CAC HE SO HOI QUI CUA NHAN BE MAT Rz: b0=28.977480 b1=8.491366 b2=-9.752964 b3=-10.234771 b12=0.217090 b13=-1.449336 b23=1.647837 b11=-0.160584 b22=-0.284649 b33=0.936091 He so xac dinh R^2= 0.775675
Gia tri ham so thong ke de kiem dinh gia thiet F0=1.921012 va gia tri p-value=0.244398
Kiem dinh ket qua CAC SO THONG KE
Phan bo Student voi muc y nghia 0.05 bac tu do 2 t_anpha=2.919986
Phan bo F voi muc y nghia 0.01 bac tu do 5 va 2 F_anpha=99.415852
Phuong sai du:0.009668
phuong sai tai sinh (binh phuong)=0.000148
t(1)=9232.840119 > t_anpha=2.919986, He so beta(1) co nghia t(2)=2184.187362 > t_anpha=2.919986, He so beta(2) co nghia t(3)=3987.194584 > t_anpha=2.919986, He so beta(3) co nghia t(4)=18690.547651 > t_anpha=2.919986, He so beta(4) co nghia t(5)=71.568653 > t_anpha=2.919986, He so beta(5) co nghia t(6)=2136.757828 > t_anpha=2.919986, He so beta(6) co nghia t(7)=3861.135494 > t_anpha=2.919986, He so beta(7) co nghia t(8)=47.489946 > t_anpha=2.919986, He so beta(8) co nghia t(9)=164.919739 > t_anpha=2.919986, He so beta(9) co nghia t(10)=9805.197496 > t_anpha=2.919986, He so beta(10) co nghia ---
---KIEM TRA y^:----
F^=S^2du/S^2ts =1.921012<F(anpha) = 99.415852, Mo hinh phu hop ---
SAU KHI MU HOA- Nham be mat co quan he theo dang:
Rz= e^(28.977480+8.491366*lnt-9.752964lnS-10.234771lnV+0.217090lnt*lnS- 1.449336lnt*lnV+1.647837lnS*lnV-0.160584lnt^2-0.284649lnS^2+0.936091lnV^2) ---