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					   VII. Factorial experiments
        VII.A         Design of factorial experiments
        VII.B         Advantages of factorial experiments
        VII.C         An example two-factor CRD experiment
        VII.D         Indicator-variable models and estimation for
                      factorial experiments
        VII.E         Hypothesis testing using the ANOVA method
                      for factorial experiments
        VII.F         Treatment differences
        VII.G         Nested factorial structures
        VII.H         Models and hypothesis testing for three-
                      factor experiments
Statistical Modelling Chapter VII                                1
        Factorial experiments
        • Often be more than one factor of interest to the
          experimenter.
        • Definition VII.1: Experiments that involve more than one
          randomized or treatment factor are called factorial
          experiments.
        • In general, the number of treatments in a factorial
          experiments is the product of the numbers of levels of the
          treatment factors.
        • Given the number of treatments, the experiment could be
          laid out as
              – a Completely Randomized Design,
              – a Randomized Complete Block Design or
              – a Latin Square
          with that number of treatments.
        • BIBDs or Youden Squares are not suitable.

Statistical Modelling Chapter VII                                      2
        VII.A Design of factorial experiments
        a) Obtaining a layout for a factorial experiment in R
        •      Layouts for factorial experiments can be
               obtained in R using expressions for the chosen
               design when only a single-factor is involved.
        •      Difference with factorial experiments is that the
               several treatment factors are entered.
                 – Their values can be generated using fac.gen.
        • fac.gen(generate, each=1, times=1,
          order="standard")
        • It is likely to be necessary to use either the
          each or times arguments to generate the
          replicate combinations.
        • The syntax of fac.gen and examples are given
          in Appendix B, Randomized layouts and sample
          size computations in R.
Statistical Modelling Chapter VII                                  3
        Example VII.1 Fertilizing oranges
        • Suppose an experimenter is interested in investigating
          the effect of nitrogen and phosphorus fertilizer on yield of
          oranges.
        • Investigate 3 levels of Nitrogen (viz 0,30,60 kg/ha) and 2
          levels of Phosphorus (viz. 0,20 kg/ha).
        • The yield after six months was measured.
        • Treatments are all possible combinations of the
          3 Nitrogen  2 Phosphorus levels: 32 = 6 treatments.
        • The treatment combinations, arranged in standard order,
          are:
                                    Treatment    N    P
                                        1        0    0
                                        2        0   20
                                        3       30    0
                                        4       30   20
                                        5       60    0
                                        6       60   20
Statistical Modelling Chapter VII                                        4
        Generating a layout in R for a CRD
        with 3 reps
        > #
                                                              Specifies units
        > # CRD                                  indexed by Seedling
        > #                                      with 18 levels.
        > n <- 18
        > CRDFac2.unit <- list(Seedling = n)
        > CRDFac2.ran <- fac.gen(list(N = c(0, 30, 60), P = c(0,
           20)), times = 3)
        > CRDFac2.lay <- fac.layout(unrandomized = CRDFac2.unit,
        +                     randomized = CRDFac2.ran, seed = 105)
        > remove("CRDFac2.unit“, "CRDFac2.ran")
                                                            Creates 3 copies of the
Does CRD                            Remove excess objects   levels combinations of N
randomization                                               and P, with 3 and 2 levels;
                                                            stores these in the
                                                            data.frame
                                                            CRDFac2.ran.
Statistical Modelling Chapter VII                                                     5
        The layout
        > CRDFac2.lay
           Units Permutation Seedling N P
        1      1           2        1 30 20
        2      2          18        2 0 0
        3      3           4        3 30 0
        4      4           5        4 30 0
        5      5           7        5 30 20
        6      6          12        6 30 0
        7      7          15        7 60 0
        8      8          13        8 0 0
        9      9           6        9 60 0
        10    10           1       10 60 0
        11    11          10       11 30 20
        12    12          16       12 60 20
        13    13           8       13 0 20
        14    14          14       14 0 20
        15    15           3       15 0 0
        16    16          11       16 60 20
        17    17           9       17 60 20
        18    18          17       18 0 20


Statistical Modelling Chapter VII             6
        What about an RCBD?
        • Suppose we decide on a RCBD with three blocks
          — how many units per block would be required?
        • Answer 6.
        • In factorial experiments not limited to two factors
        • Thus we may have looked at Potassium at 2
          levels as well. How many treatments in this
          case?
        • Answer 322 =12.



Statistical Modelling Chapter VII                               7
        VII.B Advantages of factorial
              experiments
        a) Interaction in factorial experiments
        • The major advantage of factorial experiments is
           that they allow the detection of interaction.
        • Definition VII.2: Two factors are said to interact
           if the effect of one, on the response variable,
           depends upon the level of the other.
        • If they do not interact, they are said to be
           independent.
        • To investigate whether two factors interact, the
           simple effects are computed.

Statistical Modelling Chapter VII                              8
        Effects
        • Definition VII.3: A simple effect, for the means
          computed for each combination of at least two factors, is
          the difference between two of these means having
          different levels of one of the factors but the same levels
          for all other factors.
        • We talk of the simple effects of a factor for the levels of
          the other factors.
        • If there is an interaction, compute an interaction effect
          from the simple effects to measure the size of the
          interaction
        • Definition VII.4: An interaction effect is half the
          difference of two simple effects for two different levels of
          just one factor (or is half the difference of two interaction
          effects).
        • If there is not an interaction, can separately compute the
          main effects to see how each factor affects the response.
        • Definition VII.5: A main effect of a factor is the
          difference between two means with different levels of that
          factor, each mean having been formed from all
          observations having the same level of the factor.               9
Statistical Modelling Chapter VII
        Example VII.2 Chemical reactor experiment
        • Investigate the effect of catalyst and temperature on the
          yield of chemical from a chemical reactor.
        • Table of means from the experiment was as follows:
                                               Temperature (°C)
                                                 160     180
                                           A      60      72
                                Catalyst
                                           B     52       64

        • For A the temperature effect is 72-60 = 12
        • For B the temperature effect is 64-52 = 12
        • These are called the simple effects of temperature.
        • Clearly, the difference between (effect of) the
          temperatures is independent of which catalyst is used.
        • Interaction effect: [12 - 12]/2 = 0
Statistical Modelling Chapter VII                                     10
        Illustrate using an interaction plot
                                    Effects of catalyst and temperature on yield
                                                   - no interaction

                 75

                 70

                 65
               Y
               i 60                                                                      A
               e
               l 55                                                                      B
               d
                 50

                 45

                 40
                   150              160                 170                 180    190
                                                    Temperature




        • A set of parallel lines indicates no interaction

Statistical Modelling Chapter VII                                                            11
        Interaction & independence are
        symmetrical in factors
        • Thus,
              – the simple catalyst effect at 160°C is 52-60 = -8
              – the simple catalyst effect at 180°C is 64-72 = -8
        • Thus the difference between (effect of) the
          catalysts is independent of which temperature is
          used.
        • Interaction effect is still 0 and factors are additive.



Statistical Modelling Chapter VII                                   12
        Conclusion when independent
        • Can consider each factor separately.
        • Looking at overall means will indicate what is happening in
          the experiment.
                                           Temperature (°C)
                                             160     180

                                    Mean     56       68
                                              Catalyst
                                             A         B

                                    Mean     66       58
          • So differences between the means in these tables are the
            main effects of the factors.
                – That is, the main effect of Temperature is 12 and that of Catalyst
                  is -8.
          • Having used 2-way table of means to work out that there
            is no interaction, abandon it for summarizing the results.
Statistical Modelling Chapter VII                                                      13
        Example VII.3 Second chemical reactor
        experiment

        • Suppose results from experiment with 2nd reactor as
          follows:
                                                   Temperature (°C)
                                                     160     180
                                               A      60      72
                                    Catalyst
                                               B     52       83


          • The simple temperature effect for A is 72-60 = 12
          • The simple temperature effect for B is 83-52 = 31
          • Difference between (effect of) temperatures depends on
            which catalyst is used.
          • Statement symmetrical in 2 factors — say 2 factors
            interact. (also dependent or nonadditive)
Statistical Modelling Chapter VII                                     14
        Interaction plot
                                    Effects of catalyst and temperature on yield
                                                     - interaction

                    85
                    80
                    75
                Y   70
                i   65                                                                   A
                e
                l   60                                                                   B
                d   55
                    50
                    45
                    40
                      150           160                 170                 180    190
                                                    Temperature



        • Clearly an interaction as lines have different
          slopes.
        • So cannot use overall means.
Statistical Modelling Chapter VII                                                            15
     Why using overall means is
     inappropriate
     • Overall means are:                 • Main effects:
                  Temperature (°C)            – cannot be equal to simple
                    160     180                 effects as these differ
                                              – have no practical
          Mean       56       77.5              interpretation.
                                          • Look at means for the
                       Catalyst
                      A         B
                                            combinations of the factors

          Mean       66       67.5
                                                   • Interaction effect:
                                                          [(72-60) - (83-52)]/2
                                                            = [12 - 31]/2 = -9.5
                                  Temperature (°C)
                                    160     180      or [(52-60) - (83-72)]/2
                              A      60      72             = [-8 - 9]/2 = -9.5.
               Catalyst                            • two non-interacting factors
                              B      52      83
Statistical Modelling Chapter VII
                                                     is the simpler              16
    b) Advantages over one-factor-at-
       a-time experiments
     • Sometimes suggested better to keep it simple and
       investigate one factor at a time.
     • However, this is wrong.
     • Unable to determine whether or not there is an interaction.
     • Take temperature-catalyst experiment at 2nd reactor.
                                             Experiment 1
                                            Temperature (°C)
                                              160     180
                                        A      60      ?
                Experiment 2 Catalyst
                                        B     52       83
          • WELL YOU HAVE ONLY APPLIED THREE OF THE
              FOUR POSSIBLE COMBINATIONS OF THE TWO
              FACTORS
          • Catalyst A at 180°C has not been tested but catalyst B at
              160°C has been tested twice as indicated above.
Statistical Modelling Chapter VII                                       17
        Limitation of inability to detect
        interaction
        • The results of the experiments would indicate
          that:
              – temperature increases yield by 31 gms
              – the catalysts differ by 8 gms in yield.
        • If we presume the factors act additively, predict
          the yield for catalyst A at 160°C to be:
              – 60+31 = 83 + 8 = 91.
        • This is quite clearly erroneous.
        • Need the factorial experiment to determine if
          there is an interaction.

Statistical Modelling Chapter VII                             18
        Same resources but more info
        • Exactly the same total amount of resources are involved
          in the two alternative strategies, assuming the number of
          replicates is the same in all the experiments.
        • In addition, if the factors are additive then the main
          effects are estimated with greater precision in factorial
          experiments.
        • In the one-factor-at-a time experiments
              – the effect of a particular factor is estimated as the difference
                between two means each based on r observations.
        • In the factorial experiment
              – the main effects of the factors are the difference between two
                means based on 2r observations
              – which represents a sqrt(2) increase in precision.
        • The improvement in precision will be greater for more
          factors and more levels

Statistical Modelling Chapter VII                                                  19
        Summary of advantages of factorial
        experiments
        if the factors interact, factorial experiments
             allow this to be detected and estimates of
             the interaction effect can be obtained,
             and
        if the factors are independent, factorial
             experiments result in the estimation of the
             main effects with greater precision.


Statistical Modelling Chapter VII                      20
        VII.C An example two-factor CRD
             experiment

          • Modification of ANOVA: instead of a single
            source for treatments, will have a source
            for each factor and one for each possible
            combinations of factors.




Statistical Modelling Chapter VII                    21
      a) Determining the ANOVA table for a
         two-Factor CRD
       a) Description of pertinent features of the study
        1. Observational unit
              –     a unit
        2. Response variable
              –     Y
        3. Unrandomized factors
              –     Units
        4. Randomized factors
              –     A, B
        5. Type of study
              –     Two-factor CRD

       b) The experimental structure
                Structure    Formula
                unrandomized n Units
                randomized   a A*b B
Statistical Modelling Chapter VII                          22
        c) Sources derived from the structure formulae
        • Units            = Units
        • A*B              = A + B + A#B
        d) Degrees of freedom and sums of squares
        • Hasse diagrams for this study with
                    – degrees of freedom                       Unrandomized terms          Randomized terms
                    – M and Q matrices
                                                                                                   
 Unrandomized terms                 Randomized terms                MG        MG               MG       MG


                                             
        1       1                         1       1
                                                                   Unit     U       A      A                 B      B
                                                                    s
                                                                   MU     MU-MG     MA   MA-MG               MB   MB-MG


     Unit        U         A         A                     B        B
      s
      n         n-1        a        a-1                    b       b-1                   AB             A#B
                                                                                         MAB        MAB-MA-MB+MG


                                    AB                  A#B
                                     ab               (a-1)(b-1)
Statistical Modelling Chapter VII                                                                                    23
        e) The analysis of variance table


                       Source            df         SSq
                       Units            n-1        YQUY
                          A             a-1        YQ A Y
                          B             b-1        YQB Y
                          A#B        (a-1)(b-1)   YQ ABY
                          Residual    ab(r-1)     YQURes Y




Statistical Modelling Chapter VII                             24
        f) Maximal expectation and variation models


        • Assume the randomized factors are fixed and that
          the unrandomized factor is a random factor.
        • Then the potential expectation terms are A, B and
          AB.
        • The variation term is: Units.
        • The maximal expectation model is
              – y = E[Y] = AB
        • and the variation model is
              – var[Y] = Units



Statistical Modelling Chapter VII                         25
        g) The expected mean squares
        • The Hasse diagrams, with contributions to
          expected mean squares, for this study are:

                      Unrandomized terms                     Randomized terms


                                                                          
                               1        1                              1       1




                           Unit             U      A        A                         B       B
                           s2
                             U
                                            U
                                             2
                                                 qA  ψ  qA  ψ                   qB ψ  qB  ψ 




                                                            AB                     A#B
                                                           qAB  ψ                qAB  ψ 



Statistical Modelling Chapter VII                                                                      26
        ANOVA table with E[MSq]

           Source                       df         SSq            E[MSq]
           Units                       n-1        YQUY
             A                         a-1        YQ A Y     U  qA  y 
                                                               2


             B                         b-1        YQB Y      U  qB  y 
                                                               2


             A#B                    (a-1)(b-1)   YQ ABY      U  qAB  y  
                                                               2


             Residual                ab(r-1)     YQURes Y   U
                                                              2




Statistical Modelling Chapter VII                                                27
        b) Analysis of an example
        Example VII.4 Animal survival experiment
        • To demonstrate the analysis I will use the example from
          Box, Hunter and Hunter (sec. 7.7).
        • In this experiment                         Treatment
                                              1      2       3    4
          three poisons and               I 0.31   0.82    0.43 0.45
          four treatments                   0.45   1.10    0.45 0.71
          (antidotes) were                  0.46   0.88    0.63 0.66
                                            0.43   0.72    0.76 0.62
          investigated.
        • The 12 combinations            II 0.36   0.92    0.44 0.56
          of poisons and        Poison      0.29   0.61    0.35 1.02
          treatments were                   0.40   0.49    0.31 0.71
          applied to animals                0.23   1.24    0.40 0.38
          using a CRD and the           III 0.22   0.30    0.23 0.30
          survival times of the             0.21   0.37    0.25 0.36
          animals measured                  0.18   0.38    0.24 0.31
          (10 hours).                       0.23   0.29    0.22 0.33


Statistical Modelling Chapter VII                                      28
      A. Description of pertinent features of the study
       1. Observational unit         • These are the
          – a animal                   steps that need to
       2. Response variable            be performed
          – Survival Time              before R is used
       3. Unrandomized factors         to obtain the
          – Animals                    analysis.
       4. Randomized factors         • The remaining
          – Treatments, Poisons        steps are left as
       5. Type of study                an exercise for
          – Two-factor CRD             you.
      B. The experimental structure
                Structure    Formula
                unrandomized 48 Animals
                randomized   3 Poisons*4 Treatments
Statistical Modelling Chapter VII                       29
        Interaction plot


                                                    0.9
                                                                               Treat




                                                    0.8
                                                                                       2
                                                                                       4
                                                                                       3
                                                                                       1




                                                    0.7
                                mean of Surv.Time

                                                    0.6
                                                    0.5
                                                    0.4
                                                    0.3
                                                    0.2




                                                          1   2            3

                                                                  Poison



        • There is some evidence of an interaction in that
          the traces for each level of Treat look to be
          different.
Statistical Modelling Chapter VII                                                          30
        Hypothesis test for the example
        Step 1:  Set up hypotheses
          a) H0: there is no interaction between Poison and
                 Treatment
             H1: there is an interaction between Poison and
                 Treatment

             b) H0: r1 = r2 = r3
                H1: not all population Poison means are equal

             c) H0: t1 = t2 = t3 = t4
                H1: not all population Treatment means are equal

             Set a = 0.05.

Statistical Modelling Chapter VII                                  31
        Hypothesis test for the example
        (continued)
        Step 2:    Calculate test statistics
        • The ANOVA table for a two-factor CRD, with
          random factors being the unrandomized factors
          and fixed factors the randomized factors, is:
              Source                df   SSq    MSq        E[MSq]          F    Prob
              Animals               47 3.0051
                Poison               2 1.0330 0.5165  2  q y
                                                       A    P          23.22 <.000

                Treatment            3 0.9212 0.3071    A  qT  y 
                                                         2               13.81 <.001

                Poison#Treat         6 0.2501 0.0417    A  qPT  y 
                                                         2               1.87   0.112

                Residual            36 0.8007 0.0222  A
                                                       2



Statistical Modelling Chapter VII                                                       32
        Hypothesis test for the example
        (continued)

        Step 3: Decide between hypotheses
          Interaction of Poison and Treatment is not
          significant, so there is no interaction.
          Both main effects are highly significant,
          so both factors affect the response.
          More about models soon.
        • Also, it remains to perform the usual
          diagnostic checking.


Statistical Modelling Chapter VII                      33
        VII.D Indicator-variable models and
              estimation for factorial
              experiments
        • The models for the factorial experiments will
          depend on the design used in assigning the
          treatments — that is, CRD, RCBD or LS.
        • The design will determine the unrandomized
          factors and the terms to be included involving
          those factors.
        • They will also depend on the number of
          randomized factors.
        • Let the total number of observations be n and the
          factors be A and B with a and b levels,
          respectively.
        • Suppose that the combinations of A and B are
          each replicated r times — that is, n = abr.
Statistical Modelling Chapter VII                         34
        a)            Maximal model for two-factor
                      CRD experiments
        • The maximal model used for a two-factor CRD
          experiment, where the two randomized factors A and B
          are fixed, is:
                                          
                  y AB = E  Y  = X AB a and V =  UIn
                                                     2


        where
        Y   is the n-vector of random variables for the response
            variable observations,
        (a) is the ab-vector of parameters for the A-B
                  combinations,,
        XAB       is the nab matrix giving the combinations of A and B
                  that occurred on each unit, i.e. X matrix for AB,
          U is the variability arising from different units.
           2

        • Our model also assumes Y ~ N(yAB, V)
Statistical Modelling Chapter VII                                         35
        Standard order
        • Expression for X matrix in terms of direct
          products of Is and 1s when A and B are in
          standard order.
        • Previously used standard order — general
          definition in notes.
        • The values of the k factors A1, A2, …, Ak with a1,
          a2, …, ak levels, respectively, are systematically
          ordered in a hierarchical fashion:
              – they are ordered according to A1, then A2, then A3, …
                and then Ak.
        • Suppose, the elements of the Y vector are
          arranged so that the values of the factors A, B
          and the replicates are in standard order, as for a
          systematic layout.
        • Then X AB = Ia  Ib  1r
Statistical Modelling Chapter VII                                       36
        Example VII.5 22 Factorial experiment
        • Suppose A and B have 2 levels each and that
          each combination of A and B has 3 replicates.
        • Hence, a = b = 2, r = 3 and n = 12.
                      
        • Then  a  = a11 a12 a 21 a 22 
        • Now Y is arranged so that the              X AB
          values of A, B and the reps are     A    1 1 2                              2
          in standard order — that is         B    1 2 1                              2
                                                                         1   0   0   0
                                                                         1   0   0   0
   Y = Y111 Y112 Y113 Y121 Y122 Y123 Y211 Y212 Y213 Y221 Y222 Y223    1   0   0   0
                                                                         0   1   0   0
                                                                         0   1   0   0
        • Then X AB = I2  I2  13                                       0
                                                                         0
                                                                              1   0   0
                                                                              0   1   0
                                                                         0   0   1   0
                                                                         0           0
        • so that XAB for 4 level AB is:
                                                                              0   1
                                                                         0   0   0   1
                                                                         0   0   0   1
                                                                         0           1
                                                                             0   0    
Statistical Modelling Chapter VII                                                          37
        Example VII.5 22 Factorial experiment
        (continued)
        • For the maximal model,
                                                                               a 11 
                                                                               a  
                                             1    0   0   0                       11 
                                             1    0   0   0                  a 11 
                                             1    0   0   0                  a 12 
                                             0            0
                                                                 a 11   a 12 
                                                   1   0
                                             0            0
                                             
                                                   1   0         a    a  
                                                           0
                y AB = E  Y  = X AB    
                                        a = 0
                                               0
                                                   1
                                                   0
                                                       0
                                                       1   0
                                                                      12  =       12 
                                                                 a 21   a 21 
                                             0            0
                                             0
                                                   0   1        a 22   a 21 
                                                                          
                                                           0
                                                                                a 21 
                                                   0   1
                                             0    0   0   1                          
                                             0    0   0   1                 a 22 
                                             0            1                 a  
                                                  0   0    
                                                                               a 22 
                                                                               22 
                                                                                       

        • That is, the maximal model allows for a different
          response for each combination of A and B.
Statistical Modelling Chapter VII                                                           38
        b) Alternative expectation models —
        marginality-compliant models
        Rule VII.1: The set of expectation models
          corresponds to the set of all possible
          combinations of potential expectation terms,
          subject to restriction that terms marginal to
          another expectation term are excluded from the
          model;
        • it includes the minimal model that consists of a
          single term for the grand mean.
        • For marginality of terms refer to Hasse diagrams
          and can be deduced using definition VI.9.
        • This definition states that one generalized factor
          is marginal to another if
              – the factors in the marginal generalized factor are a
                subset of those in the other and
              – this will occur irrespective of the replication of the
                levels of the generalized factors.
Statistical Modelling Chapter VII                                        39
                                     Unrandomized terms        Randomized terms

        Two-factor                                                      
                                             1       1               1       1
        CRD
                                          Unit        U    A    A                     B       B
                                           s
                                           n         n-1   a   a-1                    b       b-1


• For all randomized factors fixed, the potential
  expectation terms are A, B and AB.                          AB                  A#B
• Maximal model                                                 ab               (a-1)(b-1)

     – includes all terms: E[Y] = A + B + AB
     – However, marginal terms must be removed
     – so the maximal model reduces to E[Y] = AB
• Next model leaves out AB giving additive
  model E[Y] = A + B
     – no marginal terms in this model.
• A simpler model than this is either E[Y] = A
  and E[Y] = B.
• Only other possible model is one with neither
  A nor B: E[Y] = G.
Statistical Modelling Chapter VII                                                                 40
        Alternative expectation models in
        terms of matrices
                  y AB = X AB a               A and B interact
                                                  in effect on response   
                y A+B      = X A a  XB  
                                         A and B independently
                                         affect response                      
                    y A = X Aa                     A only affects response 
                    yB = XB                       B only affects response 
                    yG = XG                      no factors affect response 
        • Expressions for X matrices in terms of direct
          products of Is and 1s when A and B are in
          standard order.
Statistical Modelling Chapter VII                                                 41
        X matrices
        • Again suppose, the elements of the Y vector are
          arranged so that the values of the factors A, B
          and the replicates are in standard order, as for a
          systematic layout.
        • Then the X matrices can be written as the
          following direct products:

                                     X G = 1a  1b  1r = 1abr
                                     X A = Ia  1b  1r
                                     XB = 1a  Ib  1r
                                    X AB = Ia  Ib  1r

Statistical Modelling Chapter VII                                42
        Example VII.5 22 Factorial experiment
        (continued)

        • Remember A and B have two levels each and that each
          combination of A and B is replicated 3 times.
        • Hence, a = b = 2, r = 3 and n = 12. Then
                                    a = a1 a 2 
                                   =  1  2 
                                   
                                
                                a = a11 a12 a 21 a 22 

        • Suppose Y is arranged so that the values of A, B and the
          replicates are in standard order — that is
           Y = Y111 Y112 Y113 Y121 Y122 Y123 Y211 Y212 Y213 Y221 Y222 Y223 

        • Then


Statistical Modelling Chapter VII                                                43
        Example VII.5 22 Factorial experiment
        (continued)     XG     XA      XB                              X AB
                                    A         1 2                  1 1 2 2
                                    B                   1 2        1 2 1 2
                                        1   1   0   1   0   1   0   0   0
                                        1   1   0   1   0   1   0   0   0
  X G = 12  12  13 = 112              1   1   0   1   0   1   0   0   0
                                        1   1   0   0   1   0   1   0   0
  X A = I2  12  13                    1   1   0   0   1   0   1   0   0
                                        1   1   0   0   1   0   1   0   0
   XB = 12  I2  13                    1   0   1   1   0   0   0   1   0
                                        1   0   1   1   0   0   0   1   0
X AB = I2  I2  13                     1   0   1   1   0   0   0   1   0
                                        1   0   1   0   1   0   0   0   1
                                        1   0   1   0   1   0   0   0   1
                                        1
                                            0
                                                  1
                                                       0
                                                            1
                                                                 0
                                                                      0   0   1
                                                                                
        Notice, irrespective of the replication of the levels of AB ,
        • XG can be written as a linear combination of the columns
          of each of the other three
        • XA and XB can be written as linear combinations of the
          columns of XAB.
Statistical Modelling Chapter VII                                                   44
        Example VII.5 22 Factorial experiment
        (continued)
        • Marginality of indicator-variable terms (for generalized
          factors)
              – XG  XAa, XB, XAB(a).
              – XAa, XB  XAB(a).
              – More loosely, for terms as seen in the Hasse diagram, we say
                that
                    • G < A, B, AB
                    • A, B < AB
        • Marginality of models
              – yG  yA, yB, yA+B, yAB
                [yG = XG, yA = XAa, yB = XB, yA+B = XAa + XB, yAB = XAB(a)]
              – yA, yB  yA+B, yAB
                [yA = XAa, yB = XB, yA+B = XAa + XB, yAB = XAB(a)]
              – yA+B  yAB
                [yA+B = XAa + XB, yAB = XAB(a)]
              – More loosely,
                    • G < A, B, A+B, AB,
                    • A, B < A+B, AB
                    • A+B < AB.
Statistical Modelling Chapter VII                                                 45
     Estimators of the fitted values for the
     expectation models
     • They are all functions of means.
     • So can be written in terms of mean operators, Ms.
     • If Y is arranged so that the associated factors A, B and
       the replicates are in standard order, the M operators
       written as the direct product of I and J matrices:
                 Model                         Estimator
                      
          y AB = X AB a      y AB =  A  B  = MABY = r -1Ia  Ib  Jr Y
                              ˆ
          y A+B = X A a  XB y A+B = A  B - G
                              ˆ
          y A = X Aa                                 -1
                              y A = A = MA Y =  br  Ia  Jb  Jr Y
                              ˆ
          yB = XB                                  -1
                              y = B = M Y =  ar  J  I  J Y
                              ˆ  B         B             a   b     r
          yG = XG             yG = G = MGY = n -1Ja  Jb  Jr Y
                               ˆ

        G, A, B and A  B are the n-vectors of means, the latter
              for the combinations of A and B, that is for the
              generalized factor AB.
Statistical Modelling Chapter VII                                            46
        Example VII.5 22 Factorial experiment
        (continued)
        • The mean vectors, produced by an MY, are as follows:

                                    G    A        B         A Β

                                         A1     B1     A1  B1 
                                G                        A1  B1 
                                G      A1     B1 
                                         A1     B1    A B 
                                G                        1     1
                                        A      B 
                                G 
                                         A1    B2      A1  B2 
                                G                        A1  B2 
                                G      1      2
                                         A1    B2      A1  B2 
                                                         A2  B1 
                                G      A2     B1 
                                         A2     B1    A B 
                                G                        2     1
                                         A2     B1 
                                G 
                                         A2    B2      A2  B1 
                                G                        A2  B2 
                                G     A      B 
                                         A2    B2      A2  B2 
                                G 
                                       2      2      A2  B2 
                                                                   


Statistical Modelling Chapter VII                                       47
  VII.E Hypothesis testing using the ANOVA
        method for factorial experiments
          • Use ANOVA to choose between models.
          • In this section will use generic names of A, B and Units
            for the factors
          • Recall ANOVA for two-factor CRD.
           Source                       df         SSq            E[MSq]
           Units                       n-1        YQUY
              A                        a-1        YQ A Y     U  qA  y 
                                                               2


              B                        b-1        YQB Y      U  qB  y 
                                                               2


              A:B                   (a-1)(b-1)   YQ ABY      U  qAB  y  
                                                               2


              Residual               ab(r-1)     YQURes Y   U
                                                              2


Statistical Modelling Chapter VII                                                48
        a) Sums of squares for the
           analysis of variance
        • Require estimators of the following SSqs for a
          two-factor CRD ANOVA:
              – Total or Units; A; B; A#B and Residual.
        • Use Hasse diagram.




Statistical Modelling Chapter VII                          49
                                                  Unrandomized terms          Randomized terms


          Vectors                                                                     
                                                      MG        MG                MG       MG
          for sums
          of squares                                 Unit
                                                      s
                                                              U        A      A                 B      B
                                                     MU     MU-MG      MA   MA-MG               MB   MB-MG




                                                                            AB             A#B
                                                                            MAB        MAB-MA-MB+MG

Total or Units SSq:                 QUY =  MU - MG  Y = Y - G = DG
A SSq:                              Q A Y =  MA - MG  Y = A - G = A e   • All the Ms
                                                                            and Qs are
B SSq:                              QBY =  MB - MG  Y = B - G = Be        symmetric
A:B SSq:                            Q ABY =  MAB - MA - MB  MG  Y        and
                                         = A  B - A - B  G =  A  B e   idempotent.
Residual SSq:                       QURes Y =  MU - MAB  Y = Y -  A  B  = D AB

Statistical Modelling Chapter VII
                                         = Y -  A  B e - Ae - Be - G                                      50
        SSq (continued)
        • From section VII.C, Models and estimation
          for factorial experiments, we have that

                                    G = MGY = n -1Ja  Jb  Jr Y
                                                    -1
                                    A = MA Y =  br  Ia  Jb  Jr Y
                                                    -1
                                    B = MBY =  ar  Ja  Ib  Jr Y
                       A  B  = MABY = r -1Ia  Ib  Jr Y



Statistical Modelling Chapter VII                                      51
        SSq (continued)
   • So SSqs for the ANOVA are given by

                                    YQUY = D DG
                                             G
                                    YQ A Y = A Ae
                                               e
                                    YQBY = B Be
                                             e

                               YQ ABY =  A  B e  A  B e
                             YQURes Y = DABDAB




Statistical Modelling Chapter VII                                 52
 ANOVA table constructed as follows:
 Source                df             SSq              MSq                   E[MSq]            F       p
 Units               n-1             YQUY
                                    YQA Y         YQ A Y                  U  qA  y  sA sURes
  A                  a-1                                                     2             2  2        pA
                                                           = sA
                                                              2
                                                    a -1
  B                  b-1             YQB Y        YQB Y
                                                          = sB
                                                             2              U  qB  y 
                                                                             2               2  2
                                                                                            sB sURes   pB
                                                    b -1
  A#B            (a-1)(b-1) YQABY                 YQ AB Y
                                                                 = sAB
                                                                        2          2
                                                                                       
                                                                    2  U  qAB y sAB sU
                                                                                       2    pAB
                                                 a - 1 b - 1
                                                                                        Res




  Residual         ab(r-1)          YQURes Y    YQURes Y                 U
                                                                            2
                                                               =    2
                                                                   sURes
                                                 ab  r - 1

 Total              abr-1            YQUY


• Can compute the SSqs by decomposing y as follows:
            y = g  ae  be   a b e  dAB
Statistical Modelling Chapter VII                                                                           53
        d) Expected mean squares
        • The E[MSq]s involve three quadratic functions of
          the expectation vector:
                                     
                                    qA y = yQ A y  a - 1 ,

                                 qB  y  = yQBy  b - 1 ,

                                qAB  y  = yQ ABy  a - 1 b - 1 .
        • That is, numerators are SSqs of
             – QAy = (MA-MG)y,
             – QBy = (MB-MG)y and
             – QABy = (MAB-MA-MB+MG)y,
             where y is one of the models                        • Require
             – yG = XG                                           expressions
             – yA = XAa                                           for the
             – yB = XB                                           quadratic
             – yA+B = XAa + XB                                   functions
             – yAB = XAB(a)                                     under each of
Statistical Modelling Chapter VII                                  these models. 54
        Zero & nonzero quadratic functions
                                                             Source
                     Expectation                A              B                 A#B
                     model
                     yG = XG                                
                                           qA y G = 0 qB y G = 0 qAB y G = 0       
                     y Α = X Aa              qA  y Α     qB  y Α  = 0    qAB  y Α  = 0
                     yΒ = XB              qA  yB  = 0     qB  y B       qAB  yB  = 0
                     y A+B = X A a  XB    qA  y Α+B     qB  y A+B     qAB  y A+B  = 0
                     y AB = X AB a       qA  y ΑB      qB  y AB        qAB  y AB 

        • Firstly, considering the column for source A#B,
           – the only model for which qAB(y)  0 is yAB = XAB(a).
           – Consequently, A#B is significant indicates that qAB(y) > 0 and that
              the maximal model is the appropriate model.
        • Secondly, considering the column for source A,
           – qA(y)  0 implies either a model that includes XAa or the maximal
              model XAB(a):
                    if A#B is significant, know need maximal model and test for A irrelevant.
                    If A#B is not significant, know maximal model is not required and so
                       significant A indicates that the model should include XAa.
        • Thirdly for source B, provided A#B is not significant, a significant B
          indicates that the model should include XB.
Statistical Modelling Chapter VII                                                               55
        Choosing an expectation model
        for a two-factor CRD
                            pa                                        p>a
                                              A:B hypothesis
                          reject H0                                  retain H0


                 Factors interact
     in their effect on response variable.
    Use maximal model yAB= XAB(a).




                                               one or both p  a
                                                                    A and B hypotheses
                                                  reject H0(s)

                                                                                 for both p > a
                                                                                  retain all H0s


                                    Factors independent              Factors have no effect
                           in their effect on response variable.      on response variable.
                           Use a model that includes significant   Use minimal model yG= XG.
                              terms, that is a single factor or
                                       additive model

Statistical Modelling Chapter VII                                                                  56
        Nonzero quadratic functions
        • In the notes show that the non-zero q-functions are given
          by:                        a
                                                          rb a i - a. 
                                             2


                               
                       qA y Α = qA y Α+B =                 i =1
                                                                  a - 1
                                                             b
                                                          ra   j - . 
                                                                             2


                                       
                        qB yB = qB y A+B =                  j =1
                                                                  b - 1

                                                   a ij - a i. - a . j  a .. 
                                         a       b                                              2
                                        r
                          
                    qAB y AB =         i =1 j =1
                                                             a - 1 b - 1
        • So q-functions are zero when expressions in parentheses
          are zero.
        • That is when a i = a.,  j = . and a ij = a i .  a . j - a ..
        • That is equality or an additive pattern obtain.
        • These, or equivalent, expressions are given for H0.
Statistical Modelling Chapter VII                                                                   57
        e) Summary of the hypothesis test
        Step 1:  Set up hypotheses
          a) H0: there is no interaction between A and B
                                  (or model simpler than XAB(a) is
                                  adequate)
                           a    ij
                                         - a i . - a . j  a .. = 0   for all i, j   
                 H1: there is an interaction between Poison and
                     Treatment

                           a    ij
                                         - a i . - a . j  a ..  0 for some i, j        
             b) H0: a1 = a2 = … = aa (or XAa not required in model)
                H1: not all population Poison means are equal

             c) H0: 1 = 2 = … = b (or XB not required in model)
                H1: not all population Treatment means are equal
             Set a = 0.05.
Statistical Modelling Chapter VII                                                                     58
         Hypothesis test for the example
         (continued)
         Step 2:                Calculate test statistics
 Source                df             SSq              MSq                   E[MSq]          F       p
 Units                n-1            YQUY
                                    YQA Y         YQ A Y                  U  qA  y  sA sURes
  A                   a-1                                                    2             2  2      pA
                                                           = sA
                                                              2
                                                    a -1
  B                   b-1            YQB Y        YQB Y
                                                          = sB
                                                             2              U  qB  y  sB sURes
                                                                             2             2  2      pB
                                                    b -1
  A#B            (a-1)(b-1) YQABY                 YQ AB Y
                                                                 = sAB
                                                                        2          2
                                                                                     
                                                                    2  U  qAB y sAB sU
                                                                                       2    pAB
                                                 a - 1 b - 1
                                                                                        Res




  Residual         ab(r-1)          YQURes Y    YQURes Y                 U
                                                                            2
                                                               =    2
                                                                   sURes
                                                 ab  r - 1

 Total              abr-1            YQUY
Statistical Modelling Chapter VII                                                                         59
        Hypothesis test for the example
        (continued)
        Step 3:     Decide between hypotheses
          If A#B is significant, we conclude that the maximal
          model yAB = E[Y] = XAB(a) best describes the data.

             If A#B is not significant, The choice between
             these models depends on which of A and B are
             not significant. A term corresponding to the
             significant source must be included in the model.

             For example, if both A and B are significant, then
             the model that best describes the data is the
             additive model yA+B = E[Y] = XAa + XB.
Statistical Modelling Chapter VII                                 60
        f) Computation of ANOVA and
           diagnostic checking in R
        • The assumptions underlying a factorial
          experiment will be the same as for the
          basic design employed, except that
          residuals-versus-factor plots of residuals
          are also produced for all the factors in the
          experiment.




Statistical Modelling Chapter VII                        61
        Example VII.4 Animal survival
        experiment (continued)
        • Previously determined the following experimental
          structure for this experiment.
                          Structure    Formula
                          unrandomized 48 Animals
                          randomized   3 Poisons*4 Treatments


        • From this we conclude that the model to be used
          for aov function is
                Surv.Time ~ Poison * Treat +
                         Error(Animals).

Statistical Modelling Chapter VII                               62
        R instructions
        • First data entered into R data.frame Fac2Pois.dat.
             Fac2Pois.dat <- fac.gen(generate = list(Poison = 3, 4, Treat=4))
             Fac2Pois.dat <- data.frame(Animals = factor(1:48), Fac2Pois.dat)
             Fac2Pois.dat$Surv.Time <-
               c(0.31,0.82,0.43,0.45,0.45,1.10,0.45,0.71,0.46,0.88,0.63,0.66,
                 0.43,0.72,0.76,0.62,0.36,0.92,0.44,0.56,0.29,0.61,0.35,1.02,
                 0.40,0.49,0.31,0.71,0.23,1.24,0.40,0.38,0.22,0.30,0.23,0.30,
                 0.21,0.37,0.25,0.36,0.18,0.38,0.24,0.31,0.23,0.29,0.22,0.33)
             attach(Fac2Pois.dat)                          Treatment
             Fac2Pois.dat                             1    2      3    4
                                                      I   0.31   0.82   0.43   0.45
                                                          0.45   1.10   0.45   0.71
                                                          0.46   0.88   0.63   0.66
                                                          0.43   0.72   0.76   0.62

                                                     II   0.36   0.92   0.44   0.56
                                           Poison         0.29   0.61   0.35   1.02
                                                          0.40   0.49   0.31   0.71
                                                          0.23   1.24   0.40   0.38

                                                    III   0.22   0.30   0.23   0.30
                                                          0.21   0.37   0.25   0.36
                                                          0.18   0.38   0.24   0.31
Statistical Modelling Chapter VII                         0.23   0.29   0.22   0.33   63
        R output
   > Fac2Pois.dat
      Animals Poison Treat Surv.Time
    1       1      1     1      0.31   25   25   2   1   0.40
    2       2      1     2      0.82   26   26   2   2   0.49
    3       3      1     3      0.43   27   27   2   3   0.31
    4       4      1     4      0.45   28   28   2   4   0.71
    5       5      1     1      0.45   29   29   2   1   0.23
    6       6      1     2      1.10   30   30   2   2   1.24
    7       7      1     3      0.45   31   31   2   3   0.40
    8       8      1     4      0.71   32   32   2   4   0.38
    9       9      1     1      0.46   33   33   3   1   0.22
   10      10      1     2      0.88   34   34   3   2   0.30
   11      11      1     3      0.63   35   35   3   3   0.23
   12      12      1     4      0.66   36   36   3   4   0.30
   13      13      1     1      0.43   37   37   3   1   0.21
   14      14      1     2      0.72   38   38   3   2   0.37
   15      15      1     3      0.76   39   39   3   3   0.25
   16      16      1     4      0.62   40   40   3   4   0.36
   17      17      2     1      0.36   41   41   3   1   0.18
   18      18      2     2      0.92   42   42   3   2   0.38
   19      19      2     3      0.44   43   43   3   3   0.24
   20      20      2     4      0.56   44   44   3   4   0.31
   21      21      2     1      0.29   45   45   3   1   0.23
   22      22      2     2      0.61   46   46   3   2   0.29
   23      23      2     3      0.35   47   47   3   3   0.22
   24      24      2     4      1.02   48   48   3   4   0.33

Statistical Modelling Chapter VII                               64
        R instructions and output
             interaction.plot(Poison, Treat, Surv.Time, lwd=4)
             Fac2Pois.aov <- aov(Surv.Time ~ Poison * Treat +
               Error(Animals), Fac2Pois.dat)
             summary(Fac2Pois.aov)

         • Function interaction.plot to produce the plot for
           initial graphical exploration.
         • Boxplots not relevant as single factor.
        > interaction.plot(Poison, Treat, Surv.Time, lwd = 4)
        > Fac2Pois.aov <- aov(Surv.Time ~ Poison * Treat +
           Error(Animals), Fac2Pois.dat)
        > summary(Fac2Pois.aov)
        Error: Animals
                     Df Sum Sq Mean Sq F value     Pr(>F)
        Poison        2 1.03301 0.51651 23.2217 3.331e-07
        Treat         3 0.92121 0.30707 13.8056 3.777e-06
        Poison:Treat 6 0.25014 0.04169 1.8743      0.1123
        Residuals    36 0.80073 0.02224
Statistical Modelling Chapter VII                                65
        Diagnostic checking
        • As experiment was set up as a CRD, the assumptions
          underlying its analysis will be the same as for the CRD
        • Diagnostic checking the same — in particular, Tukey’s
          one-degree-of-freedom-for-nonadditivity cannot be
          computed.
        • The R output produced by the expressions that deal with
          diagnostic checking is as follows:
              >   #
              >   # Diagnostic checking
              >   #
              >   res <- resid.errors(Fac2Pois.aov)
              >   fit <- fitted.errors(Fac2Pois.aov)
              >   plot(fit, res, pch=16)
              >   plot(as.numeric(Poison), res, pch=16)
              >   plot(as.numeric(Treat), res, pch=16)
              >   qqnorm(res, pch=16)
              >   qqline(res)
Statistical Modelling Chapter VII                                   66
        Diagnostic checking (continued)                                                                                                   Normal Q-Q Plot




                                                                                                          0.4
                  0.4




                                                                                                          0.2
                  0.2




                                                                                       Sample Quantiles
            res




                                                                                                          0.0
                  0.0




                                                                                                          -0.2
                  -0.2




                          0.2    0.3    0.4     0.5         0.6    0.7    0.8    0.9                                    -2         -1               0                   1         2

                                                      fit                                                                                 Theoretical Quantiles




                                                                                                           0.4
                   0.4




                                                                                                           0.2
                   0.2




                                                                                              res
            res




                                                                                                           0.0
                   0.0




                                                                                                           -0.2
                   -0.2




                           1.0         1.5            2.0           2.5         3.0                               1.0        1.5        2.0         2.5           3.0       3.5       4.0

                                              as.numeric(Poison)                                                                              as.numeric(Treat)




        • All plots indicate a problem with the assumptions — will
          a transformation fix the problem?
Statistical Modelling Chapter VII                                                                                                                                                           67
     g) Box-Cox transformations for correcting
        transformable non-additivity
        • Box, Hunter and Hunter (sec. 7.9) describe the
          Box-Cox procedure for determining the
          appropriate power transformation for a set of
          data.
        • It has been implemented in the R function
          boxcox supplied in the MASS library that comes
          with R.
        • When you run this procedure you obtain a plot of
          the log-likelihood of l, the power of the
          transformation to be used (for l = 0 use the ln
          transformation).
        • However, the function does not work with
          aovlist objects and so the aov function must
          be repeated without the Error function.
Statistical Modelling Chapter VII                            68
        Example VII.4 Animal survival experiment
        (continued)

          > Fac2Pois.NoError.aov <- aov(Surv.Time ~ Poison *
            Treat, Fac2Pois.dat)
          > library(MASS)

            The following object(s) are masked from package:MASS
             :

                   Animals
          boxcox(Fac2Pois.NoError.aov, lambda=seq(from = -2.5,
            to = 2.5, len=20), plotit=T)

           The message reporting the masking of Animals is saying
            that there is a vector Animals that is part of the MASS
            library that is being overshadowed by Animals in
            Fac2Pois.dat.

Statistical Modelling Chapter VII                                 69
        Example VII.4 Animal survival experiment
        (continued)




                                                 30
                                                       95%




                                                 20
                                                 10
                                log-Likelihood

                                                 0
                                                 -10
                                                 -20
                                                 -30
                                                 -40




                                                             -2   -1   0   1   2




        • Output indicates that, as the log likelihood is a maximum
          around l = -1, the reciprocal transformation should be
          used.
        • The reciprocal of the survival time will be the death rate
          — the number that die per unit time
Statistical Modelling Chapter VII                                                  70
        Repeat the analysis on the
        reciprocals
        •    detach Fac2Pois.dat data.frame
        •    add Death.Rate to the data.frame
        •    reattach the data.frame to refresh info in R.
        •    repeat expressions from the original analysis with
             Surv.time replaced by Death.Rate appropriately.
                                                                                  Treat

                                                                                          1
                                                                                          3
                                                                                          4
                                                         4




                                                                                          2
                                    mean of Death.Rate

                                                         3
                                                         2




                                                             1   2            3

                                                                     Poison


        • Looking like no interaction.
Statistical Modelling Chapter VII                                                             71
        Repeat the analysis on the
        reciprocals (continued)
       > detach(Fac2Pois.dat)
       > Fac2Pois.dat$Death.Rate <- 1/Fac2Pois.dat$Surv.Time
       > attach(Fac2Pois.dat)

                    The following object(s) are masked from package:MASS :

                      Animals

       > interaction.plot(Poison, Treat, Death.Rate, lwd=4)
       > Fac2Pois.DR.aov <- aov(Death.Rate ~ Poison * Treat +
       Error(Animals), Fac2Pois.dat)
       > summary(Fac2Pois.DR.aov)

       Error: Animals
                    Df Sum Sq Mean Sq F value    Pr(>F)
       Poison         2 34.877 17.439 72.6347 2.310e-13
       Treat          3 20.414  6.805 28.3431 1.376e-09
       Poison:Treat 6 1.571     0.262 1.0904     0.3867
       Residuals    36 8.643    0.240


Statistical Modelling Chapter VII                                            72
        Repeat the analysis on the
        reciprocals (continued)                                                                                              Normal Q-Q Plot




                                                                                          1.0
              1.0




                                                                                          0.5
              0.5




                                                                       Sample Quantiles
       res




                                                                                          0.0
              0.0




                                                                                          -0.5
              -0.5




                                                                                                       -2         -1                  0                    1         2
                            2                3               4
                                                                                                                             Theoretical Quantiles
                                             fit




                                                                                          1.0
               1.0




                                                                                          0.5
               0.5




                                                                          res
        res




                                                                                          0.0
               0.0




                                                                                          -0.5
               -0.5




                      1.0       1.5          2.0           2.5   3.0                             1.0        1.5        2.0            2.5            3.0       3.5       4.0

                                      as.numeric(Poison)                                                                       as.numeric(Treat)




       • Looking good
Statistical Modelling Chapter VII                                                                                                                                              73
     Comparison of untransformed and
     transformed analyses
                                    UNTRANSFORMED                TRANSFORMED
    Source                    df     MSq    F   Prob        df   MSq    F    Prob

    Animals                    47                           47
     Poison                     2   0.5165   23.22 0.0000    2 17.439   72.63 <0.001
     Treatment                  3   0.3071   13.81 0.0000    3 6.805    28.34 <0.001
     Poison#Treat               6   0.0417    1.87 0.1112    6 0.262     1.09 0.387
     Residual                  36   0.0222                  36 0.240


    • The analysis of the transformed data indicates that there is
      no interaction on the transformed scale — confirms plot.
    • The main effect mean squares are even larger than before
      indicating that we are able to separate the treatments even
      more on the transformed scale.
    • Diagnostic checking now indicates all assumptions are met.
Statistical Modelling Chapter VII                                                      74
        VII.F Treatment differences

        • As usual the examination of treatment
          differences can be based on multiple
          comparisons or submodels.




Statistical Modelling Chapter VII                 75
        a) Multiple comparison procedures

        • For two factor experiments, there will be
          altogether three tables of means, namely
          one for each of A, B and A#B.
        • Which table is of interest depends on the
          results of the hypothesis tests outlined
          above.
        • However, in all cases Tukey’s HSD
          procedure will be employed to determine
          which means are significantly different.

Statistical Modelling Chapter VII                     76
        A#B Interaction significant
        • In this case you look at the table of means for the
          AB combinations.
                                                              A
                                         1       2     3      .     .       .     a
                                    1    x       x     x      .     .       .     x
                                    2    x       x     x      .     .       .     x
                                    .    .       .     .      .     .       .     .
                          B         .    .       .     .      .     .       .     .
                                    .    .       .     .      .     .       .     .
                                    b    x       x     x      .     .       .     x

                                        q ab,,0.05           q ab,,0.05     2
                       w  5%  =                     sxd =                 s
                                             2                     2          r
        • In this case you look at the table of means for the
          AB combinations.
Statistical Modelling Chapter VII                                                     77
        A#B interaction not significant
        • In this case examine the A and B tables of
          means for the significant lines.

                                    A                                                         B
               1      2         3   .        .       .      a           1       2         3   .       .        .        b
   Means       x      x         x   .        .       .      x   Means   x       x         x   .       .        .        x

                   q a,,0.05           q a,,0.05      2                    q b,,0.05           q b,,0.05        2
     w  5% =                  sxd =                s           w  5% =                sxd =                s
                          2                  2         rb                           2                  2           ra




        • That is, we examine each factor separately.


Statistical Modelling Chapter VII                                                                                           78
        Example VII.4 Animal survival experiment
        (continued)
            • Tables of means and studentized ranges:
            > #
            > # multiple comparisons
            > #
            > model.tables(Fac2Pois.DR.aov, type="means")
            Tables of means
            Grand mean
                                            Poison:Treat
                                                  Treat
            2.622376                       Poison 1      2     3     4
                                                 1 2.487 1.163 1.863 1.690
             Poison                              2 3.268 1.393 2.714 1.702
            Poison                               3 4.803 3.029 4.265 3.092
                                           > q.PT <- qtukey(0.95, 12, 36)
                1      2    3
                                           > q.PT
            1.801 2.269 3.797              [1] 4.93606
                                           > q.P <- qtukey(0.95, 3, 36)
             Treat                         > q.P
            Treat                          [1] 3.456758
                                           > q.T <- qtukey(0.95, 4, 36)
                1      2    3     4
                                           > q.T
            3.519 1.862 2.947 2.161        [1] 3.808798
Statistical Modelling Chapter VII                                            79
        Example VII.4 Animal survival experiment
        (continued)

        • For our example, as the interaction is not significant, the
          overall tables of means are examined.
        • For the Poison means
                                                    3.456758   0.240  2
        Poison                            w  5% =          
            1     2     3                                2        16
        1.801 2.269 3.797                         = 0.42

        • All Poison means are significantly different.
        • For the Treat means
        Treat                                       3.808798   0.240  2
            1     2     3     4           w  5% =          
                                                         2        12
        3.519 1.862 2.947 2.161                   = 0.54
        • All but Treats 2 and 4 are different.
Statistical Modelling Chapter VII                                          80
        Plotting the means in a bar chart
        >   # Plotting means
        >   #
        >   Fac2Pois.DR.tab <- model.tables(Fac2Pois.DR.aov, type="means")
        >   Fac2Pois.DR.Poison.Means <-
        +            data.frame(Poison = levels(Poison),
        +                       Death.Rate = as.vector(Fac2Pois.DR.tab$tables$Poison))
        >   barchart(Death.Rate ~ Poison, main="Fitted values for Death rate",
        +                                 ylim=c(0,4), data=Fac2Pois.DR.Poison.Means)
        >   Fac2Pois.DR.Treat.Means <-
        +              data.frame(Treatment = levels(Treat),
        +                         Death.Rate = as.vector(Fac2Pois.DR.tab$tables$Treat))
        >   barchart(Death.Rate ~ Treat, main="Fitted values for Death rate",
        +                                ylim=c(0,4), data=Fac2Pois.DR.Treat.Means)
                                                                                                Fitted values for Death rate
                                        Fitted values for Death rate




                                                                                        3
                                3




                                                                           Death.Rate
                   Death.Rate




                                2                                                       2




                                1                                                       1




                                    1                  2               3



        • Max death rate with Poison 3 and Treats 1.
                                                                                            1          2               3       4




        • Min death rate with Poison 1 and either Treats 2 or 4.
Statistical Modelling Chapter VII                                                                                                  81
    If interaction significant, 2 possibilities
        • Possible researcher’s objective(s):
                i. finding levels combination(s) of the factors that
                     maximize (or minimize) response variable or
                     describing response variable differences between all
                     levels combinations of the factors
                ii. for each level of one factor, finding the level of the
                     other factor that maximizes (or minimizes) the
                     response variable or describing the response
                     variable differences between the levels of the other
                     factor
                iii. finding a level of one factor for which there is no
                     difference between the levels of the other factor
        • For i: examine all possible pairs of differences
          between all means.
        • For ii & iii: examine pairs of mean differences
          between levels of one factor for each level of
          other factor i.e. in slices for each level of other
          factor.
Statistical Modelling Chapter VII                                        82
        Table of Poison by Treat means
        Poison:Treat
              Treat
        Poison 1                            2         3         4
             1 2.487                        1.163     1.863     1.690
             2 3.268                        1.393     2.714     1.702
             3 4.803                        3.029     4.265     3.092
                                              4.93606   0.240  2
                                    w  5%  =        
                                                   2        4
                                            = 1.21

        • Look for overall max or max in each column
        • Do not do for this example as interaction is not
          significant
Statistical Modelling Chapter VII                                       83
        b) Polynomial submodels
        • As stated previously, the formal expression for
          maximal indicator-variable model for a two-factor
          CRD experiment, where the two randomized
          factors A and B are fixed, is:
                                                
                            y AB = E  Y  = X AB a and V =  UIn
                                                               2



        •       In respect of fitting polynomial submodels, two
                situations are possible:
              i. one factor only is quantitative, or
              ii. both factors are quantitative.




Statistical Modelling Chapter VII                                    84
        One quantitative (B) and one qualitative
        factor (A)
        • Following set of models for E[Yijk] is considered:
                   E Yijk  = a ij
                                                              depends on combination of A and B

                   E Yijk  = a i  a i1 x  j  a i 2 x  j
                      
                                                                2


                                                                quadratic response to B, differing for A
                   E Yijk  = a i  a i1 x  j
                                                              linear response to B, differing for A
indicator          E Yijk  = a i   j
                                                              nonsmooth, independent response to A & B
-variable          E Yijk  = a i   1x  j   2 x  j
                                                      2
                                                                quadratic response to B, intercept differs for A
                      
models             E Yijk  = a i   1x  j                   linear response to B, intercept differs for A
                      
                   E Yijk  = a i
                                                              nonsmooth response, depends on A only
                   E Yijk  =  j
                                                              nonsmooth response, depends on B only
                   E Yijk  =    1x  j   2 x  j
                      
                                                    2
                                                                quadratic response to B, A has no effect

                   E Yijk  =    1x  j
                                                              linear response to B, A has no effect

                   E Yijk  = 
                                                              neither factor affects the response
Statistical Modelling Chapter VII                                                                               85
          Matrix expressions for models
             E  Y  = X AB a                             depends on combination of A and B
             E  Y  = X A a  X A1  a 1  X A2  a 2   quadratic response to B, differing for A
             E  Y  = X A a  X A1 a 1                   linear response to B, differing for A
             E  Y  = X A a  X B                          nonsmooth, independent response to A & B
             E  Y  = X A a  X1 1  X 2 2                quadratic response to B, intercept differs for A
             E  Y  = X A a  X1 1                         linear response to B, intercept differs for A
             E Y = XAa                                     nonsmooth response, depends on A only
             E  Y  = X B                                  nonsmooth response, depends on B only
             E  Y  = X G   X1 1  X 2 2                quadratic response to B, A has no effect
             E  Y  = X G   X1 1                         linear response to B, A has no effect
             E  Y  = XG                                   neither factor affects the response
   a  = a ij  is an ab-vector of effects              X1 is an n-vector containing the values of the levels of B

 a 1 = a i1 is an a-vector of linear coefficients X 2 is an n-vector containing the (values) of the levels of B
                                                                                                          2

                                                             X A1 is an n  a matrix whose ith column contains
 a 2 = a i 2  is an a-vector of quadratic coeffients        the values of the levels of B for just those units that
    a = a i  is an a-vector of effects                        received the ith level of A
          is a b-vector of effects
     = j                                               X A2 is an n  a matrix whose ith column contains
    =  1  2 
    2                                                           the (values)2 of the levels of B for just those units
 Statistical Modelling Chapter VII                                                                                      86
         Example VII.6 Effect of operating temperature
         on light output of an oscilloscope tube
         • Suppose an experiment conducted to investigate the effect of the
           operating temperatures 75, 100, 125 and 150, for three glass types,
           on the light output of an oscilloscope tube.
         • Further suppose that this was done using a CRD with 2 reps.
         • Then X matrices for the analysis of the experiment:
                   75                       75    0     0                                 5625     0       0 
                   75                       75    0     0                                 5265     0       0 
                  100                      100    0     0                                10000     0       0 
                  100                      100    0     0                                10000     0       0 
                  125                      125    0     0                                15625     0       0 
                  125                      125    0     0                                15625     0       0 
                  150                      150    0     0                                22500     0       0 
                  150                      150    0     0                                22500     0       0 
                   75                       0     75    0                                 0       5625     0 
                   75                       0     75    0                                 0       5625     0 
                  100                       0    100    0   a                        0      10000     0   a  
                                                        0   a 11  ,                                     0   a 12 
          X1 1 = 100   1, X A1  a 1 =  0
                                                           0   21
                                                                             X A2  a 2
                                                                                                                0   22 
                                                    100                                     = 0      10000
                   125                         0    125                                          0    15625
                  125 
                      
                                              0
                                                   125     0  a 31
                                                                      
                                                                                              0
                                                                                                     15625     0  a 32 
                                                                                                                           
                  150                       0    150    0                                 0      22500     0 
                  150                       0    150    0                                 0      22500     0 
                   75                       0      0    75                                0         0     5625 
                   75                       0      0    75                                0         0     5625 
                  100                       0      0   100                                0         0    10000 
                  100                       0      0   100                                0         0    10000 
                  125                       0      0   125                                0         0    15625 
                  125                       0      0   125                                0         0    15625 
                  150                       0      0   150                                0         0    22500 
                  150 
                                            0
                                                     0   150 
                                                                                             0
                                                                                                        0    22500 
                                                                                                                    
Statistical Modelling Chapter VII                                                                                           89
        Why this set of expectation
        models?
        • As before, s are used for the coefficients of polynomial
          terms
              – a numeric subscript for each quantitative fixed factor in the
                experiment is placed on the s to indicate the degree(s) to which
                the factor(s) is(are) raised.
        • The above models are ordered from the most complex to
          the simplest.
        • They obey two rules:
        • Rule VII.1: The set of expectation models corresponds to
          the set of all possible combinations of potential
          expectation terms, subject to restriction that terms
          marginal to another expectation term are excluded from
          the model;
        • Rule VII.2: An expectation model must include all
          polynomial terms of lower degree than a polynomial term
          that has been put in the model.
Statistical Modelling Chapter VII                                                   90
        Definitions to determine if a
        polynomial term is of lower degree
          • Definition VII.7: A polynomial term is one in
            which the X matrix involves the quantitative
            levels of a factor(s).
          • Definition VII.8: The degree for a polynomial
            term with respect to a quantitative factor is the
            power to which levels of that factor are to be
            raised in this term.
          • Definition VII.9: A polynomial term is said to be
            of lower degree than a second polynomial term
            if,
                – for each quantitative factor in first term, its degree is
                  less than or equal to its degree in the second term and
                – the degree of at least one factor in the first term is less
                  than that of the same factor in the second term.
Statistical Modelling Chapter VII                                          91
     Marginality of terms and models
     • Note that the term X11 is not marginal to X22 — the
       column X1 is not a linear combination of the column X2.
     • However,
        – the degree of X11 is less than that of X22
        – the degree rule above implies that if term X22 is included in the
          model, so must the term X11.
     • As far as the marginality of models is concerned, the
       model involving just X11 is marginal to the model
       consisting of X11 and X22
              E  Y = XG  X1 1  E  Y = XG  X1 1  X2 2
          • Also note that the term X11 is marginal to XA1(a)1 since
              X1 is the sum of the columns of XA1.
          • Consequently, a model containing XA1(a)1 will not
              contain X11.
          • In general, the models to which a particular model is
              marginal will be found above it in the list.
Statistical Modelling Chapter VII                                              92
        ANOVA table for a two-factor CRD
        with one quantitative factor
                       Source              df          SSq
                       Units              n-1        YQUY
                        A                 a-1        YQ A Y
                        B                 b-1         YQB Y
                         Linear            1         YQBL Y
                         Quadratic         1         YQBQ Y
                         Deviations       b-3       YQBDev Y
                        A#B           (a-1)(b-1)     YQ ABY
                         A#BLinear        a-1       YQ ABL Y
                         A#BQuadratic     a-1       YQ ABQ Y
                         Deviations   (a-1)(b-3)    YQ ABDev Y
                        Residual      (r-1)(ab-1)   YQURes Y

Statistical Modelling Chapter VII                                 93
        Strategy in determining models to be
        used to describe the data.
        For Deviations
        • Only if the terms to which a term is marginal are not significant then, if
          P(F  Fcalc)  a, the evidence suggests that H0 be rejected and the
          term must be incorporated in the model.
        • Deviations for B is marginal to Deviations for A#B so that if the latter is
          significant, the Deviations for B is not tested; indeed no further testing
          occurs as the maximal model has to be used to describe the data.
        For A#BLinear and A#BQuadratic
        • Only if the polynomial terms are not of lower degree than a significant
          polynomial term then, if P(F  Fcalc)  a, the evidence suggests that
          H0 be rejected and the term be incorporated in the model.
        • A#BLinear is of lower degree than to A#BQuadratic so that if the latter is
          significant, A#BLinear is not tested.
        For A, Linear for B, Quadratic for B
        • Only if the terms to which a term is marginal and the polynomial terms
          of higher degree are not significant then, if P(F  Fcalc)  a, the
          evidence suggests that H0 be rejected and the term be incorporated in
          the model.
        • For example, for the Linear term for B, it is of lower degree than the
          Quadratic term for B and it is marginal to A#BLinear so that if either of
          these is significant, Linear for B is not tested.
Statistical Modelling Chapter VII                                                  94
        Both factors quantitative
        • Example VII.7 Muzzle velocity of an antipersonnel
          weapon
        • In a two-factor CRD experiment with two replicates the
          effect of
              – Vent volume and
              – Discharge hole area
             on the muzzle velocity of a mortar-like antipersonnel
             weapon was investigated.
                          Vent              Discharge hole area
                         volume     0.016      0.03    0.048      0.062
                          0.29      294.9     295.0    270.5      258.6
                                    294.1     301.1    263.2      255.9
                           0.40     301.7     293.1    278.6      257.1
                                    307.9     300.6    267.9      263.6
                           0.59     285.5     285.0    269.6      262.6
                                    298.6     289.1    269.1      260.3
                           0.91     303.1     277.8    262.2      305.3
                                    305.3     266.4    263.2      304.9
Statistical Modelling Chapter VII                                         95
        Interaction.Plot produced
        using R
                                                                                            Hole.Area


                                                300
                                                                                                 0.062
                                                                                                 0.016
                                                                                                 0.03
                                                                                                 0.048
                                                290
                             mean of Velocity

                                                280
                                                270
                                                260




                                                      0.29   0.4   0.59              0.91

                                                                          Vent.Vol



        • Pretty clear that there is an interaction.
Statistical Modelling Chapter VII                                                                        96
        Maximal polynomial submodel, in
        terms of a single observation
               E Yijk  =    10 xai   20 xai   01x  j   02 x  j
                  
                                                2                       2


                                     11xai x  j   12 xai x  j   21xai x  j   22 xai x  j
                                                                2          2                2 2

        where
        • Yijk is the random variable representing the response
          variable for the kth unit that received the ith level of factor
          A and the jth level of factor B,
        •  is the overall level of the response variable in the
          experiment,
        • xa is the value of the ith level of factor A,
              i
        • x  is the value of the jth level of factor B,
                j
        • s are the coefficients of the equation describing the
          change in response as the levels of A and/or B changes
          with the first subscript indicating the degree with respect to
          factor A and the second subscript indicating the degree
          with respect to factor B.
Statistical Modelling Chapter VII                                                                      97
        Maximal polynomial submodel, in
        matrix terms
                                    E  Y = XG  X22
       where  =  10  20  01  02  11  12  21  22 
              22

                     X =  X10      X 20   X01 X 02   X11 X12   X 21 X 22 

        • X is an n  8 matrix whose columns are the
          products of the values of the levels of A and B as
          indicated by the subscripts in X.
        • For example
              – 3rd column consists of the values of the levels of B
              – 7th column the product of the squared values of the
                levels of A with the values of the levels of B.
Statistical Modelling Chapter VII                                             98
        Set of expectation models considered
        when both factors are quantitative
                             
            E  Y  = X AB a                    depends on combination of A and B
            E  Y  = XG   X22                smooth response in A and B
                           or some subset of X22 that obeys the degrees rule 
            E  Y  = X A a  X B               nonsmooth, independent response to A & B

non-   E  Y  = X A a  X 01 01  X 02 02 quadratic response for B, intercept differs for A
smooth E  Y  = X A a  X 01 01            linear response for B, intercept differs for A
       E Y = XAa
A
                                             nonsmooth response, depends on A only

non-   E  Y  = XB  X10 10  X 20 20        quadratic response for A, intercept differs for B
smooth E  Y  = XB  X10 10                   linear response for A, intercept differs for B
B
       E  Y  = X B                            nonsmooth response, depends on B only
         where        a  = a ij 
                         a = a i 
                                
                           = j
                        =  10
                        22             20  01  02  11  12  21  22 
Statistical Modelling Chapter VII                                                                 99
        Set of expectation models (continued)
        • Again, rules VII.1 and VII.2 were used in deriving
          this set of models.
        • Also, the subsets of terms from 22 mentioned
          above include the null subset and must conform
          to rule VII.2 so that whenever a term from X22 is
          added to the subset, all terms of lower degree
          must also be included in the subset.
              – X1111 < X1212 so model with X1212 must include X1111
              – X1212  X2121 so model with X1212 does not need
                X2121
        • Further, if for a term the Deviation for a marginal
          term is significant, polynomial terms are not
          considered for it.
Statistical Modelling Chapter VII                                      100
        Interpreting the fitted models
        • models in which there are only single-factor polynomial
          terms define
               –      a plane if both terms linear
               –      a parabolic tunnel if one term is linear and the other quadratic
               –      a paraboloid if both involve quadratic terms
        • models including interaction submodels define nonlinear
          surfaces
               –      they will be monotonic for factors involving only linear terms,
               –      for interactions involving quadratic terms, some candidate
                      shapes are:




Statistical Modelling Chapter VII                                                        101
        ANOVA table for a two-factor CRD
        with both factors quantitative
                       Source                             df          SSq
                       Units                             n-1         YQUY
                        A                                a-1         YQA Y
                           Linear                         1          YQ AL Y
                             Quadratic                    1          YQ A Y
                                                                          Q

                             Deviations                  a-3        YQBDev Y
                         B                               b-1         YQB Y
                             Linear                       1         YQBL Y
                             Quadratic                    1         YQB YQ

                             Deviations                  b-3        YQBDev Y
                         A#B                          (a-1)(b-1)     YQABY
                           ALinear#BLinear                1         YQ ALBL Y
                             ALinear#BQuadratic           1         YQ A B Y
                                                                         L Q

                             AQuadratic#BLinear           1         YQ AQBL Y
                             AQuadratic#BQuadratic        1         YQ AQBQ Y
                             Deviations              (a-1)(b-1)-4   YQ ABDev Y
                         Residual                    (r-1)(ab-1)    YQU YRes


Statistical Modelling Chapter VII                                                 102
        Step 3: Decide between hypotheses
        For Deviations
        • Only if the terms to which a term is marginal are not
          significant then, if Pr{F  F0} = p  a, the evidence
          suggests that H0 be rejected and the term must be
          incorporated in the model.
        • Deviations for A and B are marginal to Deviations for A#B
          so that if the latter is significant, neither the Deviations for
          A nor for B is tested; indeed no further testing occurs as
          the maximal model has to be used to describe the data.
        For all Linear and Quadratic terms
        • Only if the polynomial terms are not of lower degree than a
          significant polynomial term and the terms to which the
          term is marginal are not significant then, if Pr{F  F0} = p 
          a, the evidence suggests that H0 be rejected; the term and
          all polynomial terms of lower degree must be incorporated
          in the model.
        • For example, Alinear#BLinear is marginal to A#B and is of
          lower degree than all other polynomial interaction terms
          and so is not tested if any of them is significant.
Statistical Modelling Chapter VII                                       103
        Example VII.7 Muzzle velocity of an
        antipersonnel weapon (continued)
          •   Here is the analysis produced using R, where
               > attach(Fac2Muzzle.dat)
               > interaction.plot(Vent.Vol, Hole.Area, Velocity, lwd=4)
               > Vent.Vol.lev <- c(0.29, 0.4, 0.59, 0.91)
               > Fac2Muzzle.dat$Vent.Vol <-
                 ordered(Fac2Muzzle.dat$Vent.Vol, levels=Vent.Vol.lev)
               > contrasts(Fac2Muzzle.dat$Vent.Vol) <- contr.poly(4,
                 scores=Vent.Vol.lev)
               > contrasts(Fac2Muzzle.dat$Vent.Vol)
               > Hole.Area.lev <- c(0.016, 0.03, 0.048, 0.062)
               > Fac2Muzzle.dat$Hole.Area <-
                 ordered(Fac2Muzzle.dat$Hole.Area,levels=Hole.Area.lev)
               > contrasts(Fac2Muzzle.dat$Hole.Area) <- contr.poly(4,
                 scores=Hole.Area.lev)
               > contrasts(Fac2Muzzle.dat$Hole.Area

both factors are converted to ordered
and polynomial contrasts for unequally-
spaced levels obtained
Statistical Modelling Chapter VII                                    104
        Contrasts
 > contrasts(Fac2Muzzle.dat$Vent.Vol)
               .L         .Q          .C
 0.29 -0.54740790 0.5321858 -0.40880670
 0.4 -0.31356375 -0.1895091 0.78470636
 0.59 0.09034888 -0.7290797 -0.45856278
 0.91 0.77062277 0.3864031 0.08266312
 > contrasts(Fac2Muzzle.dat$Hole.Area)
               .L   .Q         .C
 0.016 -0.6584881 0.5 -0.2576693
 0.03 -0.2576693 -0.5 0.6584881
 0.048 0.2576693 -0.5 -0.6584881
 0.062 0.6584881 0.5 0.2576693
 > summary(Fac2Muzzle.aov, split = list(
 +         Vent.Vol = list(L=1, Q=2, Dev=3),
 +         Hole.Area = list(L=1, Q= 2, Dev=3),
 +         "Vent.Vol:Hole.Area" = list(L.L=1, L.Q=2, Q.L=4, Q.Q=5, Dev=c(3,6:9))))
                                    Factor                              B
  Table shows                                         Contrast    1     2     3
  numbering of                               Contrast  Label      L     Q    Dev
                                                1        L       L.L   L.Q   Dev
  contrasts                                                      (1)   (2)   (3)
  (standard order;                    A         2        Q       Q.L
                                                                 (4)
                                                                       Q.Q
                                                                       (5)
                                                                             Dev
                                                                             (6)
  by rows).                                     3       Dev      Dev   Dev   Dev
                                                                 (7)   (8)   (9)   105
Statistical Modelling Chapter VII
                                                         split used for both
        R ANOVA                                          factors and interactions
   > summary(Fac2Muzzle.aov, split = list(               in the summary function.
   +         Vent.Vol = list(L=1, Q=2, Dev=3),
   +         Hole.Area = list(L=1, Q= 2, Dev=3),
   +         "Vent.Vol:Hole.Area" = list(L.L=1, L.Q=2, Q.L=4, Q.Q=5, Dev=c(3,6:9))))

   Error: Test
                                          Df   Sum Sq Mean Sq F value    Pr(>F)
   Vent.Vol                                3    379.5   126.5  5.9541 0.0063117
     Vent.Vol: L                           1    108.2   108.2  5.0940 0.0383455
     Vent.Vol: Q                           1     72.0    72.0  3.3911 0.0841639
     Vent.Vol: Dev                         1    199.2   199.2  9.3771 0.0074462
   Hole.Area                               3   5137.2 1712.4 80.6092 7.138e-10
     Hole.Area: L                          1   4461.2 4461.2 210.0078 1.280e-10
     Hole.Area: Q                          1    357.8   357.8 16.8422 0.0008297
     Hole.Area: Dev                        1    318.2   318.2 14.9776 0.0013566
   Vent.Vol:Hole.Area                      9   3973.5   441.5 20.7830 3.365e-07
     Vent.Vol:Hole.Area:            L.L    1   1277.2 1277.2 60.1219 8.298e-07
     Vent.Vol:Hole.Area:            L.Q    1     89.1    89.1  4.1962 0.0572893
     Vent.Vol:Hole.Area:            Q.L    1   2171.4 2171.4 102.2166 2.358e-08
     Vent.Vol:Hole.Area:            Q.Q    1    308.5   308.5 14.5243 0.0015364
     Vent.Vol:Hole.Area:            Dev    5    127.2    25.4  1.1975 0.3541807
   Residuals                              16    339.9    21.2



Statistical Modelling Chapter VII                                                   106
        Analysis summary
                   Source                         df    SSq      MSq       F        p
                   Runs                      31
                    Vent.Vol                  3         379.5    126.5     5.95   0.006
                     Linear                       1     108.2    108.2     5.09   0.038
                     Quadratic                    1      72.0     72.0     3.39   0.084
                     Deviations                   1     199.2    199.2     9.38   0.007
                    Hole.Area                 3        5137.2   1712.4    80.61   0.000
                     Linear                       1    4461.2   4461.2   210.01   0.000
                     Quadratic                    1     357.8    357.8    16.84   0.001
                     Deviations                   1     318.2    318.2    14.98   0.001
                    Vent.Vol#Hole.Area        9        3973.5    441.5    20.78   0.000
                     VLinear#HLinear              1    1277.2   1277.2    60.12   0.000
                     VLinear#HQuadratic           1      89.1     89.1     4.20   0.057
                     VQuadratic#HLinear           1    2171.4   2171.4   102.22   0.000
                     VQuadratic#HQuadratic        1     308.5    308.5    14.52   0.002
                     Deviations                   5     127.2     25.4     1.20   0.354
                    Residual                 16         339.9     21.2
     • 5 interaction Deviations lines have been pooled — df and SSq have
       been added together.
     • While the Deviations for the interaction is not significant (p = 0.354),
       those for both the main effects are significant (p = 0.007 and p = 0.001).
            – Hence a smooth response function cannot be fitted.
     • Furthermore, the Vquadratic#HQuadratic source is significant (p = 0.002) so
       that interaction terms are required.
     • In this case, we must revert to the maximal model.
Statistical Modelling Chapter VII                                                         107
        Fitting these submodels in R
        • Extension of the procedure for a single factor:
        • Having specified polynomial contrasts for each
          quantitative factor, the list argument of the summary
          function is used to obtain SSqs.
        • The general form of the summary function for one factor,
          B say, quantitative is (details in Appendix C.5, Factorial
          experiments.):
        summary(Experiment.aov, split = list(
                B = list(L = 1, Q = 2, Dev = 3:(b-1)),
            "A:B" = list(L = 1, Q = 2, Dev = 3:(b-1))))
        • and for two factors, A and B say, quantitative is
        summary(Experiment.aov, split = list(
                 A = list(L = 1, Q = 2, Dev = 3:(a-1)),
                 B = list(L = 1, Q = 2, Dev = 3:(b-1)),
              "A:B" = list(L.L=1, L.Q=2, Q.L=b, Q.Q=(b+1),
                            Dev=c(3:(b-1),(b+2:(a-1)(b-1)))))
Statistical Modelling Chapter VII                                      108
        VII.G Nested factorial structures
        • Nested factorial structures commonly arise when
              – a control treatment is included or
              – an interaction can be described in terms of one cell being
                different to the others.
        • Set up
              – a factor (One say) with two levels:
                    • for the control treatment or the different cell
                    • for the other treatments or cells.
              – A second factor (Treats say) with same number of levels as there
                are treatments or cells.
        • Structure for these two factors is One/Treats
        • Terms in the analysis are One + Treats[One].
              – One compares the control or single cell with the mean of the
                others.
              – Treats[One] reflects the differences between the other treatments
                or cells.
        • Can be achieved using an orthogonal contrast, but
          nested factors is more convenient.

Statistical Modelling Chapter VII                                               109
        General nested factorial structure
        set-up

        • An analysis in which there is:
              – a term that reflects the average differences
                between g groups;
              – a term that reflects the differences within
                groups or several terms each one of which
                reflects the differences within a group.




Statistical Modelling Chapter VII                              110
        Example VII.8 Grafting experiment

        • For example, consider the following RCBD
          experiment involving two factors each at two
          levels.
        • The response is the percent grafts that take.

                                                B        1             2
                                                A    1        2    1        2
                                            I       64       23   30       15†
                                           II       75       14   50       33
                                    Block III       76       12   41       17
                                          IV        73       33   25       10
                                    †observation  missing; value inserted
                                    so that residual is zero.
Statistical Modelling Chapter VII                                                111
      Example VII.8 Grafting experiment
        (continued)
       a) Description of pertinent features of the study
        1. Observational unit
              –     a plot
        2. Response variable
              –     % Take
        3. Unrandomized factors
              –     Blocks, Plots
        4. Randomized factors
              –     A, B
        5. Type of study
              –     Two-factor RCBD

       b) The experimental structure
                Structure    Formula
                unrandomized 4 Blocks/4 Plots
                randomized   2 A*2 B
Statistical Modelling Chapter VII                      112
        R output
        > attach(Fac2Take.dat)
        > Fac2Take.dat




                                                     70
                                                                           B


           Blocks Plots A B Take                                               1
                                                                               2

        1       1     1 1 1   64




                                                     60
        2       1     2 2 1   23
        3       1     3 1 2   30




                                                     50
                                      mean of Take
        4       1     4 2 2   15
        5       2     1 1 1   75




                                                     40
        6       2     2 2 1   14
        7       2     3 1 2   50




                                                     30
        8       2     4 2 2   33
        9       3     1 1 1   76




                                                     20
        10      3     2 2 1   12            1                          2

        11      3     3 1 2   41                               A


        12      3     4 2 2   17                          An interaction
        13      4     1 1 1   73
        14      4     2 2 1   33
        15      4     3 1 2   25
        16      4     4 2 2   10
        > interaction.plot(A, B, Take, lwd=4)

Statistical Modelling Chapter VII                                                  113
        R output (continued)
        •    > Fac2Take.aov <- aov(Take ~ Blocks + A * B +
        •    +                      Error(Blocks/Plots), Fac2Take.dat)
        •    > summary(Fac2Take.aov)
        •    Error: Blocks
        •           Df Sum Sq Mean Sq
        •    Blocks 3 221.188 73.729
        •    Error: Blocks:Plots
        •              Df Sum Sq Mean Sq F value    Pr(>F)
        •    A          1 4795.6 4795.6 52.662 4.781e-05
        •    B          1 1387.6 1387.6 15.238 0.003600
        •    A:B        1 1139.1 1139.1 12.509 0.006346
        •    Residuals 9 819.6      91.1




Statistical Modelling Chapter VII                                    114
        R output (continued)
        > res <- resid.errors(Fac2Take.aov)
        > fit <- fitted.errors(Fac2Take.aov)
        > plot(fit, res, pch=16)
        > plot(as.numeric(A), res, pch=16)
        > plot(as.numeric(B), res, pch=16)
        > qqnorm(res, pch=16)
        > qqline(res)
        > tukey.1df(Fac2Take.aov, Fac2Take.dat,
        +                            error.term = "Blocks:Plots")
        $Tukey.SS
        [1] 2.879712

        $Tukey.F
        [1] 0.02820886

        $Tukey.p
        [1] 0.870787

        $Devn.SS
        [1] 816.6828
Statistical Modelling Chapter VII                                   115
        Recompute for missing value
         • Recalculate either in R or in Excel.
         • See notes for Excel details
         >   #
         >   # recompute for missing value
         >   #
         >   MSq <- c(73.729, 4795.6, 1387.6, 1139.1, 2.8797)
         >   Res <- c(rep(819.6/8, 4), 816.6828/7)
         >   df.num <- c(3,rep(1,4))
         >   df.den <- c(rep(8, 4),7)
         >   Fvalue <- MSq/Res
         >   pvalue <- 1-pf(Fvalue, df.num, df.den)
         >   data.frame(MSq,Res,df.num,df.den,Fvalue,pvalue)
                   MSq      Res df.num df.den      Fvalue       pvalue
         1     73.7290 102.4500      3      8 0.71965837 0.5677335580
         2   4795.6000 102.4500      1      8 46.80917521 0.0001320942
         3   1387.6000 102.4500      1      8 13.54416789 0.0062170009
         4   1139.1000 102.4500      1      8 11.11859444 0.0103158259
         5      2.8797 116.6690      1      7 0.02468266 0.8795959255


Statistical Modelling Chapter VII                                   116
        Diagnostic checking
                                                                                                                                        Normal Q-Q Plot




                                                                                                                 15
                  15




                                                                                                                 10
                  10




                                                                                                                 5
                  5




                                                                                              Sample Quantiles
           res




                                                                                                                 0
                  0




                                                                                                                 -5
                  -5




                                                                                                                 -10
                  -10




                              20         30   40            50     60         70         80                            -2          -1            0              1           2

                                                      fit                                                                               Theoretical Quantiles
                 15




                                                                                                                   15
                 10




                                                                                                                   10
                 5




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          res




                                                                                                         res
                 0




                                                                                                                   0
                 -5




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                 -10




                                                                                                                   -10




                        1.0        1.2        1.4            1.6        1.8        2.0                                      1.0   1.2     1.4             1.6       1.8   2.0
                                                as.numeric(A)                                                                                as.numeric(B)



Statistical Modelling Chapter VII                                                                                                                                               117
        Hypothesis test for this example
        • Step 1: Set up hypotheses
              a) H0: (a)21 - (a)11 - (a)22 + (a)12 = 0
                 H1: (a)21 - (a)11 - (a)22 + (a)12  0
              b) H0: a1 = a2
                 H1: a1 = a2
              c) H0: 1 = 2
                 H1: 1 = 2
              Set a = 0.05.


Statistical Modelling Chapter VII                            118
        Hypothesis test for this example (continued)
        Step 2: Calculate test statistics
        • The ANOVA table for the two-factor RCBD is:
                   Source            df   SSq     MSq      E[MSq]       F     Prob
                   Blocks           3     221.9   73.7  BP  4 B
                                                         2       2    0.72   0.568

                   Plots[Blocks]    12    8141.8
                     A              1     4795.6 4795.6  BP  qA  y  46.81 <0.001
                                                          2

                     B              1     1387.6 1387.6  BP  qB  y  13.54 0.006
                                                          2

                     A#B            1     1139.1 1139.1  BP  qAB  y  11.12 0.010
                                                          2

                     Residual        8†    819.6 102.4  S
                                                         2

                      Nonadditivity 1         2.9  2.9             0.02 0.880
                      Deviations     7     816.7 116.7
                   †
                    Residual degrees of freedom have been reduced by one to allow
                     for the missing observation
        Step 3: Decide between hypotheses
        • Note residuals-versus-fitted-values plot reveals nothing untoward, test
           for nonadditivity is not significant and the normal probability plot also
           appears to be satisfactory.
        • Significant interaction between A and B so fitted model is E[Y] =
           XAB(a).
Statistical Modelling Chapter VII                                                      119
        Table of means
        •    Means for combinations of A and B need to be examined.
        •    Suppose the researcher wants to determine the level of A that has the greatest take
             for each level of B.
              > #
              > # multiple comparisons
              > #
              > Fac2Take.tab <- model.tables(Fac2Take.aov,
                type="means")
              > Fac2Take.tab$tables$"A:B"
                 B
              A   1     2
                1 72.00 36.50
                2 20.50 18.75                        4.52881 102.4  2
              > q <- qtukey(0.95, 4, 8)    w  5% =        
                                                         2      4
              > q
              [1] 4.52881                          = 22.91


             • no difference between A at level two of B
             • there is an A difference at level one of B —
               level one of A maximizes.

Statistical Modelling Chapter VII                                                                  120
        Best description
              > Fac2Take.tab$tables$"A:B"
              Dim 1 : A
              Dim 2 : B
                    1     2
              1 72.00 36.50
              2 20.50 18.75
        • A and B both at level 1 different from either A or B not at
          level 1.
        • However, the results are only approximate because of the
          missing value.
        • Testing for this can be achieved by setting up a factor for
          the 4 treatments and a two-level factor that compares the
          cell with A and B both at level 1 with the remaining
          factors.
        • The four-level factor for treatments is then specified as
          nested within the two-level factor.

Statistical Modelling Chapter VII                                  121
        Re-analysis achieved in R
    >  Fac2Take.dat$Cell.1.1 <- factor(1 + as.numeric(A != "1" | B != "1"))
    >  Fac2Take.dat$Treats <- fac.combine(list(A, B))
    >  detach(Fac2Take.dat)
    >  attach(Fac2Take.dat)
    >  Fac2Take.dat
        Blocks Plots A B Take Cell.1.1 Treats
    1        1      1 1 1   64       1      1
    2        1      2 2 1   23       2      3
    3        1      3 1 2   30       2      2
    4        1      4 2 2   15       2      4
    5        2      1 1 1   75       1      1
    6        2      2 2 1   14       2      3
    7        2      3 1 2   50       2      2
    8        2      4 2 2   33       2      4
    9        3      1 1 1   76       1      1
    10       3      2 2 1   12       2      3
    11       3      3 1 2   41       2      2
    12       3      4 2 2   17       2      4
    13       4      1 1 1   73       1      1
    14       4      2 2 1   33       2      3
    15       4      3 1 2   25       2      2
    16       4      4 2 2   10       2      4


Statistical Modelling Chapter VII                                        122
        Re-analysis (continued)
        > Fac2Take.aov <- aov(Take ~ Blocks + Cell.1.1/Treats +
           Error(Blocks/Plots), Fac2Take.dat)
        > summary(Fac2Take.aov)

        Error: Blocks
               Df Sum Sq Mean Sq
        Blocks 3 221.188 73.729

        Error: Blocks:Plots
                        Df Sum Sq Mean Sq F value     Pr(>F)
        Cell.1.1          1 6556.7 6556.7 72.0021 1.378e-05
        Cell.1.1:Treats 2 765.5      382.8 4.2032    0.05139
        Residuals         9 819.6     91.1
        > # recompute for missing value
        > MSq <- c(73.729,6556.7,382.8)
        > Res <- rep(819.6/8, 3)
        > df.num <- c(3, 1, 2)
        > Fvalue <- MSq/Res
        > pvalue <- 1-pf(Fvalue, df.num, 8)
        > data.frame(MSq,Res,df.num,Fvalue,pvalue)
               MSq    Res df.num      Fvalue       pvalue
        1   73.729 102.45       3 0.7196584 5.677336e-01
        2 6556.700 102.45       1 63.9990239 4.367066e-05
        3 382.800 102.45        2 3.7364568 7.146140e-02
Statistical Modelling Chapter VII                                 123
        Revised analysis of variance table
                   Source           df    SSq       MSq        F      Prob
                   Blocks            3    221.9      73.7     0.72    0.568

                   Plots[Blocks]      12 8141.8
                     Cell 1,1 vs rest 1 6556.7 6556.7 64.00 <0.001
                     Among rest        2    765.5     382.8     3.74   0.071
                     Residual          8†   819.6     102.4
                   †
                          the Residual and Total degrees of freedom have been
                   reduced by one to allow for the missing observation


        • Appears difference between the treatments is best
          summarized in terms of this single degree of freedom
          contrast between cell1,1 and the others.
        • The mean for cell 1,1 is 72.0 and, for the other three
          treatments, the mean is 25.2, a difference of 46.8.
        • Such one-cell interactions are a very common form of
          interaction.
Statistical Modelling Chapter VII                                               124
        Example VII.9 Spraying sultanas
        • An experiment was conducted to investigate the effects of
          tractor speed and spray pressure on the quality of dried
          sultanas.
        • Response was lightness of the dried sultanas measured
          using a Hunterlab D25 L colour difference meter.
        • Three tractor speeds and two spray pressures resulting in
          6 treatment combinations which were applied to 6 plots,
          each consisting of 12 vines, using a RCBD with 3 blocks.
        • However, these 6 treatment combinations resulted in only
          4 rates of spray application as indicated in the following
          table.
                            Table of application rates for the sprayer experiment

                                                             -1
                                    Tractor Speed (km hour )
                     Pressure (kPa)       3.6             2.6                1.8
                          140            2090            2930               4120
                          330            2930            4120               5770

Statistical Modelling Chapter VII                                                   125
        Set up for analysis
        • To analyze this experiment set up:
              – a factor, Rates, with 4 levels to compare 4 rate means
              – two factors with 3 levels, Rate2 and Rate3, each of
                which compares the means of 2 treatment
                combinations with the same rate.
                   Table of factor levels for Rate2 and Rate3 in the sprayer experiment

                                                          Rate2                     Rate3
                                    -1              3.6    2.6     1.8       3.6     2.6    1.8
      Tractor Speed (km hour )
                   Pressure (kPa)
                         140                         1       2      1         1       1     2
                         330                         3       1      1         1       3     1

        • The experimental structure for this experiment is:

                   Structure    Formula
                   unrandomized 3 Blocks/ 6 Plots
                   randomized   4 Rates/(3 Rate2+3 Rate3)
Statistical Modelling Chapter VII                                                                 126
        Sources in the ANOVA table
                      Source           df          E[MSq]
                      Blocks            2    BP   B
                                              2      2


                      Plots[Blocks]    15
                        Rates           3    BP  qR  y 
                                              2


                        Rate2[Rates]    1    BP  qR2  y 
                                              2


                        Rate3[Rates]    1    BP  qR3  y 
                                              2


                        Residual       10    BP
                                              2


                      Total            17

Statistical Modelling Chapter VII                              127
        VII.H Models and hypothesis testing
              for three-factor experiments
        • Experiment with
              – factors A, B and C with a, b and c levels
              – each of the abc combinations of A, B and C replicated
                r times.
              – n = abcr observations.
        • The analysis is an extension of that for a two-
          factor CRD.
        • Initial exploration using interaction plots of two
          factors for each level of the third factor.
        • use interaction.ABC.plot from the DAE
          library.
Statistical Modelling Chapter VII                                   128
      a) Using the rules to determine the ANOVA
         table 3-factor CRD experiment
       a) Description of pertinent features of the study
        1. Observational unit
              –     a unit
        2. Response variable
              –     Y
        3. Unrandomized factors
              –     Units
        4. Randomized factors
              –     A, B, C
        5. Type of study
              –     Three-factor CRD

       b) The experimental structure
                Structure    Formula
                unrandomized n Units
                randomized   a A*b B*c C
Statistical Modelling Chapter VII                      129
        c) Sources derived from the structure formulae

        • Randomized only
          A*B*C = A + (B*C) + A#(B*C)
                = A + B + C + B#C
                     + A#B + A#C + A#B#C




Statistical Modelling Chapter VII                    130
        d) Degrees of freedom and sums of squares
        • The df can be derived by the cross product rule.
              – For each factor in the term, calculate the number of
                levels minus one and multiply these together.
     Unrandomized factors

                                                     Randomized factors
            MG       MG

                                                                    
                                                               MG       MG
    Units               U
     MU               MU-MG
                              A           A                B                 B              C        C
                              MA        MA-MG              MB           MB-MG               MC     MC-MG



                           AB          A#B          AC                     A#C           BC      B#C
                           MAB      MAB-MA-MB+MG     MAC            MAC-MA-MC+MG           MBC MBC-MB-MC+MG




                                                   ABC                         A#B#C

                                                   MABC                 MABC-MAB-MAC-MBC
                                                                          +MA+MB+MC-MG
Statistical Modelling Chapter VII                                                                        131
        e) The analysis of variance table
        • Enter the sources for the study, their degrees of
          freedom and quadratic forms, into the ANOVA
          table below.

        f) Maximal expectation and variation models

        • Given that the only random factor is Units, the
          following are the symbolic expressions for the
          maximal expectation and variation models:
              – var[Y] = Units
              – y = E[Y] = ABC



Statistical Modelling Chapter VII                             132
       g)          The expected mean squares

        Source                            df            SSq           E[MSq]
        Units                            n-1           YQUY
          A                              a-1           YQ A Y     U  qA  y 
                                                                    2


          B                              b-1           YQBY       U  qB  y 
                                                                    2


          A#B                         (a-1)(b-1)      YQ ABY      U  qAB  y 
                                                                    2


          C                              c-1           YQCY       U  qC  y 
                                                                    2


          A#C                         (a-1)(c-1)      YQ ACY      U  qAC  y 
                                                                    2

          B#C                         (b-1)(c-1)      YQBCY       U  qBC  y 
                                                                    2

          A#B#C                     (a-1)(b-1)(c-1)   YQ ABCY     U  qABC  y 
                                                                    2

          Residual                     abc(r-1)       YQURes Y   U2

        Total                           abcr-1         YQUY

Statistical Modelling Chapter VII                                                    133
  b) Indicator-variable models and
     estimation for the three-factor CRD
• The models for the expectation:                                              a  = a ijk 
 E  Y  = X ABC a  
                                                                                 a  = a ij 
 E  Y  = X AB a   X AC a   XBC   
                                                                                  =    jk 
                                        and equivalent models with a pair of 
 E  Y  = X AB a   X AC a       two-factors interactions               a  = a  
                                                                                             ik

                                           and equivalent models with two factors    a = a i 
                 
 E  Y  = X A a  XBC                 interacting and one factor independent   
                                                                                      =  j 
                                           and equivalent models with two factors     =  
                  
 E  Y  = X AB a                         interacting                                     k
                                                                                    
                                 and other models consisting of 
                          
 E  y  = X A a  XB   X C                                
                                 only main effects              
 E  Y  = XG 
• Altogether 19 different models.
Statistical Modelling Chapter VII                                                                   134
        Fitted values
        • Expressions for fitted values for each model given in
          terms of following mean vectors:
                A, B,  A  B  , C,  A  C  , Β  C  and  A  Β  C 
        • Being means vectors can be written in terms of mean
          operators, Ms.                         1 MG =   abcr
                                                                 Ja  Jb  Jc  Jr

        • Further, if Y is arranged so             MA =    1
                                                          bcr
                                                                 Ia  Jb  Jc  Jr
          that the associated factors              MB =    1     Ja  Ib  Jc  Jr
                                                          acr
          A, B, C and the replicates
          are in standard order                  MAB =      1
                                                           cr
                                                                 Ia  Ib  Jc  Jr
          M operators can be written               MC =    1
                                                          abr
                                                                 Ja  Jb  Ic  Jr
          as the direct product of I and
          J matrices as follows:                 MAC =     1
                                                           br
                                                                 Ia  Jb  Ic  Jr
                                                  MBC =     1
                                                           ar
                                                                 Ja  Ib  Ic  Jr

Statistical Modelling Chapter VII
                                                MABC =     1
                                                           r
                                                                  Ia  Ib  Ic  Jr   135
        c) Expected mean squares under
           alternative models
        • Previously given the E[MSq]s under the maximal model.
        • Also need to consider them under alternative models so
          that we know what models are indicated by the various
          hypothesis tests.
        • Basically, need to know under which models q(y) = 0.
        • From two-factor case, q(y) = 0 only when the model does
          not include a term to which the term for the source is
          marginal.
        • So, when doing the hypothesis test for a MSq for a fixed
          term,
              – provided terms to which it is marginal have been ruled out by
                prior tests,
              – it tests whether the expectation term corresponding to it is zero.
        • For example, consider the A#B mean square.
Statistical Modelling Chapter VII                                                    136
    An example: the A:B mean square
• Its expected value is  U  qAB  y 
                            2

• Now, q(y)  0 for models involving the AB term or terms
  to which the AB term is marginal.
• All models are of the form         • So q(y)  0 for the models
    E  Y  = X ABC a               E  Y  = X ABC a  

                               
    E  Y  = X AB a  X AC a  XBC                       
                                                 E  Y  = X AB a  X AC a   
    E  Y  = X AB a   X AC a                                  XBC   
    E  Y  = X A a   XBC   
                                                 E  Y  = X AB a   XBC   
    E  Y  = X AB  a 
                                                 E  Y  = X AB a   X AC a 
    E  y = X A a   XB    X C  
                                                 E  Y  = X AB a   X C 
    E  Y  = XG 
                                                 E  Y  = X AB a 
• Hence test for A#B, provides a test for whether AB should
      be included in the model, provided that the test for A#B#C
      has already indicated that ABC can be omitted.
Statistical Modelling Chapter VII                                                    137
        d) The hypothesis test
        Step 1:       Set up hypotheses                                                                    Term being tested
                     a  - a  - a  - a                      
        a)      H0:        ijk         ij .         i .k      . jk
                                                                                                               ABC
                      a   a   a  - a  = 0 for all i,j,k 
                               i ..          . j.         ..k      ...     
                     a  - a  - a  - a                        
                H1:                                                          
                            ijk          ij .         i .k     . jk
                      a   a   a  - a   0 for some i,j,k 
                               i ..          . j.         ..k       ...      
        b)          
                H0: a ij - a i . - a . j  a .. = 0 for all i,j                                     AB
                H1:  a     ij
                                    - a i . - a . j  a ..  0      for some i,j          
        c)      H0:  a ik
                                    - a i . - a .k  a .. = 0       for all i,k                        AC
                H1:  a ik
                                    - a i . - a .k  a ..  0       for some i,k          
        d)      H0:       jk
                                    -    j . -   .k    .. = 0    for all j,k                       BC
                H1:       jk
                                    -    j . -   .k    ..  0    for some j,k             
        e)    H0: a1 = a2 = ... = aa                                                                              A
              H1: not all population A means are equal
        f)    H0: 1 = 2 = ... = b                                                                              B
              H1: not all population B means are equal
        g)    H0: 1 = 2 = ... = c                                                                              C
              H1: not all population C means are equal
        Set a = 0.05.
Statistical Modelling Chapter VII                                                                                              138
        Step 2: Calculate test statistics
                  Source                  df            SSq           E[MSq]
                  Units                  n-1           YQUY
                    A                    a-1           YQ A Y     U  qA  y 
                                                                    2


                    B                    b-1           YQBY       U  qB  y 
                                                                    2


                    A#B               (a-1)(b-1)      YQ ABY      U  qAB  y 
                                                                    2


                    C                    c-1           YQCY       U  qC  y 
                                                                    2


                    A#C               (a-1)(c-1)      YQ ACY      U  qAC  y 
                                                                    2

                    B#C               (b-1)(c-1)      YQBCY       U  qBC  y 
                                                                    2

                    A#B#C           (a-1)(b-1)(c-1)   YQ ABCY     U  qABC  y 
                                                                    2

                    Residual           abc(r-1)       YQURes Y   U2

                  Total                 abcr-1         YQUY

        • MSqs would be added to this table by taking each SSq
          and dividing by its df,
        • F statistics computed by dividing all MSqs, except the
          Residual MSq, by the Residual MSq, and
        • p values obtained for each F statistic.
Statistical Modelling Chapter VII                                                    139
        Step 3: Decide between hypotheses
        For A#B#C interaction source
        • If Pr{F  F0} = p  a, the evidence suggests that
          H0 be rejected and the term should be
          incorporated in the model.
        For A#B, A#C and B#C interaction sources
        • Only if A#B#C is not significant, then if Pr{F  F0}
          = p  a, the evidence suggests that H0 be
          rejected and the term corresponding to the
          significant source should be incorporated in the
          model.
        For A, B and C source
        • For each term, only if the interactions involving
          the term are not significant, then if Pr{F  F0} = p
           a, the evidence suggests that H0 be rejected
          and the term corresponding to the significant
          source should be incorporated in the model.
Statistical Modelling Chapter VII                            140
        VII.J Exercises
        • Ex. VII-1 asks for the complete analysis of a
          factorial experiment with qualitative factors
        • Ex. VII-2 involves a nested factorial analysis –
          not examinable
        • Ex. VII-3 asks for the complete analysis of a
          factorial experiment with both factors quantitative
        • Ex. VII-4 asks for the complete analysis of a
          factorial experiment with random treatment
          factors
        • Ex. VII-5 asks for the complete analysis of a
          factorial experiment with interactions between
          unrandomized and randomized factors

Statistical Modelling Chapter VII                           141

				
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