Solution. (a) \(\bar X = 3.5\) and \(\bar Y = 11\). The deviations are \(x = (-2.5, -1.5, -0.5, 0.5, 1.5, 2.5)\) and \(y = (-5, -2, -2, 1, 3, 5)\), so \(\sum x_i^2 = 17.5\), \(\sum x_iy_i = 12.5 + 3 + 1 + 0.5 + 4.5 + 12.5 = 34\) and \(\sum y_i^2 = 68\). Hence \(b = 34/17.5 = 1.9429\) and \(a = 11 - 1.9429(3.5) = 4.2\). The fitted values are \(6.143, 8.086, 10.029, 11.971, 13.914, 15.857\), and the residuals are \(-0.143, 0.914, -1.029, 0.029, 0.086, 0.143\), which sum to zero. ESS \(= b\sum x_iy_i = 66.057\), so RSS \(= 68 - 66.057 = 1.943\) and \(R^2 = 66.057/68 = 0.9714\).

(b) \(s^2 = 1.943/4 = 0.4857\), so \(\operatorname {se}(b) = \sqrt {0.4857/17.5} = 0.1666\). Then \(t = (1.9429 - 2)/0.1666 = -0.34\). Since \(|t| < t_{0.025,4} = 2.776\), the hypothesis \(\beta = 2\) is not rejected.

(c) \(\hat Y_0 = 4.2 + 1.9429(8) = 19.743\). The variance factor is \(1 + \frac 16 + (8 - 3.5)^2/17.5 = 2.3238\), so the interval is \(19.743 \pm 2.776\sqrt {0.4857 \times 2.3238} = 19.743 \pm 2.950 = (16.79,\ 22.69)\). \(X = 8\) lies outside the observed range \(1\) to \(6\), so the prediction is an extrapolation. The interval allows for sampling error in \(a\) and \(b\), but it assumes that the straight line continues beyond the data, which nothing in the data can confirm.

Solution. With \(Y^* = 1000Y\), the deviations are \(y^* = 1000y\), so \(b^* = \sum x_iy_i^*/\sum x_i^2 = 1000b\) and \(a^* = \bar Y^* - b^*\bar X = 1000a\). The residuals are \(e^* = 1000e\), so \(s^* = 1000s\) and both standard errors scale by \(1000\). Each \(t\) ratio is a ratio of two quantities scaled by \(1000\), so it is unchanged; \(R^2 = 1 - \text {RSS}/\text {TSS}\) is a ratio of two sums each scaled by \(10^6\). If instead \(X^* = 1000X\), then \(x^* = 1000x\) and \(b^* = b/1000\), \(\operatorname {se}(b^*) = s/\sqrt {10^6\sum x_i^2} = \operatorname {se}(b)/1000\). The intercept, its standard error, the residuals, \(s\), the \(t\) ratios and \(R^2\) are all unchanged. Changing units changes the size of a coefficient but never the evidence about it.

Solution. (a) Since \(b_{YX} = \sum xy/\sum x^2\) and \(b_{XY} = \sum xy/\sum y^2\), \[ b_{YX}b_{XY} = \frac {(\sum xy)^2}{\sum x^2\sum y^2} = r^2 . \]

(b) Both lines pass through \((\bar X, \bar Y)\). In the \((X, Y)\) plane the first has slope \(b_{YX}\) and the second, written as \(Y\) against \(X\), has slope \(1/b_{XY}\). They coincide only if \(b_{YX} = 1/b_{XY}\), that is \(r^2 = 1\). When \(|r| < 1\), \(1/b_{XY} = b_{YX}/r^2\) is steeper than \(b_{YX}\). The two regressions answer different questions, minimising vertical and horizontal distances respectively, and each is the best linear predictor of its own dependent variable.

(c) \(b_{YX}b_{XY} = 0.7 \times 1.3208 = 0.9245\), which is \(R^2\) from Example 2.2.5.

Solution. (a) Minimising \(\sum (Y_i - \beta X_i)^2\) gives the single normal equation \(\sum X_i(Y_i - \tilde bX_i) = 0\), so \(\tilde b = \sum X_iY_i/\sum X_i^2\). Substituting the model, \(\tilde b = \beta + \sum X_iu_i/\sum X_i^2\), with variance \(\sigma ^2\sum X_i^2/(\sum X_i^2)^2 = \sigma ^2/\sum X_i^2\).

(b) The only normal equation is \(\sum X_ie_i = 0\). There is no equation \(\sum e_i = 0\), because there is no intercept to estimate. The proof of Theorem 2.2.3 used \(\sum e_i = 0\) to drop a cross term, so TSS \(=\) ESS \(+\) RSS fails, and RSS can exceed \(\sum y_i^2\) when the line through the origin fits worse than a horizontal line at \(\bar Y\). Software therefore reports an “uncentred” \(R^2\), \(1 - \text {RSS}/\sum Y_i^2\), for such regressions, which is not comparable with the usual one.

(c) \(\sum X_iY_i = 6 + 16 + 36 + 48 + 90 = 196\) and \(\sum X_i^2 = 220\), so \(\tilde b = 0.891\). It is larger than \(b = 0.7\) because forcing the line through the origin, when the data suggest a positive intercept, steepens it. Note also that \(\sum X_i^2 = 220 > \sum x_i^2 = 40\), so \(\tilde b\) has a smaller variance than \(b\). That is a gain if \(\alpha \) really is zero; if it is not, \(\tilde b\) is biased.

Solution. \(\check \beta = \sum c_iY_i\) with \(c_1 = -1/(X_n - X_1)\), \(c_n = 1/(X_n - X_1)\) and all other \(c_i = 0\). Then \(\sum c_i = 0\) and \(\sum c_iX_i = (X_n - X_1)/(X_n - X_1) = 1\), so it is unbiased by the argument of Theorem 2.3.4. Its variance is \(\sigma ^2\sum c_i^2 = 2\sigma ^2/(X_n - X_1)^2\). With \(X = 2, 4, 6, 8, 10\), this is \(2\sigma ^2/64 = 0.03125\sigma ^2\), against \(\Var (b) = \sigma ^2/40 = 0.025\sigma ^2\). The end-point estimator is unbiased but \(25\%\) more variable, because it throws away the three middle observations.

Solution. \(\Var (b) = \sigma ^2/\sum x_i^2\) is minimised by making \(\sum (x_i - \bar x)^2\) as large as possible, which for eight points in \([3, 4]\) means four at \(3\) and four at \(4\). That is exactly the design used, with \(\sum x_i^2 = 2\). Within the interval nothing better is possible; the second researcher did better only by sampling a much wider range. The risk is that a design with only two distinct \(x\) values cannot detect curvature: any function at all can be passed through two points. A careful researcher places a few observations in the middle of the range, accepting a slightly larger \(\Var (b)\) in exchange for being able to check linearity.

Solution. (a) \(R^2 = r^2 = 0.317^2 = 0.100\), and by Theorem 2.4.2 \[ t = \frac {r\sqrt {n-2}}{\sqrt {1 - r^2}} = \frac {0.317\sqrt {48}}{\sqrt {0.900}} = 2.32 . \] The R command summary(lm(sr ~ pop75, data = LifeCycleSavings)) confirms both, up to rounding of \(r\).

(b) pop75 is correlated \(-0.908\) with pop15. Countries with many elderly people have few children, so the two-variable regression on pop75 picks up the effect of having few young dependants. The omitted variable pop15 sits in the disturbance and is correlated with the regressor, violating (A1). The slope is a correct estimate of the best linear predictor and a biased estimate of any effect of the elderly share. Including both variables, in a multiple regression, is the remedy.