20150620 序列分析工作坊 發佈版

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2015/6/1 政政政政政政政政政政政政政政政 政政政政政政政政政政政政政 陳陳陳 [email protected] 陳陳陳陳陳陳陳陳陳 陳陳陳陳陳陳陳 1

Transcript of 20150620 序列分析工作坊 發佈版

  1. 1. 2015/6/1 [email protected] 1
  2. 2. 2
  3. 3. Experimental Group Control Group 3 ()
  4. 4. 4
  5. 5. ? 5
  6. 6. Bakeman (1997) 6
  7. 7. 1. 2. (Coding Scheme) 3. (behavioral sequences) 4. (observer agreement) 5. 6. : 1 7. (event sequences) 8. (time sequences) 9. (cross-classified events) 10. 7 Bakeman, R., & Gottman, J. M. (1997). Observing interaction: an introduction to sequential analysis. Cambridge: University Press.
  8. 8. Part 1. Part 2. Part 3. Part 4. 8
  9. 9. Part 1. 9
  10. 10. ? 10
  11. 11. 1949~1953(Bloom) 1960 2001 Anderson, L. W., & Krathwohl, D. R. (2001). A taxonomy for learning, teaching, and assessing: a revision of Blooms taxonomy of educational objectives. New York: Longman. 11
  12. 12. 12
  13. 13. B1 Remember B2 Understand B3 Apply B4 Analyze B5 Evaluate B6 Create B7 Others (Revised Blooms Taxonomy) 13
  14. 14. Gunawardena, LoweAnderson1997 (Interaction Analysis Model) IAM IAM IAM 1997IAM Gunawardena, C. N., Lowe, C. A., & Anderson, T. (1997). Analysis of a Global Online Debate and the Development of an Interaction Analysis Model for Examining Social Construction of Knowledge in Computer Conferencing. Journal of Educational Computing Research, 17(4), 397-431. 14
  15. 15. K1 / K2 K3 / K4 K5 / K6 (Interaction Analysis ModelIAM) 15
  16. 16. Hou2008 Mayer, 1992; D'Zurilla & Goldfried, 1971 16 Hou, H.-T., Chang, K.-E., & Sung, Y.-T. (2008). Analysis of Problem-Solving-Based Online Asynchronous Discussion Pattern. Educational Technology & Society, 11(1), 17-28.
  17. 17. 17 P1 P2 / P3 P4 P5
  18. 18. Part 2. 18
  19. 19. 19 Morea
  20. 20. Morae Recorder Morae Observer Morae Manager 20
  21. 21. xAPI (LMS)( )(LRS) LRSLRS 21 LMS xAPI LRS LMS xAPI LMS xAPI
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  24. 24. 1StuC 2StuB 3StuC 4StuB 5StuC 6StuB 7StuB ~~ 8StuB (1)(2) (3) (4) 9StuB 10StuB 11StuB 12StuB 13StuC ~~ 14StuB 24
  25. 25. 25 http://j.mp/20150620-coding
  26. 26. Part 3. 26
  27. 27. 27
  28. 28. 28 P3 P2
  29. 29. 29 1 2 0.72 0.08 0.80 0.18 0.02 0.20 0.90 0.10 1.00 = 0.72+0.02 1 = 0.74
  30. 30. Cohen's Kappa Cohen Kappa-1~1 30 Cohen, J. (1960). A coefficient of agreement for nominal scale. Educational and Psychological Measurement, 20(1), 3746. doi:10.1177/001316446002000104 Kappa < 0.4 0.4 ~ 0.6 0.6 ~ 0.8 > 0.8
  31. 31. = 11 + 00 (1+ +1 + 0+ +0) 1 (1+ +1 + 0+ +0) 1 2 0.72 (p11) 0.08 (p10) 0.80 (p1+) 0.18 (p01) 0.02 (p00) 0.20 (p0+) 0.90 (p+1) 0.10 (p+0) 1.00 31 2015Kappa : 2712015531 http://www.shanghaiarchivesofpsychiatry.org/cn/assets/215010cn.pdf
  32. 32. 32 http://j.mp/2015-kappa
  33. 33. Part 4. 33
  34. 34. 34
  35. 35. Code z-score A B C D 13.53 3.1 13.5 8 2.05 8.96 5.53 35
  36. 36. Sackett,G.P1974 1997BakmanGottmanObserving introduction 36
  37. 37. 1. 2. 3. 4. Adjusted Residuals Table 5. (behavioral transition graph) 37
  38. 38. N (10) ABBCBBCAAC Ns (9) AB BB BC CB BB BC CA AA AC break(1) ABBCBBCAAC 38
  39. 39. A B C A 1 (A->A) 1 (A->B) 1 (A->C) 3 B 0 2 2 4 C 1 1 0 3* AB * C1 ABBCBBCAAC 39
  40. 40. A B C A 1 1 1 3 B 0 2 2 4 C 1 1 0 3 B->C 40
  41. 41. Zero-order model A B C1/3 p(BC)exp = 1/N * 1/N = 1/3 * 1/3 = 0.11 f(BC)exp = p(BC)exp * Ns = 0.11 * 9 = 0.99 41
  42. 42. First-order model 1.f(B) = B = 4 p(B) = f(B) / N = 4 / 10 = 0.4 2.f(C) = C = 3 p(C) = f(C) / N = 3 / 10 = 0.3 3.p(BC)exp = p(B) * p(C) = 0.4 * 0.3 = 0.12 f(BC)exp = p(BC)exp * Ns = 0.12 * 9 = 1.08 42
  43. 43. x N P Q(1-P) 1.96( p < 0.05) 43 NPQ NPx z
  44. 44. 44 10.94370143 855631520.97488460 0.92 0.9504 92.0 )12.01(12.09 08.12 ))(1()( )()( expexp exp BCpBCpN BCfBCf z s obs z-score
  45. 45. 45 http://j.mp/2010-sahttp://j.mp/2015-code-to-seq
  46. 46. GSEQ(Generalized Sequential Querier) Roger Bakeman & Vicen Quera 1992 GSEQ5.1 variety frequenciesratedurations proportion(percentages)kappa 46 http://www2.gsu.edu/~psyrab/gseq/index.html
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  48. 48. 1. ex: B->B 2. 3. 48
  49. 49. o o B->CB->C 49
  50. 50. Ns = m = kL k = L = ABC2 m = 3^2 = 9 (9*(9^2)) / (9-1) = 91.125 92 50 1 9 2 m m Ns
  51. 51. Ns = m = k(k-1)L-1 k = L = ABC2 m = 3 * (3-1) ^ (2-1) = 3 * 2 = 6 (6*(6^2)) / (6-1) = 43.2 44 51 1 9 2 m m Ns
  52. 52. 52 )1( 9 PP Ns Ns = P = B->C0.6 Ns = 9 / (0.6 (1- 0.6)) = 37.5 38
  53. 53. 53
  54. 54. o o o 54 A B C D
  55. 55. o o o1 oStu1: ABBCBB / Stu2: BCBBAC : ABBCBB BCBBAC (2) o 55
  56. 56. N = 12 Ns = 10 oAB BB BC CB BB BC CB BB BA AC oB B 2 breaks ABBCBB BCBBAC 56
  57. 57. o ABBBCBB ACCBBCC oz oz A->BA- >B 57
  58. 58. 58
  59. 59. o oz-score o2 59
  60. 60. 60 B C 13.58 B C 8.93 0.53
  61. 61. 61 A 17% B 48% C 16% D 19%
  62. 62. (Passive Joint attention P)(Object attentionO) [2] P->O oP oP->Oz-score 62
  63. 63. [3] Pi->OPm->O oPPi Pm 1.Pi->OPm->OzPm->Oz-score 2.Pm->O [23] 1.P->O 2.Pm->OPi->O 63
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