91影视

$H_0:{尾}_1={尾}_2={尾}_3={尾}_4=0$

H_a: At least one of the parameters${尾}_1$,${尾}_2$,${尾}_3$, and${尾}_4$is non zero

Here, F test statistic =$\frac{SSE}{n-(k+1)}=\frac{0.41}{25-5}=0.0205$

Value of F_0.05,25,25 is 1.964

H₀ is rejected if F statistic > F_0.05,28,28. For 伪=0.05, since F < F_0.05,28,28 Not sufficient evidence to reject H_o at 95% confidence interval.

Therefore, ${尾}_1$=${尾}_2$=${尾}_3$=${尾}_4$=0 .

H₀: ${尾}_1$=0

H_a: ${尾}_1$< 0

Here, t-test statistic = $\frac{{\widehat{尾}}_1}{s{\widehat{尾}}_1}=\frac{-2.43}{1.21}=-2.008$

Value oft_0.05,25is 1.708

H₀ is rejected if t statistic > t_0.05,25. For 伪=0.05, since t < t_0.05,25 Not sufficient evidence to reject H_o at 95% confidence interval.

Therefore, ${尾}_1$=0 .

H₀:${尾}_2$= 0

H_a:${尾}_2$> 0

Here, t-test statistic = $\frac{{\widehat{尾}}_2}{s{\widehat{尾}}_2}=\frac{0.05}{0.16}=0.3125$

Value oft_0.05,25is 1.708

H₀is rejected if t statistic > t_0.05,25. For 伪 = 0.05, since t < t_0.05,31Not sufficient evidence to rejectH_oat 95% confidence interval.

Therefore, ${尾}_2$=0 .

H₀:${尾}_3$=0

H_a: ${尾}_3$鈮�0

Here, t-test statistic =$\frac{{\widehat{尾}}_3}{s{\widehat{尾}}_3}=\frac{0.62}{0.26}=2.38461$

Value oft_0.025,25is 2.060

H₀is rejected if t statistic > t_0.05,24,24. For 伪 = 0.05, since t>t_0.05,31 Sufficient evidence to rejectH_oat 95% confidence interval.

Therefore, ${尾}_3$鈮�0 .