Exam 2, PS 306, Spring 20021. We’d briefly discussed the Dutton & Aron bridge study in class. In that study, elevated physiologicalarousal of the male participants was presumed to be the root cause of a greater attraction to the youngwoman waiting at the end of the high rope bridge as compared to the perceived attraction from the non-elevated physiological arousal that followed crossing a stable concrete bridge. But maybe the perceivedattraction had to do with elements other than physiological arousal. Dutton & Aron also used only maleparticipants. As an extension of their study, suppose that male and female participants are given one ofthree levels of epinephrine [None (saline), Small, Large] to produce varying levels of physiological arousal.After 5 minutes (to allow the drug to have its effect), participants were shown a picture of the face a personwho had previously been judged by other people to be moderately attractive (4 on a 7-point scale ofattractiveness). For the male participants, the picture was of a female face. For the female participants, thepicture was of a male face. Each participant rated the attractiveness of the target face on a 7-point scale.Complete the analysis of this 2x3 independent groups design and interpret the results as completely as youcan. [10 pts]1 6.021 6.021 5.282 .0266 5.282 .6082 19.625 9.812 8.608 .0007 17.217 .9672 3.792 1.896 1.663 .2018 3.326 .31942 47.875 1.140DF Sum of Squares Mean Square F-Value P-Value Lambda PowerGenderEpinephrineGender * EpinephrineResidualANOVA Table for Attraction Score8 4.875 1.126 .3988 4.000 .926 .3278 4.500 1.195 .4238 6.250 .707 .2508 4.000 1.309 .4638 5.250 1.035 .366Count Mean Std. Dev. Std. Err.Female, LargeFemale, NoneFemale, SmallMale, LargeMale, NoneMale, SmallMeans Table for Attraction ScoreEffect: Gender * EpinephrineBecause the interaction is not significant, I would focus my attention on the twomain effects, which are significant. Gender requires no additional analysis, beyondthe computation of the means for Males (5.2) and Females (4.5). Thus, I couldconclude that males rated the pictures as significantly more attractive than didfemales. For Epinephrine, I would need to compute Tukey’s HSD = .92 (q = 3.44).Thus, I would conclude that Large doses of epinephrine lead to higher ratings (M =5.56) than Small doses (M = 4.88) or None (M = 4.00), neither of which differed.2. Power…it’s all about power! [15 pts]First of all, clearly define power.Power is the probability of correctly rejecting H0.Jacob Cohen and others have advocated working to achieve a level of power of roughly.80. What does such a level of power say about the level of Type II Error that these folksare willing to tolerate? What does that say about people’s tolerance for Type II Errorscompared to Type I Errors?Because power and Type II errors are complimentary, power of .80 means that theprobability of a Type II error would be .20. That then means that people are willingto tolerate 4x’s as much Type II as Type I error.Finally, tell me very explicitly how you would design a study to maximize its power (i.e.,talk about the aspects of the design on which you would focus to achieve the greatestpower).Examples would help, but the basic notion is to increase n (or have the largest npossible), increase treatment effect (or have the largest treatment effect you couldimagine as reasonable), and decrease the error (or design the study to have the leastindividual differences and random variability that you could achieve). Details abouthow you would accomplish those goals would be useful.3. Repeated measures designs are more powerful than independent groups designs. Theyare also more efficient. First, determine the number of participants needed to achieve aminimum of 25 scores per cell in the following 3x5 designs. Then use that information toillustrate (briefly) the efficiency of a repeated measures design. [15 pts]B1B2B3B4B5A1A2A3Completely BetweenHow many total participants would you need? 375How many total pieces of data would occur in the entire experiment? 375Completely WithinHow many total participants would you need? 30 (incomplete counterbalancing)How many total pieces of data would occur in the entire experiment? 450Mixed (A independent groups and B repeated measures)How many total participants would you need? 90 (incomplete), 360 (complete)How many total pieces of data would occur in the entire experiment? 450 OR 1800Mixed (A repeated measures and B independent groups)How many total participants would you need? 150 (complete)How many total pieces of data would occur in the entire experiment? 450What does all the above information say about efficiency?Note that for a completely between design, you would need 375 people and only endup with 375 pieces of data. However, if you could use a completely within design,you would need only 30 people and they would generate 450 pieces of data for you.The two mixed designs fall in between these two extremes.4. Dr. Reeder was interested in the extent to which people believe reports that they readin various media. To that end, he constructs three stories: Plausible, Barely Believable,and Implausible. For example, the Plausible story would describe the details of an actualevent (such as an armed robbery). The Barely Believable story would describe a situationthat might be true, but seems unlikely (such as the number of women with whom WiltChamberlain claimed to have had sexual relations). Finally, the Implausible story woulddescribe a situation that seems extraordinarily unlikely (such as the sighting of a thinElvis Presley). Each of the three stories was set up (on proper paper, with proper font,etc.) so that they appeared to have come from one of four sources: The New York Times,The National Enquirer, Newsweek, and People Magazine. Each participant read one ofthe 12 possible stories and rated its credibility on a scale from 1 (Absolutely convincedthat the story was not true) to 7 (Absolutely convinced that the story was true). Thus, thisis a 3x4 independent groups design. Complete the analysis below and tell Dr. Reederwhat he should conclude from the analyses of these data. [20 pts]3 55.333 18.444 25.441 <.0001 76.322 1.0002 130.133 65.067 89.747 <.0001 179.494 1.0006 11.467 1.911 2.636 .0273 15.816 .81248 34.800 .725DF Sum of Squares Mean Square F-Value P-Value Lambda PowerSourceStorySource * StoryResidualANOVA Table for Credibility5 2.200 1.095 .4905 1.400 .548 .2455 5.000 .707 .3165