## D-Score

##### D-Score
 Author Message Nighthawk          Group: Forum Members Posts: 15, Visits: 1 Hey there,I have a question about Scoring the IAT. I have an finished SPSS File with the IAT Reactionstime (One Line Per Subject) with the variables compatibleblock and incopatibleblock...I calculated an IAT Score by just using the mean difference over subjects between the two blocks...Can somebody tell me how I can calculate the D-Score out of it? What exactly is the advantage of the D-Score in contrast to the plain difference between the two blocks?Can you also tell me which correlation coefficient you use for correlations with explicit measures?Thanks a lot Tags Dave posted 12 Years Ago ANSWER         Group: Administrators Posts: 12K, Visits: 83K I calculated an IAT Score by just using the mean difference over subjects between the two blocks...Can somebody tell me how I can calculate the D-Score out of it? What exactly is the advantage of the D-Score in contrast to the plain difference between the two blocks?All of this is discussed at length in Greenwald et al. (2003), available here: http://faculty.washington.edu/agg/pdf/GN&B.JPSP.2003.pdf. Additionally, SPSS syntax for computing D-scores from data gathered with IAT Inquisit templates is available from the IAT download page: http://www.millisecond.com/download/samples/v3/IAT/default.aspx.Can you also tell me which correlation coefficient you use for correlations with explicit measures?The appropriate correlation coefficient would depend on a number of factors (design, scale level, etc.). Pearson's r should be okay to use in most cases. However, I suggest you look through the available research literature, find a similar case to yours and check what was used there.~Dave Nighthawk          Group: Forum Members Posts: 15, Visits: 1 Thanks for your quick answer. OK I found something in the stuff you told me for inspection. This is kind of a manual how to calculate the D-Score.COMPUTE Numerator_for_D = (Mn2 - Mn1) .COMPUTE Denominator_for_D = SQRT( ( ((N1-1) * SD1**2 + (N2-1) * SD2**2)                   + ((N1+N2) * ((Mn2-Mn1)**2) / 4) ) / (N1 + N2 - 1) ) .COMPUTE D = Numerator_for_D / Denominator_for_D .I think I try that one out. Yet I don't understand the N1 and N2 in this formula. This should be the sample size. But why is there and N1 N2.  Yes, ok the Sample Sizes you calculate the two means of but they are the same, aren't they? N1 is equal to N2 because it is a paired sample you got here. So I am not sure if I am right when inserting the same values for N1 and N2...[:^)]And I also don't understand what the double **means... Dave posted 12 Years Ago ANSWER         Group: Administrators Posts: 12K, Visits: 83K N1 is equal to N2 because it is a paired sample you got here. So I am not sure if I am right when inserting the same values for N1 and N2...Since trials with latencies greater than 10.000ms are to be discarded according to Greenwald et al.'s "improved scoring algorithm", N1 and N2 aren't necessarily equal.And I also don't understand what the double **means...Exponentiation, more commonly denoted as a^n.~Dave

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