tXt1Xt2Ytxt1xt2ytxt1

xt2

yt

xt1xt2xt1yxt2y

1855-1,4-1,3-1,81,961,693,241,822,522,34

101278+2,6+0,7+1,26,760,491,441,823,120,84

suma94636800024,420,133,610,824,816,6

media9,46,36,80002,442,013,361,082,481,66

Yt=B0+B1Xt1+B2Xt2+Et, t=1,10

Vom determina prin metoda celor mai mici patrate estimatiile coeficientilor celor mai mici patrate:1) XX =

EMBED Equation.3 2) = 24,4*20,1-10,8=373,8 3) (XX)= formula = * ( ) => ()4) Xy = formula= () 5) B= formula= () * () => 6 ) B0= formula=-3.5393206 7) Scrie ec. de regresie obtinuta: Yt= -3.5393206+0.8539317Xt1+0.3670414Xt2

8) Determinam analiza dispersionala: Q = 33.6 9) Q1 = 27.2703984 10) Q2= 6.3296016

11) Determinam estimatia variantei erorilor: s= 0.9042288

12) Determinam estimatia matricei de covarianta a lui B: V(B)= 0.9042288*()= ()13) Determinam estimatia variantei a lui B1 si B2: V(B1)= 0.9042288*0.0537720= 0.048622191

V(B2)=0.90422880*0.0652755= 0.0590239

14) Determinam estimatia variantei a lui B0: V(B0)= 0.9042288*(3.9200613+1/10)= 3.6350552

(XX) = ( ; )*(XX) *()

15) Determinam abateriel standart ale estimatiilor parametrului model:

S(B1)=0.220504401 ; S(B2)=0.2429483 : S(B0)=1.9065820

16) Testul Student. Determinam valorile statisticii student pt parametric model:

tB1=3.8726288; tB2=1.5107811: tB0=-1.8563694

17) Determinam valoarea critica: tcrit= 2.365 , Fixam pragul de semnificatie alfa = 0.05

Verificam ipoteza principala H0: B1=0 si ipoteza bilaterala de alternative H1: B10 cu pragul de semnificatie =0.05. Deoarece /tB1/=3.8726288>2.365=tcrit, ipoteza H0 se respinde si se accepta ipoteza H1. Deci parametrul B1 este semnificativ 0.

Verificam ipoteza principala H0: B2=0 si ipoteza bilaterala de alternative H1:B20 cu pragul de semnificatie =0.05. Deoarece /tB2/=1.5107811