RE: Understanding the difference between the equations cohen's d and d

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Understanding the difference between the equations cohen's d and d_biep

Understanding the difference between the equations cohen's d and...

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Julian	Julian posted 8 Years Ago Topic Details Share Topic
Group: Forum Members Posts: 3, Visits: 17	Hello all, I have been reviewing the IAT SPSS . SPSS command syntax for processing the IAT data generated by the IAT templates - See more at: http://www.millisecond.com/download/library/IAT/#sthash.AtTohJL9.dpuf . SPSS command syntax for processing the IAT data generated by the IAT templates - See more at: http://www.millisecond.com/download/library/IAT/#sthash.AtTohJL9.dpuf command syntax for processing IAT data which I downloaded here ( http://www.millisecond.com/download/library/IAT/ ). I have also been reviewing the Greenwald, Nosek, and Banaji (2003) article which outlines the syntax. Greenwald, Nosek, and Banaji (2003) state that the only difference between Cohen's d and D is that the standard deviation in the denominator of D is calculated from the scores in both conditions, ignoring condition membership. The denominator for D appears in the command as following D_asis_denom = SQRT( ( ((N1-1) * SD1*2 + (N2-1) SD2*2) + ((N1+N2) ((M1-M2)*2) / 4) ) / (N1 + N2 - 1) ) The denominator for Cohen's d (pooled within treatment standard deviation), in the same format, i believe is SQRT( ((N1-1) SD1*2 + (N2-1) SD2*2) / (N1 + N2 - 2) ) I do not understand how the differences between these two denominators can be simply explained by D ignoring condition membership, while d does not. Does including the addition of ((N1+N2) ((M1-M2)**2) / 4) and subtracting by 1 instead of 2 somehow account for ignoring condition membership? I have scoured the forums and other articles by Greenwald, Nosek, and Banaji for an explanation, but can not find one. Thanks in advance for the help! I am so grateful for all the information the forums have already provided. Greenwald, A. G., Nosek, B. A., & Banaji, M. R. (2003). - See more at: http://www.millisecond.com/download/library/IAT/#sthash.AtTohJL9.dpuf Greenwald, A. G., Nosek, B. A., & Banaji, M. R. (2003). - See more at: http://www.millisecond.com/download/library/IAT/#sthash.AtTohJL9.dpuf . SPSS command syntax for processing the IAT data generated by the IAT templates - See more at: http://www.millisecond.com/download/library/IAT/#sthash.AtTohJL9.dpuf . SPSS command syntax for processing the IAT data generated by the IAT templates - See more at: http://www.millisecond.com/download/library/IAT/#sthash.AtTohJL9.dpuf . SPSS command syntax for processing the IAT data generated by the IAT templates - See more at: http://www.millisecond.com/download/library/IAT/#sthash.AtTohJL9.dpuf(http://www.millisecond.com/download/library/IAT/ Tags Cohen's D d_biep
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Dave	Dave posted 8 Years Ago ANSWER Post Details Share Post
Group: Administrators Posts: 13K, Visits: 104K	I'm afraid I am not sure what exactly it is you don't understand. Perhaps an example that's easy enough to calculate by hand will illustrate. Suppose you have the following raw data: 5,4 are the raw values making up SET1. Thus N1 = 2, M1 = 4.5 and SD1 = 0.7071. 3,2,4,7 are the raw values making up SET2. Thus N2 = 4, M2 = 4 and SD2 = 2.1602. Now calculate the D-score nominator as well as Cohen's pooled SD: D_asis_denom = 1.7248 Cohen_pooled = 1.9039 Also calculate the SD across all six raw data points, i.e. SDALL = sd(5,4,3,2,4,7). The result is SDALL = 1.7224. You'll notice that SDALL is -- disregarding rounding errors, etc. -- identical to D_asis_denom, but noticeably different from the pooled SD for Cohen's d. In SPSS syntax: COMPUTE N1 = 2. COMPUTE M1 = mean(5,4). COMPUTE SD1=sd(5,4). COMPUTE N2 = 4. COMPUTE M2 = mean(3,2,4,7). COMPUTE SD2=sd(3,2,4,7). COMPUTE SDALL=sd(5,4,3,2,4,7). EXECUTE. COMPUTE D_asis_denom = SQRT( ( ((N1-1) * SD1*2 + (N2-1) SD2*2) + ((N1+N2) ((M1-M2)*2) / 4) ) / (N1 + N2 - 1) ). COMPUTE Cohen_pooled = SQRT( ((N1-1) SD1*2 + (N2-1) SD2**2) / (N1 + N2 - 2) ) . EXECUTE.
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Julian	Julian posted 8 Years Ago ANSWER Post Details Share Post
Group: Forum Members Posts: 3, Visits: 17	Hi Dave, Thanks for your response. That example did help explain how the differences in the calulcations in the D score that are not in Cohen's d work to ingore condition membership. If the standard deviation of all caluclation is virtually identical to the D score calculaton, then why bother with the elaborate D score calculation? Thank you!
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Dave	Dave posted 8 Years Ago ANSWER Post Details Share Post
Group: Administrators Posts: 13K, Visits: 104K	In the syntax, the denominator calculation happens at a stage where you're not operating on the full-raw data set anymore but are already working with aggregate values derived from a specific subset of the raw data.
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