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Factorial Anova – Lecture 2. REVIEW. 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.)

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## Factorial Anova – Lecture 2

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**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.**Analysis of Variance**• In our statistical analysis, we will determine whether differences among the treatment groups’ means are consistent with the null hypothesis that they differ solely because of random sampling fluctuation. • We will use the F tests to answer our question. Thus, we will compare estimates of sigma2.**Three F tests, instead of one.**• In Chapter 9, we had only two estimates of sigma2 and a single F test. In the two way factorial analysis of variance, you make four estimates of sigma2 and do three F tests. Three of the four estimates will be based on the variation of group means around the overall mean. Each of these three estimates will serve as the numerator in one of three F tests. • The other estimate is MSW, derived from the variation of scores around their own group means. It will serve as the denominator in all three F tests.**As you know…**• Sigma2 is estimated either by comparing a score to a mean (the within group estimate) or by comparing a mean to another mean. This is done by • Calculating a deviation or • Squaring the deviations. • Summing the deviations. • Dividing by degrees of freedom**H0 vs. H1**• Again, the null will predict that a ratio of two estimates of sigma2 will be about 1.00. • Again, the experimental hypothesis says the ratio will be greater than one. • But we can’t just compare MSB to MSw as we did in Chapter 9. We have a problem that needs to be solved before any estimates of sigma2 can be compared.**The Problem**• Unlike the one-way ANOVA of Chapter 9, we now have two variables that may push the means of the experimental groups apart. • Moreover, combining the two variables may have effects beyond those that would occur were each variable presented alone. We call such effects the interaction of the two variables. • Such effects can be multiplicative as opposed to additive. • Example: Moderate levels of drinking can make you high. Barbiturates can make you sleep. Combining them can make you dead. The effect (on breathing in this case) is multiplicative.**The difference between analyzing single and multifactor**designs – a review • If we simply did the same thing as in Ch. 9, our between group term could be effected by Factor 1, Factor 2, and/or their interaction as well as by sampling fluctuation. • An F test requires two estimates of sigma2, one that indexes only sampling fluctuation, the other indexes sampling fluctuation plus one (and only one) other variable or interaction of variables. • So we are going to have to do something different to get appropriate between groups terms.**We create mean squares for Factors 1 & 2 by combining the**original experimental groups into groups that differ on only one factor. • To obtain proper between groups mean squares for each independent variable, we must divide the sums of squares and df between groups (SSB and dfB) into three components. • We begin the analysis by combining the experimental groups to calculate sums of squares and df for the main effects of Factors 1 and 2 (SSIV1, SSIV2, dfIV1 and dfIV2). • We obtain the sum of squares and df for the interaction by subtraction.**Two way Anova for independent groups**• Remember, the denominator of the F ratio will be the mean square within groups (MSW) • where MSW =SSW/ n-k. (AGAIN!) • In the factorial analysis of variance, the problem is obtaining proper mean squares for the numerator.**Embarrassment and task difficulty**• Remember the experiment in your book. • Participants were either not embarrassed or exposed to mildly or highly embarrassing conditions while doing a hard or easy task. • Dependent variable was liking for the task. • Did embarrassment or task difficulty have main effects on liking for the task? • Did the two variables interact so that which task was preferred depended on how embarrassed the person was?**Computing SS for Factor 1**• Start the analysis by 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 degree to which a group is embarrassed). Create groups reflecting only differences on Factor 1. • So, when computing the main effect of Factor 1 (level of embarrassment), ignore Factor 2 (whether the task was hard or easy). Divide participants into three groups depending solely on whether they not embarrassed, mildly embarrassed, or severely embarrassed.**Computing SS for Factor 1**• Next, find the deviation of the mean of the severely, mildly, and not embarrassed participants from the overall mean. • Then square and sum those differences. • Finally, total the summed squared deviations of each person’s group’s score from the overall mean in the severely, mildly, and not embarrassed conditions. That total is the sum of squares for Factor 1. (SSIV1).**dfIV1 and MSIV1**• Compute a mean square that takes only differences on Factor 1 into account by dividing SSIV1 by dfIV1. • dfIV1= LIV1 – 1 where LIV1 equals the number of levels of the first factor (IV1). • For example, in this experiment, embarrassment was either absent, mild or high. These three ways participants are treated are called the three “levels” of Factor 1. So, dfIV1= LIV1 – 1 =3-1 = 2.**Then**• Do the same for factor 2 • Then subtract the sums of squares and df for factors 1 & 2 from the sum of squares and degrees of freedom between group to obtain SSINT & dfINT. • Once you have all the sums of squares and degrees of freedom, compute ANOVA and determine significance with the F table. • Finally, plot the means and carefully interpret the results.**Alternative Problem Formats**• Once you get down the basics of the analysis of variance for factorial designs, problems can be presented in a variety of formats. • These alternative problem formats require less computation and more understanding of what you are doing when you compute a factorial Anova than does a format in which you are given all the scores.**Format 1: Another drug study:group means and SSW given**A drug company compared three new drugs (SansCho, MuckFree, and NoBlock) for cholesterol reduction. Patients also ate either the “HeartHealthy” or “EnoughFat” diet.. There were 3 participants in each of the six conditions. The patients who ate the “HeartHealthy”diet lost an average of 12 points of cholesterol if they also took SansCho, an average of 15 points if they also took MuckFree,and an average of 18 points in the HeartHeathy diet/NoBlock condition. The patients who ate the “EnoughFat”diet lost an average of 15 points of cholesterol if they also took SansCho, an average of 21 points if they also took MuckFree,and an average of 26 points in the EnoughFat/NoBlock condition. SSW=86.00. Compute the Anova.**Start by determining df from the design and setting up the**ANOVA Table. Then see what you still need to compute.Let’s call type of diet Factor 1.Type of drug received is Factor 2.**Based on the DESIGN, what is total n? Second, what is k.**Then, what are dfB, dfW, dfIV1, dfIV2 and dfINT? • There are 2x3=6 groups. • Three participants/group. • n=18. • There are 6 groups. k = 6. • dfW=n-k=18-6=12 • dfB= k – 1 = 5 • dfIV1 =LIV1-1 = 1, dfIV2=LIV2-1=2 and dfINT=5-(1+2) =2**You already know SSW**I told you it was 86. Let’s fill in the Anova summary table on the basis of what we know from the design and what was given.**ANOVA summary table**SS df MS F p Type of Diet ? 1 Type of Drug ? 2 Interaction ? 2 Error 86.00 12 7.17**What we need**• At this point we need the sums of squares for factors 1 & 2 and for the interaction. • To find the sum of squares for the interaction, we need the sum of squares between groups. Then we can subtract the sums of squares for factors 1 & 2 and find the sum of squares for the interaction. • Best thing we can do next is set up a table of means.**Type of New Drug**SansCho MuckFree NoBlock HeartHealthy Diet Type EnoughFat Means for Cholesterol Lowering 12 15 18 15 16 21 26 21 14 18 22**Then lets compute SSB, SSIV1, SSIV2, and SSINT using the**table of means**Compare each group mean**to the overall mean. Type of Drug SansCho MuckFree NoBlock HH Task Difficulty EF Sum of Squares Between Groups (SSB)**-6**-6 -6 --3 --3 --3 0 0 0 36 36 36 9 9 9 0 0 0 -2 -2 -2 3 3 3 8 8 8 4 4 4 9 9 9 64 64 64 SSB=366.00 Sum of Squares Between Groups (SSB) 12 12 12 15 15 15 18 18 18 18 18 18 18 18 18 18 18 18 16 16 16 21 21 21 26 26 26 18 18 18 18 18 18 18 18 18**Compare each participant’s diet**group mean to the overall mean. Type of Drug Sans Cho MuckFree NoBlock HH Diet Type EF SSIV1: Main Effectof Diet Type**Type of New Drug**SansCho MuckFree NoBlock HeartHealthy Diet Type EnoughFat Means for Cholesterol Lowering 15 12 15 18 21 16 21 26 14 18 22**In this kind of problem, it is often a good idea to make a**table showing the number of participants in each group and of each combined group. • The most frequent mistake people make in solving these problems is having the wrong number of people in each of the combined groups, groups that differ on only one factor. • So, this may help. Use it if you like it, but forget it if it makes things more confusing.**Type of New Drug**SansCho MuckFree NoBlock HeartHealthy Diet Type EnoughFat n for each group in the cholesterol lowering study 9 3 3 3 9 3 3 3 6 6 6**Notice that the different factors have different n per group**• Each of the combined groups on factor 1, type of diet has n=9 • Each of the combined groups on factor 2, brand of drug, has n=6**Type of New Drug**SansCho MuckFree NoBlock HeartHealthy Diet Type EnoughFat n for each group in the cholesterol lowering study 9 3 3 3 9 3 3 3 6 6 6**-3**-3 -3 --3 --3 --3 -3 -3 -3 9 9 9 9 9 9 9 9 9 3 3 3 3 3 3 3 3 3 9 9 9 9 9 9 9 9 9 SSIV1=162.00 Sum of Squares Factor 1 (SSIV1) 15 15 15 15 15 15 15 15 15 18 18 18 18 18 18 18 18 18 21 21 21 21 21 21 21 21 21 18 18 18 18 18 18 18 18 18**ANOVA summary table**SS df MS F p Type of Diet 162.00 1 162.00 22.60 .01 Brand of Drug ? 2 Interaction ? 2 Error 86.00 12 7.17**Compare each participant’s**“Type of drug” group mean to the overall mean. Type of drug SansCho MuckFree NoBlock Heart Hlthy Diet Type Enough Fat SSIV2: Main Effectof Type of Drug**Type of New Drug**SansCho MuckFree NoBlock HeartHealthy Diet Type EnoughFat Means for Cholesterol Lowering 12 15 18 15 16 21 26 21 14 18 22**-4**-4 -4 0 0 0 4 4 4 16 16 16 0 0 0 16 16 16 -4 -4 -4 0 0 0 4 4 4 16 16 16 0 0 0 16 16 16 SSF1=192.00 Sum of Squares Factor 2 (SSIV2) 14 14 14 18 18 18 22 22 22 18 18 18 18 18 18 18 18 18 14 14 14 18 18 18 22 22 22 18 18 18 18 18 18 18 18 18**ANOVA summary table**SS df MS F p Type of Diet 162.00 1 162.00 22.60 .01 Brand of Drug 192.00 2 96.00 13.39 .01 Interaction ? 2 Error 86.00 12 7.17**Means Squares - Interaction**REARRANGE**ANOVA summary table**SS df MS F p Type of Diet 162.00 1 162.00 22.60 .01 Type of Drug 192.00 2 96.00 13.39 .01 Interaction 12.00 2 6.00 0.84 n.s. Error 86.00 12 7.17**Interpret the results.**• There was a significant main effect for type of diet [F(1,12) = 22.60, p<.01]. People who ate the EnoughFat diet lost an average of 6 points of cholesterol more than did the participants who ate the HeartHealthy diet. • There was a significant main effect for type of medication [F(2,12) = 12.70]. People who took NoBlock lost an average of 8 points more than did the participants who took SansCho, with those who took MuckFree scoring in the middle.**More interpretation**• The interaction was not significant F(2,12) = 0.84, n.s. Thus, the EnoughFat diet and NoBlock seem the best choices, resulting in the most cholesterol lowering. The effects of the drug and diet interventions seem to be additive.**Another Format**• A researcher compared responses to two doses of a new drug, ThinkRight, for schizophrenia, plus either standard hospital treatment or a behavioral innovation treatment. There were 6 participants in each group. She measures responses on a scale of normal cognition. Higher scores equal more normal functioning. The sum of squares within group is 100.00. The sum of squares between groups was 120.00. The sum of squares for Factor 1 (Drug Dose) is 60.00. The sum of squares for the interaction is 40.00. Compute the Anova.**Based on the DESIGN, what is total n? Second, what is k.**Then, what are dfB, dfW, dfIV1, dfIV2 and dfINT? • There are 2x2=4 groups. k=4 • Six participants/group. • n=24. • dfW=n-k=24-4=20 • dfB= k – 1 = 3 • dfIV1 =LIV1-1 = 1, dfIV2=LIV2-1=1 • and dfINT=3-(1+1)=1**We are given several sums of squares.**• We are given SSW (SSW=100.00) • We are given SSB (SSB=120.00) • We are given SSIV1 (SSIV1=60.00) • We are given SSINT(SSINT=40.00)**ANOVA summary table**SS df MS F p Type of Diet 60.00 1 60.00 12.00 .01 Dose of Drug ? 1 ? ? ? Interaction 40.00 1 40.00 8.00 .05 Error 100.00 20 5.00**The whole equals the sum of its parts**• SSB=SSIV1 + SSIV2 + SSINT • 120.00=60.00 +SSIV2 + 40.00 • SSIV2 = 20.00

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