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Chapter 10 - Part 1

Chapter 10 - Part 1. Factorial Experiments. Nomenclature. We will use the terms Factor and Independent Variable interchangeably. They mean the same thing.

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Chapter 10 - Part 1

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  1. Chapter 10 - Part 1 Factorial Experiments

  2. Nomenclature • We will use the terms Factor and Independent Variable interchangeably. They mean the same thing. • The term “factorial analysis of variance” simply means the analysis of variance when there are multiple factors (multiple independent variables.) • I will sometimes use the phrase Factor 1 or 2 interchangeably with Independent Variable 1 or 2. But to prevent confusion, whenever the term is abbreviated I will use IV1 or IV2, not F1 or F2.

  3. Two-way Factorial Experiments In Chapter 10, we are studying experiments with two independent variables, each of which will have multiple levels. We call each independent variable a factor. The first IV is called Factor 1 or IV1. The second IV is called Factor 2 or IV2. So, in this chapter we will study two factor, unrelated groups experiments.

  4. Conceptual Overview • In Chapter 9 you learned to do the F test comparing two estimates of sigma2, MSB and MSW. That is what you do in simple experiments, those with only one independent variable. • In the single IV, unrelated groups experiments in Ch. 9, F = MSB/MSW • That is one version of a more generic formula. The generic formula tells us what to do in the two (or more) factor case. • Here is the basis for the generic formula

  5. Generic formula for the unrelated groups F test • F = SAMP.FLUC. + (?) ONE SOURCE OF VARIANCE) (SAMPLING FLUCTUATION) • Which can also be written as • F = (ID + MP + (?) ONE SOURCE OF VARIANCE) (ID+ MP) Let me explain why.

  6. The denominator of the F test • The denominator in the F test reflects variation within each group because of random individual differences (ID) and measurement problems (MP). Since everyone in the same group is treated the same, only ID + MP can contribute to within group variation. So, MSW can not reflect the effect of any independent variable or combination of independent variables. • Random sampling fluctuation is based on random individual differences and measurement problems (ID + MP). • Thus, in the unrelated groups F and t test, our best index of the random effects of individual differences and measurement problems, the basis of random sampling fluctuation, is MSW, our least squares, unbiased estimate of sigma2.

  7. Denominator = MSW • In the F tests we are doing, F tests for unrelated groups, MSW serves as our best estimate of sigma2. • To repeat, in computing MSW, we compare each score to the mean of its own, specific group. Everyone in each specific group is treated the same. • So, the only reasons that scores differ from the other scores in their group and their own group means is that people differ from each other (ID) and there are always measurement problems (MP). • MSW = ID +MP

  8. Numerator of the F ratio: Generic formula • Numerator of the F ratio is an estimate of sigma2 that reflects sampling fluctuation + the possible effects of one difference between the groups. • In Ch. 9, there was only one difference among the ways the groups were treated, the different levels of the independent variable (IV) • MSB reflected the effects of random individual differences (there are different people in each group), random measurement problems, and the effects of the independent variable.In Ch. 9, we could write that as MSB = ID + MP + (?)IV

  9. To repeat, in the single factor analysis of variance F = (ID + MP + ?IV)(ID + MP) • Both the numerator and denominator reflect the same elements underlying sampling fluctuation • The numerator includes one, and only one, systematic source of variation not found in the denominator.

  10. This allows us to conclude that: • IF THE NUMERATOR IS SIGNIFICANTLY LARGER THAN THE DENOMINATOR, THE SIZE DIFFERENCE MUST BE MUST BE CAUSED BY THE ONE ADDITIONAL THING PUSHING THE MEANS APART, the IV. • But notice there can’t be more than one thing in the numerator that does not appear in the denominator to make that conclusion inevitable.

  11. Why we can’t use MSB as the numerator in the multifactor analysis of variance • In the two factor analysis of variance, the means can be pushed apart by: • The effects of the first independent variable (IV1). • The effects of the second independent variable (IV2) • The effects of combining IV1 and IV2 that are above and beyond the effects of either variable considered alone (INT) • Random sampling fluctuation (ID + MP)

  12. So if we compared MSW to MSB in a two factor experiment, here is what we would have. • F = (ID + MP + ?IV1 + ?IV2 + ?INT) (ID + MP) That’s not an F test. In an F test the numerator must have one and only one source of variation beyond sampling fluctuation. HERE THERE ARE THREE OF THEM! Each of these three things besides sampling fluctuation could be pushing the means apart. So, the F ratio would be meaningless.

  13. To use the F test with a two factor design, we must create 3 numerators to compare to MSW, each comprising ID + MP + one other factor.

  14. HOW? To obtain our 3 numerators for the F test, we divide (analyze) the sums of squares and degrees of freedom between groups (SSB & dfB)into component parts. Each part must contain only one factor along with ID and MP. Then each component will yield an estimate of sigma2 that can be compared to MSW in an F test.

  15. Write out the answer to these two questions without reading the answer from the slides: • Why can’t you compare MSB to MSW in the two factor, unrelated groups F test? • What must you do instead?

  16. ANSWER 1: If we compared MSW to MSB in a two factor experiment, here is what we would have. • F = (ID + MP + ?IV1 + ?IV2 + ?INT) (ID + MP) That’s not an F test. In an F test the numerator must have one and only one source of variation beyond sampling fluctuation. HERE THERE ARE THREE OF THEM! Each of these three things besides sampling fluctuation could be pushing the means apart. So, the F ratio would be meaningless.

  17. ANSWER 2:We must take apart (analyze) the sums of squares and degrees of freedom between groups (SSB & dfB)into component parts. Each part must contain only one factor along with ID and MP. Then each component will yield an estimate of sigma2 that can be compared to MSW in an F test.

  18. Here is how we divide SSB and dfB into their component parts: • First, we create a way to study the effects of factor 1 alone. • To do that, we combine groups so that the resulting, larger aggregates of participants differ only because they received different levels of the first independent variable, IV1. • Each such combined group will include an equal number of people who received the different levels of IV2. • So the groups are the same in that regard. • They differ only on how they were treated on the first independent variable, IV1.

  19. Computing MSIV1, one of the three numerators in a two factor F test • If we find the differences between each person’s combined group mean and the overall mean, square and sum them, we will have a sum of squares for the first independent variable (SSIV1). • Call the number of levels of an independent variable L. df for the combined group equals the number of levels of its IV minus one (LIV – 1). • An estimate of sigma2 that includes only ID + MP + (?) IV1 can be computed by dividing this sum of squares by its degrees of freedom, as usual. • MSIV1 = SSIV1/dfIV1 = (ID + MP + ?IV1)

  20. Once you have MSIV1, you have one of the three F tests you do in a two factor ANOVA • F = MSIV1/MSW

  21. Then you do the same thing to find MSIV2 • You combine groups so that you have groups that differ only on IV2. • You compare each person’s mean for this combined group to the overall mean, squaring athe differences for each person and then summing them for the entire sample to get SSIV2. • Degrees of freedom = the number of levels of Factor 2 minus 1 (dfIV2 = LIV2 – 1). • Then MSIV2 = SSIV2/dfIV2 • FIV2 = MSIV2/MSW

  22. What’s left is the interaction. • Remember, we are subdividing SSB and dfB into their three component parts. • We have already computed SSIV1, SSIV2, dfIV1, and dfIV2. • WHAT’S LEFT? The part of SSB and dfB that hasn’t been accounted for is the sum of squares and degrees of freedom for the interaction. • The interaction involves the means being pushed apart by the two independent variables having a different effect when present together than either has by itself alone. • For example, a moderate dose of alcohol can make you intoxicated. A moderate dose of barbituates can make you sleepy. Taken together they multiply each others’ effects and the interaction of the two drugs can easily make you dead.

  23. How to compute the interaction. • To compute the sum of squares and df for the interaction, we find that part of the sum of squares between groups and degrees of freedom between groups that are not accounted for by Factor 1 (IV1) and Factor 2 (IV2) • THAT IS, Y0U SUBTRACT THE SUMS OF SQUARES AND df YOU’VE ALREADY ACCOUNTED FOR (SSIV1, SSIV2, dfIV1, and dfIV2) FROM THE SUM OF SQUARES AND DEGREES OF FREEDOM BETWEEN GROUPS (SSB & dfB). • WHAT’S LEFT IS SSINT & dfINT, THE SUM OF SQUARES AND DEGREES OF FREEDOM FOR THE INTERACTION.

  24. Look at that another way: The whole is equal to the sum of its parts. • SSB and dfB are the between group sum of squares and degrees of freedom. Each is composed of three parts, SSIV1, SSIV2, SSINT and dfIV1, dfIV2, and dfINT. • So if we subtract the SS and df for factors 1 & 2 from SSB and dfB, what is left is the sum of squares and df for the interaction. • SSINT = SSB-(SSIV1+ SSIV2)=SSB- SSIV1- SSIV2 • dfINT = dfB-(dfIV1+ dfIV2)=dfB- dfIV1- dfIV2

  25. Designs of 2 factor studies

  26. Analysis of Variance • Each possible combination of IV1 and IV2 creates an experimental group. Participants are randomly assigned to each of the treatment groups. • Each experimental group is treated differently from all other groups in terms of one or both factors. • For example, if there are 2 levels of the first variable (Factor 1or IV1) and 2 of the second (IV2), we will need to create 4 groups (2x2). If IV1 has 2 levels and IV2 has 3 levels, we need to create 6 groups (2x3). If IV1 has 3 levels and IV2 has 3 levels, we need 9 groups. Etc.

  27. Some nomenclature • Two factor designs are identified by simply stating the number of levels of each variable. So a 2x4 design (called “a 2 by 4 design”) has 2 levels of IV1 and 4 levels of IV2. A 3x2 design has 3 levels of IV1 and 2 levels of IV2. And so on. • Which factor is called IV1 and which is called IV2 is arbitrary (and up to the experimenter).

  28. Example of 2 x 3 design To make it more concrete, assume we are testing new treatments for Generalized Anxiety Disorder. In a two factor design we examine the effects of cognitive behavior therapy vs. a social support group among GAD patients who receive Ativan, Zoloft or Placebo. Thus, IV1 has 2 levels (CBT/Social Support) while IV2 has 3 levels (Ativan/Zoloft/Placebo) So, we would form 2 x 3 = 6 groups to do this experiment. Half the patients would get CBT, the other half get social support. A third of the CBT patients and one-third of the Social Support patients also get Ativan. Another third of those who receive CBT and one-third of those who get social support also receive Zoloft. The final third in each psychotherapy condition get a pill placebo.

  29. A 2 x 3 design yields 6 groups. Let’s say you have 24 participants. Four are randomly assigned to each group. • Here are the six treatment groups: • CBT + Ativan • CBT + Zoloft • CBT + Placebo • Social support + Ativan • Social support + Zoloft • Social support + Placebo

  30. Example: Experiment Outline • Population: Outpatients with Generalized Anxiety Disorder • Subjects: 24 participants divided equally among 6 treatment groups. • Independent Variables: • Factor 1: Psychotherapy: CBT or Social Support (SoSp) • Factor 2: Medication: Ativan, Zoloft, or Placebo • Groups: 1=CBT + Ativan; 2=CBT + Zoloft; 3=CBT + Placebo; 4=SoSp + Ativan; 5=SoSp + Zoloft; 6=SoSp + Placebo. • Dependent variable: Anxiety remaining after treatment. Lower scores equal less anxiety and a better outcome.

  31. Type of drug Ativan Zoloft Placebo CBT Type of therapy SoSp A 3X2 STUDY

  32. Effects • We are interested in the main effects of type of psychotherapy and type of drug. Do participants get better with CBT and not with SoSp or the reverse? Do people get better when they get a mild tranquilizer (Ativan) an SSRI (Zoloft) or Placebo. • We are also interested in assessing how combining different levels of both factors affect the response in ways beyond those that can be predicted by considering the effects of each IV separately. So, we are interested in the interaction of the independent variables.

  33. Compare each score to the mean for its group. Drug Given Ativan Zoloft Placebo CBT Type of therapy SoSp MSW

  34. 6 6 6 6 4 4 4 4 14 14 14 14 -3 -2 2 3 3 -1 0 -2 1 0 2 -3 9 4 4 9 9 1 0 4 1 0 4 9 8 8 8 8 10 10 10 10 18 18 18 18 -4 -1 -1 -2 -4 -1 2 3 -4 0 1 3 16 1 1 4 16 1 4 9 16 0 1 9 Mean Squares Within Groups 1.1 1.2 1.3 1.4 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 3 4 8 9 7 3 4 2 15 14 16 11 4.1 4.2 4.3 4.4 5.1 5.2 5.3 5.4 6.1 6.2 6.3 6.4 12 7 7 6 6 9 12 13 14 18 19 21

  35. MSW • SSw = 132.00 • dfW = 18 • MSW = 132/18 = 7.33

  36. Then we compute a sum of squares and df between groups • This is the same as in Chapter 9 • The difference is that we are going to subdivide SSB and dfB into component parts. • Thus, we don’t use SSB and dfB in our Anova summary table, rather we use them in an intermediate calculation.

  37. Compare each group mean to the overall mean. Type of Drug Ativan Zoloft Placebo CBT Type of Therapy SoSp Sum of Squares Between Groups (SSB)

  38. Type of Drug Ativan Zoloft Placebo Cognitive BT Diet Type Social Support Means for GAD study 6 4 14 8.00 8 10 18 12.00 7.00 7.00 16.00

  39. -4 -4 -4 -4 -6 -6 -6 -6 4 4 4 4 16 16 16 16 36 36 36 36 16 16 16 16 -2 -2 -2 -2 0 0 0 0 8 8 8 8 4 4 4 4 0 0 0 0 64 64 64 64 Sum of Squares Between Groups (M=10.00, SSB=544 , dfB=5) 6 6 6 6 4 4 4 4 14 14 14 14 10 10 10 10 10 10 10 10 10 10 10 10 8 8 8 8 10 10 10 10 18 18 18 18 10 10 10 10 10 10 10 10 10 10 10 10

  40. Next, we answer the questions about each factor having an overall effect. • To get proper between groups mean squares we have to divide the sums of squares and df between groups into components for factor 1, factor 2, and the interaction. • Let’s just look at factor 1. Our question about factor 1 was “Do people undergoing different therapies have differential responses to any task?” • We can group participants into all those who were treated with CBT and those treated with Social Support.

  41. Forming Groups that Differ only on Factor 1 • Pretend that the experiment was a simple, single factor experiment in which the only difference among the groups was the first factor (that is, the type of therapy given each group). Create groups reflecting only differences on Factor 1. • So, when computing the main effect of Factor 1 (type of psychotherapy), ignore Factor 2 (type of drug). Divide participants into two groups depending solely on whether they were given CBT or Social Support. Then, find the mean of each of the two combined groups (CBT and Social Support).

  42. Computing SS for Factor 1 Next, find the deviation of the mean of the CBT participants from the overall mean. Then sum and square those differences. Then, find the deviation of the mean of the Social Support participants from the overall mean. Then sum and square those differences. The total of the summed and squared deviations the mean of each of the combined groups from M, the overall mean, is the sum of squares for Factor 1. (SSIV1).

  43. dfIV1 and MSIV1 • Compute a mean square that takes only differences on Factor 1 into account by dividing SSIV1 by dfIV1. • For example, in this experiment, type of therapy was either CBT or Social Support. The two ways participants are treated are called the two “levels” of Factor 1. • dfIV1= L1 – 1 = 2-1 = 1 [where L1 equals the number of levels (or different variations) of the first factor (IV1)].

  44. Compare each score’s therapy mean to the overall mean. Type of drug Ativan Zoloft Placebo CBT Therapy Type SoSp SSIV1: Main Effectof Therapy Type

  45. Type of New Drug Ativan Zoloft Placebo Cognitive BT Diet Type Social Support Means for GAD study 6 4 14 8.00 8 10 18 12.00 7.00 7.00 16.00

  46. CBT/SoSp Means, M=10.00Means: CBT=8.00; SoSp=12.00 CBT 1.1 1.2 1.3 1.4 2.1 2.2 2.3 2.4 3.1 3.2 3.3 3.4 group 8 8 8 8 8 8 8 8 8 8 8 8 SoSp 4.1 4.2 4.3 4.4 5.1 5.2 5.3 5.4 6.1 6.2 6.3 6.4 group 12 12 12 12 12 12 12 12 12 12 12 12

  47. SSIV1=96.00; dfIV1=2-1=1;MSIV1=96.00 8 8 8 8 8 8 8 8 8 8 8 8 10 10 10 10 10 10 10 10 10 10 10 10 -2 -2 -2 -2 -2 -2 -2 -2 -2 -2 -2 -2 4 4 4 4 4 4 4 4 4 4 4 4 12 12 12 12 12 12 12 12 12 12 12 12 10 10 10 10 10 10 10 10 10 10 10 10 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4 4 4

  48. Factor 2 Then pretend that the experiment was a single experiment with only the second factor. Proceed as you just did for Factor 1 and obtain SSIV2 and MSIV2 where dfIV2=LIV2 – 1=3-1=2.

  49. Compare each score’s Drug Type mean to the overall mean. Drug Type Ativan Zoloft Placebo CBT Therapy Type SoSp SSIV1: Main Effectof Drug Type

  50. Type of New Drug Ativan Zoloft Placebo Cognitive BT Diet Type Social Support Means for GAD study 6 4 14 8.00 8 10 18 12.00 7.00 7.00 16.00

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