We have five observations on \(x\) and \(y\). They are \(x_{i}=3,2,1,-1,0\) with corresponding \(y\) values \(y_{i}=4,2,3,1,0\). The fitted least squares line is \(\hat{y}_{i}=1.2+0.8 x_{i}\), the sum of squared least squares residuals is \(\sum_{i=1}^{5} \hat{e}_{i}^{2}=3.6, \sum_{i=1}^{5}\left(x_{i}-\bar{x}\right)^{2}=10\), and \(\sum_{i=1}^{5}\left(y_{i}-\bar{y}\right)^{2}=10\). Carry out this exercise with a hand calculator. Compute

a. the predicted value of \(y\) for \(x_{0}=4\).

b. the \(\mathrm{se}(f)\) corresponding to part (a).

c. a \(95 \%\) prediction interval for \(y\) given \(x_{0}=4\).

d. a \(99 \%\) prediction interval for \(y\) given \(x_{0}=4\).

e. a \(95 \%\) prediction interval for \(y\) given \(x=\bar{x}\). Compare the width of this interval to the one computed in part (c).