chap06 - CHAPTER 6 Exercise Answers EXERCISE 6.3(a Let the...

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29 CHAPTER 6 Exercise Answers EXERCISE 6.3 (a) Let the total variation, unexplained variation and explained variation be denoted by SST , SSE and SSR , respectively. Then, we have 42.8281 SSE 802.0243 SST 759.1962 SSR (b) A 95% confidence interval for 2 is 2 (0.975,17) 2 se( ) (0.2343,1.1639) bt b  A 95% confidence interval for 3 is 2 (0.975,17) 3 se( ) (1.3704, 2.1834) b (c) To test H 0 : 2 1 against the alternative H 1 : 2 < 1, we calculate 1.3658 t  . Since (0.05,17) 1.3658 1.740 t  , we fail to reject 0 H . There is insufficient evidence to conclude 2 1  . (d) To test 02 3 :0 H  against the alternative 12 H and/or 3 0  , we calculate 151 F . Since (0.95,2,17) 151 3.59 F  , we reject H 0 and conclude that the hypothesis 2 = 3 = 0 is not compatible with the data. (e) The t -value for testing 023 :2 H   against the alternative 123 H   is    23 2 0.37862 0.634 se 2 0.59675 bb t  Since (0.025,17) 2.11 0.634 2.11 t  , we do not reject 0 H . There is no evidence to suggest that 2  .
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Chapter 6, Exercise Answers, Principles of Econometrics, 4e 30 EXERCISE 6.5 (a) The null and alternative hypotheses are: 02 4 3 5 12 4 : and : or or both H H   (b) The restricted model assuming the null hypothesis is true is 22 14 5 6 ln( ) ( ) ( ) WAGE EDUC EXPER EDUC EXPER HRSWK e    (c) The F -value is 70.32 F .The critical value at a 5% significance level is (0.95,2,994) 3.005 F . Since the F -value is greater than the critical value, we reject the null hypothesis and conclude that education and experience have different effects on ln( ) WAGE . EXERCISE 6.10 (a) The restricted and unrestricted least squares estimates and their standard errors appear in the following table. The two sets of estimates are similar except for the noticeable difference in sign for ln( PL ). The positive restricted estimate 0.187 is more in line with our a priori views about the cross-price elasticity with respect to liquor than the negative estimate 0.583. Most standard errors for the restricted estimates are less than their counterparts for the unrestricted estimates, supporting the theoretical result that restricted least squares estimates have lower variances. CONST ln( PB ) ln( PL ) ln( PR ) ln( ) I Unrestricted 3.243 1.020 0.583 0.210 0.923 (3.743) (0.239) (0.560) (0.080) (0.416) Restricted 4.798 1.299 0.187 0.167 0.946 (3.714) (0.166) (0.284) (0.077) (0.427) (b) The high auxiliary 2 s R and sample correlations between the explanatory variables that appear in the following table suggest that collinearity could be a problem. The relatively large standard error and the wrong sign for ln( ) PL are a likely consequence of this correlation.
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chap06 - CHAPTER 6 Exercise Answers EXERCISE 6.3(a Let the...

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