Question 1 (a) (1) lot size (2) desirable location (3) lot size * desirable location You have to include the interaction term (3), otherwise no points. You may exclude (2) but not (1). Also if you used price per square meter as predictor, you got no points. Selling price is the dependent variable! (b) D, it is not given that variances are equal. (c) B. Question 2 (a) 0.430274 + 0.095687*(1000/100) = 1.387144 You got 2 points for 0.430274 + 0.095687*1000 = 96.11727 (b) Yes, p << 0.05 (c) R^2 = 0.594, so 59.4% (d) No, all coefficients are positive and TLC > 0. Question 3 (a) z= -6+0.05*40+1*3.5 = -0.5 Pr(y=1|x)= 1/(1+exp(-z)) = 1/(1+exp(0.5)) = 0.3775407 (b) The decision boundary is: -6+0.05*x_1+x_2 = 0 Filling in x_2 = 3.5 we get: -6+0.05*x_1+3.5 = 0 Solving for x_1 we find x_1 = 50, so 50 hours. (c) z = -6 + 0.05*30 + 1*4 = -0.5 d Pr(y=1|x)/d x_1 = \beta_1 * exp(-z)/(1+exp(-z))^2 = 0.05 * exp(0.5)/(1+exp(0.5))^2 = 0.01175019 Some used a "discrete approximation", for example by increasing x_1 from 30 to 31 and calculating the difference in fitted probabilities. This was awarded with 3 points. Question 4 (a) \beta_0 = -8.2 (b) -8.2+0.4*x_1+0.8*x_2 = 0 Any multiplication of this equation by a constant is also correct of course. (c) f(8,6) = -0.2 Since this is a negative number we predict the class -1. Question 5 The maximum likelihood estimate for the given sample is 5.23, and in general it is the sample maximum. Explanation: the ML estimate can't be smaller than the sample maximum: any estimate smaller than the sample maximum will assign probability zero to the sample maximum and hence to the whole set of observed data (multiplication by 0). For any estimate t of \theta that is at least as large as the sample maximum, the probability of the observed data is (1/t)^n, where n is the number of observations in the sample. This function is decreasing in t, so we should take t as small as possible, which is the sample maximum. Some people proposed 2 times the sample mean, which does have its merits (e.g. in contrast to the ML estimate it is unbiased), but it is not the ML estimate. Nevertheless you were awarded 2 points, just for proposing a reasonable estimator.