boris milašinović faculty of electrical engineering and computing university of zagreb, croatia

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Some experiences using various assessments methods with the emphasis on the automatic evaluation of (programming) assignments Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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Continuous assessment Mid term exams and final exam Periodical small tests Homeworks In class assessment Problems: Manual reviewing process is usually more precise but time consuming Lack of teaching staff Solution: Using software for automatic evaluation of assignments Complete automatic reviewing would be to strict  Mid term and final exams reviewed manually  Homeworks and quizzes (multiple choice tests) with automatic evaluation 3

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Page 1: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

Some experiences using various assessments methods with the

emphasis on the automatic evaluation of (programming)

assignments Boris Milašinović

Faculty of Electrical Engineering and Computing

University of Zagreb, Croatia

Page 2: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

2

MotivationCourses

Programming (and software engineering)Data Structures and Algorithms>= 700 students on each course

Old examination system2 mid term examsClassic written exam as an option=> no continuous assessment!

How to introduce continuous assessments?

Page 3: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

3

Continuous assessmentMid term exams and final examPeriodical small testsHomeworksIn class assessmentProblems:

Manual reviewing process is usually more precise but time consuming

Lack of teaching staffSolution:

Using software for automatic evaluation of assignments Complete automatic reviewing would be to strict

Þ Mid term and final exams reviewed manuallyÞ Homeworks and quizzes (multiple choice tests) with automatic evaluation

Page 4: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

4

Grading scheme (1 point = 1% of final points)Programming and

software engineering1st mid-term exam: 15

(points)2nd mid-term exam: 25 Final exam : 30Homeworks: 3 x 2 = 6Quizzes: 6 x 3 = 18Class activity: 6

Algorithms and Data Structures1st mid-term exam: 152nd mid-term exam: 20Final exam : 30Homeworks: 3 x 3 = 9Quizzes: 6 x 3 = 18Class activity: 8

To pass a student must collect at least 50 points and at least 8 (of 30) points on the final exam

• Grades are awarded by Gaussian distribution

Page 5: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

5

Multiple-choice tests (Quizzes)30 minutes, 12 questions, 5 possible answers (only 1 is

correct) Programming and software engineering

6 quizzes (3 points per quiz) Correct answer: 0.25 points Incorrect: -0.05

Algorithms and Data Structures: 3 quizzes (6 points per quiz) Correct answer: 0.5 points Incorrect: -0.1

1 point is equal to 1% of final score

In order to retain the initial quality new questions had to be added every year to enlarge the questions database.

Time span between the assessments should be reduced (all students should do the test in the same day)

Page 6: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

6

Multiple-choice tests (Quizzes) – Histograms of results in 2007/08

Histogram (P iP I2007Z 16v*861c)

Quiz 1 = 859*0,5*norm al(x; 2,1742; 0,6827)

-1,0 -0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Q uiz 1

0

50

100

150

200

250

300

350

No

of o

bs

His togram (PiPI2007Z 16v*861c)

Quiz 2 = 861*0,5*norm al(x; 1,858; 0,7399)

-1,0 -0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Quiz 2

0

20

40

60

80

100

120

140

160

180

200

220

240

260

No

of o

bs

Histogram (P iPI2007Z 16v*861c )

Quiz 3 = 861*0,5*norm al(x; 1,7513; 0,7998)

-1,0 -0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Q uiz 3

0

20

40

60

80

100

120

140

160

180

200

220

240

No

of o

bs

His togram (P iP I2007Z 16v*861c)

Quiz 4 = 861*0,5*norm al(x; 2,1584; 0,8334)

-1,0 -0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Q uiz 4

0

50

100

150

200

250

300

350

400

No

of o

bs

His togram (PiP I2007Z 16v*861c )

Q uiz 5 = 861*0,5*norm al(x; 1,7771; 0,8729)

-1,0 -0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Q uiz 5

0

20

40

60

80

100

120

140

160

180

200

220

240

260

280

No

of o

bs

His togram (PiPI2007Z 16v*861c )

Q uiz 6 = 861*0,5*norm al(x; 1,6478; 0,8756)

-1,0 -0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Quiz 6

0

20

40

60

80

100

120

140

160

180

200

220

240

260

No

of o

bs

His togram (ASP_2007 13v*729c )

Q uiz 1 = 729*0,5*norm al(x; 4,3102; 1,3663)

-1 0 1 2 3 4 5 6 7

Q uiz 1

0

20

40

60

80

100

120

140

160

No

of o

bs

His tog ram (ASP_2007 13v*729c )

Q uiz 2 = 729*1*norm al(x; 4,1034; 1,5903)

-2 -1 0 1 2 3 4 5 6 7

Quiz 2

0

20

40

60

80

100

120

140

160

180

200

220

240

260

No

of o

bs

His tog ram (ASP_2007 13v*729c)

Q uiz 3 = 729*0,5*norm al(x; 3,5207; 1,7958)

-1 0 1 2 3 4 5 6 7

Q uiz 3

0

20

40

60

80

100

No

of o

bs

Programming and software engineering

Algorithms and Data Structures

Page 7: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

7

Suitability of Multiple-choice tests (Quizzes)Advantages:

Simple to create and easy to runQuick (quiz lasts 30 minutes)Has good results distributionHelps achieving continuous assessment

Disadvantages: Backwash effect:

Students tend to study the matter needed to pass the test, not the matter representing the core knowledge being thought

Does not represent real knowledge of programming! Students learn how to recognize answers.

Page 8: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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HomeworksStudents are given either whole programs or

individual functions as programming assignmentsOn upload, students' code is joined with the code

previously defined by the teachers and compiled.Upon the successful compilation, program is run

against the predefined tests and its output is compared with the expected results.Tests with fixed set dataRandomly generated inputs

Time to collect the assignment: 7-10 daysTime to solve: 2 days from the time of assignment

collection

Page 9: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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Avoiding ambiguities in homeworks test definitionsSwitching to automatic evaluation can bring

problems. The lack of experience (on behalf of students and

teachers)Avoiding ambiguities in test definitionsExams for automatic evaluation must be precisely

definedFirst test with automatic evaluation:

300 received e-mails with complaints on automatic evaluation process (more than 40%)

Now: less than 40 complaints per test (5%) => mostly with request for explanation of student’s solution errors

Page 10: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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Homeworks - Histograms

His togram (ASP_20 07 13v*729 c)

Hom ework 1 = 729*0,5*norm al(x; 2,3047; 0,9892)

-0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Hom ework 1

0

50

100

150

200

250

300

350

400

450

500

No

of o

bs

His tog ram (ASP_20 07 13v*729 c)

Hom ework 2 = 729*0,5*norm al(x; 2,4412; 0,9108)

-0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Hom ework 2

0

100

200

300

400

500

600

No

of o

bs

Histog ram (ASP_20 07.sta 13v*729c)

Hom ework 3 = 729*0,5*norm al(x; 2,4295; 1,0057)

-0,5 0,0 0,5 1,0 1,5 2,0 2,5 3,0 3,5

Hom ew ork 3

0

100

200

300

400

500

600

No

of o

bs

Histogram (P iPI2007Z.s ta 16v*861c)

Hom ework 1 = 861*0,2*norm al(x; 1,7321; 0,6197)

-0,4 -0,2 0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0 2,2

Hom ework 1

0

100

200

300

400

500

600

700

800

No

of o

bs

Histogram (PiPI2007Z.s ta 16v*861c)

Hom ework 2 = 861*0,2*norm al(x; 1,6099; 0,7232)

-0, 4 -0,2 0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0 2,2

Hom ew ork 2

0

100

200

300

400

500

600

700

No

of o

bs

Histogram (PiPI2007Z.s ta 16v*861c )

Hom ew ork 3 = 861*0,2*norm al(x; 1,4473; 0,8636)

-0,4 -0,2 0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0 2,2

Hom ework 3

0

100

200

300

400

500

600

700

No

of o

bs

Page 11: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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Suitability of automatically evaluated homeworksAdvantages

Improve students programming skills More time to create assignments, but no manual review

evaluation laterDisadvantages

Minor error can lead to zero points for the assignment More tests refines the grading scale Running pre-tests examples before the final submission

Poor distribution (although not unusual for homeworks) Easy to cheat

Suspicious situations: similar solutions too short time between assignment collection and submission

Page 12: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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Correlation between results on each assignment and student’s final grade – Algorithms and Data Structures (2007/08)

1st mid-term exam

2nd mid-term exam

Final exam

Quiz 1 Quiz 2 Quiz 3 Homework 1

Homework 2

Homework 3

Classroom activity

Total Grade

1st mid-term exam   0,67 0,64 0,53 0,51 0,37 0,14 0,24 0,31 0,45 0,80 0,75

2nd mid-term exam 0,67   0,68 0,47 0,55 0,37 0,12 0,22 0,31 0,48 0,84 0,80

Final exam 0,64 0,68   0,52 0,58 0,52 0,18 0,25 0,34 0,53 0,90 0,86

Quiz 1 0,53 0,47 0,52   0,53 0,41 0,20 0,27 0,39 0,42 0,66 0,56

Quiz 2 0,51 0,55 0,58 0,53   0,49 0,19 0,28 0,40 0,47 0,72 0,62

Quiz 3 0,37 0,37 0,52 0,41 0,49   0,21 0,29 0,44 0,41 0,61 0,51

Homework 1 0,14 0,12 0,18 0,20 0,19 0,21   0,23 0,26 0,11 0,26 0,21

Homework 2 0,24 0,22 0,25 0,27 0,28 0,29 0,23   0,40 0,26 0,37 0,27

Homework 3 0,31 0,31 0,34 0,39 0,40 0,44 0,26 0,40   0,28 0,48 0,31

Classroom activity 0,45 0,48 0,53 0,42 0,47 0,41 0,11 0,26 0,28   0,66 0,62

Total 0,80 0,84 0,90 0,66 0,72 0,61 0,26 0,37 0,48 0,66   0,92

Grade 0,75 0,80 0,86 0,56 0,62 0,51 0,21 0,27 0,31 0,62 0,92  

Page 13: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

13

E1

E2

FE

Q1

Q2

Q3

Q4

Q5

Q6

H1

H2

H3

CA

T GR

1st mid-term exam 1,00 0,65 0,59 0,43 0,40 0,55 0,44 0,46 0,40 0,17 0,12 0,20 0,28 0,76 0,69

2nd mid-term exam 0,65 1,00 0,69 0,48 0,48 0,61 0,49 0,56 0,50 0,16 0,12 0,20 0,40 0,87 0,85

Final exam 0,59 0,69 1,00 0,52 0,55 0,64 0,59 0,68 0,64 0,18 0,17 0,33 0,48 0,91 0,87

Quiz 1 0,43 0,48 0,52 1,00 0,55 0,56 0,50 0,48 0,51 0,24 0,20 0,29 0,41 0,62 0,55

Quiz 2 0,40 0,48 0,55 0,55 1,00 0,59 0,57 0,55 0,55 0,21 0,22 0,34 0,50 0,64 0,57

Quiz 3 0,55 0,61 0,64 0,56 0,59 1,00 0,57 0,62 0,59 0,22 0,21 0,32 0,42 0,74 0,68

Quiz 4 0,44 0,49 0,59 0,50 0,57 0,57 1,00 0,70 0,65 0,29 0,29 0,43 0,51 0,68 0,56

Quiz 5 0,46 0,56 0,68 0,48 0,55 0,62 0,70 1,00 0,70 0,24 0,24 0,41 0,53 0,74 0,66

Quiz 6 0,40 0,50 0,64 0,51 0,55 0,59 0,65 0,70 1,00 0,23 0,25 0,42 0,51 0,70 0,59

Homework 1 0,17 0,16 0,18 0,24 0,21 0,22 0,29 0,24 0,23 1,00 0,24 0,25 0,25 0,27 0,20

Homework 2 0,12 0,12 0,17 0,20 0,22 0,21 0,29 0,24 0,25 0,24 1,00 0,22 0,24 0,24 0,16

Homework 3 0,20 0,20 0,33 0,29 0,34 0,32 0,43 0,41 0,42 0,25 0,22 1,00 0,32 0,38 0,26

Class activity 0,28 0,40 0,48 0,41 0,50 0,42 0,51 0,53 0,51 0,25 0,24 0,32 1,00 0,57 0,50

Total number of points 0,76 0,87 0,91 0,62 0,64 0,74 0,68 0,74 0,70 0,27 0,24 0,38 0,57 1,00 0,93

Grade 0,69 0,85 0,87 0,55 0,57 0,68 0,56 0,66 0,59 0,20 0,16 0,26 0,50 0,93 1,00

Programming and software engineering (2007/08)

Page 14: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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Prediction of grades after 1/3 of the semester

Classification Matrix (Algorithms and Data structures-2007/08)Rows: Grades Columns: Predicted grades

Percent 1 2 3 4 51 72,73% 112 0 38 4 02 0% 25 0 52 15 03 52,55% 23 2 103 67 14 68,78% 1 0 51 130 75 39,80% 0 0 0 59 39

Total 52,67% 161 2 244 275 47

1 homework

1 quiz1 mid-term

exam

1 homework

2 quizzes1 mid-term

exam

Classification Matrix (Programming and software enginereeing-2007/08) Rows: Grades Columns: Predicted grades

Percent 1 2 3 4 51 76,76% 218 0 60 6 02 0% 33 0 45 9 03 51% 35 0 102 63 04 73,13% 3 0 48 147 35 11,49% 0 0 2 75 10

Total 55,53% 289 0 257 300 13

Page 15: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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Prediction of grades after 2/3 of the semester

Classification Matrix (Algorithms and Data structures-2007/08)Rows: Grades Columns: Predicted grades

Percent 1 2 3 4 51 75,32% 116 19 18 1 02 17,39% 17 16 56 3 03 76,53% 4 7 150 35 04 69,31% 0 1 41 131 165 62,24% 0 0 0 37 61

Total 65,02% 137 43 265 207 77

2 homeworks

2 quizzes2 mid-term

exam

2 homeworks

4 quizzes2 mid-term

exam

Classification Matrix (Programming and software enginereeing-2007/08) Rows: Grades Columns: Predicted grades

Percent 1 2 3 4 51 86,27% 245 7 28 4 02 14,95% 29 13 44 1 03 72,5% 15 3 145 37 04 75,62% 0 0 36 152 135 68,97% 0 0 0 27 60

Total 71,59% 289 23 253 221 73

Page 16: Boris Milašinović Faculty of Electrical Engineering and Computing University of Zagreb, Croatia

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ConclusionWhy is prediction not more accurate?

Extremely poor prediction for grade 2! Reason: Significant number of students learn just to pass Gaussian distribution of grades

15%-35%-35%-15%Nevertheless: Very good prediction about the number of

students that will pass the exam

What about various assessments methodsRequires more work but covers more aspects of

assessmentEliminates backwash effectContinuous assessment