Question

. regress finaid parent hsrank male Source MS Number of obs F3, Prob >F R-squared Adj R-squared0.7493 Root MSE Model Residual 1.D785e+09 332013409 3 359495714 467217682.8 46)49.81 0.0o00 -0.7646 Total 1.4105e+0949 28785725.5 -2686.6 Einaid Std. Err [95% Conf. Interval] parent34275390315054 -10.88 0.000 hsrank 12.70553 male1570.143 784.2971-2.00 0.051-3148.851 6304.344 -.406171-.2793368 123.817 8.565037 13321.7 83.26124 20.14795 4.13 0.000 9B13.022 1743.1 5.63 0.000 1. Use the regression above to answer the following questions finaid: Financial aid in dollars. parent: Parents income in dollars hsrank; High school rank. male-1 if student is male What is the regression equation for predicted financial aid? What is the regression equation for males predicted financial aid? What is the regression equation for females predicted financial aid? What is the predicted financial aid for a female whom parents income is $20,000 a year a. b. c. d. and hsrank 3? What is the adjusted R2 for this regression? Interpret. Interpret the coefficient on male Interpret the coefficient on parent e. f. g. h. If you were to define male 0 for male students, and male-1 otherwise. How this new definition would affect the point estimates shown in the Stata output? i. Using TSS and RSS show that adjusted R2 0.7493

Answer 1

1. (a) The regression equation for predicted financial aid would be

$\dpi{80} E(finaid) = \widehat{finaid} = 9813.022 - 0.3427539*parent$$\dpi{80} + 83.26124*hsrank - 1570.143*male$.

(b) The regression equation for males predicted financial aid would be

$\dpi{80} E(finaid|male=1) = 9813.022 - 0.3427539*parent$$\dpi{80} + 83.26124*hsrank - 1570.143*1$

or $\dpi{80} E(finaid|male=1) = 8242.879 - 0.3427539*parent$ $\dpi{80} + 83.26124*hsrank$.

(c) The regression equation for females predicted financial aid would be

$\dpi{80} E(finaid|male=0) = 9813.022 - 0.3427539*parent$$\dpi{80} + 83.26124*hsrank - 1570.143*0$

or $\dpi{80} E(finaid|male=0) = 9813.022 - 0.3427539*parent$ $\dpi{80} + 83.26124*hsrank$.

(d) For the given values, we have

$\dpi{80} E(finaid|male=0) = 9813.022 - 0.3427539*20000$$\dpi{80} + 83.26124*3$

or $\dpi{80} E(finaid|male=0) = 3207.72772$ .

(e) The adjusted R-square is given as 0.7493 or 74.93%. This means that, adjusting for the degrees of freedom, the explanatory variables explains about 74.93% of the dependent variable. The adjustment is due to the fact that R-square increases as number of explanatory variables increase, and the adjusted R-square is a more dependable explanation of the model.

(f) The coefficient on male is -1570.143, meaning that the average financial aid is $1570.143 less (since negative) for male students, than the female students.

(g) The coefficient on parent is -0.3427539, meaning that for a unit increase in the parent's income, the financial aid on average decreases by $0.3427539 on average.

(h) The result would be the same if male=0 for male and male=1 other wise, but the regression output would be arithmetically different. We would have then

$\dpi{80} E(finaid) = \widehat{finaid} = 8242.879 - 0.3427539*parent$$\dpi{80} + 83.26124*hsrank + 1570.143*male$.

This would then mean the same that, the financial aid is more for non-male by $1570.143 than male students, and provided that other variables are zero, the average financial aid for males (male=0) is $8242.879, while for non-male (male=1) is $9813.022. Rest would be the same.

(i) The adjusted R-square calculation would be as below.

$\dpi{80} \bar{R^2} = 1 - \frac{RSS/(n-k)}{TSS/(n-1)}$

or $\dpi{80} \bar{R^2} = 1 - \frac{332013409/(50-4)}{1410500000/(50-1)}$ (since number of observation is 50, and the number of parameters to be estimated, including the intercept is 4)

or $\dpi{80} \bar{R^2} = 1 - \frac{332013409/46}{1410500000/49}$

or $\dpi{80} \bar{R^2} = 1 - 0.25073836$

or $\dpi{80} \bar{R^2} = 0.7493$ , as is given.