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验证儿童综合连线测试的两因素模型解释。

Confirmation of a two-factor model for interpretation of the comprehensive trail making test with children.

机构信息

Department of Educational Psychology, Texas A & M University, College Station, 77843-4225, USA.

出版信息

Arch Clin Neuropsychol. 2011 Apr;26(3):235-9. doi: 10.1093/arclin/acq105. Epub 2011 Jan 20.

DOI:10.1093/arclin/acq105
PMID:21252035
Abstract

Exploratory factor analyses (EFAs) of the Comprehensive Trail Making Test suggested a possible two-factor solution that might better reflect the differences in Trails 1-3 and Trails 4 and 5 as opposed to a single Composite Index for the total standardization sample. The purpose of this study was to conduct a confirmatory factor analysis (CFA) of the two-factor structure with a subset of the standardization sample ages 18 or younger that had completed all five-trail tasks. The sample included 251 boys and 306 girls, ages 8-18, with a mean age of 12.76 (SD = 3.07). Data were collected across 16 states with representation from all regions of the USA. Standardized scores on each of the trails (1-5) tasks were considered. The results of CFA using M-plus indicated a good fit for the two-factor model, χ(2)(4) = 18.686, p = .0009, root mean-square error of approximation = 0.081, comparative fit index = 0.986 and standardized root-mean-squared residual = 0.021. A one-factor model was not supported. As suggested by the EFA in the manual, Trails 1-3 and Trails 4 and 5, while related, appear to be different in subtle ways that may be most meaningful in conjunction with evaluation of children with neurodevelopmental differences. Implications and possible explanations for this difference are discussed.

摘要

探索性因素分析(EFAs)对综合连线测试的结果表明,可能存在一种两因素解决方案,与总标准化样本的单一综合指数相比,该方案可能更好地反映连线测试 1-3 和连线测试 4 和 5 之间的差异。本研究的目的是使用标准化样本中年龄在 18 岁以下的子样本进行两因素结构的验证性因素分析(CFA),这些子样本完成了所有五项连线任务。样本包括 251 名男孩和 306 名女孩,年龄在 8 至 18 岁之间,平均年龄为 12.76(SD=3.07)。数据来自美国 16 个州,代表了美国所有地区。考虑了每条连线(1-5)任务的标准化分数。使用 M-plus 进行 CFA 的结果表明,两因素模型拟合良好,χ(2)(4)=18.686,p=0.0009,近似均方根误差=0.081,比较拟合指数=0.986,标准化根均方残差=0.021。不支持单因素模型。正如手册中的 EFA 所建议的那样,连线测试 1-3 和连线测试 4 和 5 虽然相关,但似乎在细微方面有所不同,这在结合评估具有神经发育差异的儿童时可能具有重要意义。讨论了这种差异的含义和可能的解释。

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