CHAPTER VII
"TECHNIQUE PROCESSING OF SCORING THE LEARNING OUTCOMES"
Step of changing in raw scores into a number value in doing statistically with the producer.
1. Preparation of class interval
Suppose a physics teacher raw scores that is obtained from the test results that have been awarded to 30 students as follows.
80 36 68 58 60 54 82 72 78 60
72 76 73 54 65 70 40 64 52 60
45 84 59 60 62 70 48 78 57 73
a. Determine the maximum score and minimum scores
Minimum score of 36
Maximum score of 84
b. Looking for range (distance measurement between maximum score and minimum scores)
Range = maximum score - minimum score
Range = 84-36 = 48
c. Determine the magnitude of the interval class using Sturges formula as follows:
K = 1 + 3.3 log n
K = total of classes is sought
1 = number remains
n = total of students are tested
K = 1 + 3,3 log n
= 1 + 3,3 . 1,48 = 5,87 ~ 5
d. Determine the extent of the interval
Range 48
Formula: I = = = 9,6 ~ 10
Class i 5
e. Looking for the number of initial class that the lowest - i "which is a multiple of i and minimum scores are included in the lowest class i. In this case the initial numbers is the lowest i class of 35 which is a multiple of i. So, the lowest interval class of is 35-44.
f. Arrange class of interval until the maximum score included.
g. After class of interval the next column arranged by entering each interval score of students into class of by giving a single stroke for the students in the tally column.
h. Calculate the frequency for each class of interval by calculating the total of streaks
Table. Frequency distribution of the Physic result test
Class of interval
Tally
f
X
Fx
x
(X-)
x2
fx2
35 - 44
II
2
39,5
79
-23,67
560,27
1020,54
45 - 54
IIII
5
49,5
247,5
-13,67
186,87
934,35
55 - 64
IIII IIII
9
59,5
535,5
-3,67
13,47
121,23
65 - 74
IIII III
8
69,5
556
6,33
40,07
320,56
75 - 84
IIII I
6
79,5
477
16,33
266,67
1600,02
30
1895
Σfx2 =3996,70
2. Calculate the mean (arithmetic average)
a. Determine the central figure by adding the upper and lower bounds for each class interval and then divided by two.
35 + 44 79
X = = = 39,5
2 2
b. Calculate fx for each class interval that is calculated how much frequency is then multiplied by the middle number.
fx
c. Calculate the mean with the formula =
N
1895
M = = 63,17
30
3. Calculate the standard deviation (SD)
a. Calculate x for each class interval
x = the middle number - mean
x = X - x
b. Calculates the square of x in each class interval.
c. fX2 calculate for each class interval and Σ fX2
d. Calculate the standard deviation (SD) by the formula :
Σ fx2 3996,70
SD = = = 133,22 = 11,5
n 30
4. Determining the value of 1-10 with formula translation.
The translation formula
Calculation results
Elaboration
+ 2,25 SD = 10
+ 1,75 SD = 9
+ 1,25 SD = 8
+ 0,75 SD = 7
+ 0,25 SD = 6
- 0,25 SD = 5
- 0,75 SD = 4
- 1,25 SD = 3
- 1,75 SD = 2
- 2,25 SD = 1
63,17 + 2,25 . 11,54 = 89,14
63,17 + 1,75 . 11,54 = 83,37
63,17 + 1,25 . 11,54 = 77,59
63,17 + 0,75 . 11,54 = 71,83
63,17 + 0,25 . 11,54 = 66,06
63,17 - 0,25 . 11,54 = 60,29
63,17 - 0,75 . 11,54 = 54,52
63,17 - 1,25 . 11,54 = 48,75
63,17 - 1,75 . 11,54 = 42,98
63,17 - 2,25 . 11,54 = 37,21
Score > 89 = 10
83 - 88 = 9
78 - 82 = 8
72 - 76 = 7
66 - 71 = 6
60 - 65 = 5
55 - 59 = 4
49 - 54 = 3
43 - 48 = 2
37 - 42 = 1
< 36 = 0
With the translation of the above guidelines, teachers now live transfer or convert the raw scores obtained by each student into grades 1-10. The advantage of assessment system such as this, that the values obtained by the students truly reflect the capacity of the group. The disadvantage is that values are obtained by the system does not reflect the extent to which the achievement of the scope of materials tested lessons.
Another system that can be used in the assessment of student learning outcomes is to use percentages correction or assessment conducted on percentage. The value obtained by the students is a percentage of the overall scope of lessons to be tested.
Valuation formula is as follows:
R
S = x SM
N
Specification S = value sought
R = raw score obtained by the students
N = ideal maximum score of the test in question
SM = standard mark (desired rating scale)
For example, 1-10 or 1-100
From examples of physics data score earlier, when a student receives a score of 68 by using the percentage correction, then the value obtained is:
R 68
S = x SM = x 10 = 6,8
N 100
Assessment using the percentage correction is easily implemented by any teacher, more practical and reliable.
EXERCISE
Presented data from daily test result of physics from 10 students with a score of 80, 63, 52, 60, 76, 76, 68, 79, 58, 72. Find the value of each student by way of correction percentages, when the ideal score of 90 and a rating scale 1-10.
I'm sure you have understood the material regarding the approach in the assessment of learning outcomes. Now you can do the following exercises in the workbook and try disciplined, do not see the answers until you have finished the exercise.
EXERCISE
ANSWER KEY
Assessment with correction percentages using the formula
R
S = x SM
N
52 70
Score 52 S = x 10 = 5,7 Score 70 S = x 10 = 7,8
90 90
58 72
58 S = x 10 = 6,4 Score 72 S = x 10 = 8
90 90
60 76
60 S = x 10 = 6,7 Score 76 S = x 10 = 8,4
90 90
63 76
63 S = x 10 = 7,0 Score 76 S = x 10 = 8,4
90 90
68 80
68 S = x 10 = 7,6 Score 80 S = x 10 = 8,9
90 90
"TECHNIQUE PROCESSING OF SCORING THE LEARNING OUTCOMES"
Step of changing in raw scores into a number value in doing statistically with the producer.
1. Preparation of class interval
Suppose a physics teacher raw scores that is obtained from the test results that have been awarded to 30 students as follows.
80 36 68 58 60 54 82 72 78 60
72 76 73 54 65 70 40 64 52 60
45 84 59 60 62 70 48 78 57 73
a. Determine the maximum score and minimum scores
Minimum score of 36
Maximum score of 84
b. Looking for range (distance measurement between maximum score and minimum scores)
Range = maximum score - minimum score
Range = 84-36 = 48
c. Determine the magnitude of the interval class using Sturges formula as follows:
K = 1 + 3.3 log n
K = total of classes is sought
1 = number remains
n = total of students are tested
K = 1 + 3,3 log n
= 1 + 3,3 . 1,48 = 5,87 ~ 5
d. Determine the extent of the interval
Range 48
Formula: I = = = 9,6 ~ 10
Class i 5
e. Looking for the number of initial class that the lowest - i "which is a multiple of i and minimum scores are included in the lowest class i. In this case the initial numbers is the lowest i class of 35 which is a multiple of i. So, the lowest interval class of is 35-44.
f. Arrange class of interval until the maximum score included.
g. After class of interval the next column arranged by entering each interval score of students into class of by giving a single stroke for the students in the tally column.
h. Calculate the frequency for each class of interval by calculating the total of streaks
Table. Frequency distribution of the Physic result test
Class of interval
Tally
f
X
Fx
x
(X-)
x2
fx2
35 - 44
II
2
39,5
79
-23,67
560,27
1020,54
45 - 54
IIII
5
49,5
247,5
-13,67
186,87
934,35
55 - 64
IIII IIII
9
59,5
535,5
-3,67
13,47
121,23
65 - 74
IIII III
8
69,5
556
6,33
40,07
320,56
75 - 84
IIII I
6
79,5
477
16,33
266,67
1600,02
30
1895
Σfx2 =3996,70
2. Calculate the mean (arithmetic average)
a. Determine the central figure by adding the upper and lower bounds for each class interval and then divided by two.
35 + 44 79
X = = = 39,5
2 2
b. Calculate fx for each class interval that is calculated how much frequency is then multiplied by the middle number.
fx
c. Calculate the mean with the formula =
N
1895
M = = 63,17
30
3. Calculate the standard deviation (SD)
a. Calculate x for each class interval
x = the middle number - mean
x = X - x
b. Calculates the square of x in each class interval.
c. fX2 calculate for each class interval and Σ fX2
d. Calculate the standard deviation (SD) by the formula :
Σ fx2 3996,70
SD = = = 133,22 = 11,5
n 30
4. Determining the value of 1-10 with formula translation.
The translation formula
Calculation results
Elaboration
+ 2,25 SD = 10
+ 1,75 SD = 9
+ 1,25 SD = 8
+ 0,75 SD = 7
+ 0,25 SD = 6
- 0,25 SD = 5
- 0,75 SD = 4
- 1,25 SD = 3
- 1,75 SD = 2
- 2,25 SD = 1
63,17 + 2,25 . 11,54 = 89,14
63,17 + 1,75 . 11,54 = 83,37
63,17 + 1,25 . 11,54 = 77,59
63,17 + 0,75 . 11,54 = 71,83
63,17 + 0,25 . 11,54 = 66,06
63,17 - 0,25 . 11,54 = 60,29
63,17 - 0,75 . 11,54 = 54,52
63,17 - 1,25 . 11,54 = 48,75
63,17 - 1,75 . 11,54 = 42,98
63,17 - 2,25 . 11,54 = 37,21
Score > 89 = 10
83 - 88 = 9
78 - 82 = 8
72 - 76 = 7
66 - 71 = 6
60 - 65 = 5
55 - 59 = 4
49 - 54 = 3
43 - 48 = 2
37 - 42 = 1
< 36 = 0
With the translation of the above guidelines, teachers now live transfer or convert the raw scores obtained by each student into grades 1-10. The advantage of assessment system such as this, that the values obtained by the students truly reflect the capacity of the group. The disadvantage is that values are obtained by the system does not reflect the extent to which the achievement of the scope of materials tested lessons.
Another system that can be used in the assessment of student learning outcomes is to use percentages correction or assessment conducted on percentage. The value obtained by the students is a percentage of the overall scope of lessons to be tested.
Valuation formula is as follows:
R
S = x SM
N
Specification S = value sought
R = raw score obtained by the students
N = ideal maximum score of the test in question
SM = standard mark (desired rating scale)
For example, 1-10 or 1-100
From examples of physics data score earlier, when a student receives a score of 68 by using the percentage correction, then the value obtained is:
R 68
S = x SM = x 10 = 6,8
N 100
Assessment using the percentage correction is easily implemented by any teacher, more practical and reliable.
EXERCISE
Presented data from daily test result of physics from 10 students with a score of 80, 63, 52, 60, 76, 76, 68, 79, 58, 72. Find the value of each student by way of correction percentages, when the ideal score of 90 and a rating scale 1-10.
I'm sure you have understood the material regarding the approach in the assessment of learning outcomes. Now you can do the following exercises in the workbook and try disciplined, do not see the answers until you have finished the exercise.
EXERCISE
ANSWER KEY
Assessment with correction percentages using the formula
R
S = x SM
N
52 70
Score 52 S = x 10 = 5,7 Score 70 S = x 10 = 7,8
90 90
58 72
58 S = x 10 = 6,4 Score 72 S = x 10 = 8
90 90
60 76
60 S = x 10 = 6,7 Score 76 S = x 10 = 8,4
90 90
63 76
63 S = x 10 = 7,0 Score 76 S = x 10 = 8,4
90 90
68 80
68 S = x 10 = 7,6 Score 80 S = x 10 = 8,9
90 90
