Handout from Psych 5741/5751: Data Analysis - Two-Way ANOVA and Regression Analysis - Prof, Exams of Statistics

Download Handout from Psych 5741/5751: Data Analysis - Two-Way ANOVA and Regression Analysis - Prof and more Exams Statistics in PDF only on Docsity! Handout from Psych 5741/5751 Univ of Colorado used with Judd, C.M., & McClelland, G.H. (1989). Data Analysis: A Model Comparison Approach. HBJ. Spring Final 1996 Answers — 1 — May 13, 1996 Final Exam: Answers Spring 1996 Note: The textual and graphical answers are more detailed and complete than was expected on the exam. Question 1 A. SAS Code: data memory; infile 'memory.dat'; input mood$ word$ recall; emotvsno = (1/3)*(mood='sad') - (2/3)*(mood='neutral') + (1/3)*(mood='pleasant'); plsvssad = (-1/2)*(mood='sad') + 0*(mood='neutral') + (1/2)*(mood='pleasant'); wordemot = (1/2)*(word='emotional') - (1/2)*(word='unemotional'); inter1 = emotvsno * wordemot; inter2 = plsvssad * wordemot; run; proc reg; title 'Two-Way ANOVA of Recall Data'; model recall = emotvsno plsvssad wordemot inter1 inter2/ ss2 pcorr2; run; B. Source Table Outline Source df Model 5 Mood (Main Effect) 2 Emotional vs Neutral 1 Pleasant vs Sad 1 Word Emotionality (Main Effect) 1 Interaction: Mood x Word 2 Inter1: Em vs Neu x Word 1 Inter2: Pl vs Sad x Word 1 Error 12 Total 17 Handout from Psych 5741/5751 Univ of Colorado used with Judd, C.M., & McClelland, G.H. (1989). Data Analysis: A Model Comparison Approach. HBJ. Spring Final 1996 Answers — 2 — May 13, 1996 Question 2 A. Source Table Source b df SS MS F* PRE p Between Subjects Period 2 13957.8 6978.9 8.82 .60 .0047 Linear ** -51.30 1 13158.4 13158.4 16.63 .58 .0015 Quadratic 10.95 1 799.4 799.4 1.01 .08 .3347 Error 12 9494.0 791.2 Total Between Subj 14 23451.8 Within Subjects Time ** -11.60 1 1009.2 1009.2 11.99 .50 .0047 Time x Period 2 912.3 456.2 5.42 .48 <.05 Time x Lin ** -7.40 1 68.5 68.5 0.81 .06 .3848 Time x Quad -22.63 1 843.8 843.8 10.03 .46 .0081 Error 12 1009.6 84.1 Total Within Subj 15 2931.1 Total 29 26382.9 B. On average, clerks sorted 51.3 more dates when using four periods than when using two (F*(1,12) = 16.63, PRE = .58, p = .0015) and there was no evidence for a nonlinear effect when comparing the mean of the three-period group to the means of the other two groups (F*(1,12) = 1.01, PRE = .08, p = .33). On average, clerks were less efficient in the morning, sorting 11.6 fewer dates than in the afternoon (F*(1,12) = 11.99, PRE = .50, p = .0047). However, there was an interaction between time of day and the quadratic trend for period such that there was a quadratic effect for morning but not for afternoon (F*(1,12) = 10.03, PRE = .46, p = .008). Or, as can be seen in Figure 1, virtually all of the time-of-day effect is due to the difference for the three-period group. Handout from Psych 5741/5751 Univ of Colorado used with Judd, C.M., & McClelland, G.H. (1989). Data Analysis: A Model Comparison Approach. HBJ. Spring Final 1996 Answers — 5 — May 13, 1996 F. All outlier indices suggest that Observation #3 is not like the other data values. Its size ($575 million) is almost twice as large as the second largest company ($305 million), resulting in a very high lever of .58 (i.e., one-half of a parameter is allocated in the model to just this observation). It also has the largest residual value and the test of whether it would be useful to add an additional parameter to the model just for this observation is significant (F*(1,15) = 96.7, p < .0001). For no other observation would it be useful to add an additional parameter to the model. The combined unusualness of both the predictor (Size) and the dependent variable (Time, with respect to a model of Time) result in an enormous value of Cook's D (6.67, with the next largest being only .15). This outlier does not appear as a long tail in the squished normal-normal quantile plot produced by proc univariate, but does appear in the better plot produced by SAS/INSIGHT. The unusual observation also stands out in the plot of residuals against predicted values. Except for that point, there is an apparent trend in the residuals, but it would probably disappear in an analysis in which Observation #3 is deleted. The obvious next step is to redo the analysis either deleting #3 or, if one's advisor is cranky about deleting observations, by adding a separate parameter just for observation #3. A power transformation would not be in order until heteroscedasticity could be checked in the new analysis. However, given that the data are essentially reaction times or counts, a priori transformations using either the log or the square root would not be inappropriate. Also, Observation #4 has a lever that is a bit unusual (it is the smallest company), but it might not be so unusual once the extremely large company (#3) is removed. Observation #6 has the second most unusual adoption time relative to the model and it is for the company that adopted the innovation first. It too probably won't be so unusual in the new analysis; if it is, then it might be worth considering what special conditions might facilitate the earliest adoption; i.e., what distinguishes the first adopter from subsequent adopters? Handout from Psych 5741/5751 Univ of Colorado used with Judd, C.M., & McClelland, G.H. (1989). Data Analysis: A Model Comparison Approach. HBJ. Spring Final 1996 Answers — 6 — May 13, 1996 Question 4 A. Price reduction percentage is significantly related to the non-redemption rate (Chi-squre(1) = 151, p < .0001), such that higher price reduction percentages were associated with lower non-redemption rates. The logit of non-redemption decreased by -.1 for each 1-pt increase in the percentage price reduction. Or, the odds of non-redemption decreased by a factor of .897 for each 1-pt increase in the percentage price reduction. B. There is no suggestion of a non-linear effect of price reduction percentage; the quadratic trend is not significant (Chi-square(1) = .04, p = .84). C. loˆ g it = 2.19 − 0.11 Reduce loˆ g it = 2.19 − 0.11 (25) = −0.56 ˆ p = e loˆ g it 1 + e loˆ git = e − .56 1 + e −.56 =.36 In other words, the predicted rate of non-redemption is .36, so the predicted proportion of coupons redeemed for a 25% price reduction is 1 - .36 = .64.