Hypothesis Testing and Model Estimation in Econometrics, Study notes of French

Download Hypothesis Testing and Model Estimation in Econometrics and more Study notes French in PDF only on Docsity! 112 CHAPTER 6 Exercise Solutions Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 113 EXERCISE 6.1 (a) To compute 2R , we need SSE and SST. We are given SSE. We can find SST from the equation 2( )ˆ 13.45222 1 1 i y y y SST N N − σ = = = − − ∑ Solving this equation for SST yields 2 2ˆ ( 1) (13.45222) 39 7057.5267ySST N= σ × − = × = Thus, 2 979.8301 1 0.8612 7057.5267 SSER SST = − = − = (b) The F-statistic for testing 0 2 3: 0H β = β = is defined as ( ) ( 1) (7057.5267 979.830) / 2 114.75 ( ) 979.830 /(40 3) SST SSE KF SSE N K − − − = = = − − At 0.05α = , the critical value is (0.95, 2, 37) 3.25F = . Since the calculated F is greater than the critical F, we reject 0H . There is evidence from the data to suggest that 2 0β ≠ and/or 3 0β ≠ . Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 116 Exercise 6.3 (continued) (e) The t-statistic for testing 0 2 3: 2H β = β against the alternative 1 2 3: 2H β ≠ β is ( ) ( ) 2 3 2 3 2 se 2 b b t b b − = − For a 5% significance level we reject 0H if (0.025,17) 2.11t t< = − or (0.975,17) 2.11t t> = . The standard error is given by ( ) ( ) 2 2 3 2 3 2 3se 2 2 var( ) var( ) 2 2 cov( , ) 4 0.048526 0.03712 2 2 0.031223 0.59675 b b b b b b− = × + − × × = × + − × × − = The numerator of the t-statistic is 2 32 2 0.69914 1.7769 0.37862b b− = × − = − leading to a t-value of 0.37862 0.634 0.59675 t − = = − Since 2.11 0.634 2.11− < − < , we do not reject 0H . There is no evidence to suggest that 2 32β ≠ β . Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 117 EXERCISE 6.4 In each case we use a two-tail test with a 5% significance level. The critical values are given by (0.025,60) 2.000t = − and (0.975,60) 2.000t = . The rejection region is 2t < − or 2t > . (a) The value of the t statistic for testing the null hypothesis 0 2: 0H β = against the alternative 1 2: 0H β ≠ is 2 2 3 1.5 se( ) 4 bt b = = = Since 2 1.5 2− < < , we fail to reject H0 and conclude that there is no sample evidence to suggest that β2 ≠ 0. (b) For testing H0: β1 + 2β2 = 5 against the alternative H1: β1 + 2β2 ≠ 5, we use the statistic ( ) ( ) 1 2 1 2 2 5 se 2 b b t b b + − = + For the numerator of the t-value, we have 1 22 5 2 2 3 5 3b b+ − = + × − = The denominator is given by 1 2 1 2 1 2 1 2se( ) var( 2 ) var( ) 4 var( ) 4 cov( , ) 3 4 4 4 2 11 3.3166 b b b b b b b b+ = + = + × + × = + × − × = = Therefore, 3 0.9045 3.3166 t = = Since 2 0.9045 2− < < , we fail to reject H0. There is no sample evidence to suggest that 1 22 5β + β ≠ . Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 118 Exercise 6.4 (continued) (c) For testing 0 1 2 3: 4H β −β +β = against the alternative 1 1 2 3: 4H β −β +β ≠ , we use the statistic 1 2 3 1 2 3 ( ) 4 se( ) b b bt b b b − + − = − + Now, 1 2 3( ) 4 2 3 1 4 6b b b− + − = − − − = − and 1 2 3 1 2 3 1 2 3 1 2 1 3 2 3 se( ) var( ) var( ) var( ) var( ) 2cov( , ) 2cov( , ) 2cov( , ) 3 4 3 2 2 2 1 0 4 b b b b b b b b b b b b b b b − + = − + = + + − + − = + + + × + × − = Thus, 6 1.5 4 t − = = − Since 2 1.5 2− < − < , we fail to reject H0 and conclude that there is insufficient sample evidence to suggest that β1 − β2 + β3 = 4 is incorrect. Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 121 EXERCISE 6.7 (a) The coefficients of ln(Y), ln(K) and ln(PF) are 0.6792, 0.3503 and 0.3219, respectively. Since the model is in log-log form the coefficients are elasticities. The estimate 0.6792 is the percentage change in VC when Y changes by 1%, with the other variables held constant. Similarly, 0.3503 is the percentage change in VC when K changes by 1%, and 0.3219 is the percentage change in VC when PF changes by 1%, keeping the other variables constant in each case. (b) An increase in any one of the explanatory variables should lead to an increase in variable cost, with the exception of ln(STAGE). For a given level of output (passenger-miles) and a given level of capital stock, longer flights should be cheaper than shorter ones. Thus, positive signs are expected for all variables except ln(STAGE), whose coefficient should be negative. All coefficients have the expected signs with the exception of ln( )PM . (c) The coefficient of ln( )PM has a p-value of 0.4966 which is higher than 0.05, indicating that this coefficient is not significantly different from zero. The p-values of the other coefficients are all 0.0000, indicating that they are significant. (d) Augmenting the equation with the squares of the predictions, and squares and cubes of the predictions, yields the RESET test F-values of 3.3803 and 1.8601 with corresponding p- values of 0.0671 and 0.1577, respectively. These two p-values are higher than the conventional 0.05 level of significance indicating that the model is adequate. (e) From the middle panel of Table 6.6 the F-value for testing 0 2 3: 1H β +β = is 6.1048 with a p-value of 0.014. This p-value is less than the significance level of 0.05. We reject 0H and conclude that constant returns to scale do not exist. (f) The F-value and the p-value for testing 0 4 5 6: 1H β +β +β = can be read from the bottom panel of Table 6.6. The F value is very large and the corresponding p-value of 0.00000 is below the significance level of 0.05. We reject 0H and conclude that there is no evidence to suggest that if all input prices increase by the same proportion, variable cost will increase by the same proportion. Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 122 Exercise 6.7 (continued) (g) To test 0 2 3: 1H β +β = , the value of the t statistic is 2 3 2 3 1 0.6792 0.3503 1 2.48 se( ) 0.01187 b bt b b + − + − = = = + where the standard error is calculated from 2 3 2 3 2 3 2 3 se( ) var( ) var( ) var( ) 2cov( , ) 0.002851 0.002796 2( 0.002753) 0.011874 b b b b b b b b + = + = + + = + + − = We reject 0H because (0.975,261)2.48 1.969t> = . Note 2 2(2.48) 6.15 6.10t F= = ≈ = . The difference between 2t and F is due to rounding error. To test 0 4 5 6: 1H β +β +β = , the value of the t -statistic is 4 5 6 4 5 6 1 0.2754 0.3219 0.0683 1 8.69 se( ) 0.0542 b b bt b b b + + − + − − = = = − + + where 4 5 6 4 5 6se( ) var( ) 0.002938 0.0542b b b b b b+ + = + + = = with 4 5 6 4 5 6 4 5 4 6 5 6var( ) var( ) var( ) var( ) 2cov( , ) 2cov( , ) 2cov( , ) 0.001919 0.001303 0.010068 2 0.000088 2 0.002159 2 0.002929 0.002938 b b b b b b b b b b b b+ + = + + + + + = + + − × − × − × = We reject 0H because (0.025,261)8.69 1.969t− < = − . Note that 2 2( 8.69) 75.52t = − = which is approximately equal to 75.43F = . Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 123 EXERCISE 6.8 There are a number of ways in which the restrictions can be substituted into the model, with each one resulting in a different restricted model. We have chosen to substitute out 1β and 3β . With this in mind, we rewrite the restrictions as 3 4 1 2 3 4 1 3.8 80 6 1.9 3.61 β = − β β = − β − β − β Substituting the first restriction into the second yields 1 2 4 480 6 1.9(1 3.8 ) 3.61β = − β − − β − β Substituting this restriction and the first one 3 41 3.8β = − β into the equation 2 1 2 3 4i i i i iS P A A e= β +β +β +β + yields ( ) ( ) 2 2 4 4 2 4 480 6 1.9(1 3.8 ) 3.61 1 3.8i i i i iS P A A e= − β − − β − β +β + − β +β + Rearranging this equation into a form suitable for estimation yields ( ) ( ) ( )2 2 478.1 6 3.61 3.8i i i i i iS A P A A e− − = β − +β − + + Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 126 Exercise 6.10 (continued) (d)(e) The results for parts (d) and (e) appear in the following table. The t-values used to construct the interval estimates are (0.975, 25) 2.060t = for the unrestricted model and (0.975, 26) 2.056t = for the restricted model. The two 95% prediction intervals are (70.6, 127.9) and (59.6, 116.7). The effect of the nonsample restriction has been to increase both endpoints of the interval by approximately 10 litres. ln(Q) Q ln( )Q se( )f tc lower upper lower upper (d) Restricted 4.5541 0.14446 2.056 4.257 4.851 70.6 127.9 (e) Unrestricted 4.4239 0.16285 2.060 4.088 4.759 59.6 116.7 Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 127 EXERCISE 6.11 (a) The estimated Cobb-Douglas production function with standard errors in parentheses is ( ) ( ) ( ) ( ) ( ) ( ) 2ln 0.129 0.559ln 0.488ln 0.688 (se) 0.546 0.816 0.704 Q L K R= + + = The magnitudes of the elasticities of production (coefficients of ln(L) and ln(K)) seem reasonable, but their standard errors are very large, implying the estimates are unreliable. The sample correlation between ln(L) and ln(K) is 0.986. It seems that labor and capital are used in a relatively fixed proportion, leading to a collinearity problem which has produced the unreliable estimates. (b) After imposing constant returns to scale the estimated function is ( ) ( ) ( ) ( ) ( ) ( ) ln 0.020 0.398ln 0.602ln (se) 0.053 0.559 0.559 Q L K= + + We note that the relative magnitude of the elasticities of production with respect to capital and labor has changed, and the standard errors have declined. However, the standard errors are still relatively large, implying that estimation is still imprecise. Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 128 EXERCISE 6.12 The RESET test results for the log-log and the linear demand function are reported in the table below. Test F-value df 5% Critical F p-value Log-log 1 term 0.0075 (1,24) 4.260 0.9319 2 terms 0.3581 (2,23) 3.422 0.7028 Linear 1 term 8.8377 (1,24) 4.260 0.0066 2 terms 4.7618 (2,23) 3.422 0.0186 Because the RESET test returns p-values less than 0.05 (0.0066 and 0.0186 for one and two terms respectively), at a 5% level of significance we conclude that the linear model is not an adequate functional form for the beer data. On the other hand, the log-log model appears to suit the data well with relatively high p-values of 0.9319 and 0.7028 for one and two terms respectively. Thus, based on the RESET test we conclude that the log-log model better reflects the demand for beer. Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 131 Exercise 6.14 (continued) (e) The estimated model is: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 237.0540 2.2076 0.1688 2.6213 (se) 17.0160 1.0914 0.0444 0.7101 ( ) 2.178 2.023 3.800 3.691 HW HE HE HA t = − − + + − − ( ) ( ) ( ) ( ) 2 20.0278 7.9379 0.2443 0.0079 1.1012 3.525 7.208 HA CIT R− + = − The wage rate in large cities is, on average, $7.94 higher than it is outside those cities. (f) The p-value for 6b , the coefficient associated with CIT, is 0.0000. This suggests that 6b is significantly different from zero and CIT should be included in the equation. Note that when CIT was excluded from the equation in part (c), its omission was not picked up by RESET. The RESET test does not always pick up misspecifications. (g) From part (c), we have 1.4580 0.3022HW HE HE ∂ = − + ∂ 2.8895 0.0602HW HA HA ∂ = − ∂ and from part (f) 2.2076 0.3376HW HE HE ∂ = − + ∂ 2.6213 0.0556HW HA HA ∂ = − ∂ Evaluating these expressions for 6HE = , 15HE = , 35HA = and 50HA = leads to the following results. HW HE∂ ∂ HW HA∂ ∂ 6HE = 15HE = 35HA = 50HA = Part (c) 0.356 3.076 0.781 0.123− Part (e) 0.182− 2.855 0.678 0.156− The omitted variable bias from omission of CIT does not appear to be severe. The remaining coefficients have similar signs and magnitudes for both parts (c) and (e), and the marginal effects presented in the above table are similar for both parts with the exception of HW HE∂ ∂ for 6HE = where the sign has changed. The likely reason for the absence of strong omitted variable bias is the low correlations between CIT and the included variables HE and HA. These correlations are given by ( )corr , 0.2333CIT HE = and corr( , ) 0.0676CIT HA = . Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 132 EXERCISE 6.15 (a) The average price of a 40-year old house of size 3600 square feet is (40,3600) 1 2 33600 40PRICE = β + β + β The average price of a 5-year old house of size 1800 square feet is (5,1800) 1 2 31800 5PRICE = β + β + β The conjecture that we set up as the alternative hypothesis is 1 2 3 1 2 33600 40 2( 1800 5 )β + β + β > β + β + β Thus, after simplifying this inequality, the null and alternative hypotheses are 0 1 3: 30 0H −β + β ≤ 1 1 3: 30 0H −β + β > The test statistic for testing 0H is 1 3 1 3 30 se( 30 ) b bt b b − + = − + where 1 3 1 3 1 3se( 30 ) var( ) 900var( ) 60cov( , )b b b b b b− + = + − The values for these quantities and the test results for each house category are as follows. All houses Town houses French style 1 330b b− + 19296 169063− 291863 1var( )b 48855007 130798354 1088235489 3var( )b 19851 140902 12063976 1 3cov( , )b b 497267− 3248879− 15538629− 1 3se( 30 )b b− + 9826 21273 113482 t-value 1.964 7.947− 2.572 df 1077 67 94 5% critical value 1.646 1.668 1.661 p-value 0.0249 1.0000 0.0058 Decision Reject 0H Accept 0H Reject 0H For the all-house and French style categories, the data support the conjecture stated in the alternative hypothesis, namely, that (40,3600) (5,1800)2PRICE PRICE> × . In the case of town houses, whose estimated equation suggests that they quickly depreciate, the alternative hypothesis is not supported. Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 133 Exercise 6.15 (continued) (b) The average prices for the three houses are as follows. (i) (0,2000) 1 22000PRICE = β + β (ii) (20,2200) 1 2 32200 20PRICE = β + β + β (iii) (40,2400) 1 2 32400 40PRICE = β + β + β Setting (0,2000) (20,2200)PRICE PRICE= , gives 1 2 1 2 32000 2200 20β + β = β + β + β which can be simplified to 2 310 0β +β = Setting (0,2000) (40,2400)PRICE PRICE= , gives 1 2 1 2 32000 2400 40β + β = β + β + β which can be simplified to 2 310 0β +β = Thus, all three houses will be equally priced if 0 2 3:10 0H β +β = holds. The F-value for testing this null hypothesis against the alternative 1 2 3:10 0H β +β ≠ is 1.12F = . The corresponding p-value is 0.29. Thus 0H is not rejected. There is no evidence to suggest the houses are not equally priced. Remark: In the first printing of POE, the third house was given as 40 years old with 2300 (not 2400) square feet. In this case, the null and alternative hypotheses are 0 2 3: 0H β = β = and 1 2: 0H β ≠ and/or 3 0β ≠ . The test values are 773.6F = and p-value = 0.0000. The null hypothesis is rejected. (c) The application of RESET tests to all houses, town houses and French style homes leads to rejection of the adequacy of the model 1 2 3PRICE SQFT AGE e= β +β +β + in all cases. The model might be improved by the inclusion of more variables such as type of neighborhood, and whether the house has particular attributes such as a view, a pool and a fireplace. Also, the functional form might be inadequate. Log-log or log-linear forms or the inclusion of quadratic terms might improve the model Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 136 Exercise 6.17 (continued) (c) The delay from 3 trains is 43β . The extra time gained by leaving 5 minutes earlier is 25 5+ β . Thus, the hypotheses are 0 4 2: 3 5 5H β ≤ + β and 0 4 2: 3 5 5H β > + β The rejection region for the t-test is (0.95,227) 1.652t t> = , where the t-value is calculated as 4 2 4 2 3 5 5 3 2.7548 5 0.36923 5 1.546 se(3 5 ) 0.9174 b bt b b − − × − × − = = = − and the standard error is computed from 4 2 4 2 2 4se(3 5 ) 9 var( ) 25 var( ) 30 cov( , ) 9 0.092298 25 0.000241 30 0.000165 0.9174 b b b b b b− = × + × − × = × + × + × = Since 1.546 1.652< , we do not reject 0H at a 5% significance level. Alternatively, we do not reject 0H because the p-value = 0.0617, which is greater than 0.05. There is insufficient evidence to conclude that leaving 5 minutes earlier is not enough time. (d) The expected time taken when the departure time is 7:15AM, and no red lights or trains are encountered, is 1 245β + β . Thus, the null and alternative hypotheses are 0 1 2: 45 45H β + β ≤ and 1 1 2: 45 45H β + β > The rejection region for the t-test is (0.95,227) 1.652t t> = , where the t-value is calculated as 1 2 1 2 45 45 19.9166 45 0.36923 45 7.44 se( 45 ) 1.1377 b bt b b + − + × − = = = − + and the standard error is computed from 2 1 2 1 2 1 2se( 45 ) var( ) 45 var( ) 90 cov( , ) 1.574617 2025 0.00024121 90 0.00854061 1.1377 b b b b b b+ = + × + × = + × − × = Since 7.44 1.652− < , we do not reject 0H at a 5% significance level. Alternatively, we do not reject 0H because the p-value = 1.000, which is greater than 0.05. There is insufficient evidence to conclude that Bill will not get to the University before 8:00AM. Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 137 Exercise 6.17 (continued) (e) The predicted time it takes Bill to reach the University is 1 2 3 430 6 1 41.76TIME b b b b= + × + × + × = Using suitable computer software, the standard error of the forecast error can be calculated as se( ) 4.0704f = . Thus, a 95% interval estimate for the travel time is (0.975,227)se( ) 41.76 1.97 4.0704 (33.74,49.78)TIME t f± = ± × = Rounding this interval to 34 – 50 minutes, a 95% interval estimate for Bill’s arrival time is from 7:34AM to 7:50AM. Chapter 6, Exercise Solutions, Principles of Econometrics, 3e 138 EXERCISE 6.18 (a) We are testing the null hypothesis 0 2 3:H β = β against the alternative 1 2 3:H β ≠ β . The test can be performed with an F or a t statistic. Using an F-test, we reject 0H when (0.95,1,348)F F> , where (0.95,1,348) 3.868F = . The calculated F-value is 0.342. Thus we do not reject 0H because 0.342 3.868< . Also, the p-value of the test is 0.559, confirming non- rejection of 0H . The hypothesis that the land and labor elasticities are equal cannot be rejected at a 5% significance level. Using a t-test, we reject 0H when (0.975,348)t t> or (0.025,348)t t< where (0.975,348) 1.967t = and (0.025,348) 1.967t = − . The calculated t-value is 2 3 2 3 0.36174 0.43285 0.585 se( ) 0.12165 b bt b b − − = = = − − In this case 0H is not rejected because 1.967 0.585 1.967− < − < . The p-value of the test is 0.559. The hypothesis that the land and labor elasticities are equal cannot be rejected at a 5% significance level. (b) We are testing the null hypothesis 0 2 3 4: 1H β +β +β = against the alternative 1 2 3 4: 1H β +β +β ≠ , using a 10% significance level. The test can be performed with an F or a t statistic. Using an F-test, we reject 0H when (0.90,1,348)F F> , where (0.90,1,348) 2.72F = . The calculated F-value is 0.0295. Thus, we do not reject 0H because 0.0295 2.72< . Also, the p-value of the test is 0.864, confirming non-rejection of 0H . The hypothesis of constant returns to scale cannot be rejected at a 10% significance level. Using a t-test, we reject 0H when (0.95,348)t t> or (0.05,348)t t< where (0.95,348) 1.649t = and (0.05,348) 1.649t = − . The calculated t-value is 2 3 4 2 3 4 1 0.36174 0.43285 0.209502 1 0.172 se( ) 0.023797 b b bt b b b + + − + + − = = = + + In this case 0H is not rejected because 1.649 0.172 1.649− < < . The p-value of the test is 0.864. The hypothesis of constant returns to scale cannot be rejected at a 10% significance level.