HLM演講投影片-三星統計許健將顧問-20130329

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HLM演講投影片-三星統計許健將顧問-20130329

  1. 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
  2. 2. A simple questionHow much of the variability in pupil attainment is attributable toschools level factors and how much to pupil level factors?First of all we have to define what we mean by “variability”
  3. 3. A toy example – first we calculate the meanOverall mean(0)Two schools each with two pupils.Overall mean= (3+2+(-1)+(-4))/4=032-1-4attainmentSchool 2School 1
  4. 4. Calculating the “variance”Overall mean(0)32-1-4attainmentSchool 2School 1The total variance is the sum of the squares of the departures of theobservations around mean divided by the sample size(4) =(9+4+1+16)/4=7.5
  5. 5. The variance of the school means around the overall mean32-1-4attainmentSchool 2School 1Overall mean(0)2.5-2.5The variance of the school means around the overall mean=Total variance =7.5(2.52+(-2.5)2)/2=6.25
  6. 6. The variance of the pupils scores around their school’smean32-1-4attainmentSchool 2School 12.5-2.5The 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=6.25Total 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
  7. 7. Returning to our questionHow much of the variability in pupil attainment is attributable toschools level factors and how much to pupil level factors?In terms of our toy example we can now say6.25/7.5= 82% of the total variation of pupils attainment isattributable to school level factors1.25/7.5= 18% of the total variation of pupils attainment isattributable to pupil level factors
  8. 8. 常數項常數項ijXijYjW ijrju0ju1j1j000100111ijijjjijjijjijrXuuXWXWY1011100100 高階變項是對低階迴歸係數做解釋,而不是對依變項本身做解釋
  9. 9. 模型建構歷程Snijders and Bosker(2002):1.變項必要性與精簡性的取捨2.統計顯著性與理論性的取捨3.固定效果須有強力理論支持4.由資料內容決定隨機效果5.若交互作用達顯著,即使造成交互作用的變項之主要效果未達顯著,則該變項亦需被保留。6.若隨機效果達顯著,則其固定效果須被保留7.聚合(Aggregate)後的階層變項可能是個重要變項8.若變項間產生交互作用,則交互作用不具隨機效果
  10. 10. 模型建構歷程Hox(2002): (針對探索性研究)1.先用隨機效果變異數模型並記錄模型適配度2.將所有最低階之變項投入隨機效果共變數模型,決定需要被保留之變項3.檢視所有第二階層變項之固定效果,並比較模式間之適配度4.檢視第3步驟的變項中具有隨機效果者5.將第4步驟中具有顯著解釋力的變項,依其隨機性之有無,決定模型6.將資料折半,投入相同模型中,確認其顯著性是否與完整資料之模型相同
  11. 11. 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
  12. 12. 步驟二:共變數模型共變數模型階層一固定效果檢驗截距信度逐一剔除不顯著者檢驗離異數重新檢驗模型保留顯著者不顯著 顯著
  13. 13. 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
  14. 14. 步驟四:隨機係數模型保留顯著者逐一剔除不顯著者重新檢驗模型不顯著 顯著檢驗離異數階層二隨機效果檢驗離異數
  15. 15. 步驟五:完整模型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隨機效果步驟六:複核效化 複核效化
  16. 16. 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)
  17. 17. ‘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=71.2018.012.26/34.2112.26)(12.024.81/28.7124.81)(2222BwRRRR學校個體null model
  18. 18. ‘MathAch’ ≈ 46.32 + 1.98 x Homework(1.76) (0.93)‘MathAch’ ≈ 46.32 + 1.98 x Homework(1.76) (0.93)
  19. 19. ‘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)
  20. 20. ‘MathAch’ ≈ 48.52 -3.26 x Public + 2.30 xHomework - 0.50 x HomePublic(1.88) (3.02) (3.71) (1.59)‘MathAch’ ≈ 48.52 -3.26 x Public + 2.30 xHomework - 0.50 x HomePublic(1.88) (3.02) (3.71) (1.59)
  21. 21. ‘MathAch’ ≈ 52.72 -6.05 x Public + 0.92 xHomework - 0.68 x HomePublic(0.12) (0.38) (0.42) (0.10)
  22. 22. 課程訊息HLM階層線性模式基礎班台北場日期:2013年4月27日(六)至4月28日(日)時間:每日9:00-12:00與13:30-17:00地點:台北商業技術學院課程講師:中興大學 許健將顧問詳請請關鍵字搜索 : 三星課程網

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