Mpc 006 - 02-02 types of correlation
2.2 Special types of Correlation 2.3 Point Biserial Correlation rPB 2.3.1 Calculation of rPB 2.3.2 Significance Testing of rPB 2.4 Phi Coefficient (φ ) 2.4.1 Significance Testing of phi (φ ) 2.5 Biserial Correlation 2.6 Tetrachoric Correlation 2.7 Rank Order Correlations 2.7.1 Rank-order Data 2.7.2 Assumptions Underlying Pearson’s Correlation not Satisfied 2.8 Spearman’s Rank Order Correlation or Spearman’s rho (rs) 2.8.1 Null and Alternate Hypothesis 2.8.2 Numerical Example: for Untied and Tied Ranks 2.8.3 Spearman’s Rho with Tied Ranks 2.8.4 Steps for rS with Tied Ranks 2.8.5 Significance Testing of Spearman’s rho 2.9 Kendall’s Tau (ô) 2.9.1 Null and Alternative Hypothesis 2.9.2 Logic of Kendall’s Tau and Computation 2.9.3 Computational Alternative for Kendall’s Tau 2.9.4 Significance Testing for Kendall’s Tau
Recommended
Recommended
More Related Content
What's hot
What's hot (20)
Viewers also liked
Viewers also liked (20)
Similar to Mpc 006 - 02-02 types of correlation
Similar to Mpc 006 - 02-02 types of correlation (20)
More from Vasant Kothari
More from Vasant Kothari (17)
Recently uploaded
Recently uploaded (20)
Mpc 006 - 02-02 types of correlation
- 17. 𝑟 𝑝𝑏 = 𝐶𝑜𝑣 𝑋𝑌 𝑆 𝑋. 𝑆 𝑌
- 19. 𝑴𝒆𝒂𝒏 𝑮𝒆𝒏𝒅𝒆𝒓 = 𝟎. 𝟓𝟓 𝑴𝒆𝒂𝒏 𝑴𝒂𝒓𝒌𝒔 = 𝟔𝟒. 𝟖 𝑴𝒆𝒂𝒏 𝑴𝒂𝒓𝒌𝒔 𝑴𝒂𝒍𝒆 = 𝟔𝟎. 𝟖𝟖 𝑴𝒆𝒂𝒏 𝑴𝒂𝒓𝒌𝒔 𝑭𝒆𝒎𝒂𝒍𝒆 = 𝟔𝟖 𝑺 𝑮𝒆𝒏𝒅𝒆𝒓 = 𝟎. 𝟒𝟗𝟕 𝑺 𝑴𝒂𝒓𝒌𝒔 = 𝟗. 𝟏𝟕 𝑪𝒐𝒗 𝒙𝒚 = 𝟏. 𝟕𝟔
- 20. 𝑟 𝑝𝑏 = 𝐶𝑜𝑣 𝑋𝑌 𝑆 𝑋. 𝑆 𝑌 𝑟 𝑝𝑏 = 1.76 0.497 𝑥 9.17 𝑟 𝑝𝑏 = 0.389
- 22. 𝒕 = 𝒓 𝒑𝒃 𝒏 − 𝟐 𝟏 − 𝒓 𝒑𝒃 𝟐
- 23. 𝑡 = 𝑟𝑝𝑏 𝑛 − 2 1 − 𝑟𝑝𝑏 2 𝑡 = 0.386 20 − 2 1 − 0.3862 𝑡 = 1.775 ≤
- 28. ∅ = 𝒂. 𝒅 − 𝒃. 𝒄 𝒆. 𝒇. 𝒈. 𝒉
- 29. ∅ = 𝒂. 𝒅 − 𝒃. 𝒄 𝒆. 𝒇. 𝒈. 𝒉 = (𝟒 ∗ 𝟖) − (𝟔 ∗ 𝟏𝟏) 𝟏𝟎 ∗ 𝟏𝟗 ∗ 𝟏𝟏 ∗ 𝟖 = 𝟑𝟒 𝟏𝟗𝟗. 𝟕𝟓 = 𝟎. 𝟏𝟕
- 31. ∅ = 𝝌 𝟐 𝒏 𝝌 𝟐 = ∅ 𝟐. 𝒏 𝒅𝒇 = (𝒄 − 𝟏)(𝒓 − 𝟏)
- 32. 𝝌 𝟐 = ∅ 𝟐. 𝒏 = 𝟎. 𝟏𝟕 𝟐 ∗ 𝟐𝟗 = 𝟎. 𝟖𝟒
- 38. 𝒓 𝒃 = 𝒀 𝟏 − 𝒀 𝟎 𝑺 𝒀 𝑷 𝟎 𝑷 𝟏 𝒉 𝑊ℎ𝑒𝑟𝑒, 𝒀 𝟏 = the sample mean of the y variable for those individuals for whom x = 1 𝒀 𝟎 = the sample mean of the y variable for those individuals for whom x = 0 𝑺 𝒀 = the standard deviation of the y values 𝑷 𝟏 = the proportion of individuals with x = 1 𝑷 𝟎 = 1 − 𝑷 𝟏 = the proportion of individuals with x = 1 𝒉 = ordinate or the height of the standard normal distribution at the point which divides the proportions f P0 and P1
- 44. 𝑟 = cos 𝜃 𝑟 = cos 1800 1 + 𝑎𝑑 𝑏𝑐
- 45. 68 (a) 32 (b) 100 30 (c) 70 (d) 100 98 102 200 𝑟 = cos 1800 1 + 𝑎𝑑 𝑏𝑐 = cos 1800 1 + 68𝑥70 32𝑥30 = cos 55.7840 = 0.722
- 66. 𝝆 = 𝟏 − 𝟔 𝑫 𝟐 𝒏 𝒏 𝟐 − 𝟏 𝝆 = Spearman’s rank-order correlation D = difference between the pair of ranks of X & Y n = the number of pairs of ranks
- 68. 𝝆 = 𝟏 − 𝟔 𝑫 𝟐 𝒏 𝒏 𝟐 − 𝟏 𝜌 = 1 − 6 ∗ 32 10 100 − 1 𝜌 = 1 − 192 990 𝝆 = 𝟏 − 𝟎. 𝟏𝟗𝟒 = 𝟎. 𝟖𝟏
- 69. 𝝆 = 𝑿𝒀 − 𝑿 𝒀 𝒏 𝑿 𝟐 − 𝑿 𝟐 𝒏 𝒀 𝟐 − 𝒀 𝟐 𝒏
- 71. 𝝆 = 𝑿𝒀 − 𝑿 𝒀 𝒏 𝑿 𝟐 − 𝑿 𝟐 𝒏 𝒀 𝟐 − 𝒀 𝟐 𝒏 𝝆 = 𝟑𝟕𝟓. 𝟕𝟓 − 𝟓𝟓 ∗ 𝟓𝟓 𝟏𝟎 𝟑𝟖𝟒 − 𝟓𝟓 𝟐 𝟏𝟎 𝟑𝟖𝟑. 𝟓 − 𝟓𝟓 𝟐 𝟏𝟎 𝝆 = 𝟕𝟑. 𝟐𝟓 𝟖𝟏. 𝟐𝟓 = 𝟎. 𝟗𝟎𝟐
- 74. 𝝉 = 𝟎
- 88. 𝝉 −𝟏 < 𝝉 < +𝟏
- 89. 𝝉
- 91. 𝒏 𝒏 − 𝟏 𝟐
- 94. 𝝉 = 𝟐𝑺 𝒏 𝒏 − 𝟏 𝑊ℎ𝑒𝑟𝑒, 𝑆 = 𝑠𝑐𝑜𝑟𝑒 𝑜𝑓 𝑎𝑔𝑟𝑒𝑒𝑚𝑒𝑛𝑡 – 𝑠𝑐𝑜𝑟𝑒 𝑜𝑓 𝑑𝑖𝑠𝑎𝑔𝑟𝑒𝑒𝑚𝑒𝑛𝑡 𝑜𝑛 𝑋 𝑎𝑛𝑑 𝑌 𝑛 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝑜𝑟 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙𝑠 𝑟𝑎𝑛𝑘𝑒𝑑 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑋 𝑎𝑛𝑑 𝑌
- 95. 𝝉 = 𝟐𝑺 𝒏 𝒏 − 𝟏 − 𝑻 𝒙 . 𝒏 𝒏 − 𝟏 − 𝑻 𝒚 𝑊ℎ𝑒𝑟𝑒, 𝑆 = 𝑠𝑐𝑜𝑟𝑒 𝑜𝑓 𝑎𝑔𝑟𝑒𝑒𝑚𝑒𝑛𝑡 – 𝑠𝑐𝑜𝑟𝑒 𝑜𝑓 𝑑𝑖𝑠𝑎𝑔𝑟𝑒𝑒𝑚𝑒𝑛𝑡 𝑜𝑛 𝑋 𝑎𝑛𝑑 𝑌 𝑛 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝑜𝑟 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙𝑠 𝑟𝑎𝑛𝑘𝑒𝑑 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑋 𝑎𝑛𝑑 𝑌 𝑇 𝑥 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑒 𝑟𝑎𝑛𝑘𝑠 𝑖𝑛 𝑎 𝑡𝑖𝑒 𝑤𝑖𝑡ℎ𝑖𝑛 𝑋 = 𝑡 𝑡 − 1 𝑇 𝑦 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑛𝑠𝑒𝑐𝑢𝑡𝑖𝑣𝑒 𝑟𝑎𝑛𝑘𝑠 𝑖𝑛 𝑎 𝑡𝑖𝑒 𝑤𝑖𝑡ℎ𝑖𝑛 𝑌 = 𝑡(𝑡 − 1)
- 97. 𝑪 = 𝟑 𝑫 = 𝟎 𝑪 = 𝟏 𝑫 = 𝟏 𝑪 = 𝟏 𝑫 = 𝟎 𝚺𝑪 = 𝟓 𝚺𝑫 = 𝟏
- 99. 𝝉 𝑪 = 𝟑 𝑫 = 𝟎 𝑪 = 𝟏 𝑫 = 𝟏 𝑪 = 𝟏 𝑫 = 𝟎 𝚺𝑪 = 𝟓 𝚺𝑫 = 𝟏 𝜏 = 𝛴𝛴𝐶 − 𝛴𝛴𝐷 𝑛 𝑛 − 1 2 𝜏 = 5 − 1 4 4 − 1 2 𝜏 = 4 6 𝜏 = 0.667
- 100. 𝝉 𝜏 = 2𝑆 𝑛 𝑛 − 1 𝜏 = 2 𝑥 4 4 4 − 1 𝜏 = 8 12 𝜏 = 0.667
- 106. 𝝉 𝜏 = 2𝑆 𝑛 𝑛 − 1 𝜏 = 2 𝑥 −2 4 4 − 1 𝜏 = −4 12 𝝉 = 𝟎. 𝟑𝟑𝟑
- 108. 𝑻 𝒙 = 𝟎 𝑇 𝑦 = 𝑡 𝑡 − 1 𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 3 𝑡𝑖𝑒𝑠 𝑖𝑠 𝑦 𝑟𝑎𝑛𝑘𝑠 𝑇𝑤𝑜 𝑠𝑢𝑏𝑗𝑒𝑐𝑡𝑠 𝑎𝑟𝑒 𝑡𝑖𝑒𝑑 𝑎𝑡 1.5 𝑇𝑤𝑜 𝑠𝑢𝑏𝑗𝑒𝑐𝑡𝑠 𝑎𝑟𝑒 𝑡𝑖𝑒𝑑 𝑎𝑡 3.5 𝑇𝑤𝑜 𝑠𝑢𝑏𝑗𝑒𝑐𝑡𝑠 𝑎𝑟𝑒 𝑡𝑖𝑒𝑑 𝑎𝑡 10.5 𝑇ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑡𝑖𝑒 𝑖𝑠 𝑥 𝑟𝑎𝑛𝑘𝑠, 𝑆𝑜, 𝑇 𝑦 = 2 2 − 1 + 2 2 − 1 + 2 2 − 1 𝑻 𝒚 = 𝟔
- 109. 𝝉 = 𝟐𝑺 𝒏 𝒏 − 𝟏 − 𝑻 𝒙 . 𝒏 𝒏 − 𝟏 − 𝑻 𝒚 𝝉 = 𝟐 𝒙 𝟐𝟓 𝟏𝟐 𝟏𝟐 − 𝟏 − 𝟎 . 𝟏𝟐 𝟏𝟐 − 𝟏 − 𝟔 𝝉 = 𝟓𝟎 𝟏𝟑𝟐 𝒙 𝟏𝟐𝟔 = 𝟓𝟎 𝟏𝟐𝟖. 𝟗𝟔 𝝉 = 𝟎. 𝟑𝟖