Suppose you fit the model $y={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x_1}^2+{\beta }_3{x}_2+{\beta }_4{x}_1{x}_2+\epsilon$to n = 25 data points with the following results:

$$\begin{gathered} {\hat{\beta }}_0=1.26,{\hat{\beta }}_1=-2.43,{\hat{\beta }}_2=0.05,{\hat{\beta }}_3=0.62,{\hat{\beta }}_4=1.81 \\ s{\hat{\beta }}_1=1.21,s{\hat{\beta }}_2=0.16,s\widehat{{\beta }_3}=0.26,s{\hat{\beta }}_4=1.49 \\ SSE=0.41and{R}^2=0.83 \end{gathered}$$

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 H_a: β₁ < 0. Use α = .05.
3. Test H₀: β₂ = 0 against H_a: β₂ > 0. Use α = .05.
4. Test H₀: β₃ = 0 against H_a: β₃ ≠ 0. Use α = .05.

$$\begin{gathered} H_0:{\beta }_1={\beta }_2={\beta }_3={\beta }_4=0 \\ {H}_a:Atleastoneoftheparameters{\beta }_1,{\beta }_2,{\beta }_3,and{\beta }_{4}isnonzero \end{gathered}$$

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

Value of F_0.05,25,25 is 1.964

${\mathrm{H}}_0\mathrm{is}\mathrm{rejected}\mathrm{if}\mathrm{F}\mathrm{statistic}>{\mathrm{F}}_{0.05,28,28}.\mathrm{For}\mathrm{\alpha }=0.05,\mathrm{since}\mathrm{F}<{\mathrm{F}}_{0.05,28,28}$role="math" localid="1652120400407" $\mathrm{Not}\mathrm{sufficient}\mathrm{evidence}\mathrm{to}\mathrm{rejec}{\mathrm{H}}_{\mathrm{o}}\mathrm{at}95\mathrm{\%}\mathrm{confidence}\mathrm{interval}.$

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

$$\begin{gathered} {\mathrm{H}}_0:{\mathrm{\beta }}_1=0 \\ {\mathrm{H}}_{\mathrm{a}}:{\mathrm{\beta }}_1<0 \end{gathered}$$

Here, t-test statistic =$\frac{\widehat{{\beta }_1}}{2\widehat{{\beta }_1}}=\frac{-2.43}{1.21}=-2.008$

Value oft_0.05,25 is 1.708

$$\begin{gathered} {\text{H}}_{0}isrejectedift\mathrm{statistic}>t_{0.05,25}.\mathrm{For}\mathrm{\alpha }=0.05,\mathrm{since}t<t_{0.05,25.} \\ Notsufficientevidencetoreject{H}_{o}at95\%confidenceinterval. \\ Therefore,{\beta }_1=0 \end{gathered}$$

$$\begin{gathered} H_0:{\beta }_2=0 \\ {H}_a:{\beta }_2>0 \end{gathered}$$

Here, t-test statistic =$\frac{\widehat{{\beta }_2}}{2\widehat{{\beta }_2}}=\frac{0.05}{0.16}=0.3125$

Value oft_0.05,25 is 1.708

$$\begin{gathered} {\mathrm{H}}_0\mathrm{is}\mathrm{rejected}\mathrm{if}t\mathrm{statistic}>t_{0.05,25}.\mathrm{For}\mathrm{\alpha }=0.05,\mathrm{since}t<t_{0.05,31}. \\ \mathrm{Not}\mathrm{sufficient}\mathrm{evidence}\mathrm{to}\mathrm{reject}{\mathrm{H}}_{\mathrm{o}}\mathrm{at}95\mathrm{\%}\mathrm{confidence}\mathrm{interval}. \\ \mathrm{Therefore},{\mathrm{\beta }}_2=0. \end{gathered}$$

$$\begin{gathered} H_0:{\beta }_3=0 \\ {H}_a:{\beta }_3\ne 0 \end{gathered}$$

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

Value oft_0.025,25 is 2.060

$$\begin{gathered} H_{0}isrejectediftstatistic>t_{0.05,24,24}.For\alpha =0.05,sincet>t_{0.05,31}. \\ \mathrm{Sufficient}\mathrm{evidence}\mathrm{to}\mathrm{reject}{\mathrm{H}}_{\mathrm{o}}\mathrm{at}95\mathrm{\%}\mathrm{confidence}\mathrm{interval}. \\ \mathrm{Therefore},{\mathrm{\beta }}_3\ne 0. \end{gathered}$$