�ݺ�ߣ

PENILAIAN HASIL BELAJAR
guna memenuhi tugas mata kuliah Penilaian Hasil Belajar
Dosen pengampu : Isna Farahsanti, S.pd, M.Pd

Disusun Oleh :
1. Dewi Ria D.A.

1051500074

2. Diyah Sri Hariyanti

1051500083

3. Ernia Ardiati

1051500097

4. Siti Lestari

1051500102

5. Yuliana Asriningrum

1051500104

PROGRAM STUDI PENDIDIKAN MATEMATIKA
FAKULTAS KEGURUAN DAN ILMU PENDIDIKAN
UNIVERSITAS VETERAN BANGUN NUSANTARA SUKOHARJO
2012

No
Resp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
KJ

Nomor Butir Soal
1
A
A
B
A
B
E
A
C
D
A
A
C
D
C
A
A
A
B
A
A
A

L
W
L
W
L
W
L
W
L
W
L
W
L
W
L
W
L
W
L
W

No
Resp

2
B
B
A
B
B
C
D
B
E
B
B
B
C
E
B
A
B
A
B
A
B

3
A
C
B
A
C
D
C
C
D
A
E
C
B
C
A
B
A
B
C
C
C

4
B
D
C
B
A
B
E
B
A
A
B
A
B
A
B
A
B
C
D
A
A

5
C
A
D
C
B
C
A
C
E
A
C
E
D
C
C
B
C
D
A
C
C

6
D
E
E
D
C
D
B
A
E
E
C
E
A
B
D
C
D
A
E
E
E

7
E
E
E
A
D
E
C
B
A
B
B
A
E
A
E
D
A
E
E
B
E

8
A
A
A
E
A
A
D
C
B
C
A
E
A
B
A
A
D
A
B
A
A

9
B
B
B
B
E
B
A
D
C
B
B
A
B
B
B
E
B
D
E
B
B

10
C
D
A
D
D
C
E
A
D
C
C
B
A
D
C
B
D
D
E
D
D

11
D
E
D
C
A
D
E
E
A
D
D
C
B
D
D
E
D
B
D
E
D

12
A
C
A
A
B
A
A
A
E
A
A
B
C
B
A
A
B
C
A
D
A

13
B
C
D
B
A
B
B
B
B
B
E
A
D
C
B
B
A
B
D
B
B

14
C
D
D
A
B
C
C
C
D
C
D
E
A
D
C
C
B
A
A
D
D

15
A
E
A
B
C
A
D
D
B
A
B
B
E
A
A
D
C
B
A
B
B

16
D
A
B
C
D
D
A
A
D
D
D
D
D
D
D
A
D
C
B
A
D

17
E
B
C
D
A
E
E
E
E
E
A
E
E
A
E
E
A
D
C
B
E

18
A
C
D
A
E
A
E
A
B
A
A
A
A
B
A
A
E
A
D
C
A

19
B
D
A
E
B
B
B
E
B
B
A
A
B
C
B
B
B
E
A
D
B

20
A
A
E
A
E
A
A
B
C
A
A
B
C
D
A
A
A
D
E
A
A

21
B
E
A
C
C
B
C
C
C
B
B
C
D
A
B
A
C
C
A
E
C

22
C
E
D
C
D
C
C
C
C
C
C
D
A
E
C
C
C
A
B
C
C

23
D
A
D
D
D
D
C
D
C
D
D
A
E
D
D
D
A
B
C
D
D

24
E
D
A
E
E
E
C
A
E
E
A
E
E
E
E
A
B
C
D
E
E

25
A
C
C
C
E
A
C
C
C
B
E
D
C
C
A
B
C
D
A
C
C

26
B
B
B
C
B
B
B
E
B
B
E
B
C
B
B
C
D
A
E
B
B

27
C
E
E
A
E
E
E
D
D
C
D
E
C
E
C
D
A
E
E
A
E

28
D
D
A
D
D
D
B
D
D
D
B
E
D
D
D
A
E
D
D
A
D

29
A
A
E
E
E
A
E
E
E
B
E
D
C
E
B
E
E
E
A
E
E

Nomor Butir Soal

1

L

1
1

2
1

3
0

4
0

5
1

6
0

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
1 1 1 0 1 1 1 0 0 1 1 1 1 1 0 1 1 1 0 1 0 1 0 0

2
3
4

W
L
W

1
0
1

1
0
1

1
0
0

0
0
0

0
0
1

1
1
0

1
1
0

1
1
0

1 1
1 0
1 1

0 0
1 1
0 1

0 1 0 0 0 0 0 1 0
0 1 0 0 0 0 0 0 0
1 0 1 0 0 1 0 1 1

0 0 0 1 1
0 1 0 1 1
1 1 1 1 0

1
1
0

1 0 1
0 1 1
1 1 0

5
6

L
W

0
0

1
0

1
0

1
0

0
1

0
0

0
1

1
1

0 1
1 0

0 0
1 1

0 0 0 1 0 0 1 0 1
1 0 0 1 1 1 1 1 0

0 1 1 0 1
1 1 1 0 1

1
1

1 1 1
1 0 1

7
8
9

L
W
L

1
0
0

0
1
0

1
1
0

0
0
1

0
1
0

0
0
1

0
0
0

0
0
0

0 0
0 0
0 1

0 1
0 1
0 0

1 0 0 0 1 0 1 1 1
1 0 0 0 1 1 0 0 1
1 1 1 1 1 0 1 0 1

1 0 0 1 1
1 1 0 1 0
1 0 1 1 1

1
0
0

0 1 0
1 1 1
1 1 1

10
11
12

W
L
W

1
1
0

1
1
1

0
0
1

1
0
1

0
1
0

1
0
1

0
0
0

0
1
0

1 0
1 0
0 0

1 1
1 1
0 0

1 0 0 1 1 1 1 1 0
0 1 1 1 0 1 0 1 0
0 0 1 1 1 1 0 0 1

1 1 1 0 1
1 1 0 0 0
0 0 1 0 1

0
0
1

1 0 0
0 1 0
0 0 1

13
14

L
W

0
0

0
0

0
1

0
1

0
1

0
0

1
0

1
0

1 0
1 1

0 0
1 0

0 0 0 1 1 1 1 0 0
0 1 0 1 0 0 0 0 0

0 0 1 1 0
0 1 1 1 1

0
1

1 0 1
1 1 1

15
16

L
W

1
1

1
0

0
0

0
1

1
0

0
0

1
0

1
1

1 0
0 0

1 1
0 1

1 0 0 1 1 1 1 1 0
1 0 0 0 1 1 1 1 0

1 1 1 0 1
1 1 0 0 0

0
0

1 0 1
0 1 0

17

L

1

1

0

0

1

0

0

0

1 1

1 0

0 0 0 1 0 0 1 1 1

1 0 0 1 0

0

0 1 1

18
19
20

W
L
W

0
1
1

0
1
0

0
1
1

0
0
1

0
0
1

0
1
1

1
1
0

1
0
1

0 1
0 0
1 1

0 0
1 1
0 0

1 0 1 0 0 1 0 0 1
0 0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 1 0

0 0 0 0 0
0 0 0 0 0
1 1 1 1 1

1
1
0

1 1 1
1 0 0
0 1 0

1

30
E
C
C
D
C
C
A
C
C
D
A
C
C
C
C
D
C
C
B
A
C

Penyelesaian:

A. Indeks Kesukaran
1. Soal no 11
P=

𝐵
𝐽𝑆
9

= 20
= 0.45
2. Soal no 12
P=

𝐵
𝐽𝑆
10

= 20
= 0.5
3. Soal no 13
P=

𝐵
𝐽𝑆
11

= 20
= 0.55
4. Soal no 14
P=

𝐵
𝐽𝑆
6

= 20
= 0.3
5. Soal no 15
P=

𝐵
𝐽𝑆
6

= 20
= 0.3

2

B. Indeks Daya Beda
1. Teori Klasik
Kelompok Atas
No
Resp
15
1
6
10
20
4
9
14
2
5

L
L
W
W
W
W
L
W
W
L
Jumlah

11
1
1
1
1
0
0
0
1
0
0
5

Nomor Butir Soal
12
13
14
1
1
0
1
1
0
1
1
0
1
1
0
0
1
1
1
1
0
0
1
1
0
0
1
0
0
1
0
0
0
5
7
4

15
0
0
0
0
1
1
1
0
0
0
3

Skor
Siswa
20
19
19
18
18
17
17
16
15
15

Kelompok Bawah
No

Nomor Butir Soal

Resp
11

L

8

W

0

1

1

0

0

14

17

L
L

1

0

0

0

0

1

1

0

1

0

14
13

L

1
0

1
0

0
0

0
1

13

3

12
1

13
0

14
1

15
1

Skor Siswa

11
1

15

7
12

W

0
0

16

W

0

1

1

0

0

12

13

0

0

0

0

0

18

L
W

0

0

1

0

1

11
11

19

L

1

1

0

0

0

9

4

6

4

2

3

Jumlah

3

13

a. Soal no 11
Ba
Na

B
 Nbb

5

D=

4

= 10 - 10
= 0.1
b. Soal no 12
Ba
Na

B
 Nbb

5

D=

6

= 10 - 10
= -0.1
c. Soal no 13
Ba
Na

B
 Nbb

7

D=

4

= 10 - 10
= 0.3
d. Soal no 14
Ba
Na

B
 Nbb

4

D=

2

= 10 - 10
= 0.2
e. Soal no 15
D=
=

Ba
Na

3
10

B
 Nbb

-

3
10

=0

4

2. Koefisien Korelasi Biserial Titik
a. Soal no 11
No
Resp

X

Y

X²

Y²

XY

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

1
0
1
0
0
1
0
0
0
1
1
0
0
1
1
0
1
0
1
0
9

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

1
0
1
0
0
1
0
0
0
1
1
0
0
1
1
0
1
0
1
0
11

361
225
169
289
225
361
169
196
289
324
225
169
121
256
400
144
196
121
81
324
4645

19
0
13
0
0
19
0
0
0
18
15
0
0
16
20
0
14
0
9
0
143

D = rxy = rpbis=

N XY  ( X)( Y)

( N X 2  ( X) 2 )( N Y 2  ( Y) 2 )
20

=
20

143 – 9 ( 299 )

9 – 9 2

= 0.287142

5

20

4645 – 299 2

b. Soal no 12
No
Resp

X

Y

X²

Y²

XY

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

1
0
1
1
0
1
1
1
0
1
1
0
0
0
1
1
0
0
1
0
11

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

1
0
1
1
0
1
1
1
0
1
1
0
0
0
1
1
0
0
1
0
11

361
225
169
289
225
361
169
196
289
324
225
169
121
256
400
144
196
121
81
324
4645

19
0
13
17
0
19
13
14
0
18
15
0
0
0
20
12
0
0
9
0
169

D = rxy = rpbis=

N XY  ( X)( Y)

( N X 2  ( X) 2 )( N Y 2  ( Y) 2 )
20

=
20

169 – 11 ( 299 )

11 – 11 2

= 0.154615

6

20

4645 – 299 2

c. Soal no 13
No
Resp

X

Y

X²

Y²

XY

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

1
0
0
1
0
1
1
1
1
1
0
0
0
0
1
1
0
1
0
1
11

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

1
0
0
1
0
1
1
1
1
1
0
0
0
0
1
1
0
1
0
1
11

361
225
169
289
225
361
169
196
289
324
225
169
121
256
400
144
196
121
81
324
4645

19
0
0
17
0
19
13
14
17
18
0
0
0
0
20
12
0
11
0
18
178

D = rxy = rpbis=

N XY  ( X)( Y)

( N X 2  ( X) 2 )( N Y 2  ( Y) 2 )
20

=
20

178 – 11 ( 299 )

11 – 11 2

= 0.460447

7

20

4645 – 299 2

d. Soal no 14
No
Resp

X

Y

X²

Y²

XY

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

0
1
1
0
0
0
0
0
1
0
1
0
0
1
0
0
0
0
0
1
6

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

0
1
1
0
0
0
0
0
1
0
1
0
0
1
0
0
0
0
0
1
6

361
225
169
289
225
361
169
196
289
324
225
169
121
256
400
144
196
121
81
324
4645

0
15
13
0
0
0
0
0
17
0
15
0
0
16
0
0
0
0
0
18
94

D = rxy = rpbis=

N XY  ( X)( Y)

( N X 2  ( X) 2 )( N Y 2  ( Y) 2 )
20

=
20

6 – 6 2

= 0.158631

8

94 – 6 ( 299 )
20

4552 – 299 2

e. Soal no 15
No
Resp

X

Y

X²

Y²

XY

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

0
0
0
1
0
0
0
0
1
0
1
1
0
0
0
0
0
1
0
1
6

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

0
0
0
1
0
0
0
0
1
0
1
1
0
0
0
0
0
1
0
1
6

361
225
169
289
225
361
169
196
289
324
225
169
121
256
400
144
196
121
81
324
4645

0
0
0
17
0
0
0
0
17
0
15
13
0
0
0
0
0
11
0
18
91

D = rxy = rpbis=

N XY  ( X)( Y)

( N X 2  ( X) 2 )( N Y 2  ( Y) 2 )
20

=
20

91 – 6 ( 299 )

6 – 6 2

= 0.047958

9

20

4552 – 299 2

3. Koefisien Korelasi Biserial Titik
a. Soal no 11
No
Resp

X

Y

Y-Ῡ

( Y - Ῡ )²

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

1
0
1
0
0
1
0
0
0
1
1
0
0
1
1
0
1
0
1
0
9

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

4.05
0.05
-1.95
2.05
0.05
4.05
-1.95
-0.95
2.05
3.05
0.05
-1.95
-3.95
1.05
5.05
-2.95
-0.95
-3.95
-5.95
3.05

16.4025
0.0025
3.8025
4.2025
0.0025
16.4025
3.8025
0.9025
4.2025
9.3025
0.0025
3.8025
15.6025
1.1025
25.5025
8.7025
0.9025
15.6025
35.4025
9.3025
174.95

Ῡ1 =

19+13+19+18+15+16+20+14+9
9

= 15.88889

Ῡ

=
=

𝑌
20
299
20

= 14.95

10

( 𝑌− Ῡ )²

σY =

20
174 .95

=

20

= 2.957617
9

PX = 20
= 0.45
px
 Y Y 
D = rpbis   1 
  Y  (1 p x )

=

15.88889 − 14.95

0.45

2.957617

( 1−0.45 )

= 0.287142

b. Soal no 12
No
Resp

X

Y

Y-Ῡ

( Y - Ῡ )²

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

1
0
1
1
0
1
1
1
0
1
1
0
0
0
1
1
0
0
1
0
11

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

4.05
0.05
-1.95
2.05
0.05
4.05
-1.95
-0.95
2.05
3.05
0.05
-1.95
-3.95
1.05
5.05
-2.95
-0.95
-3.95
-5.95
3.05

16.4025
0.0025
3.8025
4.2025
0.0025
16.4025
3.8025
0.9025
4.2025
9.3025
0.0025
3.8025
15.6025
1.1025
25.5025
8.7025
0.9025
15.6025
35.4025
9.3025
174.95

11

Ῡ1 =

19+13+17+19+13+14+18+15+20+12+9
11

= 15.36364
Ῡ

=
=

𝑌
20
299
20

= 14.95
( 𝑌− Ῡ )²

σY =
=

20
174 .95
20

= 2.957617
11

PX =20
`

= 0.55
px
 Y Y 
D = rpbis   1 
 Y  (1 p x )


=

15.36364 − 14.95

0.55

2.957617

( 1−0.55)

= 0.154615

12

c. Soal no 13
No
Resp

X

Y

Y-Ῡ

( Y - Ῡ )²

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

1
0
0
1
0
1
1
1
1
1
0
0
0
0
1
1
0
1
0
1
11

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

4.05
0.05
-1.95
2.05
0.05
4.05
-1.95
-0.95
2.05
3.05
0.05
-1.95
-3.95
1.05
5.05
-2.95
-0.95
-3.95
-5.95
3.05

16.4025
0.0025
3.8025
4.2025
0.0025
16.4025
3.8025
0.9025
4.2025
9.3025
0.0025
3.8025
15.6025
1.1025
25.5025
8.7025
0.9025
15.6025
35.4025
9.3025
174.95

Ῡ1 =

19+17+19+13+14+17+18+20+12+11+18
11

= 16.1818182

Ῡ

=
=

𝑌

20
299
20

= 14.95

13

( 𝑌− Ῡ )²

σY =
=

20
174 .95
20

= 2.957617
11

PX = 20
= 0.55
px
 Y Y 
D = rpbis   1 
  Y  (1 p x )

=

16.18182 − 14.95

0.55

2.957617

( 1−0.55 )

= 0.460447

14

d. Soal no 14

No
Resp

X

Y

Y-Ῡ

( Y - Ῡ )²

1
L
2
W
3
L
4
W
5
L
6
W
7
L
8
W
9
L
10
W
11
L
12
W
13
L
14
W
15
L
16
W
17
L
18
W
19
L
20
W
Jumlah

0
1
1
0
0
0
0
0
1
0
1
0
0
1
0
0
0
0
0
1
6

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

4.05
0.05
-1.95
2.05
0.05
4.05
-1.95
-0.95
2.05
3.05
0.05
-1.95
-3.95
1.05
5.05
-2.95
-0.95
-3.95
-5.95
3.05

16.4025
0.0025
3.8025
4.2025
0.0025
16.4025
3.8025
0.9025
4.2025
9.3025
0.0025
3.8025
15.6025
1.1025
25.5025
8.7025
0.9025
15.6025
35.4025
9.3025
174.95

15+13+17+15+16+18

Ῡ1 =

6

= 15.66667
Ῡ

𝑌

=
=

20
299
20

= 14.95
σY =
=

( 𝑌− Ῡ )²
20
174 .95
20

= 2.957617
15

6

PX = 20
= 0.3
px
 Y Y 
D = rpbis   1 
  Y  (1 p x )

=

15.66667 − 14.95

0.3

2.957617

( 1−0.3 )

= 0.158631

e. Soal no 15
No
Resp
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

L
W
L
W
L
W
L
W
L
W
L
W
L
W
L
W
L
W
L
W
Jumlah

X

Y

Y-Ῡ

( Y - Ῡ )²

0
0
0
1
0
0
0
0
1
0
1
1
0
0
0
0
0
1
0
1
6

19
15
13
17
15
19
13
14
17
18
15
13
11
16
20
12
14
11
9
18
299

4.05
0.05
-1.95
2.05
0.05
4.05
-1.95
-0.95
2.05
3.05
0.05
-1.95
-3.95
1.05
5.05
-2.95
-0.95
-3.95
-5.95
3.05

16.4025
0.0025
3.8025
4.2025
0.0025
16.4025
3.8025
0.9025
4.2025
9.3025
0.0025
3.8025
15.6025
1.1025
25.5025
8.7025
0.9025
15.6025
35.4025
9.3025
174.95

16

17+17+15+13+11+18

Ῡ1 =

6

= 15.16667

Ῡ

𝑌

=
=

20
299
20

= 14.95
( 𝑌− Ῡ )²

σY =
=

20
174 .95
20

= 2.957617

6

PX = 20
= 0.3
px
 Y Y 
D = rpbis   1 
 Y  (1 p x )


=

15.16667 −14.95

0.3

2.957617

( 1−0.3 )

= 0.047958

17

Kesimpulan :

Keterangan

soal no 11

soal no 12

soal no 13

soal no 14

soal no 15

P

0.45
sedang
0.287142
jelek
buang

0.55
sedang
0.154615
jelek
buang

0.55
sedang
0.460447
baik
pakai

0.3
sedang
0.158631
jelek
buang

0.3
sedang
0.047958
jelek
buang

D
Keputusan

 Menganalisis soal no 13
Pengecoh soal no 13
A
1
2
3

B*
7
4
11

baik

Option
JA
JB
Jumlah

kunci

C
2
0
2
tidak
berfungsi

D
0
3
3

E
0
1
1

baik

baik

B diberi tanda (*) adalah kunci jawaban.
Pengecoh A, D, dan E sudah berfungsi dengan baik karena banyak pengecoh A, D
dan E tersebut sudah dipilih oleh lebih dari 5% dari jumlah responden pada
kelompok bawah. Sedangkan pengecoh C tidak berfungsi dengan baik karena
kelompok bawah tidak memilih lebih dari 5% dari jumlah responden sehingga
kelompok bawah tidak lebih banyak memilih option C dibanding kelompok atas.

18

�ݺ�ߣ

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