• 沒有找到結果。

數位學習研究的資訊議題與趨勢

N/A
N/A
Protected

Academic year: 2021

Share "數位學習研究的資訊議題與趨勢"

Copied!
72
0
0

加載中.... (立即查看全文)

全文

(1)

數位學習研究的資訊議題與

趨勢

國立臺南大學 黃國禎

數位學習科技系 教授

資訊教育研究所 所長

理工學院 院長

2006/1/7

(2)

突破時空限制

提供個別化教學指導

提高學習興趣

促成同學互動

建立完整的教學管理機制

數位學習的優點

(3)

數位學習的發展現況

行政院在九十年

NICI( 國家資訊通信發展方案 ) 計

畫中,將數位學習納入「網路社會化」的一環。

國科會通過「數位學習國家型科技計畫」的構想,

預計五年內投入四十億元進行跨部會計畫

「全民數位學習」

「縮減數位落差」

「行動學習載具與輔具─多功能電子書包」

「數位學習網路科學園區」

「前瞻數位學習技術研發」

「數位學習之學習與認知基礎研究」

「政策引導與人才培育」

(4)

數位學習研究的三大趨勢

導入資訊技術

( 人工智慧、演算法、

資料探勘、物件導向、代理人

) 提昇

系統平台功能的研究

對數位學習標準工具及延伸的探討

對行動學習

(M-Learning) 與普化學習

(U-Learning) 平台與教學策略的探討

(5)

將在數位學習的研究由應用層次

(Alessi, S.

M. & Trollip, S.R. 1991) 提昇到功能層次

應用層次

利用網路進行教材及試題的共享

利用電腦作為呈現教材及考試的媒介

功能層次

被動

(Chou, C. 1996) :記錄學習過程,統計、保

存、報告學習評量之結果

主動

(Kumar, D.D., Helgeson, S.L. & White, A.L. 1

994) :分析學習狀態、偵查學習迷思、引導學習、

實施有規劃的教學、測驗與評量

(6)

個人化學習路徑之規劃

線上學習過程之記錄

線上學習行為之分析

資訊技術應用 1 :

(7)
(8)

線上學習行為分析

藉由模糊專家系統來推論出此學習者的學習

狀態,並給予適當的幫助

學習效率

(Efficiency of Learning)

學習意願

(Willingness)

耐心度

(Patience)

專心度

(Concentration)

閒置

(Idleness)

理解度

(Comprehension)

聊天

(Chat)

(9)

(1) 學習意願分析

學生用心學習的意願

分析依據:有效登入時間 / 登入時間

模糊推理法則

If willingness is low

Then insert INT(T×0.5) corresponding willingness frames.

If willingness is average

Then insert INT(T×0.25) corresponding willingness frames

If willingness is high

Then keep the current status. 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1

low average high

degree

(10)

(2) 耐心度分析

學生瀏覽一個畫面的持續度

分析依據:畫面學習時間 / 預估學習時間

模糊推理法則

If patience is low

Then record this status and warn the student. If patience is average Then keep the current status. If patience is high

0.2 0.4 0.6 0.8

1 low average high

(11)

(3) 專心度分析

學生集中精神於瀏覽教材的程度

分析依據:回應時間

模糊推理法則

If concentration is low Then insert a corresponding concentration frame. If concentration is high Then keep the current status. If concentration is average Then keep the current status.

0 0.2 0.4 0.6 0.8 1 0 0.25 0.5 0.75 1 low no response high RT degree

(12)

(4) 聊天狀態分析

學生利用線上討論區來閒聊而不是討論課程

分析依據:學習相關比率

模糊推理法則

If chat is high

Then record this status and warn the student.

If chat is average

Then keep the current status. If chat is low

Then keep the current status 0

0.2 0.4 0.6 0.8

1 high average low

(13)

Gwo-Jen Hwang (1998), “A tutoring strategy supporting system for distance l earning on computer networks”, IEEE Transactions on Education, Vol. 41, No . 4, pp. 343. (SCI & EI)

(14)

資訊技術應用 2 :

多專家教學知識擷取與整合系統

Computer Networks

Java-based communication unit

Fuzzy Reasoning Interface Expert System Shell Knowledge (Rule) Base Interactive Knowledge Elicitation Unit Knowledge Analysis Unit Tutoring Strategy Negotiation Unit Knowledge Base Generator

(15)

模糊專家系統知識擷取

Step 1

: Elicit all of the elements (concepts to be learned)

from the domain expert.

(16)

Step 2

: Elicit attributes ( properties or fuzzy

variables).

Li K Fr F Cl I

boiling point LOW;MIDDLE;HIGH atom radius NARROW;NORMAL;WIDE

metalloid WEAK;NORMAL;STRONG negative charge WEAK;MIDDLE;STRONG

(17)

Step 3: Fill all of the [concept, attribute] entries of the grid. A 7-scale (-3 to +3) rating and the degree of certainty(“S”,”N”).

Consider the ratings of fuzzy variable ‘boiling point’:

3 means VERY HIGH,

2 means HIGH,

1 means MORE OR LESS HIGH, 0 means MIDDLE,

-1 means MORE OR LESS LOW,

-2 means LOW,

-3 means VERY LOW

‘S’ means ‘VERY SURE’, ‘N’ means ‘NOT VERY SURE’

Li K Fr F Cl I

boiling point -1/N 0/N 1/N 1/S 2/S 3/S LOW;MIDDLE;HIGH atom radius -2/S -1/S 1/N 1/S 2/S 3/S NARROW;NORMAL;WIDE

metalloid 1/S 2/S 3/S -3/S -3/S -3/S WEAK;NORMAL;STRONG negative charge -3/S -3/S -3/S 3/S 2/S 1/S WEAK;MIDDLE;STRONG

(18)

Step 4

:

the first column of the above fuzzy table is

translated to the following rule:

IF boiling point is MORE OR LESS LOW, and atom radius is NARROW, and

metalloid is MORE OR LESS STRONG, and

negative charge is VERY WEAK

THEN the element could be Li TRUTH = 0.8

TRUTH

=

# " "

(# " " # " ")

.

.

of S

(19)

多專家知識擷取與整合的問題

While elicit expertise from multiple experts, some

problems may arise:

same element or attribute different vocabularies Unified vocabularie s Different ratings Conflict happen Insist their ratings

?

??

???

(20)

A. Interactive knowledge

elicitation unit

Interactive

user

interface

Fuzzy table

editor

Membership function builder Knowledge base

(21)

NEGOTIATOR:Please give a set of elements(concepts to be learned or decisions to be made).EDUCATOR: Li, K, Fr, F, Cl, I.

. . .

NEGOTIATOR: Select a set of fuzzy values for fuzzy variable “boiling point”:

1. LOW/MIDDLE/HIGH 2. SHORT/MIDDLE/TALL 3. LIGHT/NORMAL/HEAVY 4. SMALL/MIDDLE/BIG 0. Other (user-defined)EDUCATOR: 1

NEGOTIATOR: Select a set of fuzzy values for fuzzy variable “atom radius”:

1. LOW/MIDDLE/HIGH 2. SHORT/MIDDLE/TALL 3. LIGHT/NORMAL/HEAVY 4. SMALL/MIDDLE/BIG 0. Other (user-defined)EDUCATOR: 0

NEGOTIATOR: Indicate the lower bound of the fuzzy values.EDUCATOR: NARROW

NEGOTIATOR: Indicate the middle of the fuzzy values.EDUCATOR: NORMAL

NEGOTIATOR: Indicates the upper bound of the fuzzy values.EDUCATOR: WIDE

(22)
(23)
(24)

(4) Knowledge analysis unit

Li K Fr F Cl I

boiling point -1/N 0/N 1/N 1/S 2/S 3/S LOW;MIDDLE;HIGH atom radius -2/S -1/S 1/N 1/S 2/S 3/S NARROW;NORMAL;WIDE

metalloid 1/S 2/S 3/S -3/S -3/S -3/S WEAK;NORMAL;STRONG

Check if conflict occurs and integrate tutoring

strategies.

 The contents of a fuzzy table is represented as

Fuzzy_value(Educator_ID, Object_name, Fuzzy_variable) and

Certainty_Degree (Educator_ID, Object_name, Fuzzy_variable)

for examples, the fuzzy table below can represented as

Fuzzy_value(Educator1, Li, boiling point) = -1

Certainty_Degree(Educator1, Li, boiling point) = “N” ...

(25)

Knowledge analysis rule:

Rule_analysis_02

IF (1)

Current_Phase is Knowledge_Analysis

and

(2) Fuzzy_value(Expi, Gk, Vs)

Fuzzy_value(Expj, Gk,

Vs)

< 0 and

(3) Certainty_Degree (Expi, Gk, Vs)

is

"S"

and

(4) Certainty_Degree(Expj, Gk, Vs)

is

”N”

and

THEN (a) Set

Suggested_Fuzzy_Value

be

Fuzzy_value(Expi, Gk

, Vs)

and

(b) Set

Suggested_Certainty_Degree be ”N"

and

(26)

Knowledge analysis rule:

Rule_analysis_04

IF (1)

Current_Phase is Knowledge_Analysis

and

(2) Fuzzy_value(Expi, Gk, Vs)

Fuzzy_value(Expj, Gk, Vs)

0 and

(3) Certainty_Degree (Expi, Gk, Vs)

is

"S"

and

(4) Certainty_Degree(Expj, Gk, Vs)

is

"S”

and

(5) Fuzzy_value(Expi, Gk, Vs)

Fuzzy_value(Expj, Gk, Vs)

0

THEN (a) Set

Suggested_Fuzzy_Value

be

Fuzzy_value(Expi, Gk, Vs)

and

(27)

Knowledge analysis rule:

Rule_analysis_03

IF (1)

Current_Phase is Knowledge_Analysis

and

(2) Fuzzy_value(Expi, Gk, Vs)

Fuzzy_value(Expj, G

k, Vs)

< 0 and

(3) Certainty_Degree (Expi, Gk, Vs)

is

"S"

and

(4) Certainty_Degree(Expj, Gk, Vs)

is

"S”

and

THEN (a) Set

Suggested_Fuzzy_Value

be

“Conflict”

and

(b)

Set

Current_Phase

be

Knowledge_Negotiati

on

(28)

B. Tutoring Strategy

Negotiation unit

Present suggestions by knowledge analysis unit

When a conflict occurs, experts are asked to give

suggestions.

over-gener

over-gener

al

al

ha

ha

ppen

ppen

Bear

Bear

invoke invoke Object_Specialization Object_Specialization procedure procedure An example

(29)
(30)

 Check conflict values and decide if Object_Specialization procedure should be invoked  Generate fuzzy rules.

(31)

(deffacts initial-state

(is boiling-point MORE-OR-LESS LOW) (is atom-radius NARROW)

(is metalloid MORE-OR-LESS STRONG) (is negative-charge VERY WEAK))

(defrule Rule1

?x1 <- (is ?X1 MORE-OR-LESS LOW) ?x2 <- (is ?X2 NARROW)

?x3 <- (is ?X3 MORE-OR-LESS STRONG) ?x4 <- (is ?X4 VERY WEAK)

=>

(retract ?x1 ?x2 ?x3 ?x4) (assert (is Li -1-21-3)) (assert (CF 0.8))

(printout t ”Li is -1-21-3 with CF=0.8" crlf))

C. Knowledge base generator

(export to CLIPS)

(32)

Experiment (1): Time for knowledge

elicitation and integration

Unit Group A Time (Hours) Com-pleteness Group B Time (Hours) Com-pleteness Input devices E1, E2 3.8 1 E3, E4 13.1 1 Output devices E1, E3 3.2 0.9 E2, E4 12.5 0.8 Central Processing Unit E1, E4 3.5 0.8 E2, E3 4.5 0.8 Operating systems E2, E3 3.6 1 E1, E4 4.3 1 Network Communica-tion Tools E2, E4 4.2 1 E1, E3 10.6 0.9 Multimedia Systems E , E 5.5 0.8 E , E 8.6 0.85

(33)

Experiment (2): Frequency for

correct inferences

250 test cases

Number of

successful inferences

Ratio of successful

inferences

Tutoring Rule Set

by Educator 1

217

0.868

Tutoring Rule Set

by Educator 2

224

0.896

Integrated Tutoring

Rule Set

243

0.972

Gwo-Jen Hwang (2002), “On the Development of a Cooperative Tutoring Env ironment on Computer Networks”, IEEE Transactions on System, Man and C

(34)

資訊技術應用 3 :

多目標最佳化配題機制

從大量試題中,選取符合出題條件(題數、測

試時間、概念最低配題比重...等)且鑑別

度最大的試卷

指定測驗時間範圍的試題配置問題模型

(Dedicated Range of Assessment Time Problem-DRAT)

符合期望測驗時間最高界限和最低界限的多目標配

題機制。

固定題數的試題配置問題模型

(35)

Fuzzy Art and Dynamic

Programming

Gwo-Jen Hwang (2003), “A Test Sheet Generating Algorithm for Multiple As sessment Requirements”, IEEE Transactions on Education, Vol. 46, No. 3, pp. 329-337. (SCI and EI)

(1) Fuzzy Art:

Classify test items

into groups

(2) Dynamic Programming:

Find optimal test item

composition

階段 n 決策 dn 投入狀態 sn 產出狀態s n+1 報酬函數 rn(sn,dn) 階段 n 決策 dn 投入狀態 sn 產出狀態s n+1 報酬函數 rn(sn,dn)

(36)

Heuristic Algorithms

FTF (Feasible Time First) Algorithm

Find a solution to meet

range of assessment time

Replace test items to find feasible solutions and to

maximize average discrimination degree

FNTF (Feasible Number of Test Item First)

Algorithm

Find a solution to meet

the number of test items

Replace test items to find feasible solutions and to

(37)

Genetic Algorithms

源自於 John Holland 在 1975 年出版的著作 Ad

aptation in Nature and Artificial Systems

仿效自然界生物進化過程

透過基因的選擇 (selection) 交換 (crossover) 及

突變 (mutation) 產生更好的下一代

選擇 (selection) 過程

較高合適值 (fitness value) 就有較大機會獲得保留

較低合適值的解答,可能會遭到淘汰

較不易陷入 local optimal

(38)

Population ( 族體 ):

Encoding ( 編碼 ):

Crossover ( 交配 ):

Mutation ( 突變 ):

Selection ( 適者生存 ):

Fitness Function ( 適合度公式 ):

Genetic Algorithm

(39)

基因演算法流程圖

基因演算法交配運算

基因演算法突變運算

Crossover randomly selects one-cut-point and

exchanges the right parts of two parents to

generate offspring.

Mutation alters one or more genes with a

probability equal to the mutation rate.

(40)

DRAT 目標函式:

 

Maximize Z =

DRAT 限制式: 第 i 題告第 j 個概念的關係

    

指定測驗時間範圍的試題配置問題

(Dedicated Range of Assessment Time)

n i 1

r

ij

x

i

h

j

,

j

1

,

2

,

,

m

;

n i 1

t

i

x

i

l

;

n i 1

t

i

x

i

u

;

  n i i n i i i

x

x

d

1 1

指定概念的最小

出題比重

指定測驗時間的下限

指定測驗時間的上限

指定鑑別度最大化

0

x

1

x

2

x

3

x

4

x

98

x

99

x

100

1 1 0

0 1 0

(41)

DRAT 的試題配置基因演算法 (1

/4)

概念程度下限先決基因演算法

(Concept Lower-bound First Genetic approach – CLFG)

CLFG 建立的母體 (Encoding)

X 為染色體,包含有 n 個基因

X = [x

1

, x

2

, …, x

n

]

X = [0, 0, 1, …, 0]

第 i 個試題被選取時, xi 為 1 ;否則,為 0 ;

(42)

= w

 dtu  ipt_u

dtu =

DRAT 的試題配置基因演算法 (2/

4)

適配等級 (Fitness ranking)

適配函數  v(S

k

) =

R =

K

j

=

0 if , K

j

= 1 else

= w

 dtl  ipt_l

w = (

in

d

i

x

i

) / average(u, l)

dtl =

j m j n i ij i j

r

x

K

h

)

*

(

1 1

n i

t

x

l

1

n i 1

t

i

x

i

u

 

n i i n i 1

d

i

x

i

R

1

x

in ij i j

r

x

h

1

(43)

DRAT 的試題配置基因演算法 (3

/4)

交配 (Crossover)

       A[1110011001]      A’[1110011

011

]

       B[0100100011]    

 

B’[0100100

001

]

Procedure: crossover Begin k = 0 while (k ≤ c / 2) do flag = 0 while flag = 0 do

Generate random numbers R1 and R2 from discrete interval [1,K]. If R1 ≠ R2 then flag=1 end while crossover function(R1,R2) end while Cut point

(44)

DRAT 的試題配置基因演算法 (4/

4)

突變 (Mutation)

   A[11100110

0

1]      A’[11100110

1

1]

 

P = ( 1 / n )

Procedure: mutation Begin for(i=1, i ≤ nk, i++){

Generate random number yi from discrete interval [0, 1]. Mutation function(P, yi) }

(45)

固定題數的試題配置問題

(Fixed Number of Test Items)

FNTI 目標函式:

  

Maximize Z =

FNTI 限制式:

  

x

i

≥ 1 Xi 代表題庫中的題號,最小題號為 1

x

i

n

最大題號為

n

1 ≤

i

q_num

– 1 共選出

q_num 題

num q i

d

xi _ 1

num q i j j x

h

j

m

r

i _ 1

,...,

2

,

1

,

指定鑑別度最大化

指定概念的最小出題比重

12

x

1

x

2

x

3

x

4

x

5

x

6

x

7

x

8

x

9

x

10

18 9 45 82 6 2 34 65 71

(46)

FNTI 的試題配置基因演算法 (1/

3)

試題數目先決基因演算法

(Feasible Item First Genetic approach – FIFG)

FIFG 的進行步驟

建立母體

X 為染色體,包含有 q_num 個基因

X = [x

1

, x

2

, …, x

q_num

]

 

X = [25, 118, …., 803]

基因值代表著一題試題的編號

(47)

FNTI 的試題配置基因演算法 (2/3)

交配 (Crossover)

   A[12,15, 96,112,193,243]    A’[12,15,96

,185,256,356

]

   B[3,56,108,185,256,356]    B’[3,56,108

,112,193,243

]

有兩相同基因值時,隨機更換其中一值,直到沒

有相同基因值為止

試卷中不可有二題相同的試題

Cut point

(48)

FNTI 的試題配置基因演算法 (3/

3)

突變 (Mutation)

 

A[

3,8,56,66,

256

,515

]    A’[

3,8,56,66,

346

,515

]

 

P = ( 1 / n )

Procedure: mutation Begin for (m = 1, m ≤ q_num  k, m++){

Generate random number rm from discrete interval [0, 1] Generate random number RC from discrete interval [1, n] mutation function(P, rm, RC) }

(49)

實驗題庫樣本資料

每一個情況進行二十次實驗處理後,採用平均求解時間和平均

鑑別度建立

實驗樣本

Item

Bank N Loading time(second) DiscriminationAverage

1 25 5.067 0.63267 2 30 5.308 0.65331 3 40 5.217 0.66602 4 250 8.522 0.60985 5 500 8.703 0.60920 6 1000 13.599 0.61208 7 2000 28.361 0.61339 8 4000 60.887 0.61534

(50)

CLFG 實驗結果及分析 (1/3)

l = 30

N CLFG Random Selection Optimum Solution Time(sec

) Discrimination Time(sec) Discrimination Time(min) Discrimination 25 0.13275 0.754664 0.03 0.63704 5 0.754664 30 0.14265 0.818120 0.03 0.69388 187 0.818120 40 0.27880 0.880276 0.03 0.64978 163840 0.881440 250 0.96815 0.943386 0.03 0.54248 >106 N/A 500 1.98875 0.952377 0.03 0.60500 N/A N/A 1000 3.75490 0.957359 0.03 0.69753 N/A N/A 2000 7.96650 0.956658 0.03 0.54342 N/A N/A

(51)

CLFG 實驗結果及分析 (2/3)

l = 60

N Time(sec) DiscriminationCLFG Time(sec)Random SelectionDiscrimination Time(min) DiscriminationOptimum Solution 30 0.13210 0.707622 0.03 0.64321 187 0.707622 40 0.22985 0.806201 0.03 0.63240 163840 0.806390 250 1.64565 0.924587 0.03 0.62150 >106 N/A 500 2.85260 0.942419 0.03 0.55859 N/A N/A 1000 4.36935 0.950709 0.03 0.63284 N/A N/A 2000 10.12005 0.952650 0.03 0.60746 N/A N/A 4000 27.90960 0.954701 0.03 0.62240 N/A N/A

(52)

CLFG 實驗結果及分析 (3/3)

l = 120

N CLFG Random Selection Optimum Solution Time(sec

) Discrimination Time(sec) Discrimination Time(min) Discrimination 250 2.93420 0.896015 0.03 0.59964 >106 N/A

500 4.01775 0.927922 0.03 0.66515 N/A N/A 1000 6.37270 0.940930 0.03 0.62918 N/A N/A 2000 14.80980 0.944458 0.03 0.59838 N/A N/A 4000 35.31320 0.947673 0.03 0.61402 N/A N/A

(53)

CLFG 與最佳解的實驗數據圖表

l =

30

Gwo-Jen Hwang, Bertrand M.T. Lin, Hsien-Hao Tseng, Tsung-Liang Lin (20 05), “On the Development of a Computer-Assisted Testing System with Genet ic Test Sheet-Generating Approach”, IEEE Transactions on Systems, Man, a nd Cybernetics: Part C, Vol. 35, No.4, pp. 590-594. (SCI, EI)

(54)

其他相關研究議題

Gwo-Jen Hwang, Peng-Yeng Yin and Shu-Heng Yeh (2006)

, “

A Tabu Search Approach to Generating Test Sheets for M

ultiple Assessment Criteria

”, accepted by IEEE Transactions

on Education. (SCI, EI)

Peng-Yeng Yin, Gwo-Jen Hwang, Kuang-Cheng Chang, Gw

o-Haur Hwang and Ying Chan, “

A Particle Swarm Optimiza

tion Approach to Composing Serial Test Sheets for Multiple

(55)

資訊技術應用 4 :

學習診斷與導引機制

概念影響關係圖的數學課程範例

Addition of integers Positive integers Multiplication of integers Division of integers Subtraction of integers Negative integers Zero Prime numbers

(56)

Cj C1 C2 C3 C4 C5 C6 C7 C8 Prerequisite Zero Positive integers Addition Subtrac -tion Multipli-c ation Negative integers Division Prime numbers C1 0 0 0 1 0 0 0 0 C2 0 0 1 0 0 0 0 0 C3 0 0 0 1 1 0 0 0 C4 0 0 0 0 0 0 0 0 C5 0 0 0 0 0 0 0 0 C6 0 0 0 0 0 1 1 0 C7 0 0 0 0 0 0 0 1 Ci C8 0 0 0 0 0 0 0 0 NPj 0 0 1 2 1 1 2 1

Conceptual effect table (CET)

概念影響關係表

(57)

Test item relationship table (TIRT)

試題概念關係表

Concept

C

j

Prerequisite

C1 C2 C3 C4 C5 C^ C7 C8 Q1 1 0.2 0 0 0 0 0 0 Q2 0 0.8 0.4 0 0 0 0 0 Q3 0 0 0.6 0.2 0 0 0 0 Q4 0 0 0 1 0 0 0 0 Q5 0 0 0 0 0 0 0 0 Q6 0.2 0 0 0 0.8 0.2 0 0 Q7 0 0 0 0 0 1 0 0 Q8 0 0 0 0 0 0 0.6 0.4 Q9 0 0 0 0 0.2 0 0 0 Test item

Q

i Q10 0 0 0 0 0.2 0 0.4 1

(58)

Student answer sheet table (AST)

學生答題狀況表

Test item Student Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 S1 0 0 1 0 0 1 1 0 0 0 S2 0 1 1 0 0 1 1 0 0 0 S3 0 0 0 1 0 1 1 0 0 0 S4 0 1 1 1 0 0 1 0 0 0 S5 0 0 1 0 0 0 1 1 0 0

O: The student has correctly answered the test item

1: The student failed to correctly answer the test item

(59)

建立補救學習路徑

C

1

C

3

C

6

C

1

C

3

C

7

C

1

C

4

C

8

C

10

C

2

C

4

C

8

C

10 0.16 0.6 0.66 0.16 0.6 1 0.16 0.16 0 0 0 0.16 0 0

C

2

C

5

C

9 0 0.28 0

假設

θ

值為

0.3 ,代

表我們對於某個概念

之錯誤比率的最大容

忍程度為

30 ,依

上述原則,可得到補

救學習路徑:

PATH1 : C

3

C

6

PATH2 : C

3

C

7

(60)
(61)

歸納式概念影響關係演算法

歸納式概念關係演算法產生的概念繼承關係圖

(Support = 0.1 & Belief=0.7 數學科 )

加法

角度

直角

分數

整數

分子

倍數

除法

乘數

乘法

0.78 0.7 1 0.7 0.7 1 0.7 0.84 1 0.84 0.92 0.89 0.89 0.73

加法

角度

直角

分數

整數

分子

倍數

除法

乘數

乘法

0.78 0.7 1 0.7 0.7 1 0.7 0.84 1 0.84 0.92 0.89 0.89 0.73

(62)

教師修改後的概念繼承關係圖

(Support = 0.1 & Belief=0.7 數學科 )

加法

角度

直角

整數

倍數

除法

乘數

乘法

0.78 0.7 1 0.7 0.7 1 0.7 0.84 0.92 0.89

加法

角度

直角

整數

倍數

除法

乘數

乘法

0.78 0.7 1 0.7 0.7 1 0.7 0.84 0.92 0.89

(63)
(64)
(65)

其他資訊技術在數位學習的應用

Gwo-Jen Hwang (2006), “

A Gray Forecast Approach for De

veloping Testing and Diagnostic Systems

”, accepted by

IEE

E Transactions on System, man and Cybernetic Part C

. (SC

I and EI)

Gwo-Jen Hwang, Tong C.K. Huang and Judy C.R. Tseng (2

004), “

A Group-Decision Approach for Evaluating Education

al Web Sites

”,

Computers & Education

, Vol. 42, No. 1, pp.

65-86. (SSCI)

黃國禎、陳佐霖、王姿婷、曾秋蓉、黃國豪

(2005) “

線上

自律學習輔助系統之研究與實證

, 投稿 科學教育學刊

黃國豪、曾秋蓉、黃國禎、黃繼緯、林農堯

, “

具自我調適

(66)

PART B – 數位學習標準及工

教材標準及工具的研究 (eg. SCORM)

測驗標準及工具的研究 (eg. QTI)

學習歷程標準及工具的研究

學習設計標準及工具的研究

The 2006 International Conference on SCOR

M (SCORM'2006)

Tamkang University, Taipei, Taiwan

January 17 - 19, 2006

(67)

PART C – 行動學習與普化學

記錄學生在實境學習環境中的學習歷程,分析學

生即時解決實境問題的能力,並規劃學習活動。

提出教學活動的設計策略,提高學生實境探索、

分析、解決問題的能力,並促成同儕合作。

建立教學元件及學習歷程標準規範與管理策略。

IEEE SUTC2006 International Workshop on Cont

ext Aware Ubiquitous Learning

(CAUL2006)

June 5, 2006, Taichung, Taiwan

(68)

數位學習研究的實驗設計

比較答案的品質及演算法的效率 - 以大量資料模擬

測試

瞭解系統的滿意度及使用意願 - 問卷調查

驗證系統或方法的效果 - 以實驗組及對照組進行

3-6 個月的測試及分析 ( 前測及後測 )

比較新系統 ( 方法 ) 與舊系統 ( 方法 ) 的效果 - 多

人交叉使用兩系統 ( 方法 ) 並比較結果

比較對象:舊系統 ( 方法 ) 、使用與未使用、 Rand

(69)

投稿的的期刊

國內期刊

如資訊管理學報 (TSSCI) 、師大學報 (TSSC

I) 、科學教育學刊、科學教育研究

國際期刊

SCI/SSCI: Computers & Education, IEEE Tran

sactions on Education, IEEE Transactions on S

MC part C, Educational Technologies and Soci

ety, Interacting with Computers

(70)

期刊的選擇

- 應用與教學策

Interacting with Computers

(SSCI, SCI, EI)

Computers & Education

(SSCI, SCI Expanded, EI)

Computers-Assisted Learning

(SSCI)

一般電腦應用

( 偏 Computers, 但接受教育的應用 )

電腦在教育上的應用

(Computers & Education 各半 )

電腦輔助學習

(71)

期刊的選擇

- 數位學習技術

IEEE Transactions on Systems,

Man and Cybernetics Part C

(SCI, EI)

Journal of Distance Education

Technology(EI)

Educational Technology

& Society (SSCI)

人工智慧技術的應用

( 偏技術 , 但接受教育的應用 )

資訊技術在線上學習的應用

(Computers & Education 各半 )

教育技術,但也接受資訊技

術的應用

( 偏 Education)

(72)

結論

數位學習已成為普遍性的研究

教育、資訊、工程、管理、社會

資訊科技的普及加怏數位學習發展的腳步

進入數位學習領域必須先瞭解需求

動機(需要)

+ 技術導入 + 應用及實驗

參考文獻

相關文件

In another word, the initial state description is the conjunct of the precondition of the task and the guard condition of the task’s method, and the state descriptions are

Finally, we train the SLU model by learning latent feature vectors for utterances and slot candidates through MF techniques. Combining with a knowledge graph propagation model based

- Effective use of a variety of texts (e.g. information texts) to encourage deep reading, connect students’ learning experience and broaden their knowledge base. -

stating clearly the important learning concepts to strengthen the coverage of knowledge, so as to build a solid knowledge base for students; reorganising and

Keywords: Requesting Song, Information Retrieval, Knowledge Base, Fuzzy Inference, Adaptation Recommendation System... 致

• Environmental Report 2020 of Transport Department, Hong Kong: to provide a transport system in an environmentally acceptable manner to align with the sustainable development of

● the F&amp;B department will inform the security in advance if large-scaled conferences or banqueting events are to be held in the property.. Relationship Between Food and

– Knowledge to form the basis for decision aids – Knowledge that reveals underlying skills..