Hi Gordon,
It's not easy for someone (like me) without specific knowledge about the PDDBI questionnaire to answer this question.
According to what I found on the internet, the scores should be standardized to mean=50 and SD=10*, so that the t-score should be calculated as ((raw score)-mean)/SD * 10 + 50, where the mean and SD in the formula are those of the relevant subgroup of the standardization sample (used by the developers of the inventory).
I suspect that 16.78 and 10.79 are not the applicable mean and SD values for this example (as otherwise the t-score should rather be 64). There seem to be at least two different versions of the questionnaire ("teacher" and "parent"), probably with different sets of mean and SD values.
Also, the CI must depend on a standard error of measurement (SEM) determined by the developers of the inventory based on their data. Strangely enough, the CI in your example is not centered around the t-score (unless this is due to extreme rounding errors: 59.5=(52.5+66.5)/2). I would have expected a CI like t-score +/- probit(0.95)*SEM.
The PDDBI presentation which is available from the product website reports a CI of 54-66 for what is presumably the same subject "Jack" (see slide 52). This looks more plausible. The SEM values can be found in the PDDBI presentation (slide 76). In this example ("parent" version of the questionnaire) we have SEM=3.46 for domain "SENSORY" (assuming that this corresponds to your "cluster #1"), which leads, indeed, to a CI of 54-66 (60 +/- probit(0.95)*3.46 = 60 +/- 5.69) for the t-score of 60.
I would expect that the relevant mean, SD and SEM values can be found in the PDDBI manual (and that the developers hopefully did not hide this information in their proprietary scoring software).
* (confirmed by the PDDBI presentation, slide 51)
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