Solution. \(\sum T_i = -2 - 1 + 0 + 1 + 2 = 0\), so \(\bar T = 0\) and \(T_i - \bar T = T_i\). The numerator is \(\sum (Y_i - \bar Y)T_i = \sum Y_iT_i - \bar Y\sum T_i = \sum Y_iT_i\), and the denominator is \(\sum T_i^2 = 10\). So \(b = \sum Y_iT_i/10\) (Theorem 3.1.1). The intercept is \(a = \bar Y - b\bar T = \bar Y\), the fitted value of \(Y\) at \(T = 0\), the middle year of the sample.

Solution. (a) Code the years \(T = -2, \dots , 2\). The logarithms of consumption are \[ 3.4177,\quad 4.3920,\quad 5.1654,\quad 5.9078,\quad 6.7006, \] with mean \(5.1167\). Then \[ b = \frac {-2(3.4177) - 4.3920 + 5.9078 + 2(6.7006)}{10} = \frac {8.0816}{10} = 0.8082 , \] and \(\hat g = e^{0.8082} - 1 = 1.244\). Consumption grew by about \(124\%\) a year, more than doubling annually. The instantaneous rate is \(80.8\%\).

(b) 2019 is \(T = 3\), so \(\widehat {\log Y} = 5.1167 + 3(0.8082) = 7.5413\) and \(\hat Y = e^{7.5413} = 1884\).

(c) \(e^{0.8082/12} - 1 = 0.0697\), about \(7.0\%\) a month. Compounded over twelve months, \(1.0697^{12} = 2.244\), which recovers the annual factor.

(d) By Theorem 3.2.3, \(e^{s^2/2} = e^{0.00292} = 1.0029\), so the estimated mean is \(1884 \times 1.0029 = 1890\). The correction is under \(0.3\%\), small because the growth curve fits so closely. The much larger uncertainty is whether growth of \(124\%\) a year can continue at all: the forecast assumes the 2014–2018 pattern holds for another year.

Solution. By Theorem 3.4.1, \(\log \bigl (Y/(1 - Y)\bigr ) = \alpha + \beta X\). At \(X = 0\), \(\alpha = \log (0.2/0.8) = \log 0.25 = -1.386\). At \(X = 2\), \(\log (0.8/0.2) = \log 4 = 1.386 = \alpha + 2\beta \), so \(\beta = 1.386\). \(Y = \frac 12\) where the log-odds are zero, at \(X = -\alpha /\beta = 1\). That is the midpoint, by the symmetry of the logistic curve about its point of inflection.

Solution. (a) As \(U \to \infty \), \(\hat \pi \to 1.5\): wage inflation falls towards a floor of \(1.5\%\) but never below it.

(b) At \(U = 4\), \(\hat \pi = 1.5 + 3 = 4.5\). The marginal effect is \(-12/U^2 = -0.75\) percentage points of inflation per point of unemployment. The elasticity is \(-0.75 \times 4/4.5 = -0.667\).

(c) Inflation of \(3.5\) requires \(12/U = 2\), so \(U = 6\): unemployment must rise by two points. The marginal effect of \(-0.75\) would suggest \(1.33\) points, but it applies only to small changes, and the effect weakens as \(U\) rises.

Solution. An extra year of schooling multiplies the wage by \(e^{0.12} = 1.1275\), an exact effect of \(12.7\%\); the approximation \(100b\) gives \(12\%\). Formal sector workers earn \(e^{0.35} - 1 = 41.9\%\) more than otherwise similar informal workers, against an approximation of \(35\%\). The exact figures should be reported, especially for the dummy variable: a dummy changes discretely from \(0\) to \(1\), so the derivative-based approximation has no special claim to validity, and here it understates the gap by seven percentage points.

Solution. By Theorem 3.4.1, \(dY/dX = \beta Y(1 - Y)\), so \(\eta = \beta Y(1 - Y)X/Y = \beta X(1 - Y)\). As \(Y \to 0\), \(1 - Y \to 1\) and \(\eta \to \beta X\), the elasticity of \(\log Y = \alpha + \beta X\) (Theorem 3.3.2). For small \(Y\), \(Y = e^z/(1 + e^z) \approx e^z\), because \(e^z\) is small, so the logistic is approximately the exponential curve and has the same elasticity. The factor \(1 - Y\) is the brake applied by saturation.

Solution. (a) They are measured in different units: squared units of \(Y\) for the first and squared log units for the second.

(b) Scaling \(Y\) by \(\tilde Y = 4.2\) divides every residual of the linear model by \(4.2\), so its RSS becomes \(12.6/4.2^2 = 0.714\). In the log model, \(\log (Y/\tilde Y) = \log Y - \log \tilde Y\) only shifts the intercept, so the residuals and RSS are unchanged at \(0.84\). The linear model has the smaller scaled RSS, so it is preferred, with a log-likelihood margin of \(\frac {40}{2}\log (0.84/0.714) = 3.25\).