plotting about a line whereas three are not plotting about this line. These are the points ).70338554, 2,.2079022754, ) ) 0, and.70338554, 2 30. 3 A simple fitting of a line to the eleven points, we have that θ i:a )b ) = 0.4007897323 +.539539δ i:a )b ). Plotting this line with our points, we obtain the graph in Figure 2.2.
20 Figure 2.2: Montgomery Example, Probability Plot and Fitted Line Points The three points.70338554, 2 ),.2079022754, ) 0, and.70338554, 2 30 3 ). are associated with the contrast estimators τ 2 ) 3, τ ) 2, and τ ), respectively. This plot provides evidence that there is no interaction between the two factors but there is effects due to the two factors. Example 2: Kutner, Nachtsheim, Neter, and Li 2005) on page 890 state in Exercise 20.8 that A food Technologist, testing storage capabilities for a newly developed type of imitation sausage made from soybeans, conducted an experiment to test the effects of humidity level factor A) and temperature level factor B) in the freezer compartment on color change in the sausage. Three humidity levels and four tem-
2 perature levels were considered. Five hundred sausages were stored at each of the 2 humidity-temperature combinations for 90 days. At the end of the storage period, the researcher determined the proportion of sausages for each humidity-temperature combination that exhibited color changes. The researcher transformed the data by means of the arcsine transformation to stabilize the variances. The transformed data Y = 2 arcsin Y follow. Table 2.2: Kutner s Exercise 20.8 Temperature level Humidity level j= j=2 j=3 j=4 i= 3.9 4.2 20.5 24.8 i=2 5.7.3 2.7 23. i=3 5. 5.4 9.9 2.
22 We see that 3.9 + 4.2 + 20.5 + 24.8 Y. = = 8.350; 4 5.7 +.3 + 2.7 + 23. Y 2. = = 9.325; 4 5. + 5.4 + 9.9 + 2. Y 3. = = 9.25; 4 3.9 + 5.7 + 5. Y. = = 4.9; 3 4.2 +.3 + 5.4 Y.2 = = 5.3; 3 20.5 + 2.7 + 9.9 Y.3 = = 20.7; 3 24.8 + 23. + 2. Y.4 = = 24.83; 3 Y.. = 3.9 + 4.2 + 20.5 + 24.8 + 5.7 +.3 + 2.7 + 23. + 5. + 5.4 + 9.9 + 2. = 227.2; and Y.. = 227.2 2 = 8.93 Assuming interaction between the two factors has the form τ 2 ) ij = λ τ ) i τ 2 ) j, then the least squares estimates of the parameters µ, τ ), τ ) 2,τ ) 3, τ 2 ), τ 2 ) 2, τ 2 ) 3, and τ 2 ) 4 are
23 µ = Y.. = 8.93; τ ) = Y. Y.. = 8.350 8.93 = 0.583; τ ) 2 = Y 2. Y.. = 9.325 8.93 = 0.39; τ ) 3 = Y 3. Y.. = 9.25 8.93 = 0.9; τ 2 ) = Y. Y.. = 4.9 8.93 = 4.03; τ 2 ) 2 = Y.2 Y.. = 5.3 8.93 = 3.3; τ 2 ) 3 = Y.3 Y.. = 20.7 8.93 =.7; τ 2 ) 4 = Y.4 Y.. = 24.83 8.93 = 5.9. Recall that the formula to be used to obtain an estimate of λ is 3 4 ) ) i= j= Y i. Y.. Y.j Y.. Yij λ = 3 4 ) 2 ) 2. i= j= Y i. Y.. Y.j Y..
24 It follows that 3 4 i= j= Y i. Y.. ) Y.j Y.. ) Yij = ) ) Y. Y.. Y. Y.. Y + ) ) Y. Y.. Y.2 Y.. Y2 + ) ) Y. Y.. Y.3 Y.. Y3 + ) ) Y. Y.. Y.4 Y.. Y4 + ) ) Y 2. Y.. Y. Y.. Y2 + ) ) Y 2. Y.. Y.2 Y.. Y22 + ) ) Y 2. Y.. Y.3 Y.. Y23 + ) ) Y 2. Y.. Y.4 Y.. Y24 + ) ) Y 3. Y.. Y. Y.. Y3 + ) ) Y 3. Y.. Y.2 Y.. Y32 + ) ) Y 3. Y.. Y.3 Y.. Y33 + ) ) Y 3. Y.. Y.4 Y.. Y34 = 0.583 ) 4.03 ) 3.9) + 0.583 ) 3.3 ) 4.2) + 0.583 ).7 ) 20.5) + 0.583 ) 5.9) 24.8) + 0.39 ) 4.03 ) 5.7) + 0.39 ) 3.3 ).3) + 0.39 ).7 ) 2.7) + 0.39 ) 5.9) 23.) + 0.9 ) 4.03 ) 5.) + 0.9 ) 3.3 ) 5.4) + 0.9 ).7 ) 9.9) + 0.9 ) 5.9) 2.) = 8.2709999 and
25 3 4 i= j= Y i. Y.. ) 2 Y.j Y.. ) 2 = ) 2 ) 2 ) 2 ) 2 Y. Y.. Y. Y.. + Y. Y.. Y.2 Y.. + ) 2 ) 2 ) 2 ) 2 Y. Y.. Y.3 Y.. + Y. Y.. Y.4 Y.. + ) 2 ) 2 ) 2 ) 2 Y 2. Y.. Y. Y.. + Y 2. Y.. Y.2 Y.. + ) 2 ) 2 ) 2 ) 2 Y 2. Y.. Y.3 Y.. + Y 2. Y.. Y.4 Y.. + ) 2 ) 2 ) 2 ) 2 Y 3. Y.. Y. Y.. + Y 3. Y.. Y.2 Y.. + ) 2 ) 2 ) 2 ) 2 Y 3. Y.. Y.3 Y.. + Y 3. Y.. Y.4 Y.. = 0.583 ) 2 ) 2 ) 2 ) 2 4.03 + 0.583 3.3 + 0.583 ) 2 ) 2 ) 2.7 + 0.583 5.9) 2 + 0.39 ) 2 ) 2 ) 2 ) 2 4.03 + 0.39 3.3 + 0.39 ) 2 ) 2 ) 2.7 + 0.39 5.9) 2 + 0.9 ) 2 ) 2 ) 2 ) 2 4.03 + 0.9 3.3 + 0.9 ) 2 ) 2 ) 2.7 + 0.9 5.9) 2 = 35.7500833. Thus, we have λ = 8.2709999 35.7500833 = 0.2335032.
2 It follows that the SST, SSA, SSB, and SSAB are SST = a b i= j= Yij Y.. ) 2 = 3.9 8.93 ) 2 + 4.2 8.93 ) 2 + 20.5 8.93 ) 2 ) 2 ) 2 + 24.8 8.93 + 5.7 8.93 +.3 8.93 ) 2 ) 2 ) 2 + 2.7 8.93 + 23. 8.93 + 5. 8.93 ) 2 ) 2 ) 2 + 5.4 8.93 + 9.9 8.93 + 2. 8.93 ) 2 = 20.907, SSA = a ) 2 ) 2 ) 2 ) 2 Y i. Y.. = 0.583 + 0.39 + 0.9 i= = 0.53043, SSB = b ) 2 ) 2 ) 2 ) 2 Y.j Y.. = 4.03 + 3.3 +.7 + 5.9) 2 SSAB = a j= = 7.3999999, and b i= j= λ 2 Y i. Y.. ) 2 Y.j Y.. ) 2 = 0.2335032) 2 35.7500833) =.935432. We then have SSE = SST SSA SSB SSAB = 20.907 0.53043 7.3999999.935432 = 4.027038.
27 The associated mean squares are MSA = 0.53043 3 MSB = 7.3999999 4 = 0.252083332; = 22.45; MSAB =.935432 =.935432; and MSE = 4.027038 3) 4) 3 4 = 28.225407. The observed value of F = MSAB /MSE is F observed =.935432 28.225407 = 0.078208973. We see that if λ = 0, then the probability of the random variable F,5 is greater than or equal to F observed is P F,5 F observed ) = P F,5 < F observed ) = FDist 0.078208973;, 5) = 0.8049399. These results suggest there is no two factor interaction. Lets assume the reduced model of no interaction between Factors A and B. Our model is µ ij = µ + τ ) i + τ 2 ) j.
28 The design matrix is X = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0. Our least squares estimates are X T X ) X T y = 8.93333333 0.5833333333 0.39 7 4.033 333333 3.33333333.7 7. Our total sum of squares, the sums of squares due to Factors A and B, and the
29 sum of squares due to error are SST = 2) 8.93333333) = 227.2; SSA = 0.5833333333) 2 + 0.397) 2 + 0.97) 2 = 0.53047; SSB = 4.033 333333) 2 + 3.33333333) 2 +.77) 2 + 5.9) 2 = 7.3999999; and SSE = SST SSA SSB = 59.295834. We observe that SSA/ a ) SSE/ a ) b )) SSB/ b ) SSE/ a ) b )) = 0.53047/ 3 ) 59.295834/ 3 ) 4 )) = 0.00999092203; and = 7.3999999/ 4 ) 59.295834/ 3 ) 4 )) = 0.84337388. The associated p-values are, respectively, P F 2, 0.00999092203) = FDist 0.00999092203; 2, ) = 0.99007525; and P F 3, 0.84337388) = FDist 0.84337388; 3, ) = 0.5808443. These results suggest that there is no effect due to either of the two factors. For the full model, the design matrix for the full model and our data in vector
30 form are X = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 and y = 3.9 4.2 20.5 24.8 5.7.3 2.7 23. 5. 5.4 9.9 2.. The matrix P associated with X T X ) is
3 P = 2 0 0 0 0 0 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 0 2 4 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 4 8 0 0 0 0 0 0 8 0 0 0 0 2 8 0 0 0 0 0 0 8 0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 0 0 0 0 0 0 8 3 2 8 3 8 3 8 3 2 8 3 3 2 3 0 0 3 2 3 3 3 2 3 2 2 0 3 0 8 3 3 2 3 3 2 3 3 3 2 The least squares estimates for the main effects and interaction is given by 8.93 0.583 0.39 4.03 3.3 θ = X T X ).7 X T y = 0.4 0.5 0.383 0.4083 0.083 0.083.
32 the contrasts of the main effects and interactions are 2 3θ θ2 + θ 3 ) 2 θ2 θ 3 ) 2 θ 4 + θ 5 + θ ) θ 2 4 θ ) 2 θ = P θ = θ 2 4 2θ 5 + θ ). 2 θ7 + θ 8 + θ 9 + θ 0 + θ + θ 2 ) 3 θ 7 + θ 8 + θ 9 θ 0 θ θ 2 ) 3 θ 2 7 θ 9 + θ 0 θ 2 ) θ 2 7 2θ 8 + θ 9 + θ 0 2θ + θ 2 ) 2 θ 7 θ 9 θ 0 + θ 2 ) 3 θ 7 2θ 8 + θ 9 θ 0 + 2θ θ 2 ) The coordinates of the random vector θ = P θ of estimators of the vector θ of the contrasts of the main effects and interactions are independent. The estimates for
33 these contrasts are θ = P θ = 5.5899059 0.49485534.378858223.8 7.03520254 3.53553390.520279580.77588004 0.80254038 0.4 0.3 0.34405. Removing the estimate θ = 5.5899059, we have the vector of ordered estimates of the given linear contrasts of the main effects and interactions along with the plotting positions for the corresponding normal probability plot given in the
34 following 2 matrix..5843359.8.09520 7.03520254 0.7288394047.77588004 0.49783072.378858223 0.2248908792 0.80254038 0 0.49485534 0.2248908792 0.3 0.49783072 0.34405 0.7288394047 0.4.09520.520279580.5843359 3.53553390. A plot of these points is given in the following figure.
35 Figure 2.3: Kutner Example, Probability Plot All the points seem to be plotting about a line except for the points with coordinates.5843359,.8) and.09520, 7.03520254). Using the other nine points, we estimate the line to be y = 0.5072897 + 2.774432x. A plot of this line along with our normal probability plot of the data is shown in the following figure.
3 Figure 2.4: Kutner Example, Probability Plot and Fitted Line 9 Points The point with coordinates.5843359, 3.53553390) may also be an outlier. To examine this possibility, we used the other eight points to estimate the line. This line is y = 0.54544288 +.73045299x. A plot of this line along with the normal probability plot of the data is given in the following figure.
37 Figure 2.5: Kutner Example, Probability Plot and Fitted Line 8 Points The contrast associated with the three estimates θ 4, θ 5, and θ are, respectively, 2 θ 4 + θ 5 + θ ), 2 θ 4 θ ), and 2 2 θ 4 2θ 5 + θ ). These contrast are all associated with Factor B: temperature. The plot shows no evidence there is an effect due to Factor A pressure) or interactions between Factors A and B which is expected.
38 2.7 Conclusion A method for analyzing unreplicated two factor experiments using selected contrasts has been presented. This method is based on a normal probability plot of the estimates the main effect contrasts and the interaction contrasts. This method provides the researcher a method of identifying the contrasts that are significantly different from zero. As was illustrated, each contrast is a contrast of a particular main effect or interaction. Hence, if a contrast is identified as being significantly different from zero, then it follows that the associated main effect or interaction is different from zero.
CHAPTER 3 THREE FACTOR EXPERIMENTS 3. Introduction The model for a three factor experiment k = 3) under the additive model expresses the response variable Y ijrs as Y ijrs = µ ijr + ɛ ijrs with µ ijr = µ + τ ) i + τ 2 ) j + τ 3 ) r + τ 2 ) ij + τ 3 ) ir + τ 23 ) jr + τ 23 ) ijr and ɛ ijrs iid N 0, σ 2 ) for i =,..., a, j =,..., b, r =,..., c, and s =,..., n. It is assumed that main effects and interactions are such that a b c c c i= τ ) i = 0; b j= τ 2) j = 0; c j= τ 2) ij = 0 for i =,..., a; a r= τ 3) r = 0; i= τ 2) ij = 0 for j =,..., b; τ 3) r= ir = 0 for i =,..., a; a τ 3) i= ir = 0 for r =,..., c; τ 23) r= jr = 0 for j =,..., b; b τ 23) j= jr = 0 for r =,..., c; r= τ 23) ijr = 0 for i =,..., a, j =,..., b; b j= τ 23) ijr = 0 for i =,..., a, r =,..., c; and a i= τ 23) ijr = 0 for j =,..., b, r =,..., c. This is referred to as the full model. One can reduce the model by assuming some of the interactions are zero. If this is done, we will refer to this model as the reduced model. We also assume that the ɛ ijrs s are independent and ɛ ijrs N 0, σ 2 ). We refer to these assumptions as the independent normal model. The design is an unreplicated
40 one if n =. Using matrix notation, we can write our additive model in the form Y = Xθ + ɛ, where Y is the abcn vector of observations, X is the abcn abc design matrix, θ is the abc vector of model parameters, and ɛ is the abcn vector of error terms. We are interested in studying the case in which n =. A study in which there is only one replicate per treatment does not allow one to perform an analysis of variance if the full model is assumed. For these data, there is not enough informations in the data to independently estimate the main effects and interactions and the common variance. Two methods have been suggested in the literature for analyzing the data from a design without replicates. The first of these is an extention of Tukey s method used to test for non-additivity. This is discussed in the next section. The second of these analyzes the data under a reduced model. This will be examined in Section 3. We present a third method in Section 4. In Section 5, we discuss the analysis of 3 k factorial designs. Some examples are given in Section. 3.2 Tukey s Method for Three Factors Tukey 949) method can be extended to develop tests for non-additivity for three factor experiments. In this case, one is to assume that the two and three factor interactions can be expressed in terms of the main effects and the parameters λ 2, λ 3, λ 23, and λ 23. Under Tukey s model, it is assumed that τ 2 ) ij = λ 2 τ ) i τ 2 ) j ; τ 3 ) ir = λ 3 τ ) i τ 3 ) r ; τ 23 ) jr = λ 23 τ 2 ) j τ 3 ) r ; and τ 23 ) ijr = λ 23 τ ) i τ 2 ) j τ 3 ) r.
In this model, there are a + b + c + 5 parameters including the common variance to be estimated. To determine these estimates using least squares, we define the function 4 Q = Q µ, τ ),..., τ ) a, τ 2 ),..., τ 2 ) b, τ 3 ),..., τ 3 ) c, λ 2, λ 3, λ 23, λ 23 ) = a b c i= j= r= Y ij µ τ ) i τ 2 ) j τ 3 ) r λ 2 τ ) i τ 2 ) j λ 3 τ ) i τ 3 ) r λ 23 τ 2 ) j τ 3 ) r λ 23 τ ) i τ 2 ) j τ 3 ) r ) 2 The least square estimates are the solutions to the following system of equations. Q µ = 0; Q τ ) i = 0; Q λ 2 = 0; Q λ 3 = 0; Q τ 2 ) j = 0; Q λ 23 = 0; and Q τ 3 ) j = 0; Q λ 23 = 0. It follows that the estimators for the model parameters τ ) i, τ 2 ) j, τ 3 ) r, λ 2, λ 3, λ 23, and λ 23 are τ ) i = Y i.. Y... ; τ 2 ) j = Y.j. Y... ; τ 3 ) r = Y..r Y... ; λ 2 = λ 3 = λ 23 = λ 23 = a b c i= j= r= c a b i= j= a b c i= j= r= b a c i= r= a b c i= j= r= a b c j= r= a b c i= j= r= a i= ) ) Y i.. Y... Y.j. Y... Yijr ) 2 ) 2 ; Y i.. Y... Y.j. Y... ) ) Y i.. Y... Y..r Y... Yijr ) 2 ) 2 ; Y i.. Y... Y..r Y... ) ) Y.j. Y... Y..r Y... Yijr ) 2 ) 2 ; and Y.j. Y... Y..r Y... ) ) ) Y i.. Y... Y.j. Y... Y..r Y... Yijr b c ) 2 ) 2 ) 2. j= r= Y i.. Y... Y.j. Y... Y..r Y...
42 We can now express Y ijr as Y ijr = Y... + ) ) ) Y i.. Y... + Y.j. Y... + Y..r Y... ) ) ) ) + λ 2 Y i.. Y... Y.j. Y... + λ3 Y i.. Y... Y..r Y... ) ) ) ) ) + λ 23 Y.j. Y... Y..r Y... + λ23 Y i.. Y... Y.j. Y... Y..r Y... + ɛijr. It follows that ɛ ijr can be expressed as ɛ ijr = Y ijr Y... ) ) ) Y i.. Y... Y.j. Y... Y..r Y... ) ) ) ) λ 2 Y i.. Y... Y.j. Y... λ3 Y i.. Y... Y..r Y... ) ) ) ) ) λ 23 Y.j. Y... Y..r Y... λ23 Y i.. Y... Y.j. Y... Y..r Y... The total sum of squares SST can be partitioned into the follow sums of squares. SSA = bc a SSB = ac b SSC = ab c SSAB = c a SSAC = b a SSBC = a b SSABC = a i= j= Y i.. Y... ) 2 with dfssa = a ; r= b Y.j. Y... ) 2 with dfssb = b ; Y..r Y... ) 2 with dfssc = c ; λ 2 2 i= j= c i= λ 2 3 r= c λ 2 23 j= r= b c λ 2 23 i= j= r= Y i.. Y... ) 2 Y.j. Y... ) 2 Y i.. Y... ) 2 Y..r Y... ) 2 Y.j. Y... ) 2 Y..r Y... ) 2 ; Y i.. Y... ) 2 Y.j. Y... ) 2 Y..r Y... ) 2 ; and SSE = SST SSA SSB SSC SSAB SSAC SSBC SSABC, where df SSAB = df SSAC = df SSBC = df SSABC =, df SSE = abc a b c 2, and SST = a b c ) 2 Yijr Y.... i= j= r=
43 Following the derivations in Tukey 949), one can shown that SSAB σ 2 χ 2, SSAC σ 2 χ 2, SSBC σ 2 χ 2, SSABC σ 2 χ 2,and SSE σ 2 χ 2 ab a b c 2. The observed values of the significance levels SLs) ) SSABC / SL 23 = P F,abc a b c 2 ; SSE / abc a b c 2) ) SSAB / SL 2 = P F,abc a b c 2 ; SSE / abc a b c 2) ) SSAC / SL 3 = P F,abc a b c 2 ; and SSE / abc a b c 2) ) SSBC / SL 23 = P F,abc a b c 2. SSE / abc a b c 2) are then examined. The observed significance levels OSLs) OSL 23, OSL 2, OSL 3, and OSL 23 can be used to judge if there is strong enough evidence in the data against the null hypotheses H 0 : λ 23 = 0, H 0 : λ 2 = 0, H 0 : λ 3 = 0, and H 0 : λ 23 = 0, respectively. Note that an observed significance level is commonly referred to as a p-value. The test for non-additivity has null and alternative hypotheses given by H 0 : λ 2 = λ 3 = λ 23 = λ 23 and H : H 0. The statistical test has decision rule that rejects the null hypotheis in favor of the alternative hypothesis if the observed value of MSAB MSE c AB MSAC MSE c AC MSBC MSE c BC MSABC MSE c ABC, where MSAB = SSAB, MSAC = SSAC MSABC = SSABC, and MSE =, MSBC = SSBC, SSE ab a b c 2.
44 The size of the test α is ) ) MSAB MSAC α = P c MSE AB + P c MSE AC ) ) MSBC MSABC +P c MSE BC + P c MSE ABC. If each of the critical values are selected to be the 00 γ)th percentile of the appropriate F -distribution, then α = 4γ or γ = α/4. 3.3 Analyzing a Reduced Model For the case in which n =, a reduced model can be entertained by assuming some of the parameters in the model associated with interactions are zero. Under this new assumption there is information in the data that can be used to estimate the common variance. For example, if there are no three factor interactions, our reduced model becomes µ ijr = µ + τ ) i + τ 2 ) j + τ 3 ) r + τ 2 ) ij + τ 3 ) ir + τ 23 ) jr. It follows that SSE under this reduced model is the SSABC under the full model. The SST can be partitioned as SST = SSA + SSB + SSC + SSAB + SSAC + SSBC + SSE. An ANOVA can then be used to analyze the data. There are many other possible reduced models that assumes various parameters representing interactions are zero. For example, suppose that a = 5 and b = 7.
45 3.4 Analysis of Contrasts The analysis of contrast is the same for any factorial experiment. The matrix P is determined such that X T X ) = PP T. The estimates θ of the vector of parameters θ for the full model are transformed into the vector of contrast θ = P θ. Under the independent normal model, θ N abc θ = P θ, Iσ 2). We observe that the contrasts of the estimates in the vector θ associated with a main effect or an interaction are the corresponding components of θ. Removing the contrast associate with the overall mean in θ, a normal probability plot of the remaining components can be examined. Points on the plot that provide evidence against the hypothesis θ = 0 are analyzed. These points suggest that the given parameter contrast differs from zero. 3.5 Unreplicated 3 k Factorial Designs In the analysis of a 3 k factorial experiment using contrasts, one needs the design matrix X to estimate the parameters in the full model and the matrix P such that X T X ) = PP T. However, one may have software that can be used to determine the estimates θ and then one can find θ = P θ. In what follows, we demonstrate that the matrix P has a general form.
4 Define the sequence of matrices, B 2 0, B 2,..., B 2 k as B 2 0 = 3 [], k B 2 = 3 k 2 2 = 2B 2 0 B 2 0 B 2 0 2B 2 0, 4 2 2 B 2 2 = 2 4 2 = 3 k 2B 2 B 2, 2 4 2 B 2 2B 2 2 2 4 8 4 4 2 4 2 2 4 8 2 4 2 4 2 4 2 8 4 2 4 2 B 2 3 = 2 4 4 8 2 2 4 3 k = 4 2 2 8 4 4 2 2 4 2 4 8 2 4 2 4 2 4 2 8 4 2 2 4 2 4 4 8 2B 2 2 B 2 2 B 2 2 2B 2 2, B 2 k =. 2B 2 k B 2 k B 2 k 2B 2 k. The matrix X T X ) can be expressed as block diagonal matrix with 2 k matrices on the block diagonal. The first block matrix is B 2 0, the matrix B 2 is the next ) k block matrices, the matrix B 2 2 is the next k 2) block matrices etc. It follows that P is a block diagonal matrix of the same construct as X T X ) with B2 i replaced with P 2 i for i =,..., k, where B 2 i = P 2 ip T 2 i.
47 for i = 0,, 2,..., k. 3. Example Montgomery 997) gives an example of a three factor experiment in which n =. He states in his example -3 that the process engineer can control three variables during the filling process: the percent carbonationa), the operating pressure in the filler B), and the bottles produced per minute or the line speed C). His data is presented in the following table. Table 3.: Montgomery s Example -3 Operating pressure 25psi 30psi Line speed Line speed Percent carbonation 200 250 200 250 0-4 - - 2 2 3 5 4 9 3 2
48 We see that Y.. = 4 + 2 = ; 4 Y 2.. = + 3 + 5 + = 5; 4 9 + 3 + + 2 Y 3.. = = 4.75; 4 4 + + 9 + 5 + Y.. = = 4.3; + 3 + 3 + 2 + + 2 Y.2. = = 8.; 4 + + 9 + 3 + 3 Y.. = = 3.5; + 5 + + 2 + + 2 Y..2 = = 9; Y... = 4 + + 9 + 5 + + 3 + 3 + 2 + + 2 = 75; and Y... = 75 2 =.25. Assuming interaction between the three factors has the form τ 2 ) ij = λ 2 τ ) i τ 2 ) j, τ 3 ) ij = λ 3 τ ) i τ 3 ) r, τ 23 ) ij = λ 23 τ 2 ) j τ 3 ) r, and τ 23 ) ijr = λ 23 τ ) i τ 2 ) j τ 3 ) r then the least squares estimates of the parameters µ, τ ), τ ) 2, τ ) 3, τ 2 ), τ 2 ) 2,
49 τ 3 ), and τ 3 ) 2 are µ = Y... =.25, τ ) = Y.. Y... =.25 = 7.25, τ ) 2 = Y 2.. Y... = 5.25 =.25, τ ) 3 = Y 3.. Y... = 4.75.25 = 8.5, τ 2 ) = Y.. Y... = 4.3.25 =.9, τ 2 ) 2 = Y.2. Y... = 8..25 =.9, τ 3 ) = Y.. Y... = 3.5.25 = 2.75, and τ 3 ) 2 = Y..2 Y... = 9.25 = 2.75. Recall that the formula to be used to obtain an estimate of λ is a b c ) ) i= j= r= Y i.. Y... Y.j. Y... Yijr λ 2 = c a b ) 2 ) 2, i= j= Y i.. Y... Y.j. Y... a b c ) ) i= j= r= Y i.. Y... Y..r Y... Yijr λ 3 = b a c ) 2 ) 2, i= r= Y i.. Y... Y..r Y... a b c ) ) i= j= r= Y.j. Y... Y..r Y... Yijr λ 23 = a b c ) 2 ) 2, and j= r= Y.j. Y... Y..r Y... a b c ) ) ) i= j= r= Y i.. Y... Y.j. Y... Y..r Y... Yijr λ 23 = c ) 2 ) 2 ) 2. r= Y i.. Y... Y.j. Y... Y..r Y... a b i= j=
50 It follows that 3 2 2 i= = 2 3 j= 2 i= j= r= Y i.. Y... ) Y.j. Y... ) Yijr Y i.. Y... ) Y.j. Y... ) Yijr = 2 ) ) Y.. Y... Y.. Y... Y + 2 ) ) Y.. Y... Y.. Y... Y2 + 2 ) ) Y.. Y... Y.2. Y... Y2 + 2 ) ) Y.. Y... Y.2. Y... Y22 + 2 ) ) Y 2.. Y... Y.. Y... Y2 + 2 ) ) Y 2.. Y... Y.. Y... Y22 + 2 ) ) Y 2.. Y... Y.2. Y... Y22 + 2 ) ) Y 2.. Y... Y.2. Y... Y222 + 2 ) ) Y 3.. Y... Y.. Y... Y3 + 2 ) ) Y 3.. Y... Y.. Y... Y32 + 2 ) ) Y 3.. Y... Y.2. Y... Y32 + 2 ) ) Y 3.. Y... Y.2. Y... Y322 = 2 7.25).9 ) 4) + 2 7.25).9 ) ) + 2 7.25).9 ) ) + 2 7.25).9 ) 2) + 2.25).9 ) ) + 2.25).9 ) 5) + 2.25).9 ) 3) + 2.25).9 ) ) + 2 8.5).9 ) 9) + 2 8.5).9 ) ) + 2 8.5).9 ) 3) + 2 8.5).9 ) 2) = 88.7,
5 3 2 2 i= = 2 3 j= 2 i= r= r= Y i.. Y... ) Y..r Y... ) Yijr Y i.. Y... ) Y..r Y... ) Yijr = 2 ) ) Y.. Y... Y.. Y... Y + 2 ) ) Y.. Y... Y..2 Y... Y2 + 2 ) ) Y.. Y... Y.. Y... Y2 + 2 ) ) Y.. Y... Y..2 Y... Y22 + 2 ) ) Y 2.. Y... Y.. Y... Y2 + 2 ) ) Y 2.. Y... Y..2 Y... Y22 + 2 ) ) Y 2.. Y... Y.. Y... Y22 + 2 ) ) Y 2.. Y... Y..2 Y... Y222 + 2 ) ) Y 3.. Y... Y.. Y... Y3 + 2 ) ) Y 3.. Y... Y..2 Y... Y32 + 2 ) ) Y 3.. Y... Y.. Y... Y32 + 2 ) ) Y 3.. Y... Y..2 Y... Y322 = 2 7.25) 2.75) 4) + 2 7.25) 2.75) ) + 2 7.25) 2.75) ) + 2 7.25) 2.75) 2) + 2.25) 2.75) ) + 2.25) 2.75) 5) + 2.25) 2.75) 3) + 2.25) 2.75) ) + 2 8.5) 2.75) 9) + 2 8.5) 2.75) ) + 2 8.5) 2.75) 3) + 2 8.5) 2.75) 2) = 379.5,
52 3 2 2 i= = 3 2 j= 2 j= r= r= Y.j. Y... ) Y..r Y... ) Yijr Y.j. Y... ) Y..r Y... ) Yijr = 3 ) ) Y.. Y... Y.. Y... Y + 3 ) ) Y.. Y... Y..2 Y... Y2 + 3 ) ) Y.2. Y... Y.. Y... Y2 + 3 ) ) Y.2. Y... Y..2 Y... Y22 + 3 ) ) Y.. Y... Y.. Y... Y2 + 3 ) ) Y.. Y... Y..2 Y... Y22 + 3 ) ) Y.2. Y... Y.. Y... Y22 + 3 ) ) Y.2. Y... Y..2 Y... Y222 + 3 ) ) Y.. Y... Y.. Y... Y3 + 3 ) ) Y.. Y... Y..2 Y... Y32 + 3 ) ) Y.2. Y... Y.. Y... Y32 + 3 ) ) Y.2. Y... Y..2 Y... Y322 = 3.9 ) 2.75) 4) + 3.9 ) 2.75) ) + 3.9 ) 2.75) ) + 3.9 ) 2.75) 2) + 3.9 ) 2.75) ) + 3.9 ) 2.75) 5) + 3.9 ) 2.75) 3) + 3.9 ) 2.75) ) + 3.9 ) 2.75) 9) + 3.9 ) 2.75) ) + 3.9 ) 2.75) 3) + 3.9 ) 2.75) 2) = 79.0249999,
53 3 2 2 i= j= r= Y i.. Y... ) Y.j. Y... ) Y..r Y... ) Yijr = ) ) ) Y.. Y... Y.. Y... Y.. Y... Y + ) ) ) Y.. Y... Y.. Y... Y..2 Y... Y2 + ) ) ) Y.. Y... Y.2. Y... Y.. Y... Y2 + ) ) ) Y.. Y... Y.2. Y... Y..2 Y... Y22 + ) ) ) Y 2.. Y... Y.. Y... Y.. Y... Y2 + ) ) ) Y 2.. Y... Y.. Y... Y..2 Y... Y22 + ) ) ) Y 2.. Y... Y.2. Y... Y.. Y... Y22 + ) ) ) Y 2.. Y... Y.2. Y... Y..2 Y... Y222 + ) ) ) Y 3.. Y... Y.. Y... Y.. Y... Y3 + ) ) ) Y 3.. Y... Y.. Y... Y..2 Y... Y32 + ) ) ) Y 3.. Y... Y.2. Y... Y.. Y... Y32 + ) ) ) Y 3.. Y... Y.2. Y... Y..2 Y... Y322 = 7.25).9 ) 2.75) 4) + 7.25).9 ) 2.75) ) + 7.25).9 ) 2.75) ) + 7.25).9 ) 2.75) 2) +.25).9 ) 2.75) ) +.25).9 ) 2.75) 5) +.25).9 ) 2.75) 3) +.25).9 ) 2.75) ) + 8.5).9 ) 2.75) 9) + 8.5).9 ) 2.75) ) + 8.5).9 ) 2.75) 3) + 8.5).9 ) 2.75) 2) = 8.44797,
54 2 3 2 i= j= Y i.. Y... ) 2 Y.j. Y... ) 2 = 2 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.. Y... + 2 Y.. Y... Y.. Y... + 2 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.2. Y... + 2 Y.. Y... Y.2. Y... + 2 ) 2 ) 2 ) 2 ) 2 Y 2.. Y... Y.. Y... + 2 Y 2.. Y... Y.. Y... + 2 ) 2 ) 2 ) 2 ) 2 Y 2.. Y... Y.2. Y... + 2 Y 2.. Y... Y.2. Y... + 2 ) 2 ) 2 ) 2 ) 2 Y 3.. Y... Y.. Y... + 2 Y 3.. Y... Y.. Y... + 2 ) 2 ) 2 ) 2 ) 2 Y 3.. Y... Y.2. Y... + 2 Y 3.. Y... Y.2. Y... = 2 7.25) 2.9 ) 2 + 2 7.25) 2.9 ) 2 + 2 7.25) 2.9 ) 2 + 2 7.25) 2.9 ) 2 + 2.25) 2.9 ) 2 + 2.25) 2.9 ) 2 + 2.25) 2.9 ) 2 + 2.25) 2.9 ) 2 + 2 8.5) 2.9 ) 2 + 2 8.5) 2.9 ) 2 + 2 8.5) 2.9 ) 2 + 2 8.5) 2.9 ) 2 = 374.020834,
55 2 3 2 i= r= Y i.. Y... ) 2 Y..r Y... ) 2 = 2 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.. Y... + 2 Y.. Y... Y..2 Y... + 2 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.. Y... + 2 Y.. Y... Y..2 Y... + 2 ) 2 ) 2 ) 2 ) 2 Y 2.. Y... Y.. Y... + 2 Y 2.. Y... Y..2 Y... + 2 ) 2 ) 2 ) 2 ) 2 Y 2.. Y... Y.. Y... + 2 Y 2.. Y... Y..2 Y... + 2 ) 2 ) 2 ) 2 ) 2 Y 3.. Y... Y.. Y... + 2 Y 3.. Y... Y..2 Y... + 2 ) 2 ) 2 ) 2 ) 2 Y 3.. Y... Y.. Y... + 2 Y 3.. Y... Y..2 Y... = 2 7.25) 2 2.75) 2 + 2 7.25) 2 2.75) 2 + 2 7.25) 2 2.75) 2 + 2 7.25) 2 2.75) 2 + 2.25) 2 2.75) 2 + 2.25) 2 2.75) 2 + 2.25) 2 2.75) 2 + 2.25) 2 2.75) 2 + 2 8.5) 2 2.75) 2 + 2 8.5) 2 2.75) 2 + 2 8.5) 2 2.75) 2 + 2 8.5) 2 2.75) 2 = 745.875,
5 3 2 2 j= r= Y.j. Y... ) 2 Y..r Y... ) 2 = 3 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.. Y... + 3 Y.. Y... Y..2 Y... + 3 ) 2 ) 2 ) 2 ) 2 Y.2. Y... Y.. Y... + 3 Y.2. Y... Y..2 Y... + 3 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.. Y... + 3 Y.. Y... Y..2 Y... + 3 ) 2 ) 2 ) 2 ) 2 Y.2. Y... Y.. Y... + 3 Y.2. Y... Y..2 Y... + 3 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.. Y... + 3 Y.. Y... Y..2 Y... + 3 ) 2 ) 2 ) 2 ) 2 Y.2. Y... Y.. Y... + 3 Y.2. Y... Y..2 Y... = 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 + 3.9 ) 2 2.75) 2 = 000.4025,
57 3 2 2 i= j= r= Y i.. Y... ) 2 Y.j. Y... ) 2 Y..r Y... ) 2 = ) 2 ) 2 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.. Y... Y.. Y... + Y.. Y... Y.. Y... Y..2 Y... + ) 2 ) 2 ) 2 ) 2 ) 2 ) 2 Y.. Y... Y.2. Y... Y.. Y... + Y.. Y... Y.2. Y... Y..2 Y... + ) 2 ) 2 ) 2 ) 2 ) 2 ) 2 Y 2.. Y... Y.. Y... Y.. Y... + Y 2.. Y... Y.. Y... Y..2 Y... + ) 2 ) 2 ) 2 ) 2 ) 2 ) 2 Y 2.. Y... Y.2. Y... Y.. Y... + Y 2.. Y... Y.2. Y... Y..2 Y... + ) 2 ) 2 ) 2 ) 2 ) 2 ) 2 Y 3.. Y... Y.. Y... Y.. Y... + Y 3.. Y... Y.. Y... Y..2 Y... + ) 2 ) 2 ) 2 ) 2 ) 2 ) 2 Y 3.. Y... Y.2. Y... Y.. Y... + Y 3.. Y... Y.2. Y... Y..2 Y... = 7.25) 2.9 ) 2 2.75) 2 + 7.25) 2.9 ) 2 2.75) 2 + 7.25) 2.9 ) 2 2.75) 2 + 7.25) 2.9 ) 2 2.75) 2 +.25) 2.9 ) 2 2.75) 2 +.25) 2.9 ) 2 2.75) 2 +.25) 2.9 ) 2 2.75) 2 +.25) 2.9 ) 2 2.75) 2 + 8.5) 2.9 ) 2 2.75) 2 + 8.5) 2.9 ) 2 2.75) 2 + 8.5) 2.9 ) 2 2.75) 2 + 8.5) 2.9 ) 2 2.75) 2 = 4043.428. Thus, we have λ 2 = 88.7 374.020834 = 0.02373887242, λ 3 = 379.5 745.875 = 0.049358242, λ 23 = 79.0249999 000.4025 = 0.0790538339, and λ 23 = 8.44797 4043.428 = 0.00333493.
58 It follows that the SST, SSA, SSB, and SSAB are SST = 3 2 2 i= j= r= Yijr Y... ) 2 = 4.25) 2 +.25) 2 +.25) 2 + 2.25) 2 +.25) 2 + 3.25) 2 + 5.25) 2 +.25) 2 + 9.25) 2 + 3.25) 2 +.25) 2 + 2.25) 2 = 5.25, SSA = 4 3 ) 2 Y i.. Y... = 4 7.25) 2 +.25) 2 + 8.5) 2 ) = 505.5, SSB = 2 i= j= Y.j. Y... ) 2 =.9 ) 2 +.9 ) 2) = 44.08333335, SSC = 2 ) 2 Y..r Y... = 2.75) 2 + 2.75) 2 ) = 90.75, SSAB = 3 r= 2 λ 2 2 i= j= Y i.. Y... ) 2 Y.j. Y... ) 2 = 0.02373887242) 2 374.020834/2) =.0448827, SSAC = 3 2 ) 2 ) 2 Y i.. Y... Y..r Y... i= r= λ 2 3 = 0.049358242) 2 745.875/2) = 9.4839727, SSBC = 2 2 ) 2 ) 2 Y.j. Y... Y..r Y... j= r= λ 2 23 = 0.0790538339) 2 000.4025/3) = 2.083333333, SSABC = 3 2 2 ) 2 ) 2 ) 2 Y i.. Y... Y.j. Y... Y..r Y... i= j= r= λ 2 23 = 0.00333493) 2 4043.428) = 0.0242334322.