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Examination #2

# Examination #2 - Gregory Smith Dr Cuellar Econ 317...

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Gregory Smith 11/13/2009 Dr. Cuellar – Econ 317 Examination #2 1. Construct a model to estimate the average annual growth rate in the price of all wines in the sample. Be sure to clearly define each variable. a. Ho: β1=0 (No change in Price) Ha: β1≠0 (Change in Price) Price= β0 +β1X1+Ui X1 = Time(Year – 1999) to get the average annual growth rate, we need to make price a log. . gen lnprice=ln(price) lnPrice= β0 +β1X1+Ui 2. Estimate the model constructed above and interpret your results. What is the average annual growth rate of the price of wine? Explain fully and show graphically. reg price time Source | SS df MS Number of obs = 23404 -------------+------------------------------ F( 1, 23402) = 5.41 Model | 935.797865 1 935.797865 Prob > F = 0.0200 Residual | 4046501.14 23402 172.91262 R-squared = 0.0002 -------------+------------------------------ Adj R-squared = 0.0002 Total | 4047436.94 23403 172.945218 Root MSE = 13.15 ------------------------------------------------------------------------------ price | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- time | -.0736449 .0316567 -2.33 0.020 -.135694 -.0115958 _cons | 16.50927 .1872904 88.15 0.000 16.14217 16.87637 ------------------------------------------------------------------------------ Average growth is -.0736449. . display invttail( 23402, .05/2) 1.9600654 Both P value and absolute Crit illustrate a rejection of Ho. To take the average growth rate of price: 1

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. reg lnprice time2 time Source | SS df MS Number of obs = 23404 -------------+------------------------------ F( 2, 23401) = 29.58 Model | 23.3218122 2 11.6609061 Prob > F = 0.0000 Residual | 9224.89864 23401 .394209591 R-squared = 0.0025 -------------+------------------------------ Adj R-squared = 0.0024 Total | 9248.22045 23403 .395172433 Root MSE = .62786 ------------------------------------------------------------------------------ lnprice | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- time2 | .0013123 .0005882 2.23 0.026 .0001595 .0024652 time | -.0236181 .0057992 -4.07 0.000 -.0349849 -.0122513 _cons | 2.643722 .0125756 210.23 0.000 2.619073 2.668371 ------------------------------------------------------------------------------ . display invttail(23401, .05/2) 1.9600654 The average growth rate is -.0236181 with a P value of 0 making it statistically significant. Since both the dollar and log show statically significant values, I therefore reject Ho (null hypothesis). The diminishing returns flatten out the curve. . scatter lnprice pricehat time if lnprice>2 & lnprice<3, c(. l l l) s(oh i i i) 2
2 2.2 2.4 2.6 2.8 3 1 2 3 4 5 6 7 8 1999 2008 Year lnprice Fitted values 3. To test whether the movie Sideways increased the overall demand for wine. Construct a model that compares the growth rates of the price of all wine before and after the movie’s release. Sideways was released in October 2004, so assume the impact of the movie is observed in the years 2005 through 2008. Be sure to clearly define each variable. a. Ho: β2 =0 (Sideways had no effect on price) Ha: β2 ≠ 0 (Sideways had an effect on price) X1 = time (Year – 1999) X2 = Sideways= 1 Movie has been released 0 Movie has not been released X3 = sidewaystime (Sideways*Time) X4 = time2 (time^2) X5 = sidewaystime2 (sidewaystime*time2) lnPrice= β0 +β1Time+ β2Sideways + β3sidewaystime + β4time2 + β5sidewaystime2 + Ui . gen sideways=1 if year>=2005 (11293 missing values generated) . replace sideways=0 if year<2005 (11293 real changes made) 3

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4. Estimate the model constructed above and interpret your results. What is the average annual growth rate of the price of wine before the release of Sideways ? What is the average annual growth rate of the price of wine after the release of Sideways ? . reg lnprice sideways time time2 sidewaystime sidewaystime2 Source | SS df MS Number of obs = 23404 -------------+------------------------------ F( 5, 23398) = 15.26 Model | 30.0583969 5 6.01167938 Prob > F = 0.0000 Residual | 9218.16205 23398 .393972222 R-squared = 0.0033 -------------+------------------------------ Adj R-squared = 0.0030 Total | 9248.22045 23403 .395172433 Root MSE = .62767 ------------------------------------------------------------------------------ lnprice | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- sideways | -.0839504 .2170478 -0.39 0.699 -.5093783 .3414775 time | .0243319 .0130189 1.87 0.062 -.001186 .0498497 time2 | -.0079936 .0023925 -3.34 0.001 -.0126832 -.0033041 sidewaystime | .0066897 .0439977 0.15 0.879 -.0795486 .092928 sidewaysti~2 | .0005359 .000275 1.95 0.051 -3.03e-06 .0010748 _cons | 2.610781 .0149254 174.92 0.000 2.581526 2.640036
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