ECO375: Suggested Solutions for Problem Set 1
September 27, 2009
NOTE
This assigment was marked out of 5 (since it counts for 5% of your grade). Question 2.10 was worth 2 marks (0.4 mark for each part). All other questions were worth 1 mark each. About 75
CHAPTER 2
SOLUTIONS TO PROBLEMS 2.1 (i) Income, age, and family background (such as number of siblings) are just a few possibilities. It seems that each of these could be correlated with years of education. (Income and education are probably positively co
CHAPTER 3
SOLUTIONS TO PROBLEMS 3.1 (i) hsperc is defined so that the smaller it is, the lower the students standing in high school. Everything else equal, the worse the students standing in high school, the lower is his/her expected college GPA. (ii) Jus
CHAPTER 4
SOLUTIONS TO PROBLEMS 4.1 (i) and (iii) generally cause the t statistics not to have a t distribution under H0. Homoskedasticity is one of the CLM assumptions. An important omitted variable violates Assumption MLR.3. The CLM assumptions contain
CHAPTER 5
SOLUTIONS TO PROBLEMS 5.1 Write y = 0 + 1 x1 + u, and take the expected value: E(y) = 0 + 1 E(x1) + E(u), or y =
0 + 1 x since E(u) = 0, where y = E(y) and x = E(x1). We can rewrite this as 0 = y 1 x. Now, 0 = y 1 x1 . Taking the plim of this w
CHAPTER 6
SOLUTIONS TO PROBLEMS 6.1 The generality is not necessary. The t statistic on roe2 is only about .30, which shows that roe2 is very statistically insignificant. Plus, having the squared term has only a minor effect on the slope even for large va
CHAPTER 7
SOLUTIONS TO PROBLEMS 7.1 (i) The coefficient on male is 87.75, so a man is estimated to sleep almost one and one-half hours more per week than a comparable woman. Further, tmale = 87.75/34.33 2.56, which is close to the 1% critical value agains
CHAPTER 8
SOLUTIONS TO PROBLEMS 8.1 Parts (ii) and (iii). The homoskedasticity assumption played no role in Chapter 5 in showing that OLS is consistent. But we know that heteroskedasticity causes statistical inference based on the usual t and F statistics
CHAPTER 9
SOLUTIONS TO PROBLEMS 9.1 There is functional form misspecification if 6 0 or 7 0, where these are the population parameters on ceoten2 and comten2, respectively. Therefore, we test the joint significance of these variables using the R-squared f
CHAPTER 10
SOLUTIONS TO PROBLEMS 10.1 (i) Disagree. Most time series processes are correlated over time, and many of them strongly correlated. This means they cannot be independent across observations, which simply represent different time periods. Even s
CHAPTER 11
SOLUTIONS TO PROBLEMS 11.1 Because of covariance stationarity, 0 = Var(xt) does not depend on t, so sd(xt+h) =
0 for
any h 0. By definition, Corr(xt,xt+h) = Cov(xt,xt+h)/[sd(xt) sd(xt+h)] = h /( 0 0 ) = h / 0 . 11.3 (i) E(yt) = E(z + et) = E(z
CHAPTER 12
SOLUTIONS TO PROBLEMS 12.1 We can reason this from equation (12.4) because the usual OLS standard error is an estimate of / SSTx . When the dependent and independent variables are in level (or log) form, the AR(1) parameter, , tends to be posit
CHAPTER 13
SOLUTIONS TO PROBLEMS 13.1 Without changes in the averages of any explanatory variables, the average fertility rate fell by .545 between 1972 and 1984; this is simply the coefficient on y84. To account for the increase in average education leve
CHAPTER 1
SOLUTIONS TO PROBLEMS 1.1 (i) Ideally, we could randomly assign students to classes of different sizes. That is, each student is assigned a different class size without regard to any student characteristics such as ability and family background.
APPENDIX E
SOLUTIONS TO PROBLEMS E.1 This follows directly from partitioned matrix multiplication in Appendix D. Write
x1 x 2 X = 2 , X = ( x1 x x ), and y = n x n y1 y2 y n
Therefore, XX =
xx
t =1 t
n
t
and Xy =
x y
t =1 t
n
t
. An equivalent express
CHAPTER 15
SOLUTIONS TO PROBLEMS 15.1 (i) It has been fairly well established that socioeconomic status affects student performance. The error term u contains, among other things, family income, which has a positive effect on GPA and is also very likely t
CHAPTER 16
SOLUTIONS TO PROBLEMS 16.1 (i) If 1 = 0 then y1 = 1z1 + u1, and so the right-hand-side depends only on the exogenous variable z1 and the error term u1. This then is the reduced form for y1. If 1 = 0, the reduced form for y1 is y1 = 2z2 + u2. (N
CHAPTER 17
SOLUTIONS TO PROBLEMS 17.1 (i) Let m0 denote the number (not the percent) correctly predicted when yi = 0 (so the prediction is also zero) and let m1 be the number correctly predicted when yi = 1. Then the proportion correctly predicted is (m0
CHAPTER 18
SOLUTIONS TO PROBLEMS 18.1 With zt1 and zt2 now in the model, we should use one lag each as instrumental variables, zt-1,1 and zt-1,2. This gives one overidentifying restriction that can be tested. 18.3 For , yt zt = yt zt + ( )zt, which is an
Lecture 4: Multiple Regression Analysis: Statistical Properties
Junichi Suzuki
University of Toronto
October 1st, 2009
Announcement
I I I
Important change: 2nd problem set is due October 16th. Will be available by the next lecture The rst problem set will
Lecture 5 Multiple Regression: Inference
Junichi Suzuki
University of Toronto
October 8th, 2009
Announcement
I I I
2nd problem set is due October 16th. Start early! Sacha is going to hold an o ce hour
I I
Place: Max Gluskin House GE313 Time: October 14th
APPENDIX A
SOLUTIONS TO PROBLEMS A.1 (i) $566. (ii) The two middle numbers are 480 and 530; when these are averaged, we obtain 505, or $505. (iii) 5.66 and 5.05, respectively. (iv) The average increases to $586 while the median is unchanged ($505). A.3 If
APPENDIX B
SOLUTIONS TO PROBLEMS B.1 Before the student takes the SAT exam, we do not know nor can we predict with certainty what the score will be. The actual score depends on numerous factors, many of which we cannot even list, let alone know ahead of t
APPENDIX C
SOLUTIONS TO PROBLEMS C.1 (i) This is just a special case of what we covered in the text, with n = 4: E(Y ) = and Var(Y ) = 2/4. (ii) E(W) = E(Y1)/8 + E(Y2)/8 + E(Y3)/4 + E(Y4)/2 = [(1/8) + (1/8) + (1/4) + (1/2)] = (1 + 1 + 2 + 4)/8 = , which s
APPENDIX D
SOLUTIONS TO PROBLEMS
0 1 6 2 1 7 20 D.1 (i) AB = 1 8 0 = 4 5 0 5 3 0 0 6 12 36 24
(ii) BA does not exist because B is 3 3 and A is 2 3. D.3 Using the basic rules for transpose, ( XX) = ( X)( X) = XX , which is what we wanted to show.
D.5 (i)
CHAPTER 14
SOLUTIONS TO PROBLEMS 14.1 First, for each t > 1, Var(uit) = Var(uit ui,t-1) = Var(uit) + Var(ui,t-1) = 2 u2 , where we use the assumptions of no serial correlation in cfw_ut and constant variance. Next, we find the covariance between uit and u