β
1
=
∑
i
=
1
n
(
y
i
x
i
)
-
∑
i
=
1
n
(
y
i
)
∑
i
=
1
n
(
x
i
)
n
∑
i
=
1
n
(
x
i
2
)
-
(
∑
i
=
1
n
(
x
i
)
)
2
n