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Chapter 6

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Department
Economics
Course
ECON 3210
Professor
Razvan Sufana
Semester
Fall

Description
CHAPTER 6 Exercise Solutions 178 Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 179 EXERCISE 6.1 (a) compute R 2, we need SSE and SST. We are given SSE. We can find SST from the equation  ()i 2 SST y   13.45222 N  1 Solving this equation for SST yields 2 2 SST  ˆy (N 1) (13.45222)  39 7057.5267 Thus, R  1 S 1 97.830 0.8612 SST 7057.5267 (b) The F-statistic for testin0 203  is defined as (SST  SSE) (K 1) (7057.5267 979.830) / 2 F   SSE (N  K) 979.830 / (40 3) 114.75 At  0.05 , the critical value i(0.95, 2,37)5 . Since the calculated F is greater than the critical F, we rejec0 HThere is evidence from the data to suggest t20 and/or  0 . 3 Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 180 EXERCISE 6.2 The model from Exercise 6.1 is e z x 1 23  . The SSE from estimating this model is 979.830. The model after augmen ting with the squares and the cubes of 2 3 2 3 predictions y yand e yisy z x1 2 31 2  ˆˆ . The SSE from estimating this model is 696.5375. To use the RESET, we set the null hypothesis H 0 12  . The F-value for testing this hypothesis is (S)SER USSE J (979.830 96.5375) 2 F    7.1175 SSE U (N  K ) 696.5373 (40 5) The critical value for significance level   0.05 is F(0.95,2,35)67 . Since the calculated F is greater than the criticalF we reject H 0 and conclude that the model is misspecified. Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 181 EXERCISE 6.3 (a) Let the total variation, unexplained vari ation and explained variation be denoted by SST, SSE and SSR, respectively. Then, we have SS  eiˆ N K  (20 3) 2.5193 42.8281 Also, 2 SSE R  1 SST 0.9466 and hence the total variation is SSE 42.8281 SST    802.0243 1  2 1 0.9466 and the explained variation is SSR SST SSE 80 2.0243 42.8281 759.1962 (b) A 95% confidence interval for  is 2 2(0.975,172 0.69914 2.110 0.048526 (0.2343,1.1639) A 95% confidence interval for 3s bt  b se( ) 1.7769 2.110 0.037120 (1.3704, 2.1834) 2 (0.975,173 (c) est H 0 2 1 against the alternative1H 2  < 1, we calculate b2 2 0.69914 1 t    1.3658 se2 0.048526 At a 5% significance level, we reject0Hif tt  1.740 . Since 1.3658 1.740 , (0.05,17) we fail to rejecH 0. There is insufficient evidence to conclu21 . (d) tst H 0 23 against the alternativH 1 2  and/or 3 0 , we calculate explained variationK 1 759.1962 / 2  F   151 unexplained variation K 42.8281/17 The critical value for a 5% level of significance i(0.95,2,17)59. Since 151 3.59 , we reject 0 and conclude that the hypothesis 2 = 3 = 0 is not compatible with the data. Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 182 Exercise 6.3 (continued) (e) The t-statistic for testH :2  against the alternatiH :2  is 0 2 3 1 2 3 t  2bb2 3  se  b 2 3 For a 5% significance level we reject H0if tt (0.025,17)11 or tt (0.975,17)11. The standard error is given by se 2 3  2 b2ar( )3 var( ) 223 cov( , )  4 0.048526 0.03712 2 2  .031223  0.59675 The numerator of the t-statistic is 2bb2 3 2 0 .6991 1.7769 0.37862 leading to a t-value of 0.37862  t   0.634 0.59675 Since .1 0.634 2.11 , we do not rejectH 0. There is no evidence to suggest that . 22 3 Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 183 EXERCISE 6.4 (a) The value of the t statistic for the significance tests is calculated from: bk t  se( k We reject the null hypothesis H 00k if ttc 2 . The t-values for each of the coefficients are given in the following table. Those which are significantly different from zero at an approximate 5% level are marked *. When EDUC and EDUC 2 both appear in an equation, their coefficients are not signi ficantly different from zero, with the exception of eqn (B), where EDUC 2 is significant. In addition, the interaction term between EXPER and EDUC is not significant in eqn (A). Variable t-valuesa Eqn (A) Eqn (B) Eqn (C) Eqn (D) Eqn (E) C  3.97* 6.59* 8.38*23.82* 9.42* 1 EDUC  2 1.26 0.84 1.04 15.90* 2 EDUC  3 1.89 2.12*1.73 EXPER  4 4.58* 6.28* 5.17* 6.11* 2  EXPER 5 –5.38* –5.31* –4.90* –5.13* EXPER*EDUC  –1.06 6 HRSWK  7 8.34* 8.43* 9.87*10.11* 8.71* a Note: These t-values were obtained from the computer output. Some of them do not agree exactly with the t ratios obtained using the coefficients and standard errors in Table 6.4. Rounding error discrepancies arise because of rounding in the reporting of values in Table 6.4. (b) Using the labeling of coefficients in the above table, we see that the restriction imposed on eqn (A) that gives eqn (B) is  0 . The F-test value for testing H :0 against 6 0 6 H 1 6 can be calculated from restricted and unr estricted sums of squared errors as follows: (S)SE  SSE J (222.6674 22.4166) 1 F   R U  1.120 SSEU (N  K) 222.4166 993 corhesponding p-value is 0.290. The critical valu e at the 5% significance level is F (0.95,1,993)51. Since the F-value is less than the critical value (or the p-value is greater than 0.05), we fail to reject the null hypothsis and conclude that the interaction term, EDUCEXPER is not significant in determining the wage. The t-value for testingH 0 60  against H 1 6  is –1.058. At the 5% level, its absolute value is less than the critical valut(0.975,993)62 . Thus, the t-test gives the same result. The two tests are equivalent because 1.120 1.058 and 3.8511.962 . Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 184 Exercise 6.4 (continued) (c) The restrictions imposed on eqn (A) that give eqn (C) are4 50,nd60 . Thus, we test 0d n a 0H, :0   0 4 5 6
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