Page 1 of 6 Exam1 PS 306, Spring 2008 1. (From G&W) In order to study cardiovascular responses to embarrassment, Harris (2001) had people sing the Star Spangled Banner in front of a video camera while she recorded their heart rate and blood pressure. She found that blood pressure increases steadily for two minutes before gradually returning to normal. What about the heart rate data? Below is a partially completed source table for these heart-rate data. Complete the table and analyze/interpret the results as completely as possible. Is the pattern for heart rate similar to that for blood pressure? [15 pts] Tests of Within-Subjects Effects Measure:MEASURE_1 Source Type III Sum of Squares df Mean Square F Sig. Partial Eta Squared Noncent. Parameter Observed Powera time Sphericity Assumed 1112.667 2 556.333 333.800 .000 .968 667.600 1.000 Error(time) Sphericity Assumed 36.667 22 1.667 H0: µBaseline = µ1Min = µ2Min H1: Not H0 Decision: Reject H0, p < .001 Post Hoc Test: € HSD = 3.551.66712= 1.32 Base 1Min 2Min Base ------ 1Min 12.3 ------ 2Min 1.2 11.2 ------ There is a significant effect of time of assessing heart rate, F(2,22) = 333.8, MSE = 1.667, p < .001, η2 = .968. Post hoc tests using Tukey’s HSD indicate that heart rate is significantly higher at one minute (M = 89.25) compared to Baseline (M = 76.917) or two minutes (M = 78.083). Thus, the data aren’t similar to the blood pressure data, because the hear rate data have almost returned to baseline. Even though this is a repeated measures design, no counterbalancing is possible. Briefly explain why not. Because you’re interested in the effects of time, you cannot counterbalance that variable.Page 2 of 6 2. Mook argues that external validity is not always the purpose behind psychological research. For each of the studies below, indicate why the study is not externally valid, then why it’s not a concern, given the intentions of the researcher(s). [10 pts] Answer each question using the Mook article. Study Why not externally valid Why lack of EV is not a concern Argyle (glasses and intelligence) Harlow (infant monkeys and drive reduction theory) Hecht (dark adaptation) Brown & Hanlon (parental role in grammar acquisition) 3. In the first lab, we collected a number of different academic measures from members of both sections of PS 306. Below are the results from a correlation analysis of two different SAT scores (Math and Verbal/Critical Reading). First of all, tell me what you could conclude from these results. Then, given an SAT-V score of 600, what SAT-M score would you predict using the regression equation? Given the observed correlation, if a person studied and raised her or his SAT-V score, would you expect that person’s SAT-M score to increase as well? What would you propose as the most likely source of the observed relationship? [10 pts] There is a significant positive linear relationship between SAT-M and SAT-V, r(43) = .438, p = .003. The effect size (coefficient of determination) is r2 = .192. SAT-M = (.474)(600) + 324.3 = 608.7 No, raising SAT-V would not necessarily have an impact on SAT-M. The relationship is not a causal one. A third variable, such as intelligence (IQ) or educational experience may well lead to the scores on both SAT-M and SAT-V.Page 3 of 6 4. (From G&W) Intelligence is offered as one possible explanation for why some birds migrate and others maintain year-round residency in a single location. That is, small bird brains (relative to body size) don’t have enough computational power to allow the bird to find food during the winter, so they must migrate to warmer climates where more food is available (Sol, Lefebvre, & Rodriguez-Tejeiro, 2005). On the other hand, large bird brains (relative to body size) produce sufficient computational power that their owners are more creative and can find food even when the weather turns harsh. Below is a partially completed source table consistent with the actual research results. The numbers represent the relative brain size for the individual birds in each sample. Complete the table and analyze/interpret the results as completely as possible. [15 pts] ANOVA Relative Brain Size Sum of Squares df Mean Square F Sig. Between Groups 750.000 2 375.000 14.583 .000 Within Groups 1080.000 42 25.714 Total 1830.000 44 H0: µNonMigrating = µShort Mig = µLong Mig H1: Not H0 Decision: Reject H0, because p < .001 Post Hoc Test: € HSD = 3.4425.7115= 4.50 NonMig Short Mig Long Mig NonMig ----- Short Mig 5 ----- Long Mig 10 5 ----- There is a significant effect of type of bird on brain size, F(2,42) = 14.583, MSE = 25.714, p < .001. {Note, however, that the birds are not randomly assigned to condition (type of bird is a non-manipulated characteristic of the subjects), so one cannot make a causal claim.} Post hoc tests using Tukey’s HSD indicate that the brains of non-migrating birds are significantly higher (M = 15) than those of birds with short (M = 10) or long migration patterns (M = 5). The short-distance migration birds have brains that are larger than those that migrate for long distances.Page 4 of 6 5. (From G&W) There is some evidence to suggest that high school students justify cheating in class on the basis of the teacher’s skills or stated concern about cheating (Murdock, Miller, & Kohlhardt, 2004). Thus, students appear to rationalize their illicit behavior on perceptions of how their teachers view cheating. Poor teachers are thought not to know or care whether or not students cheat, so cheating in their classes is viewed as acceptable. Good teachers, on the other hand, do care and are alert to cheating, so students tend not to cheat in their classes. Below is a partially completed source table and summary statistics that are consistent with the findings of Murdock et al. The scores represent judgments of the acceptability of cheating for students in each sample. Complete the source table below and interpret the data as completely as you can. What is your best estimate of the population variance (σ2)? [10 pts] ANOVA Acceptability of Cheating Sum of Squares df Mean Square F Sig. Between Groups 156.025 1 156.025 132.491 .000 Within Groups 44.750 38 1.178 Total 200.775 39 In order to complete this problem, you need to compute MSWithin from the supplied standard deviations. First, you need to square the standard deviations to turn them into variances (.726 and 1.629), then