Suppose you fit the model y =${\beta }_0+{\beta }_1{x}_1+{\beta }_1{x_2}^2+{\beta }_3{x}_2+{\beta }_4{x}_1{x}_2+\epsilon$to n = 25 data points with the following results:

${\hat{\beta }}_0$=1.26,${\hat{\beta }}_1$= -2.43,${\hat{\beta }}_2$=0.05,${\hat{\beta }}_3$=0.62,${\hat{\beta }}_4$=1.81s${\hat{\beta }}_1$=1.21,s${\hat{\beta }}_2$=0.16,s${\hat{\beta }}_3$=0.26, s${\hat{\beta }}_4$=1.49SSE=0.41 and R2=0.83

1. Is there sufficient evidence to conclude that at least one of the parameters b1, b2, b3, or b4 is nonzero? Test using a = .05.

2. Test H₀: β₁ = 0 against Ha: β₁ < 0. Use α = .05.

3. Test H₀: β₂ = 0 against Ha: β₂ > 0. Use α = .05.

4. Test H₀: β₃ = 0 against Ha: β₃ ≠ 0. Use α = .05.

$H_0:{\beta }_1={\beta }_2={\beta }_3={\beta }_4=0$

H_a: At least one of the parameters${\beta }_1$,${\beta }_2$,${\beta }_3$, and${\beta }_4$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 α=0.05, since F < F_0.05,28,28 Not sufficient evidence to reject H_o at 95% confidence interval.

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

H₀: ${\beta }_1$=0

H_a: ${\beta }_1$< 0

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

Value oft_0.05,25is 1.708

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

Therefore, ${\beta }_1$=0 .

H₀:${\beta }_2$= 0

H_a:${\beta }_2$> 0

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

Value oft_0.05,25is 1.708

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

Therefore, ${\beta }_2$=0 .

H₀:${\beta }_3$=0

H_a: ${\beta }_3$≠0

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

Value oft_0.025,25is 2.060

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

Therefore, ${\beta }_3$≠0 .