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Po-128
A =
n
Σy
– B
.
Σx
–1
B =
Sxx
Sxy
r
=
Sxx = Σ(x
–1
)
2
–
Syy = Σy
2
–
Sxy = Σ(x
–1
)y –
Sxx
.
Syy
Sxy
n
(Σx
–1
)
2
n
Σx
–1
.
Σy
m =
y – A
B
n = A +
x
B
n
(Σy)
2
Regressão inversa (1/X)
A = exp
( )
n
Σ
ln y – B
.
Σ ln x
B =
n
.
Σ
(
ln
x
)
2
–
(
Σ ln x
)
2
n
.
Σ
ln x ln
y
– Σ
ln x
.
Σ ln y
r
=
{ n
.
Σ
(
ln
x
)
2
–
(
Σ ln x
)
2
}{ n
.
Σ
(
ln
y
)
2
–
(
Σ ln y
)
2
}
n
.
Σ
ln x ln
y
– Σ
ln x
.
Σ ln y
m = e
B
ln y – ln A
n = Ax
B
Regressão de potência (A•X^B)
r =
{n
.
Σx
2
–
(
Σx
)
2
}{n
.
Σ
(
ln
y
)
2
–
(
Σlny
)
2
}
n
.
Σxln
y
– Σx
.
Σ
lny
m =
lnB
lny – lnA
n = AB
x