Solutions to Exercises

1 c Atkinson et al 2007, Optimum Experimental Designs, with SAS Solutions to Exercises 1. and 2. Certainly, the solutions to these questions will be different for every reader. Examples of the techniques that are required are given in the text of the book. Becoming an expert, of course, requires a lot of practice. 3. When the block effects are fixed, the condition for orthogonality defined in Chapter 15 requires the identities N 1 ij F ij T 1 Nij = N 1 F T 1 N, i = 1, 2,..., B; j = 1, 2,..., b i, (1) to be satisfied. In this case B = 2, b 1 = b 2 = 2, N = 18, N 11 = N 12 = 6, N 21 = N 22 = 3, and the columns of F correspond to the values of the regressors: the intercept, x and, when the model is second order polynomial, x 2, at each of the observations of the corresponding blocks, i.e. 1 1 1 1 1 1 1 0 0 F 11 = F 12 = 1 0 0 1 1 1 1 1 1 and F 21 = F 22 = 1 1 1 1 0 0 1 1 1 The average level of the regressors in an orthogonally blocked design is the same at each level of each blocking variable: zero for x and two thirds for x 2. Therefore the parameters corresponding to x and x 2 (if the model is of second order) will be estimated independently from the block effects when this design is used. It is easy to verify empirically the orthogonality of the design when the blocks are random. For example, the following program can be used. The blocking variables are z 1 and z 2. The sums of squares SS I and SS III are identical. Note that a block interaction cannot be estimated with this design.. data mydata; input z1 z2 x y;

2 cards; 1 1-1 5 1 1 0 3 1 1 1 2 1 1-1 2 1 1 0 7 1 1 1 4 1 2-1 2 1 2 0 1 1 2 1 2 1 2-1 4 1 2 0 6 1 2 1 2 2 1-1 7 2 1 0 2 2 1 1 9 2 2-1 3 2 2 0 3 2 2 1 6 ; proc glm; class z1 z2; model y = z1 z2 x x*x / e1; random z1 z2; 4. A possible way to analyse the data is to run the SAS program given below. data Elastomer; do Oil = 0 to 30 by 10; do Filler = A, B, C ; do Level = 0 to 60 by 12; input Viscosity @@; logv = log(viscosity); output; end; end; end; datalines; 26 28 30 32 34 37 26 38 50 76 108 157 25 30 35 40 50 60 18 19 20 21 24 24 17 26 37 53 83 124 18 21 24 28 33 41 12 14 14 16 17 17 13 20 27 37 57 87 13 15 17 20 24 29

3. 12 12 13 14 14. 15 22 27 41 63 11 14 15 17 18 25 ; proc glm data=elastomer; where (Filler= A ); model logv = Level Oil; proc glm data=elastomer; where (Filler= A ); model logv = Oil Level ; proc glm data=elastomer; where (Filler= C ); model logv = Level Oil; proc glm data=elastomer; where (Filler= C ); model logv = Oil Level ; As the sums of squares SS I and SS III are identical for Filler C, the design is orthogonal. However, this is not true for Filler A as in this case the lost observation has destroyed this property of the planned design. Whether any of the proposed experimental designs is suitable depends on whether any further changes are planned for the new experiment. Provided that there are no such changes, a 2-level factorial design would be inappropriate to use as it would not allow for estimating a second order polynomial model. The points of the composite design when the star points are equal to -1 or 1 form a useful subset of the points of the 3-level factorial design. Such a design would be a good choice, when compared to a complete 3-level or higher order factorial design because it requires a relatively small number of observations to be taken. This design does not satisfy criteria 5 and 14, but satisfies criteria 1-4, 7, 10 and 13 from the criteria listed in Chapter 6. To some extend this also applies to criteria 8, 9 and 12, though if any of these is of particular concern, it might be possible to construct a better experimental design. An algorithmic search for a design would be appropriate when the available experimental resources require a number of observations different from that of the points of the composite design. This would also be a useful approach if the experimental scenario has changed and for example blocking is required, the design region has been changed or a different model has to be estimated. 5. Suppose that a quadratic term for the ith explanatory variable is significant. Such a model cannot be fitted as the column corresponding to the quadratic term in the extended design matrix consists of ones, exactly as that for the intercept in the model. Therefore ˆβ 0 is an estimate of β 0 + β ii. A lack of fit test would not pick up this potentially serious problem.

4 6. For simplicity of notation we rename the factors A, B,..., H. In order to obtain a design with 16 observations, 4 of the factors have to confounded by the design with high order interactions. For example, we can choose: E = BCD; F = ACD; G = ABC; H = ABD. The design can then be constructed by first writing down a complete 2-level factorial design for A, B and C; the values of the remaining factors can then be obtained using the above identities. As each factor is equal to ±1, its squared value is always equal to 1. Hence, we obtain I = ABCG = ABDH = ABEF = ACDF = ACEH = ADEG = AF GH = BCDE = BCF H = BDF G = BEGH = CDGH = CEF G = DEF H The intercept in the model is confounded with fourth order interactions. Therefore the estimate of the intercept is a sum of the effects I + ABCG + ABDH + ABEF + ACDF + ACEH + ADEG + AF GH +BCDE + BCF H + BDF G + BEGH + CDGH + CEF G + DEF H. If fourth order interactions are not significant, the intercept will be unbiased. In order to obtain the alias structure the above identity can be multiplied by one or more factors. No more than 15 further identities can be obtained. Each of them corresponds to a parameter that can be included in the model. Assuming that interactions of order higher than 3 are not significant, in a similar way as above we obtain: A + BCG + BDH + BEF + CDF + CEH + DEG + F GH B + ACG + ADH + AEF + CDE + CF H + DF G + EGH C + ABG + ADF + AEH + BDE + BF H + DGH + EF G D + ABH + ACF + AEG + BCE + BF G + CGH + EF H E + ABF + ACH + ADG + BCD + BGH + CF G + DF H F + ABE + ACD + AGH + BCH + BDG + CEG + DEH G + ABC + ADE + AF H + BDF + BEH + CDH + CEF H + ABD + ACE + AF G + BCF + BEG + CDG + DEF AB + CG + DH + EF AC + BG + DF + EH AD + BH + CF + EG AE + BF + CH + DG AF + BE + CD + GH AG + BC + DE + F H

5 AH + BD + CE + F G We would not be able to estimate effects that appear on the same rows above. As AB, AC and BC are on different lines they are estimable. (b). This becomes possible when we relabel G = x 8 and H = x 7. No more interactions would be estimable as the design becomes saturated. (c). Clearly the problem becomes more complicated. We need to choose a block generator, say AB, as well as 4 design generators, similarly to (a) above: E = BCD; F = ACD; G = ABC; H = ABD. We will no longer be able to estimate the same model as in (a) and (b). Estimable effects are listed below in a similar fashion to (b): I + ABCG + ABDH + ABEF + ACDF + ACEH + ADEG + AF GH +BCDE + BCF H + BDF G + BEGH + CDGH + CEF G + DEF H Block = AB + CG + DH + EF A + BCG + BDH + BEF + CDF + CEH + DEG + F GH B + ACG + ADH + AEF + CDE + CF H + DF G + EGH C + ABG + ADF + AEH + BDE + BF H + DGH + EF G D + ABH + ACF + AEG + BCE + BF G + CGH + EF H E + ABF + ACH + ADG + BCD + BGH + CF G + DF H F + ABE + ACD + AGH + BCH + BDG + CEG + DEH G + ABC + ADE + AF H + BDF + BEH + CDH + CEF H + ABD + ACE + AF G + BCF + BEG + CDG + DEF AC + BG + DF + EH AD + BH + CF + EG AE + BF + CH + DG AF + BE + CD + GH AG + BC + DE + F H AH + BD + CE + F G (d). If the interactions between x 1, x 2 and x 3 have to be estimated, we need to label them A, C and D in order to use the same design. However, it is not possible to estimate in addition two factor interactions of x 8 with x 4, x 5, x 6 and x 7. If such interactions are deemed significant, using this design would be inappropriate. 7. If 18 observations can be collected, it would still be possible to use one of the 16-trial designs described in Exercise 6. If the factors are continuous, the extra 2 observations can be taken at the centre of the experimental region, thus providing an opportunity to test for lack of fit of the model. Alternatively, an experimental design

6 Table 1: Central composite design for 3 explanatory variables x 1 x 2 x 3-1 -1-1 1-1 -1-1 1-1 1 1-1 -1-1 1 1-1 1-1 1 1 1 1 1-1 0 0 1 0 0 0-1 0 0 1 0 0 0-1 0 0 1 with 18 observations can be constructed using one of algorithms described in Chapters 12 and 13. What would be best depends on the experiment in which the design will be used. 8. Table 1 lists the points of a central composite design for 3 explanatory variables in standard order. If the observations are taken sequentially, their order should be randomized. This design would allow a second (or lower) order polynomial model to be estimated. 9. If a polynomial model is estimable when a complete or fractional two-level factorial design ξ is used, its information matrix is diagonal so that Hence the design is orthogonal. Also, F T F = NI p. d(x, ξ) = f T (x)f(x) = p fi 2 (x). This expression is maximised and equal to p when the elements of x are -1 or 1. When the absolute value of any element of x is less that 1, p i=1 f i 2 (x) < p. Then from the General Equivalence Theorem it follows that such a design is both D- and G-optimum. 10. (a). D = 1, 0.2, 0.2, 1. For example (1.4 1.5)/0.5 = 0.2. (b). d(x) = 1 + 1.92D 2. (c). The maximum of d(x) = 2.92 > 2 (= p), achieved at D = ±1, hence this design is not D-or G-optimum. The design is orthogonal; hence the parameters in the model will be estimated independently from each other. i=1

7 Figure 1: G for different values of a (d). D = ±1 are equally good with respect to D-optimality. Non-statistical considerations can be taken into account to decide between the two doses (1 and 2 mg/bw). (e). The new dose must be 3 in order to cover the new required experimental region. The new design is D = 1, 0.6, 0.4, 0, 1. Its standardised variance of prediction is d(x) = 2.156D 2 + 0.862D + 1.086. The next dose to use is that maximizing d(x). 11. In this case Therefore F = 1 1 1 1 1 a. G = F T F = 2a 2 + 6, which is maximum when a = ±1 (see Figure 1). Also d(x, ξ) = 3(a 2 2ax + 2 + 3x 2 )/(2a 2 + 6). Figure 2 shows plots of d(x) for a = 0, 0.5 and 1. When a = 1, the design is both D- and G-optimum. 12. In this case F = 1 1 1 1 1 1 1 1 Therefore the standardized variance is [ 1/4 0 d(x) = 4(x 1 x 2 ) 0 1/4. ] ( x1 x 2 ) = x 2 1 + x 2 2

8 Figure 2: d(x) for different values of a and has a maximum at x 1 = ±1 and x 2 = ±1 equal to the number of the parameters 2. Hence, the design is D- and G-optimum. Similarly, it is easy to see that a design including the points (1,1) and (-1,1) is both D- and G-optimum as d(x) takes exactly the same form. The optimum design is not unique and alternative symmetric arrangements of the these two points are also D-optimum. Certainly such designs are not useful because one of the explanatory variables is not varied in the design. 13. The General Equivalence Theorem makes it easy to check whether or not an experimental design is D- and G-optimum. It is sufficient to verify that the maximum standardised variance of the prediction is equal to the number of model parameters at the design points, and less elsewhere in the design region. (a) and (b). d(x) = 1 + x 2. The design is D- and G-optimum as x 1. (c) - (e). The designs are not D- and G-optimum: (c) d(x) = 1 + 1.5x 2 ; (d) d(x) = 1 + 2.25x 2 ; (e) d(x) = 4(1 x + x 2 )/3. The next point to be added to the design that maximally increases the determinant of the information matrix is -1 or 1 for all designs except for (e), for which it is -1. 14. As defined in Section 10.1, an experimental design ξ is A-optimum if p i=1 λ 1 i is minimized, where λ i is the ith eigenvalue of M(ξ). Using a Lagrange multiplier to constrain the optimization so that p i=1 λ i = p, it becomes clear that the minimum is achieved when all eigenvalues are equal to 1. The D-optimality of any two-level factorial or fractional factorial design used to estimate a first-order polynomial model, which may also include some interactions, was shown in Exercise 9. As M(ξ) = n 1 F T F = I p all its eigenvalues λ i, i = 1,..., p, are equal to 1, hence these designs are also A-optimum. In general an experimental design would be both D- and A-optimum if its information matrix is proportional to the identity matrix.

9 15. Orthogonal coding of the list of candidate points ensures that F T c F c = N c I, where F c is the extended design matrix for all candidate points and N c is their number. Minimizing the average standardised variance of prediction over the candidate points is equivalent to minimizing trace[(f T F) 1 ] required by the A-optimality criterion because Nc 1 trace[f c (F T F) 1 F T c ] = N 1 c trace[(f T F) 1 F T c F c ] = trace[(f T F) 1 ], where F is the extended design matrix. 16. Omitting explanatory variables in the analysis of data may lead to wrong results, and is inferior to analysing all data, even if the two sets of variables (those that are considered important and those that are not) are orthogonal. In addition to estimation of the parameters of the linear model, estimation of the error variance is usually important. Omission of important terms in the linear model will lead to heterogeniety of the error variance and underestimation of significance. Similarly, omitting variables in designing an experiment would lead to an inferior design as the estimates of the effects of both types of variable are not independent of each other. In fact, the volume of the confidence ellipsoid for the parameters of interest is proportional to M 11 (ξ) M 12 (ξ)m 1 22 (ξ)m T 12(ξ) = M(ξ) M 22 (ξ), not M 11, provided that the partitioning of the information matrix is done in such a way that the set of the important explanatory variables is first, while the remaining parameters are in the second set. 17. Simplex centroid designs include all possible mixtures in which the components which are included in the mixture have equal proportions. For a Scheffé polynomial of degree d the simplex lattice designs include all possible mixtures in which each component takes values 0, 1/d,..., and 1. In this exercise d = 3. The Simplex Lattice Design and the simplex centroid design for a mixture with components A, B, C and D are given in Tables 2 and 3, respectively. The simplex lattice design is saturated for the Scheffé polynomial of third degree and has 20 observations, while the Simplex Centroid Design has only 15 observation and cannot be used to fit the same model. However, it can be used to estimate the 10 parameters of the second-order Scheffé polynomial model. Hence, if the third-order Scheffé polynomial has to be estimated, the simplex centroid design has to be used. It is recommended to add a few more points to this design in order to test the model for lack of fit. If the observations are collected sequentially, it is recommended that the order of the observations be randomised. 18. (a). In this case F = 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1/2 1/2 0 1/4 0 0 1/2 0 1/2 0 1/4 0 0 1/2 1/2 0 0 1/4,

10 Table 2: Simplex Lattice Design A B C D 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.67 0.33 0.00 0.00 0.67 0.00 0.33 0.00 0.67 0.00 0.00 0.33 0.00 0.67 0.33 0.00 0.00 0.67 0.00 0.33 0.00 0.00 0.67 0.33 0.33 0.67 0.00 0.00 0.33 0.00 0.67 0.00 0.33 0.00 0.00 0.67 0.00 0.33 0.67 0.00 0.00 0.33 0.00 0.67 0.00 0.00 0.33 0.67 0.33 0.33 0.33 0.00 0.33 0.33 0.00 0.33 0.33 0.00 0.33 0.33 0.00 0.33 0.33 0.33

11 Table 3: Simplex Centroid Design A B C D 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.50 0.50 0.00 0.00 0.50 0.00 0.50 0.00 0.50 0.00 0.00 0.50 0.00 0.50 0.50 0.00 0.00 0.50 0.00 0.50 0.00 0.00 0.50 0.50 0.33 0.33 0.33 0.00 0.33 0.33 0.00 0.33 0.33 0.00 0.33 0.33 0.00 0.33 0.33 0.33 0.25 0.25 0.25 0.25 where the first three columns coincide with the design. Then (F T F ) 1 = 1 0 0 2 2 0 0 1 0 2 0 2 0 0 1 0 2 2 2 2 0 24 4 4 2 0 2 4 24 4 0 2 2 4 4 24. It is now easy to check that the standardised variance is equal to the number of model parameters (i.e. NN 1 i = 6/1 = 6) at the designs points and less elsewhere. (b). The estimators for the parameters of a second-order Scheffé polynomial are: ˆβ i = y i ˆβ ij = 4y ij 2y i 2y j, where the subscripts i and j correspond to the case when the corresponding mixture components are used. For example, y ij is the response when a mixture of equal amounts of components i and j is used. If replications are available y i and y ij are replaced by the averages of the responses at the corresponding experimental conditions.

12 (c). In this case (F T F ) 1 = 1 0 0 2 2 0 0 1 0 2 0 2 0 0 1 0 2 2 2 2 0 40/3 4 4 2 0 2 4 24 4 0 2 2 4 4 24. The standardised variance at the nonreplicated points is NN 1 i = 8/1 = 8, while at the replicated points it is NN 1 i = 8/3 = 2.667. You can use the program below to verify this result. proc iml; F = { 1 0 0 0 0 0, 0 1 0 0 0 0, 0 0 1 0 0 0, 0.5 0.5 0 0.25 0 0, 0.5 0.5 0 0.25 0 0, 0.5 0.5 0 0.25 0 0, 0.5 0 0.5 0 0.25 0, 0 0.5 0.5 0 0 0.25}; G = F *F; ginv = inv(g); print ginv; dx = nrow(f) # F * ginv * F ; print dx;} 19. The information matrix of any minimum support design with n i observations at its ith support point is G = F T F = N f T (x i )f(x i ) = i=1 p n i f T (x i )f(x i ) = FS T W F S, where F S is the design matrix for the corresponding saturated design and W is a diagonal matrix with diagonal elements n i, i = 1,..., p. If G is nonsigular the standardized variance of the prediction at the design points i=1 d(x, ξ) = NF T S (F T S W F S ) 1 F S = NW 1 as F S is a square matrix. This completes the proof. 20. (a). Here is an example of a SAS program that can be used to analyse the data. data e20; input dose y; datalines; 0.02 76 0.02 47

13 0.06 97 0.06 107 0.11 123 0.11 139 0.22 159 0.22 152 0.56 191 0.56 201 1.10 207 1.10 200 proc nlin data=e20; parms a = 210 b = 0.1; pred = a*dose/(b+dose); model y = pred; (b). F = [ x 1 β+x 1 ax1 (β+x 1) 2 x 2 β+x 2 ax2 (β+x 2) 2 ]. (c). The result follows from matrix multiplication. (d). As [ 1.25 414.65 G = 414.65 687745.84 then and d(x) = [ G 1 = ], 1 0.000602915 0.000602915 0.000001817532 2x 2 (0.06412 + x) 2 0.51296x2 (0.06412 + x) 3 + 0.164455x2 (0.06412 + x) 4. The design is locally D-optimum as the maximum of d(x) is equal to the number of parameters in the model, i.e. 2, and this maximum is reached at the design points (by the General Equivalence Theorem - see Figure 3). (e). When x = β, E(Y ) = α/2, i.e. half of the maximum possible response. This value is often referred to as ED50 or IC50, depending on the nature of the experiment. (f). Similar calculations show that d(x) = 3.07808x2 (0.06412 + x) 2 1.070722x2 (0.06412x) 3 + 0.21824x 2 (0.164455 + x) 4. The plot (see Figure 3) suggests that this design is locally D-optimum for the design region where x (0, 1). However, if x > 1 then var(ŷ) exceeds the number of parameters 2 and therefore the design is no longer locally D-optimum. 21. (a). This can be seen by substituting δ = 0 and α = 1 in Model 26.1. (b). Tedious algebra shows that d(x) = x2 (302.52 287.7ln(x) + 156.18(ln(x)) 2 ) (x + 2.5119) 4. ]

14 Figure 3: d(x) for the design of Exercise 20(d) Figure 4: d(x) for the design of Exercise 20(f)

15 Figure 5: d(x) for the design of Exercise 21(b) Figure 5 illustrates that the maximum of d(x) is equal to the number of the parameters in the model, i.e. 2, and that this maximum is reached at the design points. Hence, by the General Equivalence Theorem, this design is locally D-optimum. The interpretation of β is the same as in Exercise 20.