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.