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

=

15.88...
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

= 2.95...
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
...
( 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 )

=

16.181...
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...
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 )

= 0.15863...
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 ...
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
...
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Contoh tugas penilaian hasil belajar

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Contoh tugas penilaian hasil belajar

  1. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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

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