Exam 2 - 1 of 5 Exam 2 PS 217, Spring 2010 1. In PS 306 this semester, we conducted a lab in which subjects served as mock eyewitnesses. Even though they hadn’t actually observed a crime, they could read descriptions from eyewitnesses (see below) and then rate the similarity of each of the six pictures in a photo-array (see below) to that description. Think about it. If the police put together an unbiased photo-array, what should happen? Right! People should rate all the faces as equally similar to the eyewitness description. In other words, if the photo-array was fair, an analysis of the data would retain H0: µFace1 = µFace2 = µFace3 = µFace4 = µFace5 = µFace6. If the similarity ratings (made on a 7-pt scale, from 1 = bad match to 7 = great match) differ for the faces, it would indicate that the photo-array is biased. Complete the analysis below and interpret the results as completely as you can. (A1F1 means Array 1 Face 1, etc.) [N.B. The photos were presented simultaneously, so there was no counterbalancing.] [15 pts] African-American male in his early 20’s with dark hair, an oval face and broad forehead. Small, dark eyes and thin eyebrows. A wide nose, thick lips and small, protruding ears. Source Type III Sum of Squares df Mean Square F Sig. Partial Eta Squared Observed Powera face Sphericity Assumed 579.1 5 115.82 71.9 .000 .450 1.000 Error(face) Sphericity Assumed 707.9 440 1.61 Note that although this is a repeated measures design, there is no counterbalancing, because all six faces are presented simultaneously. Were they presented sequentially, then you would need to counterbalance. H0: µFace1 = µFace2 = µFace3 = µFace4 = µFace5 = µFace6 H1: Not H0 Decision: Reject H0, p < .05 Post Hoc Test: € HSD = 4.071.6189= .55 Face1 Face2 Face3 Face4 Face5 Face6 Face1 --- Face2 1.13 --- Face3 .46 .66 --- Face4 1.08 2.20 1.54 --- Face5 2.32 1.19 1.85 3.39 --- Face6 .22 .91 .24 1.29 2.1 --- Thus Face5 is judged to be a significantly better match (M = 5.787) than any of the other faces. Face2 is judged to be a significantly better match (M = 4.596) than any of the remaining faces. Face3 (M = 3.933), Face6 (M = 3.685), and Face 1 (M = 3.472) did not differ from one another, but all three were judged to be better matches than Face4 (M = 2.393).Exam 2 - 2 of 5 2. Some questions primarily related to repeated measures designs: a. In a single-factor repeated measures design with 7 levels of the factor (A), how many participants would you need to conduct the study if you wanted a minimum of 15 scores per condition (cell)? [2 pts] You’d use incomplete counterbalancing, so with 7 levels of the factor (conditions), you’d need 14 orders (2x7). That would only give you 14 scores per condition, so you’d need to use each order twice…meaning that you’d need 28 participants. b. Suppose that you conduct the study (as in a above, with k = 7 and n = ?). Complete the resulting source table below. [5 pts] Source SS df MS F Between (A) 120 6 20 10 Within 567 189 Subject 243 27 Error 324 162 2 Total 687 195 c. Use the above source table to illustrate why it is that the repeated measures analysis will typically be more powerful than the independent groups analysis. That is, what component of the source table is largely responsible for making the repeated measures analysis more powerful? (Show why.) [2 pts] You will typically get a smaller error term with a repeated measures design compared to an independent groups design (even with the impact of counterbalancing). That will typically occur because you remove the SSSubject, which will be relatively large. When that’s not the case, then your F may actually decrease relative to an independent groups ANOVA. For instance, in this case, if SSSubject = 81, then MSError = 3 and F = 6.67. Obviously, were SSSubject to further decrease, your F would get smaller still. d. Why is it that you typically cannot conduct a repeated measures study for a characteristic of a person (e.g., gender, IQ)? [1 pt] Because, for example, a person cannot be both low IQ and high IQ. e. Under what other circumstances would you not be able to conduct an experiment as a repeated measures design? [1 pt] When the impact of your treatment is permanent, you cannot use a repeated measures design. For example, if you’re cutting into two different areas of the brain, you can’t rewire the severed areas from the first cut before doing the second cut. f. In an independent groups ANOVA, what is the best estimate of population variance (σ2)? [2 pts] MSWithin g. Why is it that MSError in a repeated measures ANOVA is not a good estimate of population variance? [2 pts] Because population variance is due to both individual differences and random variability, but MSError in a repeated measures ANOVA reflects only random variability. 3. Although psychologists do not completely understand the phenomenon of dreaming, it does appear that people need to dream. One experiment demonstrating this fact shows that people who are deprived of dreaming one night will tend to have extra dreams the following night, as if they were trying to make up for the lost dreams. In a typical version of this experiment, the psychologist first records the number of dreams (by monitoring rapid eye movements [REM]) during a normal night's sleep. The next night, each subject is prevented from dreaming by being awakened as soon as she or he begins a dream. During the third night, the psychologist once again records the number of dreams. Hypothetical data from this experiment are as follows:Exam 2 - 3 of 5 First Night Night After Deprivation P S1 4 7 11 S2 5 5 10 S3 4 8 12 S4 6 7 13 S5 4 10 14 S6 5 7 12 S7 4 7 11 S8 4 6 10 Sum 36 57 93 Sum of squared scores (i.e., ΣX2) 166 421 587 SS 4 14.875 18.875 M 4.5 7.125 Interpret these data as completely as you can. [15 pts] First, note that there is no counterbalancing…even though with two conditions, the researcher could have done so using 8 participants. However, if you read the description of the study, all participants get no disruption on the first night and then disruption on the second night. That’s a big no-no…no counterbalancing. H0: µFirst = µDeprived H1: Not H0 Source SS df MS F Between 27.57 1 27.57 16.18 Within 18.875 14 Subject 6.94 7 Error 11.93 7 1.70 Total 46.44 15 With FCritical(1,7) = 5.59, you would reject H0 (FObtained ≥ FCritical). Thus, the number of