What Factors Affect House Prices? Suppose you have 254 observations on house sales in Seoul. Consider the linear regression model: Y₁ = B₁ + B₂X2i + B3X3i + B4D₁X2i+ei, for i=1,...,254 Y Sales Price (in million won) Size (in m²) Number of rooms in a house 1 if house is in the South otherwise 2 where X2 X3i D₁ Location dummy = Assume that the errors are homoskedastic and uncorrelated. : : (a) Compare the expected value of house sales price, E (Y₂), in the South with one in elsewhere. Briefly interpret the difference. (b) You want to calculate the change of sales price of a house in the South when a 10m²-room is added to a house. How much will you expect to get more on average (in terms of expected value), assuming everything else being held constant? (c) You estimate the regression equation as Î₁ = 12 + 20 X2i+ 10 X3i+ 5 D;X2i (0.10) (1.5) (3.5) (0.5) n = 254 RSS=5,000 ESS = 15,000 (standard errors in parentheses) Now obtain the estimate of change the sales price of a house in the South when a 10m²-room is added to a house. (d) Based on the estimates, test the significance of 34 with 5% significance level. (Note: please refer to the probability table below.) (e) Report R² of this regression. (f) Report the adjusted R² (=R²) of this regression. Is it greater than R²? (g) Test the overall significance of the model based on the estimates. You should clarify what is a test statistic and its distribution. (Note: please refer to the probability table below.) (h) Calculate the OLS estimator of o² using the above regression output. table: - P (F (1, +200) ≥ 3.84) = 0.05, P (F (1,+200) ≥ 6.63) = 0.01 for F distribution with degrees of (1, +200). (+200 indicates for more than 200 degrees of freedom.) = free dom - P(F (2, +200) ≥ 3.00) 0.05, P (F (2,+200) ≥ 4.61) = 0.01 for F distribution with degrees of freedom (2, +200). (+200 indicates for more than 200 degrees of freedom.) - P(F (3, +200) ≥ 2.60) 0.05, P (F (3,+200) ≥ 3.78) = 0.01 for F distribution with degrees of free dom (3, +200). (+200 indicates for more than 200 degrees of freedom.) = - P(Z > 1.28) = 0.1, P(Z ≥ 1.645) = 0.05, P (Z≥ 1.96) = 0.025, P(Z > 2.58) = 0.005 where Z is a standard normal random variable. • Probability