SPEARMAN BROWN
S-B = rxx’ = 2 (ry1y2) / (1+ry1y2)
rxx’ = koefisien reliabilitas spearman brown
ry1y2 = koefisien korelasi antara skor kedua belahan
|
Suby
|
Nomor Item
|
Belahan
|
X
|
||||||||||||
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
y1
|
y2
|
||
|
A
|
1
|
1
|
2
|
2
|
1
|
0
|
1
|
1
|
2
|
2
|
2
|
2
|
9
|
8
|
17
|
|
B
|
0
|
0
|
2
|
1
|
0
|
0
|
2
|
2
|
1
|
2
|
0
|
1
|
5
|
6
|
11
|
|
C
|
0
|
0
|
2
|
2
|
0
|
0
|
1
|
0
|
2
|
2
|
0
|
0
|
5
|
4
|
9
|
|
D
|
2
|
2
|
2
|
2
|
1
|
0
|
0
|
2
|
2
|
2
|
1
|
0
|
8
|
8
|
16
|
|
E
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
1
|
1
|
0
|
0
|
8
|
9
|
17
|
|
F
|
0
|
0
|
2
|
2
|
2
|
2
|
1
|
1
|
1
|
1
|
2
|
1
|
8
|
7
|
15
|
|
G
|
1
|
0
|
1
|
1
|
2
|
2
|
0
|
1
|
0
|
0
|
0
|
0
|
4
|
4
|
8
|
|
H
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
3
|
2
|
5
|
|
I
|
2
|
2
|
2
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
2
|
12
|
11
|
23
|
|
J
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
4
|
3
|
7
|
y1 = jumlah skor pada item nomor 1+3+5+7+9+11
y2 = jumlah skor pada item nomor 2+4+6+8+10+12
X = jumlah skor pada keseluruhan item, X = y1+y2
Koefisien korelasi ry1y2 = 0,957 (berasal dari rumus korelasi product
moment terhadap skor kedua belahan).
Jadi, koefisien reliabilitas tes adalah:
S-B = rxx’ = 2 (0,957) / (1+0,957)
S-B = rxx’ = 0,978
FORMULA ALPHA (α)
rxx’ ≥ (α) = 2 {1 –
(sy1²+sy2²) / sx²}
sy1 dan sy2²
= varians skor belahan 1 dan belahan 2
sx² =
varians skor tes
|
Suby
|
Nomor Item
|
Belahan
|
X
|
||||||||||||
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
y1
|
y2
|
||
|
A
|
1
|
1
|
2
|
2
|
1
|
0
|
1
|
1
|
2
|
2
|
2
|
2
|
9
|
8
|
17
|
|
B
|
0
|
0
|
2
|
1
|
0
|
0
|
2
|
2
|
1
|
2
|
0
|
1
|
5
|
6
|
11
|
|
C
|
0
|
0
|
2
|
2
|
0
|
0
|
1
|
0
|
2
|
2
|
0
|
0
|
5
|
4
|
9
|
|
D
|
2
|
2
|
2
|
2
|
1
|
0
|
0
|
2
|
2
|
2
|
1
|
0
|
8
|
8
|
16
|
|
E
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
1
|
1
|
0
|
0
|
8
|
9
|
17
|
|
F
|
0
|
0
|
2
|
2
|
2
|
2
|
1
|
1
|
1
|
1
|
2
|
1
|
8
|
7
|
15
|
|
G
|
1
|
0
|
1
|
1
|
2
|
2
|
0
|
1
|
0
|
0
|
0
|
0
|
4
|
4
|
8
|
|
H
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
3
|
2
|
5
|
|
I
|
2
|
2
|
2
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
2
|
12
|
11
|
23
|
|
J
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
4
|
3
|
7
|
Varians belahan y1 adalah sy1² = 7,24
Varians belahan y2 adalah sy2² = 7,56
Varians skor total X adalah sx² = 28,96
k1 = k2 = 6
Jadi, koefisien alpha untuk data ini adalah:
α = 2 {1
– (7,24+7,56) / 28,96}
α = 0,978
KOEFISIEN ALPHA PADA TES BELAH TIGA
rxx’ ≥ (α) = (3/2)
{1 – (sy1²+sy2²+SY3²) / sx²}
sy1, sy2²,
SY3² = varians skor belahan 1, 2, dan 3
sx² =
varians skor tes
|
Suby
|
Nomor
Item
|
Belahan
|
X
|
|||||||||||||
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
y1
|
y2
|
y3
|
||
|
A
|
1
|
1
|
2
|
2
|
1
|
0
|
1
|
1
|
2
|
2
|
2
|
2
|
6
|
5
|
6
|
17
|
|
B
|
0
|
0
|
2
|
1
|
0
|
0
|
2
|
2
|
1
|
2
|
0
|
1
|
5
|
2
|
4
|
11
|
|
C
|
0
|
0
|
2
|
2
|
0
|
0
|
1
|
0
|
2
|
2
|
0
|
0
|
5
|
0
|
4
|
9
|
|
D
|
2
|
2
|
2
|
2
|
1
|
0
|
0
|
2
|
2
|
2
|
1
|
0
|
6
|
6
|
4
|
16
|
|
E
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
1
|
1
|
0
|
0
|
7
|
5
|
5
|
17
|
|
F
|
0
|
0
|
2
|
2
|
2
|
2
|
1
|
1
|
1
|
1
|
2
|
1
|
4
|
5
|
6
|
15
|
|
G
|
1
|
0
|
1
|
1
|
2
|
2
|
0
|
1
|
0
|
0
|
0
|
0
|
2
|
3
|
3
|
8
|
|
H
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
1
|
2
|
2
|
5
|
|
I
|
2
|
2
|
2
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
2
|
8
|
7
|
8
|
23
|
|
J
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
3
|
2
|
2
|
7
|
y1 = jumlah skor pada item nomor
1+4+7+10
y2 = jumlah skor pada item nomor
2+5+8+11
y3 = jumlah skor pada item nomor
3+6+9+12
X = jumlah skor pada keseluruhan item, X = y1+y2+y3
Varians belahan y1 adalah sy1² = 4,899
Varians belahan y2 adalah sy2² = 4,899
Varians belahan y3 adalah sy3² = 3,599
Varians skor total X adalah sx² = 32,178
k1 = k2 = k3 = 4
Jadi, koefisien alpha untuk data ini adalah:
α = (3/2) {1
– (4,899+4,899+ 3,599) / 32,178}
α = 0,875
FORMULA RULON
rxx’ = 1 - sd²/sx²
sd² = varians perbedaan skor kedua belahan
sx² = varians skor tes
d = perbedaan skor kedua belahan
|
Suby
|
Nomor Item
|
Belahan
|
d =
(y1-y2)
|
X
|
||||||||||||
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
y1
|
y2
|
|||
|
A
|
1
|
1
|
2
|
2
|
1
|
0
|
1
|
1
|
2
|
2
|
2
|
2
|
9
|
8
|
1
|
17
|
|
B
|
0
|
0
|
2
|
1
|
0
|
0
|
2
|
2
|
1
|
2
|
0
|
1
|
5
|
6
|
-1
|
11
|
|
C
|
0
|
0
|
2
|
2
|
0
|
0
|
1
|
0
|
2
|
2
|
0
|
0
|
5
|
4
|
1
|
9
|
|
D
|
2
|
2
|
2
|
2
|
1
|
0
|
0
|
2
|
2
|
2
|
1
|
0
|
8
|
8
|
0
|
16
|
|
E
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
1
|
1
|
0
|
0
|
8
|
9
|
-1
|
17
|
|
F
|
0
|
0
|
2
|
2
|
2
|
2
|
1
|
1
|
1
|
1
|
2
|
1
|
8
|
7
|
1
|
15
|
|
G
|
1
|
0
|
1
|
1
|
2
|
2
|
0
|
1
|
0
|
0
|
0
|
0
|
4
|
4
|
0
|
8
|
|
H
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
3
|
2
|
1
|
5
|
|
I
|
2
|
2
|
2
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
2
|
12
|
11
|
1
|
23
|
|
J
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
4
|
3
|
1
|
7
|
y1 = jumlah skor pada item omor 1+3+5+7+9+11
y2 = jumlah skor pada item omor 2+4+6+8+10+12
d = perbedaan skor y1 dan y2
X = jumlah skor pada keseluruhan item, X = y1+y2
Varians skor d menghasilkan sd² = 0,711
Varians skor X sudah diketahui sx² = 32,178
Jadi, koefisien reliabilitas untuk data ini adalah:
rxx’ = 1 - 0,711 / 32,178
rxx’ = 0,978
FORMULA
KUDER-RICHARDSON
KR-20 = k 1 - Æ©p (1 - p)
(k - 1) sx²
sx²
= varians skor tes
k
= banyaknya item dalam tes
p
= proporsi subyek yang mendapat angka
1 pada suatu item
|
Suby
|
Nomor Item
|
X
|
|||||||||||
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
||
|
A
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
1
|
0
|
0
|
0
|
0
|
7
|
|
B
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
8
|
|
C
|
1
|
1
|
1
|
1
|
0
|
0
|
1
|
0
|
0
|
0
|
0
|
0
|
5
|
|
D
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
1
|
0
|
0
|
1
|
0
|
7
|
|
E
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
9
|
|
F
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
10
|
|
G
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
8
|
|
H
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
11
|
|
I
|
1
|
1
|
1
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
1
|
0
|
6
|
|
J
|
0
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
5
|
|
p
|
0,9
|
0,9
|
1
|
0,8
|
0,8
|
0,6
|
0,7
|
0,8
|
0,4
|
0,3
|
0,3
|
0,1
|
|
|
1-p
|
0,1
|
0,1
|
0
|
0,2
|
0,2
|
0,4
|
0,3
|
0,2
|
0,6
|
0,7
|
0,7
|
0,9
|
|
|
p(1-p)
|
0,09
|
0,09
|
0
|
0,16
|
0,16
|
0,24
|
0,21
|
0,16
|
0,24
|
0,21
|
0,21
|
0,09
|
|
Æ©p(1-p) = 1,86
sx²
= 3,64
k
= 12
KR-20 = [12
/ (12 - 1)][1 – (1,86 / 3,64)]
=
0,5345
KR-21 = k 1 - ká¹— (1 - á¹—)
(k - 1) sx²
k =
12
á¹— =
0,633
KR-21 = [12 /
(12 - 1)][1 – {12(0,633)(1 - 0,633) / 3,64}]
= 0,2554
FORMULA BELAH-DUA
DENGAN PANJANG BERBEDA (FELDT)
rxx’ = 4 (sy1y2)
sx² - [(sy1² - sy2²) / sx]
sy1²
= varians skor pada belahan 1
sy2²
= varians skor pada belahan 2
sy1y2 =
kovarians skor pada belahan 1 dan 2
= E(y1y2)
– E(x) . E(y)
E = nilai
rata-rata
sx =
deviasi standar skor tes
|
Suby
|
Nomor Item
|
Belahan
|
X
|
||||||||||||
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
y1
|
y2
|
||
|
A
|
1
|
1
|
2
|
2
|
1
|
0
|
1
|
1
|
2
|
2
|
2
|
2
|
9
|
8
|
17
|
|
B
|
0
|
0
|
2
|
1
|
0
|
0
|
2
|
2
|
1
|
2
|
0
|
1
|
5
|
6
|
11
|
|
C
|
0
|
0
|
2
|
2
|
0
|
0
|
1
|
0
|
2
|
2
|
0
|
0
|
5
|
4
|
9
|
|
D
|
2
|
2
|
2
|
2
|
1
|
0
|
0
|
2
|
2
|
2
|
1
|
0
|
8
|
8
|
16
|
|
E
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
1
|
1
|
0
|
0
|
8
|
9
|
17
|
|
F
|
0
|
0
|
2
|
2
|
2
|
2
|
1
|
1
|
1
|
1
|
2
|
1
|
8
|
7
|
15
|
|
G
|
1
|
0
|
1
|
1
|
2
|
2
|
0
|
1
|
0
|
0
|
0
|
0
|
4
|
4
|
8
|
|
H
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
3
|
2
|
5
|
|
I
|
2
|
2
|
2
|
2
|
2
|
2
|
2
|
1
|
2
|
2
|
2
|
2
|
12
|
11
|
23
|
|
J
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
0
|
0
|
0
|
0
|
4
|
3
|
7
|
y1 = jumlah
skor pada item nomor 1+3+5+7+9
y2 = jumlah
skor pada item nomor 2+4+6+8+10=11+12
X = jumlah
skor pada keseluruhan item, X = y1+y2
sy1² = 7,24
sy2² = 7,56
sy1y2 =
7,08
sx² = 28,96
sx = 5,38
k1 = 5
k2 = 7
rxx’ =
4 (7,08) / [28,96 - {(7,24 – 7,56)
/ 5,38}]
rxx’ =
0,9759
FORMULA KRISTOF UNTUK
BELAH TIGA
rxx’ = st²
sx²
st ² = sy1y2 . sy1y3 + sy1y2 . sy2y3 + sy1y3 . sy2y3 + 2 (sy1y2 + sy1y3 + sy2y3)
sy2y3 sy1y3 sy1y2
sy1y2
= kovarians belahan y1 dan belahan y2
sy1y3
= kovarians belahan y1 dan belahan y3
sy2y3 =
kovarians belahan y2 dan belahan y3
k1 = 3, k2= 4, k3 = 5
sy1y2
= 2,15
sy1y3
= 2,7
sy2y3 =
3,28
sx² =
28,96
|
Subyek
|
Belahan
|
X
|
||
|
y1
|
y2
|
y3
|
||
|
A
|
4,0
|
5,0
|
8,0
|
17
|
|
B
|
3,0
|
2,0
|
6,0
|
11
|
|
C
|
3,0
|
0,0
|
6,0
|
9
|
|
D
|
4,0
|
6,0
|
6,0
|
16
|
|
E
|
6,0
|
5,0
|
6,0
|
17
|
|
F
|
3,0
|
5,0
|
7,0
|
15
|
|
G
|
2,0
|
3,0
|
3,0
|
8
|
|
H
|
1,0
|
2,0
|
2,0
|
5
|
|
I
|
6,0
|
7,0
|
10,0
|
23
|
|
J
|
3,0
|
2,0
|
2,0
|
7
|
y1 adalah item nomor 1, 4, 7
y2 adalah item nomor 2, 5, 8, 11
y3 adalah item nomor 3, 6, 9, 10, 12
X adalah jumlah skor pada keseluruhan item, X =
y1+y2+y3
st
² = (2,15)(2,7) / 3,28 + (2,15)(3,28) / 2,7 + (2,7)(3,28) / 2,15
+
2 (2,15 + 2,7 + 3,28)
st
² = 24,7606
rxx’ = 24,7606/28,96
rxx’ = 0,85499
