Page 121 - Metode Penelitian Kuantitatif, Kualitatif, dan Penelitian Gabungan by Muri Yusuf (z-lib.org)
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BAGIAN KEDUA:GIAN KEDUA: METODE PENELITIAN KUANTITATIFMETODE PENELITIAN KUANTITATIF
B
A
∑ ∑
7
y
,
9
0
)
∑ ∑ y == ∑ ∑ y y −− ( ( y) 2 2 = = 9 96,4751−− 3 30,79 2 2 = = 1 1,67269
,
2 2
2 2
9
6
2
6
6
1
,
5
4
7
7
y
N N 10
1
0
(∑∑
∑ ∑
x
)
x
)
8
3
6
7
4
2
x
1
1 1 2 ∑
∑ ∑ x x x == ∑ x x x −− ( ( x )( x ) = = 8 87658 −− 1273 x 684 = = 5 584,8
2 2
1 1
8
x
6
5
x
7
4
8
,
8
2
1 1 2
2
N N 1 0
10
∑ ∑
(∑∑
)
)
x
y
2
7
7
9
3
x
3
1
,
0
∑ ∑ x x y == ∑ ∑ x x y −− ( ( x )( y) = = 3 3949,85−− 1273 x 30,79 = = 3 30,283
9
1 1
4
y
y
8
5
9
,
8
0
3
2
,
1 1
1 1
N N 10
0
1
(∑∑
∑ ∑
x
y
)
3
0
4
,
9
6
7
8
x
∑ ∑ x x y == ∑ ∑ x x y −− ( ( x ))( y) = = 2 2135,22−− 684 x 30,79 = = 2 29,184
1
2 2
3
,
4
2
2
8
5
9
y
1
y
,
2 2
2 2
1
0
N N 10
n
a
k
t
t
a
a
a
h
r
m
g
a
u
k
u
k
s
y
a
n
n
m
a
a
a
n
pe
s
r
i
n
s
m
l
t
a
u
m
e
k
n
e
d
e
n
t
u
m
u
a
a
l
u
S
a
j
e
n
Selanjutnya masuk ke dalam persamaan simultan untuk menentukan harga a
l
1 1
:
a
n
a
d
dan a :
2 2
∑
x
∑
∑
x
y
a
=
2 2
x
.
+
a
1. ∑x y = a ∑x + a ∑x x
1
x
1 1 1 1 1 1 2 2 1 1 2 2
y
a
=
a
+
x
2. ∑x y = a ∑x x + a ∑xx 2 2
∑
∑
2
x
x
.
∑
2 2 1 1 1 1 2 2 2 2 2
2
=
3
,
8
a
8
1
a
3 0 , 2 8 3 = 832,1 a + 584,8 a ( D i b a g i 5 8 4 , 8 )
8
(Dibagi 584,8)
30,283
,
5
+
4
2
1 1 2 2
(Dibagi 730,4)
29,184
= 584,8 a + 730,4 a
2 9 , 1 8 4 = 5 8 4 , 8 a + 7 3 0 , 4 a ( D i b a g i 7 3 0 , 4 )
1 1 2 2
0 , 0 5 1 7 8 3 5 1 5 = 1 , 1 4 2 2 8 7 9 6 1 a + a
0,051783515 = 1,142287961 a + a
1 1 2 2
6
8
0
0
,
a
4
=
0,03995619
1
0 , 0 3 9 9 5 6 1 9 = 0,800657174 a + a
7
5
0
a
7
+
1 1 2 2 – –
a
0
7
3
0
6
,
0
1
8
0,011827325 = 0,341630787 a
7
3
,
3
1
8
0
2
7
5
=
1
4
2
1 1
a a = 0 , 0 0 3 4 6 2 0 1 9 6
= 0,0034620196
1 1
= 0,051783515 – (0,034620196 x 1,42287961 = 0,012237281
a a = 0 , 0 5 1 7 8 3 5 1 5 – ( 0 , 0 3 4 6 2 0 1 9 6 x 1 , 4 2 2 8 7 9 6 1 = 0 , 0 1 2 2 3 7 2 8 1
2 2
x
x
a
=
y y = a x + a x
a
+
1 1 1 2 2 2
2
1
0
(
X
1
Y
X
0
X
–
–
,
1
)
=
0
,
0
+
8
2
)
7
2
2
3
)
6
–
Y
(
3
9
X
0
4
(
6
(Y – Y) = 0,034620196 (X – X ) + 0,012237281 (X – X )
2
1
1 1 1 1 2 2 2 2
Y = 0 , 0 3 4 6 2 0 1 9 6 X – 4 . 4 0 7 1 5 0 9 5 1 + 0 , 0 1 2 2 3 7 2 8 1 X – 0 , 8 3 7 0 3 0 0 2
Y = 0,034620196 X – 4.407150951 + 0,012237281 X – 0,83703002
1 1 2 2
+ 3 , 0 7 9
+ 3,079
,
4
X
–
6
1
7
2
9
0
=
7
Y
5
2
0
2
,
1
1
3
X
,
2
1
3
0
Y = 0,034620196 X + 0,012237281 X – 2,165180971
6
6
8
+
2
0
9
0
1
0
1
8
1 1 2 2
Y = 0,034620196 X + 0,012237281 X – 2,165181
Y = 0 , 0 3 4 6 2 0 1 9 6 X + 0 , 0 1 2 2 3 7 2 8 1 X – 2 , 1 6 5 1 8 1
1 1 2 2
(dibulatkan).
( d i b u l a t k a n ) .
a
d
l
r
d
X
a
n
o
a
a
X
d
Adapun koefisien korelasi antara Y dan X dan X adalah:
e
a
a
a
e
pu
h
Y
s
n
t
e
i
a
:
n
r
i
f
k
s
A
d
l
n
n
a
a
k
i
o
1 1 2 2
0,034620196 x 30,283 0,012237281 x 29,184
0 , 0 3 4 6 2 0 1 9 6 x 3 0 , 2 8 3++ 0 , 0 1 2 2 3 7 2 8 1 x 2 9 , 1 8 4
=
R R (1,2) =
,
(
1
2
)
a
a
k y y 1 , 6 7 2 6 9
k
1,67269
a
t
t a
s
s
u
u
1,408950225
p
p 1 , 4 0 8 9 5 0 2 2 5
3
0
0
8
9
4
4
,
8
2
2
a
i a = = = = 0,840284932
i
s 1 , 6 7 2 6 9
s
1,67269
e
e
n
n
o
o
0,9166770
d = = 0 , 9 1 6 6 7 7 0
d
n
i n
i
/
/
m
m
o
o 2 2
Jadi, R (1,2)= 0,92 (dibulatkan), dan R (1,2)= 0,840.
c
c J a d i , R y y ( 1 , 2 ) = 0 , 9 2 ( d i b u l a t k a n ) , d a n Ryy ( 1 , 2 ) = 0 , 8 4 0 .
. .
