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

##### Understanding the difference between the equations cohen's d and...
 Author Message Julian posted 4 Years Ago ANSWER         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 isSQRT( ((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.dpufGreenwald, 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 Dave posted 4 Years Ago ANSWER         Group: Administrators Posts: 10K, Visits: 51K 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 = 1.7248Cohen_pooled = 1.9039Also calculate the SD across all six raw data points, i.e. SDALL = sd(5,4,3,2,4,7). The result isSDALL = 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. Julian posted 4 Years Ago ANSWER         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! Dave posted 4 Years Ago ANSWER         Group: Administrators Posts: 10K, Visits: 51K 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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