ST4241 Design and Analysis of Clinical Trials Lecture 4: 2 2 factorial experiments, a special cases of parallel groups study

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1 ST4241 Design and Analysis of Clinical Trials Lecture 4: 2 2 factorial experiments, a special cases of parallel groups study Chen Zehua Department of Statistics & Applied Probability 8:00-10:00 am, Tuesday, August 23, 2016

2 Outline Factorial experiments for investigation of interaction

3 Factorial experiments When more than one methods can be used simultaneously for the treatment of patients, there is a concern whether there is any interaction effects between the different methods, good or bad. A factorial experiment can investigate interactions. Besides, it has the advantage of investigating more than one methods in a single experiment. 2 2 factorial experiment In a 2 2 factorial experiment, there are two factors (e.g., two medical methods), each factor has two levels (e.g., absent or present). The treatments are the combinations of the levels of the two factors. A 4-treatment parallel groups design can be applied to a 2 2 factorial experiment.

4 Example A study is caried out to investiage whether there is an interaction effect between two operations, castration and adrenalectomy, on reducing thymus gland weight of mice. Male mice were randomly assigned to receive one of four treatments: non operation, castration only, adrenalectomy only, both castration and adrenalectomy. Each animal was sacrificed at a fixed time after the operations and its thymus gland was weighed. The data is summarized below: Group Castration Adrenalectomy n ij X ij s ij µ ij 1 no no µ 11 2 yes no µ 21 3 no yes µ 12 4 yes yes µ 22

5 Measurement of interaction Effect of Castration without adrenalectomy: E C Ā = µ 21 µ 11. Effect of Castration with adrenalectomy: E C A = µ 22 µ 12. Effect of adrenalectomy without Castration: E A C = µ 12 µ 11. Effect of adrenalectomy with Castration: Interaction effect: E A C = µ 22 µ 21. E AC = E C A E = E C Ā A C E A C = µ 22 µ 12 µ 21 + µ 11.

6 Inference on interaction Let µ ij be estimated by X ij. Then the interaction is estimated by Ê AC = X 22 X 12 X 21 + X 11. Test statistic: L AC = Ê AC, Var(Ê AC ) where Var(Ê AC ) = s 2 i,j 1, s 2 i,j = (n ij 1)sij 2 n ij i,j (n ij 1). Under the hypothesis of no interaction, L AC has a t-distribution with df n 4. The interaction effect is claimed significant at level α if L AC t n 4,α/2.

7 Inference on interaction in the example Ê C Ā = X 21 X 11 = = Ê C A = X 22 X 12 = = Ê AC = Ê C A Ê C Ā = 0.3. s 2 = = L AC = ( ) = The p-value P( t ) 0.94 The interaction effect is not significant by any standard.

8 Inference on main effects The main effect of a factor is defined as its averaged within-level effects. For example, the main effect of castration is E C = E C A + E C Ā. 2 The concept of main effect is defined when more than one factors are involved. The main effect of a factor is, in general, not the pure effect of that factor when it is administered alone. The main effect is equal to the pure effect only when the interaction effect is zero. Analysis of main effect when interaction effect is non-zero is meaningless. It only makes sense when the interaction is found non-significant.

9 Inference on main effects when interaction is zero Intuitively, main effect can be estimated by the simple average of the estimated within-level effects. Due to different sample sizes, a more efficient estimate is given by the weighted average of the two estimated within-level effects, with weight being inversely proportional to their variances. Since Var(Ê ) = C Ā σ2 ( ), n 11 n 21 Var(Ê C A ) = σ 2 ( ). n 22 n 12 the weights are determined as w C Ā = n 11n 21 n 11 + n 21, w C A = n 12n 22 n 12 + n 22.

10 Inference on main effects when interaction is zero (cont.) The estimate of the main effect E C is then given by Ê C = w C ĀÊC Ā + w C AÊC A w C Ā + w. C A s 2 Var(Ê C ) = w C Ā + w. C A The test statistic for testing E C is L C = Ê C. Var(Ê C ) L C t n 4 under the null hypothesis of no main effect of C. The main effect is claimed as significant at level α if L C t n 4,α/2.

11 Inference on the main effect of castration in the Example Computation: w C Ā = = 6.74, w C A = = 4.44, Ê C = = 11.08, Var(Ê C ) = = , L C = = The p-value P( t ) = e 07. The effect of castration is significant. Main effect of Adrenalectomy can be similarly inferenced (left as excercise).

12 A wrong practice of inferece on interaction Usually, one based their inference on the ANAVA table of experimental data. The ANOVA table of the Example is as follows: Source d.f. SS MS F p-value Effect of C 1 1, , e-07 Effect of A Interaction Within 41 1, However, since the sample sizes are unequal, the ANOVA table is not valid for the analysis. The SS of interaction and main effects do not necessarily provide valid measurement for the effect they intend to measure.

13 SS of interaction in a 2 2 factorial experiment The SS of interaction is given by SS I = 2 2 n ij ( X ij X i X j + X ) 2. i=1 j=1 where X 1 = n 11 X 11 + n 12 X12, X2 = n 21 X 21 + n 22 X22, n 11 + n 12 n 21 + n 22 X 1 = n 11 X 11 + n 21 X 21 n 11 + n 21, X 2 = n 121 X 12 + n 22 X 22 n 12 + n 22, X = n 11 X 11 + n 12 X 12 + n 21 X 21 + n 22 X 22 n 11 + n 12 + n 21 + n 22.

14 Expression of a particular term in the SS of interaction X 11 X 1 X 1 + X = X 11 n 11 X 11 + n 12 X 12 n 11 X 11 + n 21 X 21 + n 11 + n 12 n 11 + n 21 n 11 X 11 + n 12 X 12 + n 21 X 21 + n 22 X 22. n 11 + n 12 + n 21 + n 22 n 11 n 21 n 11 = (1 + ) X 11 n 11 + n 12 n 11 + n 21 n 11 + n 12 + n 21 + n 22 n 12 n 12 ( ) X 12 n 11 + n 12 n 11 + n 12 + n 21 + n 22 n 21 n 21 ( ) X 21 + n 11 + n 21 n 11 + n 12 + n 21 + n 22 n 22 X 22. n 11 + n 12 + n 21 + n 22 It is not an estimate of µ 11 µ 12 µ 21 + µ 22 in general.

15 SS of interaction is not an estimation of interaction with unequal sample sizes X 11 X 1 X 1 + X is not an estimate of µ 11 µ 12 µ 21 + µ 22 when the sample sizes are unequal. Only when the sample sizes are equal, it reduces to 1 4 ( X 11 X 12 X 21 + X 22 ), which gives an unbiased estimate of 4E AC and SS I = n 4 Ê 2 AC, SS I s 2 = L2 AC. Warning When the group sizes are unequal, ANOVA table should not be used for the inference on interaction.

16 The regression model The 2 2 factorial design data can be validly analyzed by a linear regression approach whether or not the sample sizes are equal. The procedure is as follows: For each factor, create a dummy variable as follows: { 1, if factor 1 is at level 2; x 1 = 0, otherwise. { 1, if factor 2 is at level 2; x 2 = 0, otherwise. The data of the 2 2 experiment can be expressed as y ijk = β 0 + β 1 x 1ijk + β 2 x 2ijk + ξ 12 x 1ijk x 2ijk + ɛ ijk, i, j = 1, 2; k = 1,..., n ij.

17 Procedure of inference The inference on interaction reduces to test H 0 : ξ 12 = 0. When the hypothesis of no interaction is not rejected, refit the model y ijk = β 0 + β 1 x 1ijk + β 2 x 2ijk + ɛ ijk. The inference on main effects reduces to test whether or not β 1 and β 2 equal to zero. The test procedures for multple regression analysis are then used for the above tests.

18 Implementation using R Create observation vector y, which is a single vector consisting of all y ijk measurements. Create two factor vectors, x1 and x2, which are the vector of the dummy variables Use the R function lm as follows fitted = lm(y~x1*x2) Using the results contained in the object fitted to do inference. Remark: To use the linear regression approach, the original raw data must be available.