2.10. In the above table, $x_i$ is the length of the femur and $y_i$ is the length of the humerus taken from five dinosaur fossils (Archaeopteryx) that preserved both bones. See Moore (2000, p. 99).
a) Complete the table and find the least squares estimators $\hat{\beta}_1$ and $\hat{\beta}_2$.
b) Predict the humerus length if the femur length is 60 .

2.11. Suppose that the regression model is $Y_i=7+\beta X_i+e_i$ for $i=$ $1, \ldots, n$ where the $e_i$ are iid $N\left(0, \sigma^2\right)$ random variables. The least squares criterion is $Q(\eta)=\sum_{i=1}^n\left(Y_i-7-\eta X_i\right)^2$.
a) What is $E\left(Y_i\right)$ ?
b) Find the least squares estimator $\beta$ of $\beta$ by setting the first derivative $\frac{d}{d \eta} Q(\eta)$ equal to zero.
c) Show that your $\hat{\beta}$ is the global minimizer of the least squares criterion $Q$ by showing that the second derivative $\frac{d^2}{d \eta^2} Q(\eta)>0$ for all values of $\eta$.

2.12. The location model is $Y_i=\mu+e_i$ for $i=1, \ldots, n$ where the $e_i$ are iid with mean $E\left(e_i\right)=0$ and constant variance $\operatorname{VAR}\left(e_i\right)=\sigma^2$. The least squares estimator $\hat{\mu}$ of $\mu$ minimizes the least squares criterion $Q(\eta)=\sum_{i=1}^n\left(Y_i-\eta\right)^2$. To find the least squares estimator, perform the following steps.

a) Find the derivative $\frac{a}{d \eta} Q$, set the derivative equal to zero and solve for
$\eta$. Call the solution $\hat{\mu}$.
b) To show that the solution was indeed the global minimizer of $Q$, show that $\frac{d^2}{d \eta^2} Q>0$ for all real $\eta$. (Then the solution $\hat{\mu}$ is a local min and $Q$ is convex, so $\hat{\mu}$ is the global min.)

2.14. Suppose that the regression model is $Y_i=10+2 X_{i 2}+\beta_3 X_{i 3}+e_i$ for $i=1, \ldots, n$ where the $e_i$ are iid $N\left(0, \sigma^2\right)$ random variables. The least squares criterion is $Q\left(\eta_3\right)=\sum_{i=1}^n\left(Y_i-10-2 X_{\mathrm{i} 2}-\eta_3 X_{i 3}\right)^2$. Find the least squares estimator $\hat{\beta}_3$ of $\beta_3$ by setting the first derivative $\frac{d}{d \eta_3} Q\left(\eta_3\right)$ equal to zero. Show that your $\hat{\beta}_3$ is the global minimizer of the least squares criterion $Q$ by showing that the second derivative $\frac{d^2}{d \eta_3^2} Q\left(\eta_3\right)>0$ for all values of $\eta_3$.