1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. ( 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

13. 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

14. 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

15. ( 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

16. 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

17. 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

18. 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

19. 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