1 Lecture 18 The Two-way Anova We have learned how to test for the effects of independent variables considered one at a time. However, much of human behavior is determined by the influence of several variables operating at the same time. Sometimes these variables influence each other. We need to test for the independent and combined effects of multiple variables.

2 Lecture 18 The Two-way Anova Definitions: Main effect  effect of one treatment variable considered in isolation (ignoring other variables in the study) Interaction  when the effect of one variable is different across levels of one or more other variables

3 Lecture 18 Interaction of variables In order to detect interaction effects, we must use “factorial” designs.  In a factorial design each variable is tested at every level of all of the other variables. A 1 A 2 B 1 III B 2 IIIIV

4 Lecture 18 Interaction of variables I vs III Effect of B at level A 1 of variable A II vs IV Effect of B at A 2 I vs II Effect of A at B 1 III vs IV Effect of A at B 2  Two variables A & B interact if the effect of A varies at levels of B (equivalently, if the effect of B varies at levels of A).

5 Lecture 18 A 1 A 2 B1B2B1B2 B1B2B1B2 In the graphs above, the effect of A varies at levels of B, and the effect of B varies at levels of A. How you say it is a matter of preference (and your theory). In each case, the interaction is the whole pattern. No part of the graph shows the interaction. It can only be seen in the entire pattern showing all 4 data points.

6 Lecture 18 Interaction of variables  In order to test the hypothesis about an interaction, you must use a factorial design.  The designs shown on the previous slide (2 X 2s) are the simplest possible factorial designs.  We frequently use 3 and 4 variable designs, but beware: it’s almost impossible to interpret an interaction among more than 4 variables!

7 Lecture 18 Two-way Anova - Computation Raw Scores (X i ) A 1 A 2 A 3 B 1 B 2 B 1 B 2 B 1 B

8 Lecture 18 Cell Totals (T ij ) and marginal totals (T i & T j ) A 1 A 2 A 3 T j B B T i

9 Lecture 18 ΣX i 2 = … = 2427 (ΣX i ) 2 /N = (147) 2 /18 = Σ(T i ) 2 /n i = (21) 2 + (21) 2 + (105) 2 = Σ(T j ) 2 /n i = = Σ(T ij ) 2 /n ij = … =

10 Lecture 18 SS A = Σ(T i ) 2 – CM n i SS B = Σ(T j ) 2 – CM n j SS Tota l = ΣX i 2 – CM SS E = ΣX i 2 – Σ(T ij ) 2 n ij SS AB = Σ(T ij ) 2 – Σ(Tj) 2 – Σ(Ti) 2 + (ΣXi) 2 n ij n j n i N

11 Lecture 18 We now compute: SS A = – = 784 SS B = 1381 – = SS Total = 2427 – = SS E = 2427 – 2337 = 90 SS AB = 2337 – 1381 – = 172

12 Lecture 18 SourcedfSSMSF Aa-1 = Bb-1 = AB (a-1)(b-1) = ErrorN-ab = TotalN-1 =

13 Lecture 18 Two-way anova – hypothesis test for A H 0 : No difference among means for levels of A H A : At least two A means differ significantly Test statistic:F = MSA MSE Rej. region:F obt < F(2, 12,.05) = 3.89 Decision: Reject H 0 – variable A has an effect.

14 Lecture 18 Two-way anova – hypothesis test for B H 0 : No difference among means for levels of B H A : At least two B means differ significantly Test statistic:F = MSB MSE Rej. region:F obt < F(1, 12,.05) = 4.75 Decision: Reject H 0 – variable B has an effect.

15 Lecture 18 Two-way anova – hypothesis for AB H 0 : A & B do not interact to affect mean response H A : A & B do interact to affect mean response Test statistic:F = MS(AB) MSE Rej. region:F obt < F(2, 12,.05) = 3.89 Decision: Reject H 0 – A & B do interact...

16 Lecture 18 Example 1 – hypothesis test for A (illumination) H 0 : No difference among means for levels of A H A : At least two A means differ significantly Test statistic:F = MSA MSE Rej. region:F obt < F(1, 24,.05) = 4.26

17 Lecture 18 Example 1 – hypothesis test for B (type size) H 0 : No difference among means for levels of B H A : At least two B means differ significantly Test statistic:F = MSB MSE Rej. region:F obt < F(2, 24,.05) = 3.40

18 Lecture 18 Example 2 – hypothesis test for AB interaction H 0 : A & B do not interact to affect means H A : A & B do interact to affect means Test statistic:F = MS(AB) MSE Rej. region:F obt < F(2, 24,.05) = 3.40

19 Lecture 18 Two-way Anova – Example 1 Compute: CM = (12375) 2 = SS A = – CM =

20 Lecture 18 Two-way Anova – Example 1 SSB = – CM 10 = Σ(T ij ) 2 = … n ij =

21 Lecture 18 Two-way Anova – Example 1 SS AB = – – = SS Total = ΣX i 2 – CM = – = ,573.5 SS E = SS Total – SS A – SS B – SS AB =

22 Lecture 18 Two-way Anova – Example 1 SourcedfSSMS F A * B * AB Error Total * Reject H 0.

23 Lecture 18 Two-way Anova – Example 2 A researcher is interested in comparing the effectiveness of 3 different methods of teaching reading, and also in whether the effectiveness might vary as a function of the reading ability of the students. Fifteen students with high reading ability and fifteen students with low reading ability were divided into three equal-sized group and each group was taught by one of these methods. Listed on the next slide are the reading performance scores for the various groups at school year-end.

24 Lecture 18 Two-way Anova – Example 2 Teaching Method AbilityABC HighX s LowX s

25 Lecture 18 Two-way Anova – Example 2 (a) Do the appropriate analysis to answer the questions posed by the researcher (all αs =.05) (b) The London School Board is currently using Method B and, prior to this experiment, had been thinking of changing to Method A because they believed that A would be better. At α =.01, determine whether this belief is supported by these data.

26 Lecture 18 Example 2 – hypothesis test for A H 0 : No difference among means for levels of A H A : At least two A means differ significantly Test statistic:F = MSA MSE Rej. region:F obt < F(2, 12,.05) = 3.89 Decision: Reject H 0 – variable A has an effect.

27 Lecture 18 Two-way Anova – Example 1