Psych 610 Handout #8, p. 1Prof. MooreDerivation of coefficients for Trend Analysis(1) µi = β1(a1 + Xj) + β2(a2 + b2Xj + Xj2) + β3(a3 + b3Xj + c3Xj2 + Xj3) + etc.-- Assume an experiment with 4 age groups – 8, 9, 10, and 11-year-olds. We want to do trend analysis on Age. So the ages are used as the values of X.For linear: cj = a1 + Xj (from Eq. 1). Plug in the age as Xj:c1 = a1 + 8 c1 = -9.5 + 8 = -1.5Add c2 = a1 + 9 c2 = -9.5 + 9 = -.5these c3 = a1 + 10 c3 = -9.5 + 10 = .5c4 = a1 + 11 c4 = -9.5 + 11 = 1.5____________∑cj = 4a1 + 380 = 4a1 + 38; because ∑cj = 0-38 = 4a1a1 = -9.5 Now plug in –9.5 for a1-- So, coefficients for linear are –1.5, -.5, .5, 1.5. If we multiply by 2, we get the values from Keppel Table A –4, -3, -1, 1, 3.-- Next we want to find the coefficients for quadratic trend. We want these coefficients to be orthogonal to the linear coefficients.For quadratic: cj = a2 + b2Xj + Xj2 (from Eq. 1). Plug in the age as Xj:c1 = a2 + (b2)(8) + 82c1 = 89 + (-19)(8) + 64 = 1Add c2 = a2 + (b2)(9) + 92c2 = 89 + (-19)(9) + 81 = -1these c3 = a2 + (b2)(10) + 102c3 = 89 + (-19)(10) + 100 = -1c4 = a2 + (b2)(11) + 112c4 = 89 + (-19)(11) + 121 = 1__________________∑cj = 4a2 + 38b2 + 366Now apply orthogonality constraint. ∑cjcj′ = 0.Handout #8, p. 2So we multiply linear cj by the quad cj :(linear cj) (quad cj) (-3) (a2 + 8b2 + 64)Add (-1) (a2 + 9b2 + 81)these (1) (a2 + 10b2 + 100) (3) (a2 + 11b2 + 121)___________________________∑cjcj′ = 0a2 + 10b2 + 190-- Now we have 2 equations in 2 unknowns:(2) 0 = 4a2 + 38b2 + 3660 = 0a2 + 10b2 + 190; or, b2 = -19Put b2 = -19 into Eq. 2: 0 = 4a2 + (38)(-19) + 366a2 = 89-- Now plug a2 = 89 and b2 = -19 into quadratic expressions above and calculate cj for quad. This yields the values in Keppel’s table.-- This method works for unevenly spaced independent variables but will yield different cj from those in the table in Keppel, of
View Full Document
Unlocking...