SlideShare a Scribd company logo
1 of 16
Download to read offline
HASIL TES DARI MAHASISWA PRODI MATEMATIKA
ANGKATAN 2015
TES PERSEPSI GAMBAR
RUANG DAN BIDANG
N = 20
∑ X = 78
∑ X2
= 404
No X X2
1 10 100
2 4 16
3 2 4
4 3 9
5 8 64
6 5 25
7 4 16
8 6 36
9 3 9
10 2 4
11 3 9
12 2 4
13 4 16
14 4 16
15 2 4
16 1 1
17 6 36
18 1 1
19 3 9
20 5 25
Jumlah 78 404
Mean =
∑X
N
=
78
20
= 3,9
SD =
√N.∑(X2)−(∑X)2
N(N−1)
=
√20.404−(78)2
20(20−1)
=
√8080−6084
20.19
=
√1996
380
= √5,25
= 2,29
SKALA 5
BS = 3,9 + (1,8 x 2,29) = 9,12 ke atas
B = 3,9 + (0,6 x 2,29) = 5,274
C = 3,9 - (0,6 x 2,29) = 2,526
K = 3,9 - (1,8 x 2,29) = -0,22
KS = Dibawah -0,22
DISTRAKTOR DAN VALIDITAS
Distraktor
Rangking 1-10
subyek 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A D B E D D A A D D C D D B D C D A C B A
E E E E D E B A E D D C C B A B B C D D A
H A E B E E D A D A D A D D D C D D D A D
Q B E D B D A C D A B D D A D B E E B B A
F E B C B D A B A B C B B C D C A A D B A
T C D C D D B A E D A B B C E D B C B B A
B A E D D C B D C A C D A C A B D C A A C
G B E B B B A A C B C D D A D A A A C C A
M C C E D D A E A D C A A B B E E C E B A
N C C E D A A B E D A B A E E B D C B B A
Rangking 11-20
subyek 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
D E B E D E D C B C A A A B A D B C D D B
I E B E A B B A E D C A D D D A D D C A B
K A B E D A D A B E A D D D C B D B A A C
S B C E D D E C A E C A B B C E D B B A E
C A B E D D C C D A B D C C C D D B B B D
J B E E C E B A E D D D A D C A D C C C A
L A C E D A D D C C B B E D C B B B D E E
O A B E B D C A B E D A A E B E D B A E B
P C E D A B A E D C B B A D A B E C D D B
R A C D C A E C B D D A B B E D B C D A B
Subyek
Nomer item
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A D B E D D A A D D C D D B D C D A C B A
B A E D D C B D C A C D A C A B D C A A C
C A B E D D C C D A B D C C C D D B B B D
D E B E D E D C B C A A A B A D B C D D B
E E E E D E B A E D D C C B A B B C D D A
F E B C B D A B A B C B B C D C A A D B A
G B E B B B A A C B C D D A D A A A C C A
H A E B E E D A D A D A D D D C D D D A D
I E B E A B B A E D C A D D D A D D C A B
J B E E C E B A E D D D A D C A D C C C A
K A B E D A D A B E A D D D C B D B A A C
L A C E D A D D C C B B E D C B B B D E E
M C C E D D A E A D C A A B B E E C E B B
N C C E D A A B E D A B A E E B D C B B A
O A B E B D C A B E D A A E B E D B A E A
P C E D A B A E D C B B A D A B E C D D B
Q B E D B D A C D A B D D A D B E E B B A
R A C D C A E C B D D A B B E D B C D A B
S B C E D D E C A E C A B B C E D B B A E
T C D C D D B A E D A B B C E D B C B B A
Kunci B E E D E B A E D C D D A D A/C A A/D C C A
1. DISTRAKTOR
1
KET A B* C D E
Atas 2 2 3 1 2
Bawah 5 2 1 0 2
2
KET A B C D E*
Atas 0 2 2 1 5
Bawah 0 5 3 O 2
3
KET A B C D E*
Atas 0 2 2 2 4
Bawah 0 0 0 2 8
4
KET A B C D* E
Atas 0 2 0 6 1
Bawah 2 1 2 5 0
5
KET A B C D E*
Atas 1 1 1 5 2
Bawah 3 2 0 3 2
6
KET A B* C D E
Atas 6 3 0 1 0
Bawah 1 2 2 3 2
7
KET A* B C D E
Atas 5 2 1 1 1
Bawah 4 0 4 1 1
8
KET A B C D E*
Atas 2 0 2 3 3
Bawah 1 4 1 2 2
9
KET A B C D* E
Atas 3 2 0 4 0
Bawah 1 0 3 3 3
10
KET A B C* D E
Atas 2 1 5 2 0
Bawah 2 3 2 3 0
11
KET A B C D* E
Atas 2 3 1 4 0
Bawah 5 2 0 3 0
12
KET A B C D* E
Atas 3 2 3 4 0
Bawah 3 2 1 2 1
13
KET A* B C D E
Atas 2 3 3 0 1
Bawah 0 3 1 5 1
14
KET A B C D* E
Atas 2 4 0 5 2
Bawah 2 1 5 1 1
15
KET A* B C* D E
Atas 1 4 3 1 1
Bawah 2 3 0 3 2
16
KET A* B C D E
Atas 2 2 0 3 2
Bawah 0 3 0 6 1
17
KET A* B C D* E
Atas 3 0 5 1 1
Bawah 0 5 4 1 0
18
KET A B C* D E
Atas 1 3 2 2 1
Bawah 2 2 2 4 0
19
KET A B C* D E
Atas 2 6 1 1 0
Bawah 4 1 1 2 2
20
KET A* B C D E
Atas 8 0 1 1 0
Bawah 1 5 1 1 2
Kategori soal
P di atas 0,5 maka soal dikatakan mudah
P sama dengan 0,5 maka soal dikatakan sedang
P di bawah 0,5 maka soal dikatakan sulit
P1=
𝑩
𝐓
=
𝟒
𝟐𝟎
= 0,2 (Sukar) P8=
𝑩
𝐓
=
𝟓
𝟐𝟎
= 0,25 (Sukar) P15=
𝑩
𝐓
=
𝟔
𝟐𝟎
= 0,3 (Sukar)
A = Berfungsi D = Berfungsi A = Berfungsi C = Berfungsi B = Berfungsi E = Berfungsi
C = Berfungsi E = Berfungsi B = Berfungsi D = Berfungsi D = Berfungsi
P2=
𝑩
𝐓
=
𝟕
𝟐𝟎
= 0,35 (sukar) P9=
𝑩
𝐓
=
𝟕
𝟐𝟎
= 0,35 (Sukar) P16=
𝑩
𝐓
=
𝟐
𝟐𝟎
= 0,1 (Sukar)
A = Revisi C = Berfungsi A = Berfungsi C = Berfungsi B = Berfungsi D = Berfungsi
B = Berfungsi D = Berfungsi B = Berfungsi E = Berfungsi C = Revisi E = Berfungsi
P3=
𝑩
𝐓
=
𝟏𝟐
𝟐𝟎
= 0,6 (Sedang) P10=
𝑩
𝐓
=
𝟕
𝟐𝟎
= 0,35 (Sukar) P17=
𝑩
𝐓
=
𝟓
𝟐𝟎
= 0,25 (Sukar)
A = Revisi C = Berfungsi A = Berfungsi D = Berfungsi B = Berfungsi E = Berfungsi
B = Berfungsi D= Berfungsi B = Berfungsi E =Revisi C = Berfungsi
P4=
𝑩
𝐓
=
𝟏𝟏
𝟐𝟎
= 0,55 (Sedang) P11=
𝑩
𝐓
=
𝟕
𝟐𝟎
= 0,35 (Sukar) P18=
𝑩
𝐓
=
𝟒
𝟐𝟎
= 0,2 (Sukar)
A = Berfungsi C = Berfungsi A = Berfungsi C = Berfungsi A = Berfungsi D = Berfungsi
B = Berfungsi E = Berfungsi B = Berfungsi E =Revisi B = Berfungsi E = Berfungsi
P5=
𝑩
𝐓
=
𝟒
𝟐𝟎
= 0,2 (Sukar) P12=
𝑩
𝐓
=
𝟔
𝟐𝟎
= 0,3 (Sukar) P19=
𝑩
𝐓
=
𝟐
𝟐𝟎
= 0,1 (Sukar)
A = Berfungsi C = Berfungsi A = Berfungsi C = Berfungsi A = Berfungsi D = Berfungsi
B = Berfungsi D = Berfungsi B = Berfungsi E = Berfungsi B = Berfungsi E = Berfungsi
P6=
𝑩
𝐓
=
𝟓
𝟐𝟎
= 0,25 (Sukar) P13=
𝑩
𝐓
=
𝟐
𝟐𝟎
= 0,1 (Sukar) P20=
𝑩
𝐓
=
𝟗
𝟐𝟎
= 0,45 ( Sukar)
A = Berfungsi D = Berfungsi B = Berfungsi D = Berfungsi B = Berfungsi D = Berfungsi
C = Berfungsi E = Berfungsi C = Berfungsi E = Berfungsi C = Berfungsi E = Berfungsi
P7=
𝑩
𝐓
=
𝟗
𝟐𝟎
= 0,45 (sukar) P14=
𝑩
𝐓
=
𝟔
𝟐𝟎
= 0,3 (Sukar)
B = Berfungsi D = Berfungsi A = Berfungsi C = Berfungsi
C = Berfungsi E = berfungsi B = Berfungsi E = Berfungsi
2. Validasi
Item 1
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.15−(4)(78)
√{[20.(4)− (4)2][20.(404)−(78)2]}
=
300−312
√{[80− 16][8080−6084]}
=
12
√{[64][1996]}
=
12
√127744
=
12
357,41
= 0,33
Item 2
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.29−(7)(78)
√{[20.(7)− (7)2][20.(404)−(78)2]}
=
580−546
√{[140− 49][8080−6084]}
=
34
√{[91][1996]}
=
34
√18636
=
52
426,19
= 0,122
S X Y X2
Y2
X.Y
A 0 10 0 100 0
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 0 5 0 25 0
G 1 4 1 16 4
H 0 6 0 36 0
I 0 3 0 9 0
J 1 2 1 4 2
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 1 6 1 36 6
R 0 1 0 1 0
S 1 3 1 9 3
T 0 5 0 25 0
∑ 4 78 4 404 15
S X Y X2
Y2
X.Y
A 0 10 0 100 0
B 0 4 0 16 0
C 0 2 0 4 0
D 1 3 1 9 3
E 1 8 1 64 8
F 1 5 1 25 5
G 0 4 0 16 0
H 0 6 0 36 0
I 1 3 1 9 3
J 0 2 0 4 0
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 1 1 1 1 1
Q 1 6 1 36 6
R 0 1 0 1 0
S 1 3 0 9 3
T 0 5 0 25 0
∑ 7 78 6 404 29
Item 3
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.44−(12)(78)
√{[20.(12)− (12)2][20.(404)−(78)2]}
=
880−936
√{[240− 144][8080−6084]}
=
−56
√{[96][1996]}
=
−56
√191616
=
−56
437,73
=-0,127
Item 4
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.48−(11)(78)
√{[20.(11)− (11)2][20.(404)−(78)2]}
=
960−858
√{[220− 121][8080−6084]}
=
102
√{[99][1996]}
=
102
√197604
=
102
444,52
= 0,22
S X Y X2
Y2
X.Y
A 1 10 100 100 10
B 0 4 0 16 0
C 1 2 1 4 2
D 1 3 1 9 3
E 1 8 1 64 8
F 0 5 0 25 0
G 0 4 0 16 0
H 0 6 0 36 0
I 1 3 1 9 3
J 1 2 1 4 2
K 1 3 1 9 3
L 1 2 1 4 2
M 1 4 1 16 4
N 1 4 1 16 4
O 1 2 1 4 2
P 1 1 0 1 1
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 12 78 11 404 44
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 1 4 1 16 4
C 1 2 1 4 2
D 1 3 1 9 3
E 1 8 1 64 8
F 0 5 0 25 0
G 0 4 0 16 0
H 0 6 0 36 0
I 0 3 0 9 0
J 0 2 0 4 0
K 1 3 1 9 3
L 1 2 1 4 2
M 1 4 1 16 4
N 1 4 1 16 4
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 1 3 1 9 3
T 1 5 1 25 5
∑ 11 78 11 404 48
Item 5
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.19−(4)(78)
√{[20.(4)− (4)2][20.(404)−(78)2]}
=
380−312
√{[80− 16][8080−6084]}
=
68
√{[64][1996]}
=
68
√127744
=
68
357,41
= 0,21
Item 6
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.20−(5)(78)
√{[20.(5)− (5)2][20.(404)−(78)2]}
=
400−390
√{[100− 25][8080−6084]}
=
10
√{[75][1996]}
=
10
√127744
=
10
357,41
= 0,27
S X Y X2
Y2
X.Y
A 0 10 0 100 0
B 0 4 0 16 0
C 0 2 0 4 0
D 1 3 1 9 3
E 1 8 1 64 8
F 0 5 0 25 0
G 0 4 0 16 0
H 1 6 1 36 6
I 0 3 0 9 0
J 1 2 1 4 2
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 4 78 4 404 19
S X Y X2
Y2
X.Y
A 0 10 0 100 0
B 1 4 1 16 4
C 0 2 0 4 0
D 0 3 0 9 0
E 1 8 1 64 8
F 0 5 0 25 0
G 0 4 0 16 0
H 0 6 0 36 0
I 1 3 1 9 3
J 1 2 1 4 2
K 1 3 0 9 3
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 5 78 4 404 20
Item 7
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.41−(9)(78)
√{[20.(9)− (9)2][20.(404)−(78)2]}
=
820−702
√{[180−81][8080−6084]}
=
118
√{[99][1996]}
=
118
√197604
=
118
444,52
= 0,26
Item 8
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.22−(5)(78)
√{[20.(5)− (5)2][20.(404)−(78)2]}
=
440−390
√{[100−25][8080−6084]}
=
50
√{[75][1996]}
=
50
√149700
=
50
386,91
= 0,129
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 1 8 1 64 8
F 0 5 0 25 0
G 1 4 1 16 4
H 1 6 1 36 6
I 1 3 1 9 3
J 1 2 1 4 2
K 1 3 1 9 3
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 1 2 1 4 2
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 1 3 0 9 3
T 0 5 0 25 0
∑ 9 78 8 404 41
S X Y X2
Y2
X.Y
A 0 10 0 100 0
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 1 8 1 64 8
F 0 5 0 25 0
G 0 4 0 16 0
H 0 6 0 36 0
I 1 3 1 9 3
J 1 2 1 4 2
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 1 4 1 16 4
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 1 5 1 25 5
∑ 5 78 5 404 22
Item 9
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.36−(7)(78)
√{[20.(7)− (7)2][20.(404)−(78)2]}
=
720−546
√{[140−49][8080−6084]}
=
174
√{[91][1996]}
=
174
√181636
=
174
426,21
= 0,41
Item 10
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.31−(7)(78)
√{[20.(7)− (7)2][20.(404)−(78)2]}
=
620−546
√{[140−49][8080−6084]}
=
74
√{[91][1996]}
=
74
√181636
=
74
426,21
= 0,12
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 1 8 1 64 8
F 0 5 0 25 0
G 0 4 0 16 0
H 0 6 0 36 0
I 1 3 1 9 3
J 1 2 1 4 2
K 0 3 0 9 0
L 0 2 0 4 0
M 1 4 1 16 4
N 1 4 1 16 4
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 1 1 0
S 0 3 0 9 0
T 1 5 1 25 5
∑ 7 78 8 404 36
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 1 4 1 16 4
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 1 5 1 25 5
G 1 4 1 16 4
H 0 6 0 36 0
I 1 3 1 9 3
J 0 2 0 4 0
K 0 3 0 9 0
L 0 2 0 4 0
M 1 4 1 16 4
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 1 1 1 1 1
S 0 3 1 9 0
T 0 5 0 25 0
∑ 7 78 8 404 31
Item 11
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.31−(7)(78)
√{[20.(7)− (7)2][20.(404)−(78)2]}
=
620−546
√{[140−49][8080−6084]}
=
74
√{[91][1996]}
=
74
√181636
=
74
426,198
= 0,23
Item 12
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.32−(6)(78)
√{[20.(6)− (6)2][20.(404)−(78)2]}
=
640−468
√{[120−36][8080−6084]}
=
172
√{[84][1996]}
=
172
√167664
=
172
409,46
= 0,42
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 1 4 1 16 4
C 1 2 1 4 2
D 0 3 0 9 0
E 0 8 0 64 0
F 0 5 0 25 0
G 1 4 1 16 4
H 0 6 0 36 0
I 0 3 0 9 0
J 1 2 1 4 2
K 1 3 1 9 3
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 1 6 1 36 6
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 7 78 7 404 31
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 0 5 0 25 0
G 1 4 1 16 4
H 1 6 1 36 6
I 1 3 1 9 3
J 0 2 0 4 0
K 1 3 1 9 3
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 1 6 1 36 6
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 6 78 6 404 32
Item 13
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.10−(2)(78)
√{[20.(2)− (2)2][20.(404)−(78)2]}
=
200−156
√{[40−4][8080−6084]}
=
44
√{[160][1996]}
=
44
√319360
=
44
565,11
= 0,077
Item 14
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.34−(6)(78)
√{[20.(6)− (6)2][20.(404)−(78)2]}
=
680−468
√{[120−36][8080−6084]}
=
212
√{[84][1996]}
=
212
√167664
=
212
409,51
= 0,52
S X Y X2
Y2
X.Y
A 0 10 0 100 0
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 0 5 0 25 0
G 1 4 1 16 4
H 0 6 0 36 0
I 0 3 0 9 0
J 0 2 0 4 0
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 1 6 1 36 6
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 2 78 2 404 10
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 1 5 1 25 5
G 1 4 1 16 4
H 1 6 1 36 6
I 1 3 1 9 3
J 0 2 0 4 0
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 1 6 1 36 6
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 6 78 6 404 34
Item 15
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.30−(6)(78)
√{[20.(6)− (6)2][20.(404)−(78)2]}
=
600−468
√{[120−36][8080−6084]}
=
132
√{[84][1996]}
=
132
√167664
=
132
409,58
= 0,322
Item 16
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.9−(2)(78)
√{[20.(2)− (2)2][20.(404)−(78)2]}
=
180−156
√{[400−4][8080−6084]}
=
24
√{[396][1996]}
=
24
√790416
=
24
889,1
= 0,27
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 1 5 1 25 5
G 1 4 1 16 4
H 1 6 1 36 6
I 1 3 1 9 3
J 1 2 1 4 2
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 6 78 6 404 30
S X Y X2
Y2
X.Y
A 0 10 0 100 0
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 1 5 1 25 5
G 1 4 1 16 4
H 0 6 0 36 0
I 0 3 0 9 0
J 0 2 0 4 0
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 2 78 2 404 9
Item 17
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.28−(5)(78)
√{[20.(5)− (5)2][20.(404)−(78)2]}
=
560−390
√{[100−25][8080−6084]}
=
170
√{[75][1996]}
=
170
√149700
=
170
386,91
= 0,43
Item 18
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.19−(4)(78)
√{[20.(4)− (4)2][20.(404)−(78)2]}
=
380−312
√{[80−16][8080−6084]}
=
68
√{[64][1996]}
=
68
√127744
=
68
357,41
= 0,22
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 1 5 1 25 5
G 1 4 1 16 4
H 1 6 1 36 6
I 1 3 1 9 3
J 0 2 0 4 0
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 5 78 5 404 28
S X Y X2 Y2 X.Y
A 1 10 1 100 10
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 0 5 0 25 0
G 1 4 1 16 4
H 0 6 0 36 0
I 1 3 1 9 3
J 1 2 1 4 2
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 4 78 4 404 19
Item 19
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.6−(2)(78)
√{[20.(2)− (2)2][20.(404)−(78)2]}
=
120−156
√{[120−4][8080−6084]}
=
−36
√{[116][1996]}
=
−36
√231536
=
−36
481,181
= -0,07
Item 20
r.xy=
N.∑xy−(∑x)(∑y)
√{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]}
=
20.46−(9)(78)
√{[20.(9)− (9)2][20.(404)−(78)2]}
=
920−702
√{[180−81][8080−6084]}
=
218
√{[99][1996]}
=
218
√197604
=
218
444,52
= 0,49
S X Y X2
Y2
X.Y
A 0 10 0 100 0
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 0 8 0 64 0
F 0 5 0 25 0
G 1 4 1 16 4
H 0 6 0 36 0
I 0 3 0 9 0
J 1 2 1 4 2
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 0 4 0 16 0
O 0 2 0 4 0
P 0 1 0 1 0
Q 0 6 0 36 0
R 0 1 0 1 0
S 0 3 0 9 0
T 0 5 0 25 0
∑ 2 78 2 404 6
S X Y X2
Y2
X.Y
A 1 10 1 100 10
B 0 4 0 16 0
C 0 2 0 4 0
D 0 3 0 9 0
E 1 8 1 64 8
F 1 5 1 25 5
G 1 4 1 16 4
H 0 6 0 36 0
I 0 3 0 9 0
J 1 2 1 4 2
K 0 3 0 9 0
L 0 2 0 4 0
M 0 4 0 16 0
N 1 4 1 16 4
O 1 2 1 4 2
P 0 1 0 1 0
Q 1 6 1 36 6
R 0 1 0 1 0
S 0 3 0 9 0
T 1 5 1 25 5
∑ 9 78 9 404 46
KUNCI JAWABAN TES PERSEPSI GAMBAR RUANG DAN BIDANG
1. B 11. D
2. E 12. D
3. E 13. A
4. D 14. D
5. E 15. A/C
6. B 16. A
7. A 17. A/D
8. E 18. C
9. D 19. C
10. C 20. A

More Related Content

What's hot

To mat diknas 1112 02
To mat diknas 1112 02To mat diknas 1112 02
To mat diknas 1112 02
Tri Bagus
 
Fungsi eksponen-dan-logaritma
Fungsi eksponen-dan-logaritmaFungsi eksponen-dan-logaritma
Fungsi eksponen-dan-logaritma
Arya Ananda
 
Ma5 vector-u-s54
Ma5 vector-u-s54Ma5 vector-u-s54
Ma5 vector-u-s54
S'kae Nfc
 
Potencias y radicales resueltos 1-5
Potencias y radicales resueltos 1-5Potencias y radicales resueltos 1-5
Potencias y radicales resueltos 1-5
Educación
 
Math quota-cmu-g-455
Math quota-cmu-g-455Math quota-cmu-g-455
Math quota-cmu-g-455
Rungroj Ssan
 

What's hot (17)

Function problem p
Function problem pFunction problem p
Function problem p
 
To mat diknas 1112 02
To mat diknas 1112 02To mat diknas 1112 02
To mat diknas 1112 02
 
Trial terengganu 2014 spm add math k1 skema [scan]
Trial terengganu 2014 spm add math k1 skema [scan]Trial terengganu 2014 spm add math k1 skema [scan]
Trial terengganu 2014 spm add math k1 skema [scan]
 
Real problem2 p
Real problem2 pReal problem2 p
Real problem2 p
 
IIT-JEE Mains 2017 Online Mathematics Previous Paper Day 1
IIT-JEE Mains 2017 Online Mathematics Previous Paper Day 1IIT-JEE Mains 2017 Online Mathematics Previous Paper Day 1
IIT-JEE Mains 2017 Online Mathematics Previous Paper Day 1
 
Estadistica U4
Estadistica U4Estadistica U4
Estadistica U4
 
E1 f8 bộ binh
E1 f8 bộ binhE1 f8 bộ binh
E1 f8 bộ binh
 
Fungsi eksponen-dan-logaritma
Fungsi eksponen-dan-logaritmaFungsi eksponen-dan-logaritma
Fungsi eksponen-dan-logaritma
 
Ma5 vector-u-s54
Ma5 vector-u-s54Ma5 vector-u-s54
Ma5 vector-u-s54
 
Potencias y radicales resueltos 1-5
Potencias y radicales resueltos 1-5Potencias y radicales resueltos 1-5
Potencias y radicales resueltos 1-5
 
Trabajo matemáticas 7
Trabajo matemáticas 7Trabajo matemáticas 7
Trabajo matemáticas 7
 
Math quota-cmu-g-455
Math quota-cmu-g-455Math quota-cmu-g-455
Math quota-cmu-g-455
 
Arithmetic Progressions and the Construction of Doubly Even Magic Squares
Arithmetic Progressions and the Construction of Doubly Even Magic SquaresArithmetic Progressions and the Construction of Doubly Even Magic Squares
Arithmetic Progressions and the Construction of Doubly Even Magic Squares
 
ゲーム理論BASIC 第15回 -展開形ゲームにおける戦略と期待利得-
ゲーム理論BASIC 第15回 -展開形ゲームにおける戦略と期待利得-ゲーム理論BASIC 第15回 -展開形ゲームにおける戦略と期待利得-
ゲーム理論BASIC 第15回 -展開形ゲームにおける戦略と期待利得-
 
Algebra and Trigonometry 9th Edition Larson Solutions Manual
Algebra and Trigonometry 9th Edition Larson Solutions ManualAlgebra and Trigonometry 9th Edition Larson Solutions Manual
Algebra and Trigonometry 9th Edition Larson Solutions Manual
 
Radicales dobles racionalizacion widmar aguilar
Radicales dobles racionalizacion widmar aguilarRadicales dobles racionalizacion widmar aguilar
Radicales dobles racionalizacion widmar aguilar
 
IIT-JEE Mains 2017 Online Mathematics Previous Paper Day 2
IIT-JEE Mains 2017 Online Mathematics Previous Paper Day 2IIT-JEE Mains 2017 Online Mathematics Previous Paper Day 2
IIT-JEE Mains 2017 Online Mathematics Previous Paper Day 2
 

Viewers also liked

Output uji validitas dan reliabilitas spss
Output uji validitas dan reliabilitas spssOutput uji validitas dan reliabilitas spss
Output uji validitas dan reliabilitas spss
Chenk Alie Patrician
 
CONTOH BLUE PRINT DAN RUMUS EXCEL VALIDITAS
CONTOH BLUE PRINT DAN RUMUS EXCEL VALIDITAS CONTOH BLUE PRINT DAN RUMUS EXCEL VALIDITAS
CONTOH BLUE PRINT DAN RUMUS EXCEL VALIDITAS
Nur Arifaizal Basri
 

Viewers also liked (20)

V&P Advogados
V&P AdvogadosV&P Advogados
V&P Advogados
 
Material didactico(1)
Material didactico(1)Material didactico(1)
Material didactico(1)
 
Campaign for child labor free brgys provincial 2016
Campaign for child labor  free brgys provincial 2016Campaign for child labor  free brgys provincial 2016
Campaign for child labor free brgys provincial 2016
 
Kualitas alat ukur
Kualitas alat ukurKualitas alat ukur
Kualitas alat ukur
 
Ppt sociopolitica
Ppt sociopoliticaPpt sociopolitica
Ppt sociopolitica
 
Agroecologia
AgroecologiaAgroecologia
Agroecologia
 
Aterros sanitarios apostila
Aterros sanitarios apostilaAterros sanitarios apostila
Aterros sanitarios apostila
 
Output uji validitas dan reliabilitas spss
Output uji validitas dan reliabilitas spssOutput uji validitas dan reliabilitas spss
Output uji validitas dan reliabilitas spss
 
Corrossion focus on polarization
Corrossion focus on polarizationCorrossion focus on polarization
Corrossion focus on polarization
 
Validitas dan reliabilitas
Validitas dan reliabilitasValiditas dan reliabilitas
Validitas dan reliabilitas
 
Capitulo1
Capitulo1Capitulo1
Capitulo1
 
CONTOH BLUE PRINT DAN RUMUS EXCEL VALIDITAS
CONTOH BLUE PRINT DAN RUMUS EXCEL VALIDITAS CONTOH BLUE PRINT DAN RUMUS EXCEL VALIDITAS
CONTOH BLUE PRINT DAN RUMUS EXCEL VALIDITAS
 
Validitas dan realibilitas
Validitas dan realibilitasValiditas dan realibilitas
Validitas dan realibilitas
 
Tentang Uji Validitas dan Reliabilitas
Tentang Uji Validitas dan ReliabilitasTentang Uji Validitas dan Reliabilitas
Tentang Uji Validitas dan Reliabilitas
 
Patología: linfoma, sarcoma
Patología: linfoma, sarcomaPatología: linfoma, sarcoma
Patología: linfoma, sarcoma
 
STOPAH
STOPAHSTOPAH
STOPAH
 
La abdicación de la izquierda dani rodrik
La abdicación de la izquierda dani rodrikLa abdicación de la izquierda dani rodrik
La abdicación de la izquierda dani rodrik
 
Presentación nuestro blog.
Presentación nuestro blog.Presentación nuestro blog.
Presentación nuestro blog.
 
Prepositions
PrepositionsPrepositions
Prepositions
 
Sales Playbook Rapid Scale strategies video simulation
Sales Playbook Rapid Scale strategies video simulationSales Playbook Rapid Scale strategies video simulation
Sales Playbook Rapid Scale strategies video simulation
 

Similar to VALIDITAS NILAI TES

ข้อสอบเมทริกซ์
ข้อสอบเมทริกซ์ข้อสอบเมทริกซ์
ข้อสอบเมทริกซ์
K'Keng Hale's
 
Ciclotron dynamic 12000 h 2ω
Ciclotron   dynamic 12000 h 2ωCiclotron   dynamic 12000 h 2ω
Ciclotron dynamic 12000 h 2ω
Muniz Rodrigues
 
Copy of mpc 22-04-2018 jee adv-p-1 _ 2_cc_ans _ sol
Copy of mpc 22-04-2018 jee adv-p-1 _ 2_cc_ans _ solCopy of mpc 22-04-2018 jee adv-p-1 _ 2_cc_ans _ sol
Copy of mpc 22-04-2018 jee adv-p-1 _ 2_cc_ans _ sol
RAVIPUROHIT22
 
Sesión de aprendizaje de Radicación Algebra pre u ccesa007
Sesión de aprendizaje de Radicación  Algebra pre u  ccesa007Sesión de aprendizaje de Radicación  Algebra pre u  ccesa007
Sesión de aprendizaje de Radicación Algebra pre u ccesa007
Demetrio Ccesa Rayme
 

Similar to VALIDITAS NILAI TES (20)

AR勉強会第4回part1
AR勉強会第4回part1AR勉強会第4回part1
AR勉強会第4回part1
 
Salesforce Big Object 最前線
Salesforce Big Object 最前線Salesforce Big Object 最前線
Salesforce Big Object 最前線
 
ข้อสอบเมทริกซ์
ข้อสอบเมทริกซ์ข้อสอบเมทริกซ์
ข้อสอบเมทริกซ์
 
IFIR法による逆回復特性測定回路図
IFIR法による逆回復特性測定回路図IFIR法による逆回復特性測定回路図
IFIR法による逆回復特性測定回路図
 
セオリー・オブ・チェンジ(ToC)とは(2020.6.9. 神戸大学経営学部内田ゼミ)
セオリー・オブ・チェンジ(ToC)とは(2020.6.9. 神戸大学経営学部内田ゼミ)セオリー・オブ・チェンジ(ToC)とは(2020.6.9. 神戸大学経営学部内田ゼミ)
セオリー・オブ・チェンジ(ToC)とは(2020.6.9. 神戸大学経営学部内田ゼミ)
 
Theory and Methods for Unsupervised Anomaly Detection in Sounds Based on Deep...
Theory and Methods for Unsupervised Anomaly Detection in Sounds Based on Deep...Theory and Methods for Unsupervised Anomaly Detection in Sounds Based on Deep...
Theory and Methods for Unsupervised Anomaly Detection in Sounds Based on Deep...
 
ブロックチェーン: 「 書き換え不可能な記録」によって 社会はどう変化するか?
ブロックチェーン: 「書き換え不可能な記録」によって社会はどう変化するか? ブロックチェーン: 「書き換え不可能な記録」によって社会はどう変化するか?
ブロックチェーン: 「 書き換え不可能な記録」によって 社会はどう変化するか?
 
OSC 2018 Nagoya rsyncやシェルでバックアップするよりも簡単にOSSのBaculaでバックアップしてみよう
OSC 2018 Nagoya rsyncやシェルでバックアップするよりも簡単にOSSのBaculaでバックアップしてみようOSC 2018 Nagoya rsyncやシェルでバックアップするよりも簡単にOSSのBaculaでバックアップしてみよう
OSC 2018 Nagoya rsyncやシェルでバックアップするよりも簡単にOSSのBaculaでバックアップしてみよう
 
Blockchain economy
Blockchain economyBlockchain economy
Blockchain economy
 
Prelude to halide_public
Prelude to halide_publicPrelude to halide_public
Prelude to halide_public
 
Ejercicios prueba de algebra de la UTN- widmar aguilar
Ejercicios prueba de algebra de la UTN-  widmar aguilarEjercicios prueba de algebra de la UTN-  widmar aguilar
Ejercicios prueba de algebra de la UTN- widmar aguilar
 
Ciclotron dynamic 12000 h 2ω
Ciclotron   dynamic 12000 h 2ωCiclotron   dynamic 12000 h 2ω
Ciclotron dynamic 12000 h 2ω
 
Db2 Warehouse v3.0 SMP 導入ガイド 20190104 Db2 Warehouse SMP v3.0 configration Ins...
Db2 Warehouse v3.0 SMP 導入ガイド 20190104 Db2 Warehouse SMP v3.0 configration Ins...Db2 Warehouse v3.0 SMP 導入ガイド 20190104 Db2 Warehouse SMP v3.0 configration Ins...
Db2 Warehouse v3.0 SMP 導入ガイド 20190104 Db2 Warehouse SMP v3.0 configration Ins...
 
Project management
Project managementProject management
Project management
 
Copy of mpc 22-04-2018 jee adv-p-1 _ 2_cc_ans _ sol
Copy of mpc 22-04-2018 jee adv-p-1 _ 2_cc_ans _ solCopy of mpc 22-04-2018 jee adv-p-1 _ 2_cc_ans _ sol
Copy of mpc 22-04-2018 jee adv-p-1 _ 2_cc_ans _ sol
 
Sesión de aprendizaje de Radicación Algebra pre u ccesa007
Sesión de aprendizaje de Radicación  Algebra pre u  ccesa007Sesión de aprendizaje de Radicación  Algebra pre u  ccesa007
Sesión de aprendizaje de Radicación Algebra pre u ccesa007
 
On Repetitive Right Application of B-terms (for PPL 2019)
On Repetitive Right Application of B-terms (for PPL 2019)On Repetitive Right Application of B-terms (for PPL 2019)
On Repetitive Right Application of B-terms (for PPL 2019)
 
ゲーム理論BASIC 演習37 -3人ゲームの混合戦略ナッシュ均衡を求める-
ゲーム理論BASIC 演習37 -3人ゲームの混合戦略ナッシュ均衡を求める-ゲーム理論BASIC 演習37 -3人ゲームの混合戦略ナッシュ均衡を求める-
ゲーム理論BASIC 演習37 -3人ゲームの混合戦略ナッシュ均衡を求める-
 
Kertas Percubaan Matematik Tambahan Kedah Skema K2
Kertas Percubaan Matematik Tambahan Kedah Skema K2Kertas Percubaan Matematik Tambahan Kedah Skema K2
Kertas Percubaan Matematik Tambahan Kedah Skema K2
 
Key pat1 1-53
Key pat1 1-53Key pat1 1-53
Key pat1 1-53
 

More from Nur Arifaizal Basri

MODEL LAYANAN BK SMA guru penggerak kurikulum meredeka belajar
MODEL LAYANAN BK SMA guru penggerak kurikulum meredeka belajarMODEL LAYANAN BK SMA guru penggerak kurikulum meredeka belajar
MODEL LAYANAN BK SMA guru penggerak kurikulum meredeka belajar
Nur Arifaizal Basri
 
FORMAT LAPORAN ALAT PERAGA BK DENGAN PANDUAN BIMBINGAN KARIER.docx
FORMAT LAPORAN ALAT PERAGA BK DENGAN PANDUAN BIMBINGAN KARIER.docxFORMAT LAPORAN ALAT PERAGA BK DENGAN PANDUAN BIMBINGAN KARIER.docx
FORMAT LAPORAN ALAT PERAGA BK DENGAN PANDUAN BIMBINGAN KARIER.docx
Nur Arifaizal Basri
 
Pengembangan Buku Panduan Bimbingan Karier Berdasarkan Teori Trait and factor
Pengembangan Buku Panduan Bimbingan Karier Berdasarkan Teori Trait and factorPengembangan Buku Panduan Bimbingan Karier Berdasarkan Teori Trait and factor
Pengembangan Buku Panduan Bimbingan Karier Berdasarkan Teori Trait and factor
Nur Arifaizal Basri
 
Laporan hasil tindak lanjut analisis pelaksanaan program bk
Laporan hasil tindak lanjut analisis pelaksanaan program bkLaporan hasil tindak lanjut analisis pelaksanaan program bk
Laporan hasil tindak lanjut analisis pelaksanaan program bk
Nur Arifaizal Basri
 
cognitive behavioral therapy for social anxiety disorder (CBT)
 cognitive behavioral therapy for social anxiety disorder (CBT) cognitive behavioral therapy for social anxiety disorder (CBT)
cognitive behavioral therapy for social anxiety disorder (CBT)
Nur Arifaizal Basri
 

More from Nur Arifaizal Basri (20)

CONTOH RPL KLASIKAL
CONTOH RPL KLASIKALCONTOH RPL KLASIKAL
CONTOH RPL KLASIKAL
 
contoh RPL BIMBINGAN KELOMPOK.pdf
contoh RPL  BIMBINGAN KELOMPOK.pdfcontoh RPL  BIMBINGAN KELOMPOK.pdf
contoh RPL BIMBINGAN KELOMPOK.pdf
 
contoh RPL konseling individu.pdf
contoh RPL konseling individu.pdfcontoh RPL konseling individu.pdf
contoh RPL konseling individu.pdf
 
Permendikbud No 15 Tahun 2018.pdf
Permendikbud No 15 Tahun 2018.pdfPermendikbud No 15 Tahun 2018.pdf
Permendikbud No 15 Tahun 2018.pdf
 
UU ASN NO. 5 TH. 2014
UU ASN NO. 5 TH. 2014UU ASN NO. 5 TH. 2014
UU ASN NO. 5 TH. 2014
 
program kerja BK 2022-2023.pdf
program kerja BK 2022-2023.pdfprogram kerja BK 2022-2023.pdf
program kerja BK 2022-2023.pdf
 
MODEL LAYANAN BK SMA guru penggerak kurikulum meredeka belajar
MODEL LAYANAN BK SMA guru penggerak kurikulum meredeka belajarMODEL LAYANAN BK SMA guru penggerak kurikulum meredeka belajar
MODEL LAYANAN BK SMA guru penggerak kurikulum meredeka belajar
 
FORMAT LAPORAN ALAT PERAGA BK DENGAN PANDUAN BIMBINGAN KARIER.docx
FORMAT LAPORAN ALAT PERAGA BK DENGAN PANDUAN BIMBINGAN KARIER.docxFORMAT LAPORAN ALAT PERAGA BK DENGAN PANDUAN BIMBINGAN KARIER.docx
FORMAT LAPORAN ALAT PERAGA BK DENGAN PANDUAN BIMBINGAN KARIER.docx
 
Pengembangan Buku Panduan Bimbingan Karier Berdasarkan Teori Trait and factor
Pengembangan Buku Panduan Bimbingan Karier Berdasarkan Teori Trait and factorPengembangan Buku Panduan Bimbingan Karier Berdasarkan Teori Trait and factor
Pengembangan Buku Panduan Bimbingan Karier Berdasarkan Teori Trait and factor
 
Laporan hasil tindak lanjut analisis pelaksanaan program bk
Laporan hasil tindak lanjut analisis pelaksanaan program bkLaporan hasil tindak lanjut analisis pelaksanaan program bk
Laporan hasil tindak lanjut analisis pelaksanaan program bk
 
self control
self controlself control
self control
 
Carl gustav jung psychology and the occult
Carl gustav jung psychology and the occultCarl gustav jung psychology and the occult
Carl gustav jung psychology and the occult
 
kepercayan diri
kepercayan dirikepercayan diri
kepercayan diri
 
self-efficacy, and self-esteem
self-efficacy, and self-esteemself-efficacy, and self-esteem
self-efficacy, and self-esteem
 
mengenal kecemasan komunikasi
mengenal kecemasan komunikasimengenal kecemasan komunikasi
mengenal kecemasan komunikasi
 
KECEMASAN KOMUNIKASI
KECEMASAN KOMUNIKASIKECEMASAN KOMUNIKASI
KECEMASAN KOMUNIKASI
 
cognitive behavioral therapy for social anxiety disorder (CBT)
 cognitive behavioral therapy for social anxiety disorder (CBT) cognitive behavioral therapy for social anxiety disorder (CBT)
cognitive behavioral therapy for social anxiety disorder (CBT)
 
EXPLORING CAREERS WITH TYPOLOGY (JOHN HOLLAND)
EXPLORING CAREERS WITH TYPOLOGY (JOHN HOLLAND)EXPLORING CAREERS WITH TYPOLOGY (JOHN HOLLAND)
EXPLORING CAREERS WITH TYPOLOGY (JOHN HOLLAND)
 
VOCATIONAL INDECISION (JOHN HOLLAND)
VOCATIONAL INDECISION (JOHN HOLLAND)VOCATIONAL INDECISION (JOHN HOLLAND)
VOCATIONAL INDECISION (JOHN HOLLAND)
 
PERSONALITY AND VOCATIONAL John holland 1993
PERSONALITY AND VOCATIONAL John holland 1993PERSONALITY AND VOCATIONAL John holland 1993
PERSONALITY AND VOCATIONAL John holland 1993
 

Recently uploaded

Making and Justifying Mathematical Decisions.pdf
Making and Justifying Mathematical Decisions.pdfMaking and Justifying Mathematical Decisions.pdf
Making and Justifying Mathematical Decisions.pdf
Chris Hunter
 
Beyond the EU: DORA and NIS 2 Directive's Global Impact
Beyond the EU: DORA and NIS 2 Directive's Global ImpactBeyond the EU: DORA and NIS 2 Directive's Global Impact
Beyond the EU: DORA and NIS 2 Directive's Global Impact
PECB
 
The basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptxThe basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptx
heathfieldcps1
 

Recently uploaded (20)

Making and Justifying Mathematical Decisions.pdf
Making and Justifying Mathematical Decisions.pdfMaking and Justifying Mathematical Decisions.pdf
Making and Justifying Mathematical Decisions.pdf
 
Unit-IV- Pharma. Marketing Channels.pptx
Unit-IV- Pharma. Marketing Channels.pptxUnit-IV- Pharma. Marketing Channels.pptx
Unit-IV- Pharma. Marketing Channels.pptx
 
psychiatric nursing HISTORY COLLECTION .docx
psychiatric  nursing HISTORY  COLLECTION  .docxpsychiatric  nursing HISTORY  COLLECTION  .docx
psychiatric nursing HISTORY COLLECTION .docx
 
Grant Readiness 101 TechSoup and Remy Consulting
Grant Readiness 101 TechSoup and Remy ConsultingGrant Readiness 101 TechSoup and Remy Consulting
Grant Readiness 101 TechSoup and Remy Consulting
 
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
 
Nutritional Needs Presentation - HLTH 104
Nutritional Needs Presentation - HLTH 104Nutritional Needs Presentation - HLTH 104
Nutritional Needs Presentation - HLTH 104
 
ICT role in 21st century education and it's challenges.
ICT role in 21st century education and it's challenges.ICT role in 21st century education and it's challenges.
ICT role in 21st century education and it's challenges.
 
Web & Social Media Analytics Previous Year Question Paper.pdf
Web & Social Media Analytics Previous Year Question Paper.pdfWeb & Social Media Analytics Previous Year Question Paper.pdf
Web & Social Media Analytics Previous Year Question Paper.pdf
 
Class 11th Physics NEET formula sheet pdf
Class 11th Physics NEET formula sheet pdfClass 11th Physics NEET formula sheet pdf
Class 11th Physics NEET formula sheet pdf
 
Holdier Curriculum Vitae (April 2024).pdf
Holdier Curriculum Vitae (April 2024).pdfHoldier Curriculum Vitae (April 2024).pdf
Holdier Curriculum Vitae (April 2024).pdf
 
INDIA QUIZ 2024 RLAC DELHI UNIVERSITY.pptx
INDIA QUIZ 2024 RLAC DELHI UNIVERSITY.pptxINDIA QUIZ 2024 RLAC DELHI UNIVERSITY.pptx
INDIA QUIZ 2024 RLAC DELHI UNIVERSITY.pptx
 
PROCESS RECORDING FORMAT.docx
PROCESS      RECORDING        FORMAT.docxPROCESS      RECORDING        FORMAT.docx
PROCESS RECORDING FORMAT.docx
 
ICT Role in 21st Century Education & its Challenges.pptx
ICT Role in 21st Century Education & its Challenges.pptxICT Role in 21st Century Education & its Challenges.pptx
ICT Role in 21st Century Education & its Challenges.pptx
 
Micro-Scholarship, What it is, How can it help me.pdf
Micro-Scholarship, What it is, How can it help me.pdfMicro-Scholarship, What it is, How can it help me.pdf
Micro-Scholarship, What it is, How can it help me.pdf
 
Beyond the EU: DORA and NIS 2 Directive's Global Impact
Beyond the EU: DORA and NIS 2 Directive's Global ImpactBeyond the EU: DORA and NIS 2 Directive's Global Impact
Beyond the EU: DORA and NIS 2 Directive's Global Impact
 
Unit-V; Pricing (Pharma Marketing Management).pptx
Unit-V; Pricing (Pharma Marketing Management).pptxUnit-V; Pricing (Pharma Marketing Management).pptx
Unit-V; Pricing (Pharma Marketing Management).pptx
 
Food Chain and Food Web (Ecosystem) EVS, B. Pharmacy 1st Year, Sem-II
Food Chain and Food Web (Ecosystem) EVS, B. Pharmacy 1st Year, Sem-IIFood Chain and Food Web (Ecosystem) EVS, B. Pharmacy 1st Year, Sem-II
Food Chain and Food Web (Ecosystem) EVS, B. Pharmacy 1st Year, Sem-II
 
The basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptxThe basics of sentences session 2pptx copy.pptx
The basics of sentences session 2pptx copy.pptx
 
Python Notes for mca i year students osmania university.docx
Python Notes for mca i year students osmania university.docxPython Notes for mca i year students osmania university.docx
Python Notes for mca i year students osmania university.docx
 
How to Give a Domain for a Field in Odoo 17
How to Give a Domain for a Field in Odoo 17How to Give a Domain for a Field in Odoo 17
How to Give a Domain for a Field in Odoo 17
 

VALIDITAS NILAI TES

  • 1. HASIL TES DARI MAHASISWA PRODI MATEMATIKA ANGKATAN 2015 TES PERSEPSI GAMBAR RUANG DAN BIDANG N = 20 ∑ X = 78 ∑ X2 = 404 No X X2 1 10 100 2 4 16 3 2 4 4 3 9 5 8 64 6 5 25 7 4 16 8 6 36 9 3 9 10 2 4 11 3 9 12 2 4 13 4 16 14 4 16 15 2 4 16 1 1 17 6 36 18 1 1 19 3 9 20 5 25 Jumlah 78 404
  • 2. Mean = ∑X N = 78 20 = 3,9 SD = √N.∑(X2)−(∑X)2 N(N−1) = √20.404−(78)2 20(20−1) = √8080−6084 20.19 = √1996 380 = √5,25 = 2,29 SKALA 5 BS = 3,9 + (1,8 x 2,29) = 9,12 ke atas B = 3,9 + (0,6 x 2,29) = 5,274 C = 3,9 - (0,6 x 2,29) = 2,526 K = 3,9 - (1,8 x 2,29) = -0,22 KS = Dibawah -0,22
  • 3. DISTRAKTOR DAN VALIDITAS Distraktor Rangking 1-10 subyek 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A D B E D D A A D D C D D B D C D A C B A E E E E D E B A E D D C C B A B B C D D A H A E B E E D A D A D A D D D C D D D A D Q B E D B D A C D A B D D A D B E E B B A F E B C B D A B A B C B B C D C A A D B A T C D C D D B A E D A B B C E D B C B B A B A E D D C B D C A C D A C A B D C A A C G B E B B B A A C B C D D A D A A A C C A M C C E D D A E A D C A A B B E E C E B A N C C E D A A B E D A B A E E B D C B B A Rangking 11-20 subyek 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 D E B E D E D C B C A A A B A D B C D D B I E B E A B B A E D C A D D D A D D C A B K A B E D A D A B E A D D D C B D B A A C S B C E D D E C A E C A B B C E D B B A E C A B E D D C C D A B D C C C D D B B B D J B E E C E B A E D D D A D C A D C C C A L A C E D A D D C C B B E D C B B B D E E O A B E B D C A B E D A A E B E D B A E B P C E D A B A E D C B B A D A B E C D D B R A C D C A E C B D D A B B E D B C D A B Subyek Nomer item 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A D B E D D A A D D C D D B D C D A C B A B A E D D C B D C A C D A C A B D C A A C C A B E D D C C D A B D C C C D D B B B D D E B E D E D C B C A A A B A D B C D D B E E E E D E B A E D D C C B A B B C D D A F E B C B D A B A B C B B C D C A A D B A G B E B B B A A C B C D D A D A A A C C A H A E B E E D A D A D A D D D C D D D A D I E B E A B B A E D C A D D D A D D C A B J B E E C E B A E D D D A D C A D C C C A K A B E D A D A B E A D D D C B D B A A C L A C E D A D D C C B B E D C B B B D E E M C C E D D A E A D C A A B B E E C E B B N C C E D A A B E D A B A E E B D C B B A O A B E B D C A B E D A A E B E D B A E A P C E D A B A E D C B B A D A B E C D D B Q B E D B D A C D A B D D A D B E E B B A R A C D C A E C B D D A B B E D B C D A B S B C E D D E C A E C A B B C E D B B A E T C D C D D B A E D A B B C E D B C B B A Kunci B E E D E B A E D C D D A D A/C A A/D C C A
  • 4. 1. DISTRAKTOR 1 KET A B* C D E Atas 2 2 3 1 2 Bawah 5 2 1 0 2 2 KET A B C D E* Atas 0 2 2 1 5 Bawah 0 5 3 O 2 3 KET A B C D E* Atas 0 2 2 2 4 Bawah 0 0 0 2 8 4 KET A B C D* E Atas 0 2 0 6 1 Bawah 2 1 2 5 0 5 KET A B C D E* Atas 1 1 1 5 2 Bawah 3 2 0 3 2 6 KET A B* C D E Atas 6 3 0 1 0 Bawah 1 2 2 3 2 7 KET A* B C D E Atas 5 2 1 1 1 Bawah 4 0 4 1 1 8 KET A B C D E* Atas 2 0 2 3 3 Bawah 1 4 1 2 2 9 KET A B C D* E Atas 3 2 0 4 0 Bawah 1 0 3 3 3 10 KET A B C* D E Atas 2 1 5 2 0 Bawah 2 3 2 3 0 11 KET A B C D* E Atas 2 3 1 4 0 Bawah 5 2 0 3 0 12 KET A B C D* E Atas 3 2 3 4 0 Bawah 3 2 1 2 1 13 KET A* B C D E Atas 2 3 3 0 1 Bawah 0 3 1 5 1 14 KET A B C D* E Atas 2 4 0 5 2 Bawah 2 1 5 1 1 15 KET A* B C* D E Atas 1 4 3 1 1 Bawah 2 3 0 3 2 16 KET A* B C D E Atas 2 2 0 3 2 Bawah 0 3 0 6 1
  • 5. 17 KET A* B C D* E Atas 3 0 5 1 1 Bawah 0 5 4 1 0 18 KET A B C* D E Atas 1 3 2 2 1 Bawah 2 2 2 4 0 19 KET A B C* D E Atas 2 6 1 1 0 Bawah 4 1 1 2 2 20 KET A* B C D E Atas 8 0 1 1 0 Bawah 1 5 1 1 2 Kategori soal P di atas 0,5 maka soal dikatakan mudah P sama dengan 0,5 maka soal dikatakan sedang P di bawah 0,5 maka soal dikatakan sulit P1= 𝑩 𝐓 = 𝟒 𝟐𝟎 = 0,2 (Sukar) P8= 𝑩 𝐓 = 𝟓 𝟐𝟎 = 0,25 (Sukar) P15= 𝑩 𝐓 = 𝟔 𝟐𝟎 = 0,3 (Sukar) A = Berfungsi D = Berfungsi A = Berfungsi C = Berfungsi B = Berfungsi E = Berfungsi C = Berfungsi E = Berfungsi B = Berfungsi D = Berfungsi D = Berfungsi P2= 𝑩 𝐓 = 𝟕 𝟐𝟎 = 0,35 (sukar) P9= 𝑩 𝐓 = 𝟕 𝟐𝟎 = 0,35 (Sukar) P16= 𝑩 𝐓 = 𝟐 𝟐𝟎 = 0,1 (Sukar) A = Revisi C = Berfungsi A = Berfungsi C = Berfungsi B = Berfungsi D = Berfungsi B = Berfungsi D = Berfungsi B = Berfungsi E = Berfungsi C = Revisi E = Berfungsi P3= 𝑩 𝐓 = 𝟏𝟐 𝟐𝟎 = 0,6 (Sedang) P10= 𝑩 𝐓 = 𝟕 𝟐𝟎 = 0,35 (Sukar) P17= 𝑩 𝐓 = 𝟓 𝟐𝟎 = 0,25 (Sukar) A = Revisi C = Berfungsi A = Berfungsi D = Berfungsi B = Berfungsi E = Berfungsi B = Berfungsi D= Berfungsi B = Berfungsi E =Revisi C = Berfungsi P4= 𝑩 𝐓 = 𝟏𝟏 𝟐𝟎 = 0,55 (Sedang) P11= 𝑩 𝐓 = 𝟕 𝟐𝟎 = 0,35 (Sukar) P18= 𝑩 𝐓 = 𝟒 𝟐𝟎 = 0,2 (Sukar) A = Berfungsi C = Berfungsi A = Berfungsi C = Berfungsi A = Berfungsi D = Berfungsi B = Berfungsi E = Berfungsi B = Berfungsi E =Revisi B = Berfungsi E = Berfungsi P5= 𝑩 𝐓 = 𝟒 𝟐𝟎 = 0,2 (Sukar) P12= 𝑩 𝐓 = 𝟔 𝟐𝟎 = 0,3 (Sukar) P19= 𝑩 𝐓 = 𝟐 𝟐𝟎 = 0,1 (Sukar) A = Berfungsi C = Berfungsi A = Berfungsi C = Berfungsi A = Berfungsi D = Berfungsi B = Berfungsi D = Berfungsi B = Berfungsi E = Berfungsi B = Berfungsi E = Berfungsi P6= 𝑩 𝐓 = 𝟓 𝟐𝟎 = 0,25 (Sukar) P13= 𝑩 𝐓 = 𝟐 𝟐𝟎 = 0,1 (Sukar) P20= 𝑩 𝐓 = 𝟗 𝟐𝟎 = 0,45 ( Sukar) A = Berfungsi D = Berfungsi B = Berfungsi D = Berfungsi B = Berfungsi D = Berfungsi C = Berfungsi E = Berfungsi C = Berfungsi E = Berfungsi C = Berfungsi E = Berfungsi P7= 𝑩 𝐓 = 𝟗 𝟐𝟎 = 0,45 (sukar) P14= 𝑩 𝐓 = 𝟔 𝟐𝟎 = 0,3 (Sukar) B = Berfungsi D = Berfungsi A = Berfungsi C = Berfungsi C = Berfungsi E = berfungsi B = Berfungsi E = Berfungsi
  • 6. 2. Validasi Item 1 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.15−(4)(78) √{[20.(4)− (4)2][20.(404)−(78)2]} = 300−312 √{[80− 16][8080−6084]} = 12 √{[64][1996]} = 12 √127744 = 12 357,41 = 0,33 Item 2 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.29−(7)(78) √{[20.(7)− (7)2][20.(404)−(78)2]} = 580−546 √{[140− 49][8080−6084]} = 34 √{[91][1996]} = 34 √18636 = 52 426,19 = 0,122 S X Y X2 Y2 X.Y A 0 10 0 100 0 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 0 5 0 25 0 G 1 4 1 16 4 H 0 6 0 36 0 I 0 3 0 9 0 J 1 2 1 4 2 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 1 6 1 36 6 R 0 1 0 1 0 S 1 3 1 9 3 T 0 5 0 25 0 ∑ 4 78 4 404 15 S X Y X2 Y2 X.Y A 0 10 0 100 0 B 0 4 0 16 0 C 0 2 0 4 0 D 1 3 1 9 3 E 1 8 1 64 8 F 1 5 1 25 5 G 0 4 0 16 0 H 0 6 0 36 0 I 1 3 1 9 3 J 0 2 0 4 0 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 1 1 1 1 1 Q 1 6 1 36 6 R 0 1 0 1 0 S 1 3 0 9 3 T 0 5 0 25 0 ∑ 7 78 6 404 29
  • 7. Item 3 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.44−(12)(78) √{[20.(12)− (12)2][20.(404)−(78)2]} = 880−936 √{[240− 144][8080−6084]} = −56 √{[96][1996]} = −56 √191616 = −56 437,73 =-0,127 Item 4 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.48−(11)(78) √{[20.(11)− (11)2][20.(404)−(78)2]} = 960−858 √{[220− 121][8080−6084]} = 102 √{[99][1996]} = 102 √197604 = 102 444,52 = 0,22 S X Y X2 Y2 X.Y A 1 10 100 100 10 B 0 4 0 16 0 C 1 2 1 4 2 D 1 3 1 9 3 E 1 8 1 64 8 F 0 5 0 25 0 G 0 4 0 16 0 H 0 6 0 36 0 I 1 3 1 9 3 J 1 2 1 4 2 K 1 3 1 9 3 L 1 2 1 4 2 M 1 4 1 16 4 N 1 4 1 16 4 O 1 2 1 4 2 P 1 1 0 1 1 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 12 78 11 404 44 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 1 4 1 16 4 C 1 2 1 4 2 D 1 3 1 9 3 E 1 8 1 64 8 F 0 5 0 25 0 G 0 4 0 16 0 H 0 6 0 36 0 I 0 3 0 9 0 J 0 2 0 4 0 K 1 3 1 9 3 L 1 2 1 4 2 M 1 4 1 16 4 N 1 4 1 16 4 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 1 3 1 9 3 T 1 5 1 25 5 ∑ 11 78 11 404 48
  • 8. Item 5 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.19−(4)(78) √{[20.(4)− (4)2][20.(404)−(78)2]} = 380−312 √{[80− 16][8080−6084]} = 68 √{[64][1996]} = 68 √127744 = 68 357,41 = 0,21 Item 6 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.20−(5)(78) √{[20.(5)− (5)2][20.(404)−(78)2]} = 400−390 √{[100− 25][8080−6084]} = 10 √{[75][1996]} = 10 √127744 = 10 357,41 = 0,27 S X Y X2 Y2 X.Y A 0 10 0 100 0 B 0 4 0 16 0 C 0 2 0 4 0 D 1 3 1 9 3 E 1 8 1 64 8 F 0 5 0 25 0 G 0 4 0 16 0 H 1 6 1 36 6 I 0 3 0 9 0 J 1 2 1 4 2 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 4 78 4 404 19 S X Y X2 Y2 X.Y A 0 10 0 100 0 B 1 4 1 16 4 C 0 2 0 4 0 D 0 3 0 9 0 E 1 8 1 64 8 F 0 5 0 25 0 G 0 4 0 16 0 H 0 6 0 36 0 I 1 3 1 9 3 J 1 2 1 4 2 K 1 3 0 9 3 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 5 78 4 404 20
  • 9. Item 7 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.41−(9)(78) √{[20.(9)− (9)2][20.(404)−(78)2]} = 820−702 √{[180−81][8080−6084]} = 118 √{[99][1996]} = 118 √197604 = 118 444,52 = 0,26 Item 8 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.22−(5)(78) √{[20.(5)− (5)2][20.(404)−(78)2]} = 440−390 √{[100−25][8080−6084]} = 50 √{[75][1996]} = 50 √149700 = 50 386,91 = 0,129 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 1 8 1 64 8 F 0 5 0 25 0 G 1 4 1 16 4 H 1 6 1 36 6 I 1 3 1 9 3 J 1 2 1 4 2 K 1 3 1 9 3 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 1 2 1 4 2 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 1 3 0 9 3 T 0 5 0 25 0 ∑ 9 78 8 404 41 S X Y X2 Y2 X.Y A 0 10 0 100 0 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 1 8 1 64 8 F 0 5 0 25 0 G 0 4 0 16 0 H 0 6 0 36 0 I 1 3 1 9 3 J 1 2 1 4 2 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 1 4 1 16 4 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 1 5 1 25 5 ∑ 5 78 5 404 22
  • 10. Item 9 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.36−(7)(78) √{[20.(7)− (7)2][20.(404)−(78)2]} = 720−546 √{[140−49][8080−6084]} = 174 √{[91][1996]} = 174 √181636 = 174 426,21 = 0,41 Item 10 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.31−(7)(78) √{[20.(7)− (7)2][20.(404)−(78)2]} = 620−546 √{[140−49][8080−6084]} = 74 √{[91][1996]} = 74 √181636 = 74 426,21 = 0,12 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 1 8 1 64 8 F 0 5 0 25 0 G 0 4 0 16 0 H 0 6 0 36 0 I 1 3 1 9 3 J 1 2 1 4 2 K 0 3 0 9 0 L 0 2 0 4 0 M 1 4 1 16 4 N 1 4 1 16 4 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 1 1 0 S 0 3 0 9 0 T 1 5 1 25 5 ∑ 7 78 8 404 36 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 1 4 1 16 4 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 1 5 1 25 5 G 1 4 1 16 4 H 0 6 0 36 0 I 1 3 1 9 3 J 0 2 0 4 0 K 0 3 0 9 0 L 0 2 0 4 0 M 1 4 1 16 4 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 1 1 1 1 1 S 0 3 1 9 0 T 0 5 0 25 0 ∑ 7 78 8 404 31
  • 11. Item 11 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.31−(7)(78) √{[20.(7)− (7)2][20.(404)−(78)2]} = 620−546 √{[140−49][8080−6084]} = 74 √{[91][1996]} = 74 √181636 = 74 426,198 = 0,23 Item 12 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.32−(6)(78) √{[20.(6)− (6)2][20.(404)−(78)2]} = 640−468 √{[120−36][8080−6084]} = 172 √{[84][1996]} = 172 √167664 = 172 409,46 = 0,42 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 1 4 1 16 4 C 1 2 1 4 2 D 0 3 0 9 0 E 0 8 0 64 0 F 0 5 0 25 0 G 1 4 1 16 4 H 0 6 0 36 0 I 0 3 0 9 0 J 1 2 1 4 2 K 1 3 1 9 3 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 1 6 1 36 6 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 7 78 7 404 31 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 0 5 0 25 0 G 1 4 1 16 4 H 1 6 1 36 6 I 1 3 1 9 3 J 0 2 0 4 0 K 1 3 1 9 3 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 1 6 1 36 6 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 6 78 6 404 32
  • 12. Item 13 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.10−(2)(78) √{[20.(2)− (2)2][20.(404)−(78)2]} = 200−156 √{[40−4][8080−6084]} = 44 √{[160][1996]} = 44 √319360 = 44 565,11 = 0,077 Item 14 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.34−(6)(78) √{[20.(6)− (6)2][20.(404)−(78)2]} = 680−468 √{[120−36][8080−6084]} = 212 √{[84][1996]} = 212 √167664 = 212 409,51 = 0,52 S X Y X2 Y2 X.Y A 0 10 0 100 0 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 0 5 0 25 0 G 1 4 1 16 4 H 0 6 0 36 0 I 0 3 0 9 0 J 0 2 0 4 0 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 1 6 1 36 6 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 2 78 2 404 10 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 1 5 1 25 5 G 1 4 1 16 4 H 1 6 1 36 6 I 1 3 1 9 3 J 0 2 0 4 0 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 1 6 1 36 6 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 6 78 6 404 34
  • 13. Item 15 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.30−(6)(78) √{[20.(6)− (6)2][20.(404)−(78)2]} = 600−468 √{[120−36][8080−6084]} = 132 √{[84][1996]} = 132 √167664 = 132 409,58 = 0,322 Item 16 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.9−(2)(78) √{[20.(2)− (2)2][20.(404)−(78)2]} = 180−156 √{[400−4][8080−6084]} = 24 √{[396][1996]} = 24 √790416 = 24 889,1 = 0,27 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 1 5 1 25 5 G 1 4 1 16 4 H 1 6 1 36 6 I 1 3 1 9 3 J 1 2 1 4 2 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 6 78 6 404 30 S X Y X2 Y2 X.Y A 0 10 0 100 0 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 1 5 1 25 5 G 1 4 1 16 4 H 0 6 0 36 0 I 0 3 0 9 0 J 0 2 0 4 0 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 2 78 2 404 9
  • 14. Item 17 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.28−(5)(78) √{[20.(5)− (5)2][20.(404)−(78)2]} = 560−390 √{[100−25][8080−6084]} = 170 √{[75][1996]} = 170 √149700 = 170 386,91 = 0,43 Item 18 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.19−(4)(78) √{[20.(4)− (4)2][20.(404)−(78)2]} = 380−312 √{[80−16][8080−6084]} = 68 √{[64][1996]} = 68 √127744 = 68 357,41 = 0,22 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 1 5 1 25 5 G 1 4 1 16 4 H 1 6 1 36 6 I 1 3 1 9 3 J 0 2 0 4 0 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 5 78 5 404 28 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 0 5 0 25 0 G 1 4 1 16 4 H 0 6 0 36 0 I 1 3 1 9 3 J 1 2 1 4 2 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 4 78 4 404 19
  • 15. Item 19 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.6−(2)(78) √{[20.(2)− (2)2][20.(404)−(78)2]} = 120−156 √{[120−4][8080−6084]} = −36 √{[116][1996]} = −36 √231536 = −36 481,181 = -0,07 Item 20 r.xy= N.∑xy−(∑x)(∑y) √{[N.(∑X2)− (∑X)2][N.(∑Y2)−(∑Y)2]} = 20.46−(9)(78) √{[20.(9)− (9)2][20.(404)−(78)2]} = 920−702 √{[180−81][8080−6084]} = 218 √{[99][1996]} = 218 √197604 = 218 444,52 = 0,49 S X Y X2 Y2 X.Y A 0 10 0 100 0 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 0 8 0 64 0 F 0 5 0 25 0 G 1 4 1 16 4 H 0 6 0 36 0 I 0 3 0 9 0 J 1 2 1 4 2 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 0 4 0 16 0 O 0 2 0 4 0 P 0 1 0 1 0 Q 0 6 0 36 0 R 0 1 0 1 0 S 0 3 0 9 0 T 0 5 0 25 0 ∑ 2 78 2 404 6 S X Y X2 Y2 X.Y A 1 10 1 100 10 B 0 4 0 16 0 C 0 2 0 4 0 D 0 3 0 9 0 E 1 8 1 64 8 F 1 5 1 25 5 G 1 4 1 16 4 H 0 6 0 36 0 I 0 3 0 9 0 J 1 2 1 4 2 K 0 3 0 9 0 L 0 2 0 4 0 M 0 4 0 16 0 N 1 4 1 16 4 O 1 2 1 4 2 P 0 1 0 1 0 Q 1 6 1 36 6 R 0 1 0 1 0 S 0 3 0 9 0 T 1 5 1 25 5 ∑ 9 78 9 404 46
  • 16. KUNCI JAWABAN TES PERSEPSI GAMBAR RUANG DAN BIDANG 1. B 11. D 2. E 12. D 3. E 13. A 4. D 14. D 5. E 15. A/C 6. B 16. A 7. A 17. A/D 8. E 18. C 9. D 19. C 10. C 20. A