Dian Normalitasari Purnama, S. Hadi, S. Sukirno, H. Retnawati, Rizki Nor Amelia
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引用次数: 0
摘要
评估可以采用不同的测试,但这些测试不一定是平行的。能力较差的学生可能得分较高,而能力较强的学生得分较低。这种情况造成的测量误差需要进行测验等效。有几种计算机程序,包括 Bilog 和 R 程序,可用于测验等化。每种程序的准确度不同,等分结果的准确度会影响等分的标准误差。本研究旨在找出最准确的等差数列测试方法,以及估计 BILOG 和 R 程序的准确性。本研究采用了两组等效组设计的测试。根据均方根偏差(RMSD)值确定最准确的等效方法。用 BILOG 程序估计平均值和西格玛法将测试包 A 等同于测试包 B,结果 RMSD 值为 0.320。在平均值和均值法中,RMSD 值为 0.250。同时,使用 R 程序将软件包 A 等同于软件包 B 的均值和西格玛法的 RMSD 值为 0.300,均值和均值法的 RMSD 值为 0.272。均值和均值法得出的 RMSD 值小于均值和西格玛法。因此,均值和均值法更为精确和适用。此外,BILOG 程序的估计结果更为准确,可用于测试等值。
Which one is more accurate, BILOG or R program? (a comparison for score test equating)
Evaluation may be carried out using different tests that are not necessarily parallel. Students with lower abilities may get higher scores while those with higher abilities get lower scores. Measurement errors caused by this condition require test equating. Several computer programs, including Bilog and the R program, can be used for test equating. Each program has a different level of accuracy, and the accuracy of the equating results will affect the standard errors of equating. This study aimed to find out the most accurate equating test method and the accuracy of the estimated BILOG and R program. This research used two sets of tests with equivalent group designs. The determination of the most accurate equating method was based on the root mean square deviation (RMSD) value. Equating test packages, A to package B with BILOG program estimation on the mean and Sigma method resulted in RMSD value of 0.320. In the mean and mean method, RMSD value is 0.250. Meanwhile, equating package A to package B using R program on the mean and sigma resulted in RMSD value of 0.300, and the mean and mean method with the RMSD value is 0.272. The mean and mean yield RMSD values smaller than the mean and sigma methods. Therefore, the mean and mean method is more accurate and applicable. Moreover, the estimation results of the BILOG program are more accurate and can be used in test equating.