Name_____________________________ Ψ 420 Ainsworth 1Psy 420 – Midterm 1 Part 1 – In class (50 points total) True or False (circle T or F) (1 point each) 1. T F With more than 2-groups, performing multiple T-tests would be the same as doing a one-way ANOVA. 2. T F In a one-way ANOVA, total variance is the sum of within and between groups variance. 3. T F In ANOVA, your sample must have a normal distribution? Multiple Choice (circle the best answer) (1 points each) 4. If an IV has no effect on subject scores you would expect the F-ratio for that effect in an ANOVA to be: A) 0 B) 1 C) 1.96 D) cannot be determined 5. An effect you’re studying has a Cohen’s D of .5. According to Cohen, the size of this effect is: A) Large B) Moderate C) Small D) not worth studying 6. IV A and IV B interact significantly, what does this mean? A) A causes B B) A and B have the same effect on the DV C) The effect of A depends on B D) Subjects in A got to know subjects in B 7. If you have a 2 x 2 x 2 ANOVA, this means that: A) You have 2 IVs with 3 levels each B) You have 2 IVs and 3 DVs C) You have 3 IVs with 2 levels each D) You have 3 IVs and 2 subjects in each IV 8. ( )2i GMY Y−∑ is the deviation form of: A) SST B) SSBG C) SSWG D) MSWGName_____________________________ Ψ 420 Ainsworth 2Fill in the blank and short-answer 9. According to the simplest form of the GLM, a subject’s score is said to be the sum of µ (grand mean), α (effect of the IV) and ε (error). (3 points) 10. In regression terms, the residual is the difference between _____Y____ and ____Y’______. (2 points) 11. When performing an ANOVA, why do we assume homogeneity of variance? (2 points) Because we are assuming that all of the subjects in each sample come from the same population. Because we need to calculate an average within groups variance using each of the samples and you can’t do that if the variances are way off. 12. Describe 1 scenario where the assumption of independence of errors would be violated. (2 points) Repeated measures Grouped dataName_____________________________ Ψ 420 Ainsworth 313. Under what circumstance is trend analysis recommended? (2 points) Anytime you have an IV that increases quantitatively (in amount) 14. You have an IV with four groups (groups 1, 2, 3 and 4) and there is a significant quadratic trend; what might your data look like (draw a graph) (2 points)? Any picture that looks like a bowl or a mound with labels for the four groups across the bottom. 15. If you have a three-way ANOVA (IVs A, B and C), the between groups sums of squares is broken down into what effects (just list the labels of the effects, not the formulas)? (1 point each) A B C AB AC BC ABC If they put anything else, deduct half a point eachName_____________________________ Ψ 420 Ainsworth 4One-Way ANOVA Square BNL DMB John Mayer 9 7 5 3 10 5 6 4 9 7 3 4 8 8 3 4 8 6 5 5 Sum 44 33 22 20 T = 119 Mean 8.8 6.6 4.4 4 2Y∑ = 799 SD 0.84 1.14 1.34 0.71 20 subjects were randomly selected to rate 4 different recording artists (Square, Bare Naked Ladies, Dave Mathews Band and John Mayer) on a ten point scale (higher scores mean they liked the artist more). 16. Do an omnibus ANOVA (show your work) and decide if there is a significant difference between groups (5 points): 2 2 2 2 2/44 33 22 20 119781.8 708.05 73.755 20799 781.8 17.2799 708.05 90.95AS ATSSSSSS+ + += − = − == − == − = Fcrit(3, 16) = 3.24 17. Test for the homogeneity of variance assumption (2 points). 2 21.34 /1.14 1.7956 /1.2996 1.38= = It’s less than 10 so it’s fine. 18. Calculate η2 for the effect (2 points): 73.75/90.95=.81 19. Do a single test comparing Square to all the other artists combined and test the comparison using a Scheffé adjustment (3 points): 22 2 2 25[3(8.8) 1(6.6) 1(4.4) 1(4)]54.153 ( 1) ( 1) ( 1)50.141.08 1.08F− − −+ − + − + −= = = Source SS df MS F Band 73.75 3 24.58 22.76 S/Band 17.2 16 1.08 Total 90.95 19Name_____________________________ Ψ 420 Ainsworth 53-way ANOVA 20. Based on the data above, fill in the missing pieces of the source table on the next page (6 points). Use the space below for calculations, make sure you show all work. 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 256 56 61 77 69 59 94 81 60 173 205 235 227 206 180 6139 27 27 814,795.67 4,710.33 4,680.19 4,639.12 44.27ACSS+ + + + + + + + + + + += − − + =− − + = Or they can figure out the S/ABC and subtract everything to get SSAC c1 c2 c3 a1 a2 a3 a1 a2 a3 a1 a2 a3 b1 6 7 6 6 6 6 7 6 7 A1 = 173 A2 = 205 A3 = 235 6 6 7 7 6 7 7 6 7 B1 = 174 B2 = 194 B3 = 245 6 7 7 7 6 6 7 6 6 C1 = 227 C2 = 206 C3 = 180 b2 6 7 10 6 7 6 6 6 7 7 9 11 6 6 7 7 6 6 T = 613 6 8 12 6 7 7 7 7 8 ΣY2 = 5069 b3 7 11 14 6 11 13 6 7 7 6 11 13 6 10 15 7 7 6 6 11 14 6 10 14 7 8 6 A x B x C cell totals c1 c2 c3 a1 a2 a3 a1 a2 a3 a1 a2 a3 b1 18 20 20 20 18 19 21 18 20 b2 19 24 33 18 20 20 20 19 21 b3 19 33 41 18 31 42 20 22 19 2-way interaction totals a1 a2 a3 a1 a2 a3 c1 c2 c3 b1 59 56 59 c1 56 77 94 b1 58 57 59 b2 57 63 74 c2 56 69 81 b2 76 58 60 b3 57 86 102 c3 61 59 60 b3 93 91 61Name_____________________________ Ψ 420 Ainsworth 6Source Table Source SS df MS F A 71.21 2 35.61 103.00 B 99.28 2 49.64 143.61 C 41.06 2 20.53 59.39 AB 61.61 4 15.40 44.55 AC BC 52.20 4 13.05 37.75 ABC 41.58 8 5.20 15.04 S/ABC 54 .35 Total 430.11 80 21. Calculate partial Omega squared for the AB effect (3 points): 24(15.4 .35) 4(15.05).743 .74381 81 .684(15.4 .35) 4(15.05).743 .35 1.093.35 .3581 81Partialϖ−= = = = ≈−+   + +      ) 22. If you were to do this ANOVA through regression how many predictors (Xs) would you need to code the ABC interaction? (1 point) You’d need