Unformatted text preview: 73261 Econometrics October 7, 2009 Midterm 1
Student name:______________________________ 50 minutes. 50 points. Feel free to detach the table, and use it's back side as scratch paper. 1. Short/Multiple choice questions [5 points] You estimated the equation y = xβ + u, and you want to make a prediction for x = x0. ˆ ˆ The estimate of β is β , the estimate of V(ux) is σ 2 , ˆˆ ˆ ˆ and you define y0 = x0β + u, y = x β , e = y − y .
0 0 1.a Which of the following could be an actual set of estimates: (pick one answer) ˆ ˆ ˆ Case σ s.e.(e) s.e.( y ) ˆ ˆ ˆ 3 5 4 A. [ s.e.(e) ]2 = [ s.e.( y ) ]2 + [ σ ]2 B. 3 4 5 C. 5 4 3 D. 4 3 5 1.b Which s.e. would you use in constructing confidence interval for y0? ˆ A. σ ˆ ˆ B. s.e.(e) : CI is being built around y ˆ C. s.e.( y ) ˆ D. s.e.( β ) 1.c Which distribution would you use to compute the critical value for that CI? A. Chisquared B. Student's t C. Fdistribution D. Poisson
(pick one answer) (pick one answer) 1.d You have 15 observations and 2 variables, and you want a 99% confidence interval. What is the appropriate critical value c? (write your answer in the provided space) 1 c = t (0.005;12) = 3.055 Partial credit (0.5) for c = 3.012 (13 d.f.'s) or c= 2.681 (singletailed test) 1.e Given the s.e. from 1.b, and critical value c from 1.d, what is the formula for the confidence interval? ˆ A. y ± c * se
2 ˆ B. y ± c * (se ) (pick one answer) C. y0 ± c * (se ) ˆ D. y ± c * se 2 2. [15 points] You are trying to estimate the following equation: y = β0 + u You have N observations of x and y, denoted as Xi and Yi respectively. 2.a [5 points] Derive the estimator of β0:  write out the sum of squared residuals Σi (Yi − β0)2  Compute its derivative and set it equal to zero ˆ 2Σi (Yi − β 0 ) = 0 ˆ  solve for β 0 ˆ β 0 = Σ i Y i /N = Y
2.b [3 points] What assumption is necessary to make your estimate unbiased? Prove it. Assumption is: E[u] = 0 ˆ Proof: β 0 = Σi Yi/N = Σi (β0 + ui)/N = β0 + Σi ui/N ˆ E[ β ] = β0 + Σi E[ui]/N = β0 + E[u] = β0
0 Mean of the estimate is equal to the estimated parameter, which is definition of unbiased estimate 2.c [5 points] Assuming that V[ux] = σ 2, derive the population variance of your estimate. ˆ Recall that β 0 = β0 + Σiui/N Since ui 's are independent, we have: ˆ V[ β 0 ] = ΣiV[ui/N] = ΣiV[ui]/N2 = σ 2/N 2.d [2 points] You have the following data: X = 4.9, Y = 2.4, X 2 = 31, Y 2 = 7, XY = 15 Compute the estimate of β0. ˆ As we found in 2.a, β 0 = Y = 2.4 3. [13 points] You decided to check whether controlling for economic growth eliminates the link between global warming and pirate activity, but your dog has chewed up the output, and you're left with: Dependent Variable: TEMPERATURE Included observations: 15 Variable Coefficient Std. Error tStatistic Prob. C Log GDP Piracy Rsquared 5 0.65 4 0.4857 1.666 0.25 3 3 2.6 1.356 0.01 0.02 0.2 Adjusted Rsquared 0.4 3.a [2 points] Is correlation between piracy and global warming eliminated? Yes, pvalue is 0.2 which is higher than any significance level that we used (110%) Alternatively, one can say that it is almost eliminated as 20% significance is still used sometimes. 3.b [7 points] Fill in the blank cells. Done above, 1 point for each of seven numbers. Values are rounded up, sufficiently close numbers get full credit. tstat on Piracy must use the correct degrees of Freedom 3.c [2 point] Somebody suggests that your R2 could be higher, and you might improve the fit by adding add a variable measuring heat output of the sun. Name one test or criterion that you would use to check whether adding the variable leads to a meaningful improvement. Valid answers are: • AIC & SIC criteria, • ttest on newly added variable or equivalent Ftest for excluding it R2 or Adjusted R2 are not valid criteria as they can be increased by insignificant variables 3.d [2 points] Same somebody also suggests that increasing R2 will reduce standard errors. What is the basis for this claim, i.e. what is the quantity that affects both R2 and standard errors? Both depend on RSS: • R2 = 1 − RSS/TSS • s.e. are part of variancecovariance matrix that is (X'X)1 multiplied by: ˆ σ 2 = RSS /( N − k − 1) 4. [17 points] You have the following regression output:
Dependent Variable: LWAGE Included observations: 526 Variable Coefficient S. Error tStatistic Prob. … β5: MALE*MARRIED 0.292091 0.055389 5.273454 0 β6: FEMALE*(1MARRIED) 0.09675 0.057463 1.68368 0.0928 β7: FEMALE*MARRIED 0.12024 0.057984 2.07358 0.0386 and a subset of the variancecovariance matrix:
MALE*MARRIED FEMALE * (1MARRIED) FEMALE*MARRIED MALE*MARRIED 0.003068 0.001917 0.002101 FEMALE*(1MARRIED) FEMALE*MARRIED 0.001917 0.002101 0.003302 0.001947 0.001947 0.003362 Male, female and married are "dummy" variables, i.e. they are equal to 1 if individual is male, female or married (respectively), and 0 otherwise. 4.a [2 points] What is the approximate wage difference between single male and married male? Specify whether it is absolute or relative. As the dependent variable is log(wage), we can only compute relative change. We know that male = 1 and ∆married = 1 (from 0 to 1), so in terms of regression above, we the only difference between married and single males is that male * married is 0 in one case, and 1 in another. Let β5 be the coefficient on (male*married), and note that we essentially have a loglin model, so: ˆ ˆ ∆y/y = β 5 ∗∆(male*married) = β 5 = 0.292091 = 29.21% 4.b [5 points] What is the exact wage difference between single male and married male? Here, we need to plug in (male*married) = x5 = 0,1 and compute: ˆ ˆ ∆y = y(x5 = 1) − y(x5 = 1) = exp{1* β 5 + …} − exp{0* β 5 + …} ˆ If we treat single male as base case (y = exp{0* β + …}), we have:
5 ˆ ˆ ∆y/y = exp{ β 5 } − 1 = 0.3392 = 33.92%, ∆y = y(exp{ β 5 } − 1) => i.e. married males receive 33.92% higher wage than single one. If we treat married male as base case, similar calculation tell that single males get wage that is 25.33% lower.
(continued on next page) 4.c [6 points] Test whether there is a wage difference between married and unmarried females, vs. twotailed alternative, at 10% significance level. Compute the test statistic, describe its distribution (t, F, χ2, Normal, etc.) and d.f.'s, report critical value and your conclusion. ˆ ˆ ˆ ˆ ˆ Test statistic is: t = β 6 − β 7 s.e. β 6 − β 7 = 0.4462, where: )( ) ˆ ˆ ˆ ˆ ˆˆ s.e.(β − β ) = V (β ) + V (β ) − 2cov(β , β )
6 7 6 6 6 7 ( = 0.00330 + 0.00336 − 2 * 0.00195 = 0.05263 Distribution is Student's t with 519 d.f., critical value is 1.645 (using ∞), and since our tstatistic below that, we reject null hypothesis, i.e. female wage is not correlated with marriage 4.d [4 points] On the basis of this regression, can you claim that getting married would increase the wage of a given male? Why? Use back side of this page if you need more space. NO, correlation does not imply causation. It could be that males with higher income are more likely to be married. ...
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 Fall '09
 Kyrkv
 Statistics, Econometrics, Normal Distribution, Statistical hypothesis testing, critical value

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