HLM演講投影片-三星統計許健將顧問-20130329
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This document summarizes a lecture on hierarchical linear modeling (HLM). It discusses how to use HLM to determine how much variability in student achievement is attributable to school-level factors versus student-level factors. A toy example is provided to illustrate calculating variance between and within schools. The document also provides an overview of the process for building an HLM model, including determining necessary and significant variables and effects. Information is given about an upcoming basic HLM modeling course in Taipei.
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HLM演講投影片-三星統計許健將顧問-20130329
- 1. HLM階層線性模型演講 講師 : 中興大學 許健將顧問 2013/03/29 關鍵字請搜尋 : 三星課程網
- 15. A simple question How much of the variability in pupil attainment is attributable to schools level factors and how much to pupil level factors? First of all we have to define what we mean by “variability”
- 16. A toy example – first we calculate the mean Overall mean(0) Two schools each with two pupils. Overall mean= (3+2+(-1)+(-4))/4=0 3 2 -1 -4 attainment School 2School 1
- 17. Calculating the “variance” Overall mean(0) 3 2 -1 -4 attainment School 2School 1 The total variance is the sum of the squares of the departures of the observations around mean divided by the sample size(4) = (9+4+1+16)/4=7.5
- 18. The variance of the school means around the overall mean 3 2 -1 -4 attainment School 2School 1 Overall mean(0) 2.5 -2.5 The variance of the school means around the overall mean= Total variance =7.5 (2.52+(-2.5)2)/2=6.25
- 19. The variance of the pupils scores around their school’s mean 3 2 -1 -4 attainment School 2School 1 2.5 -2.5 The variance of the pupils scores around their school’s mean= The variance of the school means around overall mean = (2.52+(- 2.5)2)/2=6.25 Total variance =7.5=6.25+1.25 ((3-2.5)2 + (2-2.5)2 + (-1-(-2.5))2 + (-4-(-2.5))2 )/4 =1.25
- 20. Returning to our question How much of the variability in pupil attainment is attributable to schools level factors and how much to pupil level factors? In terms of our toy example we can now say 6.25/7.5= 82% of the total variation of pupils attainment is attributable to school level factors 1.25/7.5= 18% of the total variation of pupils attainment is attributable to pupil level factors
- 31. 重新檢驗資料 檢驗ICC 一般迴歸分析 共變數模型 零模型 檢驗截距信度 不可接受 可接受 不顯著 顯著 步驟一：零模型
- 33. 步驟三：固定效果模型 不進一步分析 檢驗截距信度 保留顯著者逐一剔除不顯著者 重新檢驗模型 檢驗離異數 不顯著 顯著 檢驗離異數 階層二固定效果 係數低 係數高
- 36. ICC=26.12/26.12+81.24=0.24ICC=26.12/26.12+81.24=0.24 ‘MathAch’ ≈50.76 (1.15) ‘MathAch’ ≈50.76 (1.15)
- 37. ‘MathAch’ ≈ 46.36 + 2.40 x Homework (1.15) (0.28) ‘MathAch’ ≈ 46.36 + 2.40 x Homework (1.15) (0.28) ∆(Deviance)=3798.68 – 3727.48=71.20∆(Deviance)=3798.68 – 3727.48=71.20 18.012.26/34.2112.26)( 12.024.81/28.7124.81)( 22 22 B w RR RR 學校 個體 null model
- 38. ‘MathAch’ ≈ 46.32 + 1.98 x Homework (1.76) (0.93) ‘MathAch’ ≈ 46.32 + 1.98 x Homework (1.76) (0.93)
- 39. ‘MathAch’ ≈ 49.05 + 1.98 x Homework ‐4.06 x Public (2.18) (0.92) (1.98) ‘MathAch’ ≈ 49.05 + 1.98 x Homework ‐4.06 x Public (2.18) (0.92) (1.98)
- 40. ‘MathAch’ ≈ 48.52 -3.26 x Public + 2.30 x Homework - 0.50 x HomePublic (1.88) (3.02) (3.71) (1.59) ‘MathAch’ ≈ 48.52 -3.26 x Public + 2.30 x Homework - 0.50 x HomePublic (1.88) (3.02) (3.71) (1.59)
- 41. ‘MathAch’ ≈ 52.72 -6.05 x Public + 0.92 x Homework - 0.68 x HomePublic (0.12) (0.38) (0.42) (0.10)