Wzory Na I Sprawdzian Z Pis

Wzory na I sprawdzian z pis.

12 S W := 2 3 , S := n (k − k )

k

n

−

∑ (∑ xij − x) 2 , x =1 j =1

f ( t ) = α 0 + α 1t , a = ( X′X ) X′y

Ve =

se 2 ,ϕ = y

∑( y t =1 n

t

∑( y t =1

t

− yˆ t ) − y)

P T

n

j= 1 i =1

ij

, χ 2 = Wn(k − 1)

1 n ( yt − yˆ t ) 2 , se = s = ∑ n − 2 t =1

a  =  0,  a1 

2 e

se2

,

2

(i ) −1 R 2 = 1 − ϕ 2 , var( a ) = se2 ( X′X ) , t =

, 2

a  yTP = ca = [1 T ] ⋅  0 , sD = T  a1 

(y

k

∑∑x k

i =1 −1

n

_

[

]

s 1 + c( X′X ) c′ 2 e

)

−1

, V DT =

s DT

ai , D ( ai )

, VI TP =

y TP

dI P T

,

P T

y

− u ⋅ s DT , yTP + u ⋅ s DT . ~2

yt = α 0 ⋅ α1 ⋅ e , ln yt = ln α 0 + t ln α1 + ξ t , s e = ξt

t

[

~

]

1 n ( ln yt − ln yˆ t ) 2 , ∑ n − 2 t =1

~

α1 ξ t −1 s DT = se2 1 + c( X′X ) c′ , s DT = s DT ⋅ yTP , yt = α 0 + α1 ln t + ξ t , yt = α 0 t e ,

1 ln yt = ln α 0 + α1 ln t + ξ t , yt = α 0 + α1 + ξ t , t y tlj

= f (T ) + g lj

^

j

+ ξ tij , et = y t − y t , g ′j = ij ij ij

1 N

N

∑ et , g j = g ′j − g ′ , g ′ = lj

l =1

1 m ∑ g ′j , m j =1

1 ytlj = f ( tlj ) w j + ξ tij , u t = yt / y t , w' j = ∑ u t , w j = w' j / w' , w' = ∑ w' j , N l =1 m j =1 N

^

ij

u=

∑(

1 yt − ytP m t∈I ep

)

, Vu =

ij

u y t∈I ep

ij

_

lj

_

1

m

m

∑g j =1

m

∑w j =1

j

j

=0 =m

sP 1 P 2 , s p = m ∑ ( yt − yt ) , V s P = y t∈I ep t∈I ep

n/2  2 n ~ 2 n ~  2π   2π   2π   2π  y t = f (t ) + ∑ α i sin  it  + β i cos it  , a i = ∑ y t sin  it , bi = ∑ y t cos it , n t =1 n t =1  n   n   n   n  i =1  a i2 + bi2 a n2 / 2 + bn2/ 2 1 n ~ 2 2 a n / 2 = 0 , bn / 2 = ∑ y t cos(π t ) , Ai = a i + bi , wi = , wn / 2 = , 2 sY2 sY2 n t =1