Abstract
This article offers a new methodological approach to investigate the degree of fit between an independent sample and 2 existing sets of norms. Specifically, with a new adaptation of a Bayesian method, we developed a user-friendly procedure to compare the mean values of a given sample to those of 2 different sets of Rorschach norms. To illustrate our technique, we used a small, U.S. community sample of 80 adults and tested whether it resembled more closely the standard Comprehensive System norms (CS 600; Exner, 2003), or a recently introduced, internationally based set of Rorschach norms (Meyer, Erdberg, & Shaffer, Citation2007). Strengths and limitations of this new statistical technique are discussed.
Acknowledgment
Thanks to Susan Huffaker who helped to process these data and to code the Rorschachs.
Notes
1 Additional information on these international data can be found in the 2007 Special Issue of the Journal of Personality Assessment devoted to International Reference Samples for the Rorschach Comprehensive System.
2 When this study was initiated, many clinicians were still using the DSM–III, despite the fact that the fourth edition (DSM–IV; American Psychiatric Association, Citation1994) had already been published.
3 Five records were missing age information but are known to be adults.
4 For the t tests, when homoscedasticity could not be assumed, the Welch–Satterthwaite method was used to adjust degrees of freedom. Also, years of education and distribution of gender and race were not reported in Meyer et al. (Citation2007).
5 In calculating the JZS B, the experimenter has to define a scale factor related to prior probabilities, which is denoted by r. Rouder et al.'s (2009) recommended setting is r = .5 in situations where small differences are important. Because small differences in sets of norms are likely to be interpretatively important we set r at .5 (for details, see Rouder et al., Citation2009).
6 XA% was also considered. It is essentially the complement of X–%, and in our sample was correlated with X–% at –.97, p < .01. Thus, it is redundant with X–% so it was not included.