Question

Salary GPA Instructions: 3.4 In this problem you will characterize the predictive relationship between 3.8 2.8 starting salary (dependent or response variable, in tens of thousands) and GPA (independent or predictor variable) using regression analysis of the 3.2 hypothetical data in the table to the right. Do things by hand (except for part g and perhaps part c. and use of a calculator or software for arithmetic) and show your work The data set is tiny to make the calculations easy. The point is to make sure you know what statistical routines are doing with real datasets. a) Write the equation for the sum of squared residuals as a function of the unknown regression coefficient estimates, B, and (with the data, not in summation notation). b) Take the partial derivatives with respect to the two unknown coefficient estimates, set them to zero, and solve the resulting first order conditions for the estimates. c) Plot the data and the regression line. (Use excel or other software if you like.) d) Calculate the standard error of ß, assuming homoscedasticity. e) Calculate the standard error of Busing Stata's small sample corrected version the Eicker- Huber-White sandwich estimator. f) Test the null hypothesis that B. = 0 (explain your choice of standard error). g) Run the regression in Stata to verify your calculations are correct. h) Test the null hypothesis that B. = 1 (explain your choice of standard error).

Answer 1

a) To find Residual Sum of Squares we need to identify the alpha and beta.

	X	Y	X^2	Y^2	XY
	3.4	5	11.56	25	17
	3.8	6	14.44	36	22.8
	2.8	4	7.84	16	11.2
	3.2	6	10.24	36	19.2
Sum	13.2	21	44.08	113	70.2

From this table, n=4, Sum X = 13.2, Sum Y=21, Sum X^2 = 44.08, Sum Y^2 = 113, Sum XY = 70.2

beta = n SumXY - (Sum X Sum Y) / n Sum X^2 - (Sum X)^2

beta = 1.7307

alpha = Ybar - Beta Xbar

Ybar = Sum Y / n = 21/4 = 5.25

Xbar = Sum X / n = 13.2 /4 = 3.3

Now, alpha = 5.25 - (1.7307 x 3.3) = -0.46131

Thus regression equation is, Y = -0.46131 + 1.7307 X

Now RSS can be calculated using the formula,

Unfortunately Latex equation is not working due to some technical problem, I will write the formula as,

RSS = Summation {from 1 to n} (y_i - (alpha + beta x_i))^2

= [5 - (-0.46131 + (1.7307 * 3.4))]^2 + [6 - (-0.46131 + (1.7307 * 3.8))]^2 + [4 - (-0.46131 + (1.7307 * 2.8))]^2 + [6 - (-0.46131 + (1.7307 * 3.2))]^2

= 1.1923

c) Plot the regression line:

Y - Dependent 2.8 3.1 3.7 3.4 X-Independent - Regression Line (ỹ = 1.731 - 0.45)

d) Standard Error of Slope = Sqrt (Sum (y_i - y(hat))^2/ n-2 ) / Sqrt (Sum (x_i - x)^2)

Which is nothing but, beta divided by the t-statistic. = 1.7307/1.347 = 1.2847

Excel Output:

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.802955069
R Square	0.644736842
Adjusted R Square	0.289473684
Standard Error	0.973328527
Observations	3
ANOVA
	df	SS	MS	F	Significance F
Regression	1	1.719298	1.719298246	1.814814815	0.406519728
Residual	1	0.947368	0.947368421
Total	2	2.666667
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-0.684210526	4.502077	-0.151976635	0.903983399	-57.88852368	56.520103	-57.888524	56.52010263
	1.842105263	1.367409	1.347150628	0.406519728	-15.53246751	19.216678	-15.532468	19.21667803
RESIDUAL OUTPUT					PROBABILITY OUTPUT
Observation	Predicted 5	Residuals	Standard Residuals		Percentile	5
1	6.315789474	-0.31579	-0.458831468		16.66666667	4
2	4.473684211	-0.47368	-0.688247202		50	6
3	5.210526316	0.789474	1.147078669		83.33333333	6