第五章 結論與未來研究
5.2 未來研究
55
56
參考文獻
中文部份
[1] 陳慶帆、許意苹、林敏慧，MOODLE 教學帄台多媒體評量編輯系統之探 討，台灣網際網路研討會，台中，2005 年。
[2] 顏春煌，數位學習與資訊素養補充教材，空大學訊，第 385 期，第 91-96 頁，2007 年。
[3] 李坤崇，多元化教學評量，心理，1999 年。
[4] 余民寧，教育測驗與評量—成就測驗與教學評量(第二版)，台北：心理出 版社，2002 年。
[5] 歐滄和，教育測驗與評量，心理，2002 年。
[6] 王文中、呂金燮，教育測驗與評量，五南，2004 年。
[7] 沈中偉，多媒體電腦輔助學習的學習理論基礎，教學科技與媒體，第 16 期，
第 16-25 頁，1995 年。
[8] 陳品仲，網路學習標準的分析與比較，國立中山大學資訊管理研究所碩士 論文，2001 年。
[9] 經濟部，學習管理帄台(LMS)及學習物件管理帄台(LCMS)整合介面研究報 告，數位學習產業推動與發展計畫，2004 年。
[10] 李長峰，合乎國際標準的學習教材管理系統與點對點搜尋工具，國立中正大 學資訊工程研究所碩士論文，2002 年。
[11] 數位學習國家型科技計畫標準推動網站，標準說明，Content Package，
http://elnpweb.ncu.edu.tw/web-e/b3.htm.
57
[12] Moodle 中文加油站, http://moodle.club.tw/moodle/.
[13] Moodle 1.5.3 架站錄影教學 , http://tpc.k12.edu.tw/1001216482/index.html.
[14] 黃信義、沈慶珩，網路同儕互評在 MOODLE 系統上的應用，教育資料與 圖書館學，第 43 期，第 267-284 頁，2006 年。
[15] 廖元偉、趙 銘，以 QTI 為標準引入 IRT 之詴題編輯工具，逢甲大學資訊 工程學系碩士班碩士論文，2006 年。
[16] 楊蹕齊、趙銘，以 MOODLE 為帄台之國中數學適性測驗工具，逢甲大學 資訊工程學系碩士班碩士論文，2006 年。
[17] 歐展嘉、Shally，MOODLE 線上教學帄台 - 松崗文魁，2005 年。
[18] 歐展嘉、Shally，MOODLE 數位學習課程管理帄台- 松崗文魁，2006 年。
[19] 羅夢娜、江仲翔，高級中學教師自編測驗評量方式對探討，國立中山大學 應用數學研究所碩士學位論文，2003 年。
[20] 陳敏彥、林原宏
，
應用 S-P 表與次序理論分析原住民學生在分數乘法之認 知診斷，
2007 年。英文部份
[21] Wen-Chih Chang and Timothy K. Shih, ―Using SPC Table to Strengthen SCORM Compliant Assessment‖, Department of Information Management, Chung Hua University, Department of Information Engineering and Computer Science, Tamkang University, 21st International Conference on Advanced Networking and Applications(AINA'07), pp. 825-830, 2007.
58
[22] HSIN-YI WU,―Software Based on S-P Chart Analysis and Its Applications‖, Proc. Natl. Sci. Counc. ROC(D),Vol. 8, No. 3, 1998. pp. 108-120,1999.
[23] Jonassen, D. H. & Reeves, T. C.,―Learning with technology: Using computer as cognitive tools‖, In D.H. Jonassen, Ed., Handbook of research on educational communication and technology, Scholastic Press pp. 693-719. NY.,1996.
[24] Dodds, P.V.W., "Current State of The SCORM", Retrieved from the World Wide Web, October 2001.
http://www.adlnet.org/library/documents/plugfest/plugfest5/Plugfest5state.ppt.
[25] IMS Global Learning Consortium, Inc., http://www.imsglobal.org/.
[26] Ariadne Foundation, http://www.ariadne-eu.org/.
[27] AICC - Aviation Industry CBT Committee, http://www.aicc.org/.
[28] Deng-Jyi Chen, Ah-Fur Lai and I-Chang Liu, ―The Design and Implementation of a Diagnostic Test System Based on the Enhanced S-P Model‖Journal of Information Science and Engineering, Vol. 21 No. 5, pp. 1007-1030, 2005.
[29] MOODLE，"Moodle documentation" ,http://docs.moodle.org/.
[30] IMS Global Learning Consortium, "IMS Question & Test Interoperability: An Overview, http://www.imsglobal.org/question/qtiv1p2/imsqti_oviewv1p2.html.
[31] Advanced Distributed Learning, ―ADL SCORM 2004 2nd Edition Overview‖,
http://www.adlnet.org.
59
附錄一 作業系統詴題分析
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
1 作業系統利用行程控制表(Process
Control Block)來管理行程(process)，
以 Unix 為例，請問下列哪一類資訊不 會紀錄在該表中?
分頁表資訊 8/55 (15%) 36 0.4855 0.436 0.32
行程優先順序 6/55 (11%)
程式計數器 21/55 (38%)
以上皆非 20/55 (36%)
2 相關暫存器找出一個頁數所花的百分
率稱為命中率，搜尋相關暫存器 20ns，存取記憶體 100ns，有頁數在相 關暫存器作對應記憶體的存取時間 120ns。如果命中率是 80%，則有效存 取時間是多少?
220ns 8/55 (15%) 15 0.3558 0.205 0.248
140ns 25/55 (45%)
120ns 12/55 (22%)
100ns 10/55 (18%)
3 電腦的多工作業系統必頇避面程式執
行時發生死結(deadlock)，請問下列哪 一項不是造成死結的必要條件?
Mutual exclusion 1/55 (2%) 96 0.1889 0.974 0.366
Hold and Wait 0/55 (0%)
No priority 53/55 (96%)
No preemption 1/55 (2%)
60
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
4 下列何者不為發生死結(deadlock)的
必要條件?
互斥(Mutual exclusion) 1/55 (2%) 95 0.2292 0.897 0.032
不可搶奪(No preemption) 0/55 (0%)
佔有與等待(Hold and wait) 2/55 (4%)
有限等待(Bounded waiting) 48/55 (87%) Cache 記憶體大小會影響電腦執
行程式效率
4/55 (7%)
5 在行程執行時，當所欲執行的指令或
存取的資料不在記憶體時，會發生分 頁錯誤(page fault)，是一種
中斷(interrupt) 44/55 (80%) 80 0.4037 0.846 0.238
例外(exception) 7/55 (13%)
系統呼叫(system call) 2/55 (4%)
訊號(signal) 1/55 (2%)
行程間通訊(interprocess communication)
0/55 (0%)
6 要解決臨界區間問題(critical section problem)，下列何者不為必需滿足的 條件?
互斥(Mutual exclusion) 1/55 (2%) 45 0.5025 0.538 0.358
進行(Progress) 6/55 (11%)
佔有與等待(Hold and wait) 25/55 (45%)
有限等待(Bounded waiting) 18/55 (33%)
以上皆非 5/55 (9%)
61
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
7 在 LRU 近似換頁法中，額外的參考位
元法則，下列何者會先被換出
0 43/55 (78%) 78 0.4168 0.795 0.277
11111111 0/55 (0%)
11000100 0/55 (0%)
1110111 12/55 (22%)
8 Second-chance page-replacement algorithm 的基本演算法是下列哪一 項演算法?
FIFO 52/55 (95%) 95 0.2292 0.974 0.496
OPT 2/55 (4%)
LRU 1/55 (2%)
9 Enhanced Second-chance 換頁演算法 運用下列哪些位元? (A)reference bit (B)dirty bit (C)modify bit (D)valid bit
A and B 33/55 (60%) 60 0.4944 0.692 0.379
B and C 17/55 (31%)
A and D 3/55 (5%)
C and D 2/55 (4%)
10 處理分頁錯誤的步驟有下列幾個，請
找出正確的順序。(A)參用(B)載入想 要的那一頁資料(C)某頁在備用儲存 體中(D)插斷(E)重新啟動指令(F)重新 設定分頁表
D, C, B, F, E, A 0/55 (0%) 89 0.3146 0.974 0.514
A, D, C, B, F, E 49/55 (89%)
B, C, F, E, A, D 2/55 (4%)
62
A, C, D, B, F, E 4/55 (7%)
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
11 關於 first-fit 的描述，哪一個正確? 將最小的一個區間分配給行程 1/55 (2%) 98 0.1348 0.974 0.481
將最大的一個區間分配給行程 0/55 (0%)
將最適合的一個區間分配給行程 0/55 (0%)
將第一個的一個區間分配給行程 54/55 (98%)
12 First-fit 與 best-fit 會導致下面哪一種 情況?
external fragmentation 46/55 (84%) 84 0.3734 0.974 0.629
internal fragmentation 4/55 (7%)
dynamic fragmentation 4/55 (7%)
compaction 1/55 (2%)
13 下列哪一個資源配置串列會出現死
結?
P1->R1->P2->R3->P3->R2->P1 31/55 (56%) 93 0.2621 0.897 0.063 P2->R3->P3->R2->P2 7/55 (13%)
P1->R1->P3->R2 3/55 (5%) P1->R1->P2->R3->P3->R2->P1->
R4
1/55 (2%)
63
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
14 下列哪一句話的描述是正確的? reader-writer 問題，可以允許兩個
writer 一起動作
0/55 (0%) 98 0.1348 0.974 0.444
Peterson's solution 不滿足 bounded waiting
0/55 (0%)
binary semaphore 的數值可以是 1, 2
1/55 (2%)
哲學家問題，無死結的解答並不 能消除飢餓的可能性
54/55 (98%)
15 死結的必要條件有哪些? (A)互斥(B)
進行(C)佔用與等候(D)不可搶先(E)循 環式等候(F)限制性等待 (G)競爭情 況
A, B, C, D 0/55 (0%) 98 0.1348 0.949 0.181
C, D, E, F 1/55 (2%)
B, G, D, E 0/55 (0%)
A, C, D, E 54/55 (98%)
16 下列哪一個演算法實做上很大的困
難?
FIFO 0/55 (0%) 89 0.3146 0.949 0.546
OPT 49/55 (89%)
LRU 5/55 (9%)
64
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
17 關於 RAID 的定義哪一個是正確的? 全名 Redundant Array of
Inexpensive Disk
12/55 (22%) 73 0.4495 0.872 0.533
RAID 需要的硬碟是很貴重的磁 碟,才能正常運作
0/55 (0%)
Mirroring 是 RAID 中最昂貴的技 術
40/55 (73%)
Hamming code 在每一個 RAID 中 都運用到
3/55 (5%)
18 資料位元 11110000，則計算出來的偶
同位元是
4 3/55 (5%) 93 0.2621 0.923 0.295
2 1/55 (2%)
1 0/55 (0%)
0 0/55 (0%)
19 下列哪一種 RAID 沒有重複的資料? RAID 0 51/55 (93%) 93 0.2621 1 0.604
RAID 1 2/55 (4%)
RAID 2 0/55 (0%)
RAID 3 1/55 (2%)
RAID 4 0/55 (0%)
RAID 5 1/55 (2%)
RAID 6 0/55 (0%)
65
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
20 下列哪一種 RAID 適合高性能應用，
且不重視資料損失?
RAID 0 45/55 (82%) 82 0.3892 0.974 0.653
RAID 1 1/55 (2%)
RAID 2 0/55 (0%)
RAID 3 4/55 (7%)
RAID 4 0/55 (0%)
RAID 5 3/55 (5%)
RAID 6 2/55 (4%)
21 下列哪一種 RAID 適合儲存大量資
料?
RAID 0 0/55 (0%) 75 0.4396 0.897 0.627
RAID 1 2/55 (4%)
RAID 2 1/55 (2%)
RAID 3 7/55 (13%)
RAID 4 1/55 (2%)
RAID 5 41/55 (75%)
RAID 6 3/55 (5%)
22 下列哪一種 RAID 具備最佳重建性? RAID 0 0/55 (0%) 78 0.4168 0.872 0.435
RAID 1 43/55 (78%)
RAID 2 5/55 (9%)
RAID 3 5/55 (9%)
RAID 4 0/55 (0%)
RAID 5 1/55 (2%)
66
RAID 6 1/55 (2%)
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
23 最早使用同位元發展 RAID 的是哪一
個?
RAID 0 10/55 (18%) 73 0.4495 0.795 0.352
RAID 1 2/55 (4%)
RAID 2 40/55 (73%)
RAID 3 2/55 (4%)
RAID 4 0/55 (0%)
RAID 5 0/55 (0%)
RAID 6 1/55 (2%)
24 Thrashing 的解釋，下列哪一個最接
近?
找出最近幾組常用的頁，集合起 來的集合
3/55 (5%) 87 0.3363 0.974 0.613
頻繁的分頁替換 48/55 (87%)
行程從自己的區域換出分頁 4/55 (7%)
25 以加強第二次機會替換法，我們透過
參考位元與修改位元來協助實做，下 列哪一個是屬於近來被使用過但未被 修改過?
(0,0) 1/55 (2%) 95 0.2292 0.949 0.275
(0,1) 2/55 (4%)
(1,0) 52/55 (95%)
(1,1) 0/55 (0%)
67
題號 題目 題目的選項 答對數 答對% 答對百分比難
易度
標準差 鑑別度
指數
鑑別度 係數
26 clock algorithm 用於下列哪一個演算
法?
FIFO 5/55 (9%) 89 0.3146 0.923 0.288
OPT 0/55 (0%)
LRU 1/55 (2%)
Second-chance 49/55 (89%)
27 配置演算法中，利用同等分配(equal
allocation)，有 93 個欄，有 5 個行程。
請問每個行程可以分配到多少欄?
16 0/55 (0%) 96 0.1889 0.974 0.392
17 1/55 (2%)
18 53/55 (96%)
19 1/55 (2%)
68
附錄二 英文論文
發表於「INTERNATIONAL CONFERENCE ON USABILITY AND LEARNING TECHNOLOGY, Chueh-sheng International Conference Hall, Tamkang University, May 2, 2008, pp. 108-120」
Items Analysis Tool Supporting S-P Chart and Bloom Cognition Taxonomy
Hsuan-Che Yang, *Wen-Chih Chang, *Chih-Huang Hsu, *Tsung-Pu Lee
Department of Computer Science and Information Science Tamkang University, Taipei, Taiwan
*Department of Information Management Chung Hua University, Hsin-Chu, Taiwan Arthur@mail.mine.tku.edu.tw, Earnest@chu.edu.tw
Abstract: E-learning provides people a convenient and efficient way for learning.
However E-learning didn‘t provide well methods for assessing learners‘ abilities.
In this paper, Sato‘s Student-Problem Chart is applied to integrate with our proposed on-line assessment system. Teachers are able to analyze learners with it easily and efficiently. In addition, Bloom Taxonomy of Educational Objective supports each item in our assessment management system during the authoring time. In our proposed system, it provides three groups of function for student, teacher and system administrator. According to the S-P Chart analysis and Bloom taxonomy of items, we can divide all items in an assessment into four types, and divide all students into six types. With these four types of diagnosis analysis chart of items, teacher can modify or delete the items which are not proper. With these six types of diagnosis analysis chart of students, teachers can realize learners‘
learning situation easily and efficiently.
Keywords: S-P Chart, Bloom Taxonomy of Educational Objective, E-Learning, On-line Assessment Tool.
Introduction
E-learning provides people a convenient and efficient way for learning. E-learning, especially web based learning is one of the popular way for people to access learning material at any time and in any place. E-learning supports teachers teaching and students learning easily with internet and multimedia. However E-learning didn‘t provide well methods for assessing learners‘ abilities. In traditional education, the teacher can change his/her lecturing style or content flexibly to maximize the teaching quality with students‘ response face to face. In E-learning era, teachers recorded and prepared their teaching materials before classes begin then it will be published on the internet.
Teachers have less time to predict students‘ learning ability thoroughly before the web class begins.
Briefly, it is hard for teachers to modify the learning content or style immediately and flexibly in the web based learning environment. How to assist the instructor estimate the learner‘s ability and analyze the learning records accurately, which provides precious information to adjust the learning contents or learning sequencing more appropriately. Assessment measures and analyzes student performance and learning skill. It also replies feedback to the teacher and student which documents growth or provides directives to improve future performance, is significant to learning and development. Formative assessment plays the role to guide student instruction and learning, diagnose
69
skill or knowledge gaps, measure progress and evaluate instruction. In daily use, teachers apply formative assessment to determine what concepts require more teaching and what teaching techniques or strategies require modification. After a period of learning days, teachers use the result to evaluate instruction strategies and curriculum. Teachers can make some adjustments for better student performance. Assessment focuses on the gap between students performance and instruction goal. Formative assessment which is beneficial to apply on E-learning to gather the learning information could adapt the teaching or the learning to meet the needs of the learner.
In this study, we mainly use a tool called S-P Chart (Student-Problem Chart) integrated in our proposed on-line assessment system. Teachers are able to analyze learners with it easily and efficiently. There are two main purposes in our study. First one is that teachers can measure and understand learners‘ further learning performances with caution index for course for students (CS) provided in S-P Chart. The Second one is that teachers are able to observe items quality with caution index for problems (CP) provided in S-P Chart with this integrated system. Bloom Taxonomy of Educational Objective is commonly used for categorizing cognitive domains. In our research, we also integrate Bloom taxonomy to assist categorizing our test items. Thus, after the assessments were held, teachers are not only able to evaluate the learners‘ ability, but also to point out the deficiency of learners in the assessments.
The rest sessions of this paper are briefly summarized as follows. The related work section describes Sato‘s S-P Chart analysis and Bloom Taxonomy of Educational Objective; some related studies are discussed as well. After related work section, the system structure and system flow chart are presented, Following is the main method of our system, it delineates how to integrate S-P Chart to diagnose students and estimate their learning abilities. An assist system is discussed in proposed system section. Finally, a brief conclusion and future work is drawn.
Related Works
The diagnosis systems play a very important role in the area of E-learning. Teachers use diagnosis systems to evaluate learners and learning objects after learning [4]. After evaluating learners‘ ability and learning objects, teachers are able to provide learning materials for learners more proper according the diagnosing results. In our study, we integrate S-P Chart and Bloom‘s taxonomy into our proposed diagnosis system. Following is the introduction of these two theories.
Sato’s Student-Problem Chart (S-P Chart)
The Student-Problem Chart (S-P Chart) is proposed by a Japanese scholar Takahiro Sato in ‗70s [1]. The S-P Chart is a graphical analysis tool to represent the relationship between students and the response situations of answering some problems by these students. The main purpose of the S-P Chart is to obtain the diagnosis materials of each student when they study. Teachers can provide better study counseling to each student according to these analyzed materials. There are numbered indices used in S-P chart. Including disparity index, homogeneity index, caution index for student (CS), and caution index for problem (CP). With these indices, teachers are able to diagnose students‘ learning conditions, instructive achievement, and problem quality [2] [3].
In our research, we use caution index for problems (CP) and students (CS) to analyze the test items (problem) and students. These indices can be used for judging whether students or problems have any unusual phenomenon when the assessment was made. New coefficient formulas can modify by these two attention coefficients as shown below. [6]
70
■ Caution Index for Problems (CP)
students the of
score average The
j"
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item for test
takers test of all for
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■ Caution Index for Students (CS)
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curve S the
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Bloom Taxonomy of Educational Objective
The Taxonomy of Educational Objectives, often called Bloom's Taxonomy, is a classification of the different objectives and skills that educators set for students. The taxonomy was proposed in 1956 by Benjamin Bloom, an educational psychologist at the University of Chicago. In 1956, Bloom created the six cognition levels from the simple recall or recognition of facts. From easy to complex, the six levels are knowledge, comprehension, application, analysis, synthesis and evaluation. The sequence of six levels is also list through more complex and more abstract mental levels. Learner has to learn the front level (ex. knowledge) before later level (ex. comprehension). Learner who is not understand lower level (ex. knowledge), cannot forward to higher level (ex. comprehension) [10]. In 2001, Anderson et al. proposed a revised version of cognition level which is composed of cognitive process dimension and knowledge dimension. The cognitive process dimension emphasizes on how to assist learner‘s knowledge of retention and transfer. The knowledge dimension focuses on inform teachers how to teach. Knowledge dimension involves factual knowledge, conceptual knowledge, procedural knowledge and meta-cognitive knowledge. Cognitive process dimension includes remember, understand, apply, analyze, evaluate and create [10].