contoh tugas penilaian hasil belajar
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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
1
No
Resp
Nomor Butir Soal
1 2 3 4 5 6 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 L A B A B C D E A B C D A B C A D E A B A B C D E A B C D A E
2 W A B C D A E E A B D E C C D E A B C D A E E A D C B E D A C
3 L B A B C D E E A B A D A D D A B C D A E A D D A C B E A E C
4 W A B A B C D A E B D C A B A B C D A E A C C D E C C A D E D
5 L B B C A B C D A E D A B A B C D A E B E C D D E E B E D E C
6 W E C D B C D E A B C D A B C A D E A B A B C D E A B E D A C
7 L A D C E A B C D A E E A B C D A E E B A C C C C C B E B E A
8 W C B C B C A B C D A E A B C D A E A E B C C D A C E D D E C
9 L D E D A E E A B C D A E B D B D E B B C C C C E C B D D E C
10 W A B A A A E B C B C D A B C A D E A B A B C D E B B C D B D
11 L A B E B C C B A B C D A E D B D A A A A B C D A E E D B E A
12 W C B C A E E A E A B C B A E B D E A A B C D A E D B E E D C
13 L D C B B D A E A B A B C D A E D E A B C D A E E C C C D C C
14 W C E C A C B A B B D D B C D A D A B C D A E D E C B E D E C
15 L A B A B C D E A B C D A B C A D E A B A B C D E A B C D B C
16 W A A B A B C D A E B E A B C D A E A B A A C D A B C D A E D
17 L A B A B C D A D B D D B A B C D A E B A C C A B C D A E E C
18 W B A B C D A E A D D B C B A B C D A E D C A B C D A E D E C
19 L A B C D A E E B E E D A D A A B C D A E A B C D A E E D A B
20 W A A C A C E B A B D E D B D B A B C D A E C D E C B A A E A
KJ A B C A C E E A B D D A B D B D E A B A C C D E C B E D E C
No
Resp
Nomor Butir Soal
1 2 3 4 5 6 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 L 1 1 0 0 1 0 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 W 1 1 1 0 0 1 1 1 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 0 1
3 L 0 0 0 0 0 1 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 1 0 1 1 1 0 1 1
4 W 1 1 0 0 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 1 1 1 1 1 1 0 0 1 1 0
5 L 0 1 1 1 0 0 0 1 0 1 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 1 1 1 1
6 W 0 0 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 1 0 1 1 1 0 1 1 1 0 1
7 L 1 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 0 1 0
8 W 0 1 1 0 1 0 0 0 0 0 0 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 0 1 1 1
9 L 0 0 0 1 0 1 0 0 0 1 0 0 1 1 1 1 1 0 1 0 1 1 0 1 1 1 0 1 1 1
10 W 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 1 1 1 1 1 0 1 1 1 0 1 0 1 0 0
11 L 1 1 0 0 1 0 0 1 1 0 1 1 0 1 1 1 0 1 0 1 0 1 1 0 0 0 0 0 1 0
12 W 0 1 1 1 0 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 1
13 L 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 1 0 1
14 W 0 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1
15 L 1 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 1 0 1 1 1 0 1 0 1 0 1
16 W 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0 1 1 1 1 0 1 1 0 0 0 0 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 W 0 0 0 0 0 0 1 1 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 0 0 0 1 1 1 1
19 L 1 1 1 0 0 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0
20 W 1 0 1 1 1 1 0 1 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 1 1 1 0 0 1 0
2
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
3
B. Indeks Daya Beda
1. Teori Klasik
Kelompok Atas
No Nomor Butir Soal Skor
Siswa Resp 11 12 13 14 15
15 L 1 1 1 0 0 20
1 L 1 1 1 0 0 19
6 W 1 1 1 0 0 19
10 W 1 1 1 0 0 18
20 W 0 0 1 1 1 18
4 W 0 1 1 0 1 17
9 L 0 0 1 1 1 17
14 W 1 0 0 1 0 16
2 W 0 0 0 1 0 15
5 L 0 0 0 0 0 15
Jumlah 5 5 7 4 3
Kelompok Bawah
No Nomor Butir Soal Skor Siswa
Resp 11 12 13 14 15
11 L 1 1 0 1 1 15
8 W 0 1 1 0 0 14
17 L 1 0 0 0 0 14
3 L 1 1 0 1 0 13
7 L 0 1 1 0 0 13
12 W 0 0 0 0 1 13
16 W 0 1 1 0 0 12
13 L 0 0 0 0 0 11
18 W 0 0 1 0 1 11
19 L 1 1 0 0 0 9
Jumlah 4 6 4 2 3
4
a. Soal no 11
D = b
b
a
a
N
B
N
B
= 5
10 -
4
10
= 0.1
b. Soal no 12
D = b
b
a
a
N
B
N
B
= 5
10 -
6
10
= -0.1
c. Soal no 13
D = b
b
a
a
N
B
N
B
= 7
10 -
4
10
= 0.3
d. Soal no 14
D = b
b
a
a
N
B
N
B
= 4
10 -
2
10
= 0.2
e. Soal no 15
D = b
b
a
a
N
B
N
B
= 3
10 -
3
10
= 0
5
2. Koefisien Korelasi Biserial Titik
a. Soal no 11
No X Y X² Y² XY
Resp
1 L 1 19 1 361 19
2 W 0 15 0 225 0
3 L 1 13 1 169 13
4 W 0 17 0 289 0
5 L 0 15 0 225 0
6 W 1 19 1 361 19
7 L 0 13 0 169 0
8 W 0 14 0 196 0
9 L 0 17 0 289 0
10 W 1 18 1 324 18
11 L 1 15 1 225 15
12 W 0 13 0 169 0
13 L 0 11 0 121 0
14 W 1 16 1 256 16
15 L 1 20 1 400 20
16 W 0 12 0 144 0
17 L 1 14 1 196 14
18 W 0 11 0 121 0
19 L 1 9 1 81 9
20 W 0 18 0 324 0
Jumlah 9 299 11 4645 143
D = rxy = pbisr =
)Y)(YN)(X)(XN(
Y)X)((XYN
2222
= 20 143 – 9 ( 299 )
20 9 – 9 2 20 4645 – 299 2
= 0.287142
6
b. Soal no 12
No X Y X² Y² XY
Resp
1 L 1 19 1 361 19
2 W 0 15 0 225 0
3 L 1 13 1 169 13
4 W 1 17 1 289 17
5 L 0 15 0 225 0
6 W 1 19 1 361 19
7 L 1 13 1 169 13
8 W 1 14 1 196 14
9 L 0 17 0 289 0
10 W 1 18 1 324 18
11 L 1 15 1 225 15
12 W 0 13 0 169 0
13 L 0 11 0 121 0
14 W 0 16 0 256 0
15 L 1 20 1 400 20
16 W 1 12 1 144 12
17 L 0 14 0 196 0
18 W 0 11 0 121 0
19 L 1 9 1 81 9
20 W 0 18 0 324 0
Jumlah 11 299 11 4645 169
D = rxy = pbisr =
)Y)(YN)(X)(XN(
Y)X)((XYN
2222
= 20 169 – 11 ( 299 )
20 11 – 11 2 20 4645 – 299 2
= 0.154615
7
c. Soal no 13
D = rxy = pbisr =
)Y)(YN)(X)(XN(
Y)X)((XYN
2222
= 20 178 – 11 ( 299 )
20 11 – 11 2 20 4645 – 299 2
= 0.460447
No X Y X² Y² XY
Resp
1 L 1 19 1 361 19
2 W 0 15 0 225 0
3 L 0 13 0 169 0
4 W 1 17 1 289 17
5 L 0 15 0 225 0
6 W 1 19 1 361 19
7 L 1 13 1 169 13
8 W 1 14 1 196 14
9 L 1 17 1 289 17
10 W 1 18 1 324 18
11 L 0 15 0 225 0
12 W 0 13 0 169 0
13 L 0 11 0 121 0
14 W 0 16 0 256 0
15 L 1 20 1 400 20
16 W 1 12 1 144 12
17 L 0 14 0 196 0
18 W 1 11 1 121 11
19 L 0 9 0 81 0
20 W 1 18 1 324 18
Jumlah 11 299 11 4645 178
8
d. Soal no 14
No X Y X² Y² XY
Resp
1 L 0 19 0 361 0
2 W 1 15 1 225 15
3 L 1 13 1 169 13
4 W 0 17 0 289 0
5 L 0 15 0 225 0
6 W 0 19 0 361 0
7 L 0 13 0 169 0
8 W 0 14 0 196 0
9 L 1 17 1 289 17
10 W 0 18 0 324 0
11 L 1 15 1 225 15
12 W 0 13 0 169 0
13 L 0 11 0 121 0
14 W 1 16 1 256 16
15 L 0 20 0 400 0
16 W 0 12 0 144 0
17 L 0 14 0 196 0
18 W 0 11 0 121 0
19 L 0 9 0 81 0
20 W 1 18 1 324 18
Jumlah 6 299 6 4645 94
D = rxy = pbisr =
)Y)(YN)(X)(XN(
Y)X)((XYN
2222
= 20 94 – 6 ( 299 )
20 6 – 6 2 20 4552 – 299 2
= 0.158631
9
e. Soal no 15
D = rxy = pbisr =
)Y)(YN)(X)(XN(
Y)X)((XYN
2222
= 20 91 – 6 ( 299 )
20 6 – 6 2 20 4552 – 299 2
= 0.047958
No X Y X² Y² XY
Resp
1 L 0 19 0 361 0
2 W 0 15 0 225 0
3 L 0 13 0 169 0
4 W 1 17 1 289 17
5 L 0 15 0 225 0
6 W 0 19 0 361 0
7 L 0 13 0 169 0
8 W 0 14 0 196 0
9 L 1 17 1 289 17
10 W 0 18 0 324 0
11 L 1 15 1 225 15
12 W 1 13 1 169 13
13 L 0 11 0 121 0
14 W 0 16 0 256 0
15 L 0 20 0 400 0
16 W 0 12 0 144 0
17 L 0 14 0 196 0
18 W 1 11 1 121 11
19 L 0 9 0 81 0
20 W 1 18 1 324 18
Jumlah 6 299 6 4645 91
10
3. Koefisien Korelasi Biserial Titik
a. Soal no 11
No X Y Y - Ῡ ( Y - Ῡ )²
Resp
1 L 1 19 4.05 16.4025
2 W 0 15 0.05 0.0025
3 L 1 13 -1.95 3.8025
4 W 0 17 2.05 4.2025
5 L 0 15 0.05 0.0025
6 W 1 19 4.05 16.4025
7 L 0 13 -1.95 3.8025
8 W 0 14 -0.95 0.9025
9 L 0 17 2.05 4.2025
10 W 1 18 3.05 9.3025
11 L 1 15 0.05 0.0025
12 W 0 13 -1.95 3.8025
13 L 0 11 -3.95 15.6025
14 W 1 16 1.05 1.1025
15 L 1 20 5.05 25.5025
16 W 0 12 -2.95 8.7025
17 L 1 14 -0.95 0.9025
18 W 0 11 -3.95 15.6025
19 L 1 9 -5.95 35.4025
20 W 0 18 3.05 9.3025
Jumlah 9 299 174.95
Ῡ1 = 19+13+19+18+15+16+20+14+9
9
= 15.88889
Ῡ = 𝑌
20
= 299
20
= 14.95
11
σY = ( 𝑌− Ῡ )²
20
= 174.95
20
= 2.957617
PX = 9
20
= 0.45
D = )xp1(
xp
Y
Y1Ypbisr
= 15.88889 − 14.95
2.957617
0.45
( 1−0.45 )
= 0.287142
b. Soal no 12
No X Y Y - Ῡ ( Y - Ῡ )²
Resp
1 L 1 19 4.05 16.4025
2 W 0 15 0.05 0.0025
3 L 1 13 -1.95 3.8025
4 W 1 17 2.05 4.2025
5 L 0 15 0.05 0.0025
6 W 1 19 4.05 16.4025
7 L 1 13 -1.95 3.8025
8 W 1 14 -0.95 0.9025
9 L 0 17 2.05 4.2025
10 W 1 18 3.05 9.3025
11 L 1 15 0.05 0.0025
12 W 0 13 -1.95 3.8025
13 L 0 11 -3.95 15.6025
14 W 0 16 1.05 1.1025
15 L 1 20 5.05 25.5025
16 W 1 12 -2.95 8.7025
17 L 0 14 -0.95 0.9025
18 W 0 11 -3.95 15.6025
19 L 1 9 -5.95 35.4025
20 W 0 18 3.05 9.3025
Jumlah 11 299 174.95
12
Ῡ1 = 19+13+17+19+13+14+18+15+20+12+9
11
= 15.36364
Ῡ = 𝑌
20
= 299
20
= 14.95
σY = ( 𝑌− Ῡ )²
20
= 174.95
20
= 2.957617
PX =11
20
` = 0.55
D = )xp1(
xp
Y
Y1Ypbisr
= 15.36364 − 14.95
2.957617
0.55
( 1−0.55)
= 0.154615
13
c. Soal no 13
No X Y Y - Ῡ ( Y - Ῡ )²
Resp
1 L 1 19 4.05 16.4025
2 W 0 15 0.05 0.0025
3 L 0 13 -1.95 3.8025
4 W 1 17 2.05 4.2025
5 L 0 15 0.05 0.0025
6 W 1 19 4.05 16.4025
7 L 1 13 -1.95 3.8025
8 W 1 14 -0.95 0.9025
9 L 1 17 2.05 4.2025
10 W 1 18 3.05 9.3025
11 L 0 15 0.05 0.0025
12 W 0 13 -1.95 3.8025
13 L 0 11 -3.95 15.6025
14 W 0 16 1.05 1.1025
15 L 1 20 5.05 25.5025
16 W 1 12 -2.95 8.7025
17 L 0 14 -0.95 0.9025
18 W 1 11 -3.95 15.6025
19 L 0 9 -5.95 35.4025
20 W 1 18 3.05 9.3025
Jumlah 11 299 174.95
Ῡ1 = 19+17+19+13+14+17+18+20+12+11+18
11
= 16.1818182
Ῡ = 𝑌
20
= 299
20
= 14.95
14
σY = ( 𝑌− Ῡ )²
20
= 174.95
20
= 2.957617
PX = 11
20
= 0.55
D = )xp1(
xp
Y
Y1Ypbisr
= 16.18182 − 14.95
2.957617
0.55
( 1−0.55 )
= 0.460447
15
d. Soal no 14
Ῡ1 = 15+13+17+15+16+18
6
= 15.66667
Ῡ = 𝑌
20
= 299
20
= 14.95
σY = ( 𝑌− Ῡ )²
20
= 174.95
20
= 2.957617
No X Y Y - Ῡ ( Y - Ῡ )²
Resp
1 L 0 19 4.05 16.4025
2 W 1 15 0.05 0.0025
3 L 1 13 -1.95 3.8025
4 W 0 17 2.05 4.2025
5 L 0 15 0.05 0.0025
6 W 0 19 4.05 16.4025
7 L 0 13 -1.95 3.8025
8 W 0 14 -0.95 0.9025
9 L 1 17 2.05 4.2025
10 W 0 18 3.05 9.3025
11 L 1 15 0.05 0.0025
12 W 0 13 -1.95 3.8025
13 L 0 11 -3.95 15.6025
14 W 1 16 1.05 1.1025
15 L 0 20 5.05 25.5025
16 W 0 12 -2.95 8.7025
17 L 0 14 -0.95 0.9025
18 W 0 11 -3.95 15.6025
19 L 0 9 -5.95 35.4025
20 W 1 18 3.05 9.3025
Jumlah 6 299 174.95
16
PX = 6
20
= 0.3
D = )xp1(
xp
Y
Y1Ypbisr
= 15.66667 − 14.95
2.957617
0.3
( 1−0.3 )
= 0.158631
e. Soal no 15
No X Y Y - Ῡ ( Y - Ῡ )²
Resp
1 L 0 19 4.05 16.4025
2 W 0 15 0.05 0.0025
3 L 0 13 -1.95 3.8025
4 W 1 17 2.05 4.2025
5 L 0 15 0.05 0.0025
6 W 0 19 4.05 16.4025
7 L 0 13 -1.95 3.8025
8 W 0 14 -0.95 0.9025
9 L 1 17 2.05 4.2025
10 W 0 18 3.05 9.3025
11 L 1 15 0.05 0.0025
12 W 1 13 -1.95 3.8025
13 L 0 11 -3.95 15.6025
14 W 0 16 1.05 1.1025
15 L 0 20 5.05 25.5025
16 W 0 12 -2.95 8.7025
17 L 0 14 -0.95 0.9025
18 W 1 11 -3.95 15.6025
19 L 0 9 -5.95 35.4025
20 W 1 18 3.05 9.3025
Jumlah 6 299 174.95
17
Ῡ1 = 17+17+15+13+11+18
6
= 15.16667
Ῡ = 𝑌
20
= 299
20
= 14.95
σY = ( 𝑌− Ῡ )²
20
= 174.95
20
= 2.957617
PX = 6
20
= 0.3
D = )xp1(
xp
Y
Y1Ypbisr
= 15.16667 −14.95
2.957617
0.3
( 1−0.3 )
= 0.047958
18
Kesimpulan :
Keterangan soal no 11 soal no 12 soal no 13 soal no 14 soal no 15
P 0.45 0.55 0.55 0.3 0.3
sedang sedang sedang sedang sedang
D 0.287142 0.154615 0.460447 0.158631 0.047958
jelek jelek baik jelek jelek
Keputusan buang buang pakai buang buang
Menganalisis soal no 13
Pengecoh soal no 13
Option A B*
C D E
JA 1 7 2 0 0
JB 2 4 0 3 1
Jumlah 3 11 2 3 1
baik kunci tidak
berfungsi 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.
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