ARE 106 HW 6

ARE 106 HW 6 - Jeff Phang 4-5 Tuesday March 3, 2009 ARE 106...

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Jeff Phang 4-5 Tuesday March 3, 2009 ARE 106 HW #6 1. Run a regression relating house prices to a constant, lajolla, bedrms, rooms, baths, firepl, garage, pool, yard, and view (in this order, for ease in grading). Report the results in standard format. Model 1: OLS estimates using the 59 observations 1-59 Dependent variable: price coefficient std. error t-ratio p-value ----------------------------------------------------------- const -31.3355 45.5026 -0.6887 0.4943 lajolla 132.235 19.1877 6.892 9.76E-09 *** bedrms -1.35268 15.2037 -0.08897 0.9295 rooms 4.09467 11.5367 0.3549 0.7242 baths 36.6244 15.8076 2.317 0.0247 ** firepl 34.7332 17.7819 1.953 0.0565 * garage 11.3213 15.8706 0.7133 0.4790 pool 30.6849 22.3680 1.372 0.1764 yard 0.00325721 0.00185451 1.756 0.0853 * view 23.1777 18.9386 1.224 0.2269 Mean of dependent variable = 228.629 Standard deviation of dep. var. = 112.436 Sum of squared residuals = 214106 Standard error of the regression = 66.1022 Unadjusted R-squared = 0.70799 Adjusted R-squared = 0.65436 F-statistic (9, 49) = 13.2006 (p-value < 0.00001) Log-likelihood = -325.52 Akaike information criterion (AIC) = 671.039 Schwarz Bayesian criterion (BIC) = 691.815 Hannan-Quinn criterion (HQC) = 679.149 Excluding the constant, p-value was highest for variable 5 (bedrms) Price bar = -31.3355 + 132.235(la jolla)+ -1.35268(bedrooms) + (-0.6887) (6.892) (-0.08897) 4.09467(rooms) + 36.6244(baths) + 34.7332(firepl) + ( 0.3549) (2.317) (1.953) 11.3213(garage) + 30.6849(pool) + 0.00325721(yard) + (0.7133) (30.6849) (1.756) 23.1777(view) ( 1.224)
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R 2 = 0.71 σ hat = 66.10 Write the null and alternate hypotheses tested in the "goodness of fit" F test, and formally draw a conclusion. Special wald test. Ho: β2 = β 3 = β4 = β 5 = β 6 = β 7 = β 8 = β 9 = β 10 = 0 Ha: not Ho. F stat = [(ESS r – ESS u ) / q]/ [(ESS u )/ T-K] Analysis of Variance: Sum of squares df Mean square Regression 519120 9 57680 Residual 214106 49 4369.51 Total 733226 58 14963.8 R^2 = 519120 / 733226 = 0.707995 F(9, 49) = 57680 / 4369.51 = 13.2006 [p-value 2.14e-010] F c (59-10) = 2.08 The probability of Ho being true when if we reject it is 2.14e-010. Since F* = 13.2006 is greater than F c = 2.08 then we reject the null. Very briefly examine the model for plausibility. It is difficult to isolate the effect that each of the variables has on price because there would be little to no houses without a bedroom or a bathroom. It’s difficult to find a house that has no bedrooms, no bathrooms but a pool. It then becomes very difficult to separate the effects of a bedroom or bathroom from a garage, pool or any other variable. This validity of this model is questionable. Are the signs of the variables as expected? The signs of the variables are as expected. Holding all others constant, an increase in the # of bedrooms with the same
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This note was uploaded on 04/11/2009 for the course ARE 106 taught by Professor Havenner during the Winter '09 term at UC Davis.

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ARE 106 HW 6 - Jeff Phang 4-5 Tuesday March 3, 2009 ARE 106...

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