38  LM5. F-test

Consider a dataset for a chemical reaction, where \(Y\) = percent of unchanged starting material, \(X_1\) = temperature, \(X_2\) = concentration of a reagent, and \(X_3\) = time of reaction.

\[Y_i=\beta_0+\beta_1 X_{i1}+\beta_2 X_{i2}+\beta_3 X_{i3}+\epsilon_i,\ i=1, \cdots, n=19\]

where \(\mathbf{\epsilon}=(\epsilon_1,\cdots,\epsilon_n)^T \sim N(\mathbf{0},\sigma^2\mathbf{I})\).

After fitting the model, the following results are obtained:

\[\hat{\mathbf{\beta}}=\begin{pmatrix} 332.11\\ -1.55\\ -1.42\\ -2.24 \end{pmatrix} ,\ \hat{Var}(\hat{\mathbf{\beta}})=\begin{pmatrix} 349.43 & -1.81 &-1.67 &-0.11 \\ -1.81& 0.0098 & 0.0068 &-0.0023 \\ -1.67 & 0.0068 & 0.022& -0.0094\\ -0.11 & -0.0023 & -0.0094 & 0.12 \end{pmatrix} \]

and \(MSE = 5.34\), \(R^2 = 0.96\).

1. Compute the F test statistic for the hypothesis \(H_0:\beta_1=\beta_2=\beta_3=0\), and give the reference distribution against which this test statistic should be compared for a test of \(H_0\).
2. Express the hypothesis \(H_0: 2\beta_1=\beta_3=-3\) in the form of the general linear hypothesis.
3. Compute the F test statistic for \(H_0\) from part (b) and give the reference distribution for this test statistic.

38.1 Linear model. Solutions

\[F=\frac{R^2/p}{(1-R^2)/(n-p-1)}=\frac{0.96/3}{0.04/(19-3-1)}=120\]

Reference distribution \(F(3,15)\)

\(H_0:C\beta =t\) where \[\begin{equation*} C=\begin{pmatrix} 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 1\\ \end{pmatrix},\ t=\begin{pmatrix} -3 \\ -3 \\ \end{pmatrix},\ \beta=(\beta_0,\beta_1,\beta_2,\beta_3)^T \end{equation*}\]

\[F=\frac{[(C \hat{\beta}-t)^T[C(X^T X)^{-1}C^T]^{-1}(C\hat{\beta}-t)]/2}{MSE} \]

\(\hat Var(\hat \beta) = MSE (X^TX)^{-1} \rightarrow (X^T X)^{-1} = \frac{1}{MSE} \begin{pmatrix} 349.43 & -1.81 &-1.67 &-0.11 \\ -1.81& 0.0098 & 0.0068 &-0.0023 \\ -1.67 & 0.0068 & 0.022& -0.0094\\ -0.11 & -0.0023 & -0.0094 & 0.12 \end{pmatrix}\)

\[C(X^T X)^{-1}C^T= \frac{1}{MSE} \begin{pmatrix} 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 1\\ \end{pmatrix} \begin{pmatrix} 349.43 & -1.81 &-1.67 &-0.11 \\ -1.81& 0.0098 & 0.0068 &-0.0023 \\ -1.67 & 0.0068 & 0.022& -0.0094\\ -0.11 & -0.0023 & -0.0094 & 0.12 \end{pmatrix} \begin{pmatrix} 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 1\\ \end{pmatrix}^T = \frac{1}{MSE} \begin{pmatrix} 0.0392 & -0.0046\\ -0.0046& 0.12 \\ \end{pmatrix} \]

\[[C(X^T X)^{-1}C^T]^{-1}=\frac{MSE}{(0.0392)(0.12)-(0.0046)^2}\begin{pmatrix} 0.12 & 0.0046\\ 0.0046& 0.392 \\ \end{pmatrix} \]

\[C\hat{\beta}-t=\begin{pmatrix} -0.1 \\ 0.76 \end{pmatrix} \] \[F=\frac{(-0.1, 0.76) \begin{pmatrix} 0.12 & 0.0046\\ 0.0046& 0.392 \\ \end{pmatrix} \begin{pmatrix} -0.1 \\ 0.76 \end{pmatrix} / 2 }{0.0392(0.12)-(0.0046)^2}=2.57 \] Reference distribution \(F(2,15)\)