### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# 5.solutions chapter 5

### View Full Document

## About this Document

## 6

## 0

## Popular in

## Popular in Department

This 21 page Bundle was uploaded by yeshuahanotzrivemelejhayehu Notetaker on Friday September 16, 2016. The Bundle belongs to at 1 MDSS-SGSLM-Langley AFB Advanced Education in General Dentistry 12 Months taught by in Fall 2016. Since its upload, it has received 6 views.

## Reviews for 5.solutions chapter 5

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/16/16

CHAPTER 5 Exercise Solutions 91 Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 92 EXERCISE 5.1 (a) y =1=,x 2 3 0, 0 xi2 xi3 yi 0 1 0 1 − 2 1 2 1 2 − 2 0 − 2 1 − 1 − 1 − 2 − 1 − 2 0 1 1 − 1 1 0 1 0 1 (b) yx* =13=, =x =2 16,2 yx** 4, x 10 ∑ ∑∑ii2 2 3i 3 i i i ** 2 * * * * (∑∑ ii∑2 3 i * 33 2i i i ix ) 13 ×−×4 0 = (c) b2 = = 2 2 2 2 0.8125 (( ) )(∑∑ ∑i ii 3i2 *− * 16×−0 0 2 (∑∑ ii∑3 2 i* 23 2i i * *i ix ) 4×−×0 = b3 = = 2 2 0.4 (( ) )(∑∑ i i3i 3i2 − * * 16 ×− 0 3 32 1 2− − = 1 (d) e =( 0.4, 0)9875−, 0.02−5, 0.3−75, 1.4125, 0.025, 0.6, 0.4125, 0.1875 ∑ eˆ2 3.8375 (e) σ=2 = = 0.6396 NK − − 93 ∗ ∗ (∑ ∑(xi2 i 2 3 3 i xxi 23 (f) r23= = 2 2 2 2 ∗ ∗ 0 ∑ ∑(i2 2i∑ x3i 3 i x x 2 3 2 (g) se(b b= = var( ) = = σ 0.6396 0.1999 2 2 6 )11 ∑( (xx − −r 2 2 i2 2 3 2 2 2 (h) SSE = =∑ ∑i 3i8375 = −=SST (y y ) 16, 2 SSR 12.1625 SSR = −SST SSE 12.16=25 = R= 0.7602 SST 16 Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 93 EXERCISE 5.2 (a) A 95% confidence interval for β is 2 = bt±± = b se( ) 0.8125 2.447 0.1999 (0.3233,1.3017) ) 6 2, 5 7 9 . 02 ( (b) The null and alternative hypotheses are H,0 2:1≠1= 1 2 cTlhelated t-value is b2−1 0.8125 1− t= = − = 0.9377 se(b2) 0.1999 At a 5% significance level, we reject H if tt> = 2.447 . Since −0<.9377 2.447 , 0 (0.975,6) we do not reject H . 0 Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 94 EXERCISE 5.3 b1 0.0091 (a) (i) The t-statistic for1b is = = 0.476 . se(1 ) 0.0191 0.0276 (ii) The standard error forb2bi2 se( ) = = 0.00418 . 6.6086 (iii) The estimate forβ3 is b3= 0.0002×− ( 6.9624=− 0.0014 . (iv) To compute R , we need SSE and SST. From the output, SSE = 5.752896. To find SST, we use the result SST σy N −1 0.0633 2 which gives SST =1×518 (= 0.0633) 6.08246. 2 SSE 5.75290 4 5Thus, R 1 1. − 0=− = SST 6.08246 (v) The estimated error standard deviation is σ= SSE = 5.752896= 0.061622 (NK − −) 1519 4 (b) Tvhelue b = 0.0276 implies that if ln(TOTEXP)increases by 1 unit the alcohol share 2 will increase by 0.0276. The change in the alcohol share from a 1-unit change in total expenditure depends on the level of total expenditure. Specifically, d(WALC) d(TOTEXP) = 0.0276 TOTEXP . A 1% increase in total expenditure leads to a 0.000276 increase in the alcohol share of expenditure. vaTue b3= −0.0014 suggests that if the age of the household head increases by 1 year the share of alcohol expenditure of that household decreases by 0.0014. vaTue b = −0.0133 suggests that if the household has one more child the share of the 4 alcohol expenditure decreases by 0.0133. (c) A 95% confidence interval for β3is − − = × ) 5 31 5 1 , 5 7 9 .3 0b se( ) 0.0014 1.96 0.0002 ( 0.0018, 0.0010) This interval tells us that, if the age of the household head increases by 1 year, the share of the alcohol expenditure is estimated to decrease by an amount between 0.0018 and 0.001. Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 95 Exercise 5.3 (Continued) (d) The null and alternative hypotheses are H,0 40≠= β1 4 . b cTlhelated t-value is − = = 4 4.075 se( 4 At a 5% significance level, we reject H if tt> = 1.96 . Since −4>.075 1.96 , we 0 (0.975,1515) reject H and conclude that the number of children in the household influences the budget 0 proportion on alcohol. Having an additional child is likely to lead to a smaller budget share for alcohol because of the non-alcohol expenditure demands of that child. Also, perhaps households with more children prefer to drink less, believing that drinking may be a bad example for their children. Chapter 5, Exercise Solutions, Principles of Econometrics,96e EXERCISE 5.4 (a) The regression results are: n 2 WTRANS =− 0.0315+ 0.0414ln T(TEXP− 0.)001AGE − 0.0130 NK R= 0.0247 ()e (0.0322) (0.0071) (0.0004) (0.0055) (b) Tvalue b2= 0.0414 suggests that as ln(TOTEXP ) increases by 1 unit the budget proportion for transport increases by 0.0414. Alternatively, one can say that a 10% increase in total expenditure will increase the budget proportion for transportation by 0.004. (See Section A.4.8 of Appendix A.) The positive sign of2is according to our expectation because as households become richer they tend to use more luxurious forms of transport and the proportion of the budget for transport increases. vTlue b3= −0.0001 implies that as the age of the head of the household increases by 1 year the budget share for transport decreases by 0.0001. The expected signb is not 3 clear. For a given level of total expenditure and a given number of children, it is difficult to predict the effect of age on transport share. valuee b4= −0.0130 implies that an additional child decreases the budget share for transport by 0.013. The negative sign means that adding children to a household increases expenditure on other items (such as food and clothing) more than it does on transportation. Alternatively, having more children may lead a household to turn to cheaper forms of transport. (c) The p-value for testingH 0 3 = against the alternativH 1 3 ≠ where β3 is the coefficient of AGE is 0.869, suggesting that AGE could be excluded from the equation. Similar tests for the coefficients of the other two variables yield p-values less than 0.05. (d) The proportion of variation in the budget proportion allocated to transport explained by this equation is 0.0247. (e) For a one-child household: n WTRANS =−00.0315+ 0.0414ln(TOTEXP − 0) 0 0001AGE− 0.013NK =− 0.0315+ 0.041× ln(98.7−) 0.000×1 −6 0.0×13 1 = 0.1420 For a two-child household: WTRANS =−0 0.0315+0.0414ln( TOTEXP − ) 0.0001 AGE− 0.013NK 0 0 0 =− 0.0315+ 0.041× ln(98.7−) 0.000×1 −6 0.0×13 2 = 0.1290 Chapter 5, Exercise Solutions,Principles of Econometrics, 3e 97 EXERCISE 5.5 (a) The estimated equation is VALUE = 2 −8.4067 −0.1834CRI +ME 22.8−109NITOX 6.3715ROOMS 0.0478AGE (se) (5.3659) (0.0365) (4.1607) (0.3924) (0.0141) −1+.3353DIST− 0.2723ACCESS 0.0126TAX 1.1768PTRATIO (0.2001) (0.0723) (0.0038) (0.1394) The estimated equation suggests that as the per capita crime rate increases by 1 unit the home value decreases by $183.4. The higher the level of air pollution the lower the value of the home; a one unit increase in the nitric oxide concentration leads to a decline in value of $22,811. Increasing the average number of rooms leads to an increase in the home value; an increase in one room leads to an increase of $6,372. An increase in the proportion of owner-occupied units built prio r to 1940 leads to a decline in the home value. The further the weighted distances to the five Boston employment centers the lower the home value by $1,335 for every unit of weighted distance. The higher the tax rate per $10,000 the lower the home value. Finally, the higher the pupil-teacher ratio, the lower the home value. (b) A 95% confidence interval for the coefficient of CRIME is − − = × ± − bt± = b se( ) 0.1834 1.965 0.0365 ( 0.255, 0.112) . 2 (0.975,497)2 A 95% confidence interval for the coefficient of ACCESS is = × bt7 ± (0.975,497)7 0.2723 1.965 0.0723 (0.130, 0.414) (c) We want to test H 0Hβr1oom 7 against≠ : room 7 . The value of the t statistic is b −7− 6.3715 7 t=−= roo=s 1.6017 se(brooms 0.3924 At α= 0.05 , we reject H if the absolute calculated t is greater than 1.965. Since 0 −1.6017 1.965 , we do not reject H 0 The data is consistent with the hypothesis that increasing the number of rooms by one increases the value of a house by $7000. (d) We want to test H :Hβ≥ 1−a<ainst : 1. The value of the t statistic is 0 p1tratio ptratio −1+.1768 1 − t= = 0.1394 1.2683 At a significance level of α= 0.05, we rejectH if the calculated t is less than the critical 0 value t (0.05,497).648 . Since −1.2683 >− 1.648, we do not reject H 0. We cannot conclude that reducing the pupil-teacher ratio by 10 will increase the value of a house by more than $10,000. Chapter 5, Exercise Solutions,Principles of Econometrics, 3e 98 EXERCISE 5.6 The EViews output for verifying the answers to Exercise 5.1 is given below. Method: Least Squares Dependent Variable: Y Method: Least Squares Included observations: 9 Coefficient Std. Error t-Statistic Prob. X1 1.000000 0.266580 3.751221 0.0095 X2 0.812500 0.199935 4.063823 0.0066 X3 0.400000 0.252900 1.581654 0.1648 R-squared 0.760156 Mean dependent var 1.000000 Adjusted R-squared 0.680208 S.D. dependent var 1.414214 S.E. of regression 0.799740 Akaike info criterion 2.652140 Sum squared resid 3.837500 Schwarz criterion 2.717882 Log likelihood -8.934631 Hannan-Quinn criter. 1.728217 Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 99 EXERCISE 5.7 (a) Estimates, standard errors and p-values for each of the coefficients in each of the estimated share equations are given in the following table. Explanatory Dependent Variable Variables Food Fuel Clothing Alcohol Transport Other Constant Estimate 0.8798 0.3179 −0.2816 0.0149 − 0.0191 0.0881 0.0ro2r S00265 0.0510 0.0370 00..00553762 p-value 0.0000 0.0000 0.0000 0.6878 0.7382 0.1006 ln(TOTEXP) Estimate − 0.1477 −0.0560 0.0929 0.0327 0.0321 0.0459 0.0ro3r S00058 0.0112 0.0082 00..00111286 p-value 0.0000 0.0000 0.0000 0.0001 0.0111 0.0001 AGE Estimate 0.00227 0.00044 −0.00056 −0.00220 0.00077 −0.00071 0.00ro5r 0St0029 0.00055 0.00040 00..0000005682 p-value 0.0000 0.1245 0.3062 0.0000 0.2167 0.2242 NK Estimate 0.0397 0.0062 −0.0048 − 0.0148 − 0.0123 − 0.0139 0.0ro4r S00044 0.0084 0.0061 00..00008984 p-value 0.0000 0.1587 0.5658 0.0152 0.1921 0.1157 An increase in total expenditure leads to decreases in the budget shares allocated to food and fuel and increases in the budget shares of the commodity groups clothing, alcohol, transport and other. Households with an older household head devote a higher proportion of their budget to food, fuel and transport and a lower proportion to clothing, alcohol and other. Having more children means a higher proportion spent on food and fuel and lower proportions spent on the other commodities. The coefficients of ln(TOTEXP) are significantly different from zero for all commodity groups. At a 5% significance level, age has a significant effect on the shares of food and alcohol, but its impact on the other budget sh ares is measured less precisely. Significance tests for the coefficients of the number of children yield a similar result. NK has an impact on the food and alcohol shares, but we can be less certain about the effect on the other groups. To summarize, ln(TOTEXP) has a clear impact in all equations, but the effect ofAGE and NK is only significant in the food and alcohol equations. Chapter 5, Exercise Solutions, Principles of Econometrics, 3e100 Exercise 5.7 (continued) (b) The t-values and p-values for testing H0 2β ≤ 0βagain1 2 : 0 are reported in the table below. Using a 5% level of significance, the critical value for each test is t(0.95,496)648 . t-value p-value decision WFOOD − 13.083 1.0000 Do not rejecH 0 WFUEL −9.569 1.0000 Do not rejecH 0 WCLOTH 8.266 0.0000 Reject H 0 WALC 4.012 0.0000 Reject H 0 WTRANS 2.548 0.0056 Reject H 0 WOTHER 3.884 0.0001 Reject H 0 Those commodities which are regarded as necessities (b2< 0) are food and fuel. The tests suggest the rest are luxuries. While alcohol, transportation and other might be luxuries, it is difficult to see clothing categorized as a luxury. Perhaps a finer classification is necessary to distinguish between basic and luxury clothing. Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 101 EXERCISE 5.8 (a) The expected sign for β 2s negative because, as the number of grams in a given sale increases, the price per gram should decrease, implying a discount for larger sales. We expect β to be positive; the purer the cocaine, the higher the price. The sign for β will 3 4 depend on how demand and supply are changing over time. For example, a fixed demand and an increasing supply will lead to a fall in price. A fixed supply and increased demand would lead to a rise in price. (b) The estimated equation is: n 2 PRICE = 9 −0.8467 0 +.0600Q− UANT 0.1162Q= UAL 2.3546TREND R 0.5097 (se) (8.5803) (0.0102) (0.2033) (1.3861) (t) (10.588) ( −5.892) (0.5717) ( −1.6987) The estimated values for β2 3 an4 are −0.0600, 0.1162 and −2.3546, respectively. They imply that as quantity (number of gram s in one sale) increases by 1 unit, the price will go down by 0.0600. Also, as the quality increases by 1 unit the price goes up by 0.1162. As time increases by 1 year, the price decreases by 2.3546. All the signs turn out according to our expectations, with β4 implying supply has been increasing faster than demand. (c) The proportion of variation in cocaine price explained by the variation in quantity, quality and time is 0.5097. (d) For this hypothesis we test H<0 2 ≥ 0βagains1 2 : 0 . The calculated t-value is −5.892 . We reject H i0 the calculated t is less than the critical ( )95,52− 1.675. Since the calculated t is less than the critical t value, we rejec0 H and conclude that sellers are willing to accept a lower price if they can make sales in larger quantities. (e) We want to test H 0 3≤ 0 against1 3 : 0. The calculated t-value is 0.5717. At α= 0.05 we reject H 0 if the calculated t is greater than 1.675. Since for this case, the calculated t is not greater than the critical t, we do not rejHc0. We cannot conclude that a premium is paid for better quality cocaine. (f) The average annual change in the cocaine price is given by the value of b4=− 2.3546 . It has a negative sign suggesting that the price decreases over time. A possible reason for a decreasing price is the development of improved technology for producing cocaine, such that suppliers can produce more at the same cost. Chapter 5, Exercise Solutions, Principles of Econometrics, 3102 EXERCISE 5.9 (a) Theefficients, β3, β4and β5are expected to be positive, whereas 2 should be negative. The dependent variable is the log of per capita consumption of beef. As the price of beef declines, consumption of beef should increase (2< 0). Also, as the prices of lamb and pork increase, the consumption of beef should increase 3 β > 0 and4β > 0). If per capita disposable income increases, the consumption of beef should increase5(0). (b) The least squares estimates and standard errors are given by ln(QB) = −.6726 0.+8266ln ( ) 0.1997ln( ) ()( 1.()96 0)1826 ( 0)2127 +0+.4371ln( ) 0.1017ln ( ) R 2 0.7609 ( ) ( ) 0.2940 The estimated coefficients are elasticities, impl ying that a 1% increase in the price of beef leads to an 0.83% decrease in the quantity of beef consumed. Likewise, 1% changes in the prices of lamb and pork will cause 0.20% and 0.44% changes, respectively, in the quantity of beef consumed. The estimates seem reasonable. They all have the expected signs and the price of beef has the greatest effect on the quantity of beef consumed, a logical result. However, the standard errors of the coefficient estimates for the lamb and pork prices, and for income, are relatively large. (c) The estimated covariance matrix is given by ⎛ 2.7⎞42 0.1087 − − −36 0.2353 0.3314 ⎜ 0.1087 0.0334 0.0040 −0− .0107 0.0326 m ⎜ ⎟ cov( 1 2 3 4 5 = ⎜−0.0−36 0.0040 0.0453 0.0439 0.1661 ⎜−0−.353 0.0107− 0.0439 0.1472 0.0303 ⎜ ⎟ ⎝−0.3⎠14 0.0326 0.1661 0.0303 0.0864 and the standard errors, reported in part (b), are the square roots of the diagonal of this matrix. The matrix contains estimates of the variances and covariances of the least squares estimators 1 2 3 4 and b 5 The variances and covariances are measures of how the least squares estimates will vary and "covary" in repeated samples. (d) Using t(0.975,12)179 , the 95% confidence intervals are as follows. β 1 β 2 β3 β4 β5 lower limit 1.0563 −1.2246 −0.2638 −0.3989 −0.5389 upper limit 8.2888 −0.4286 0.6632 1.2732 0.7423 Chapter 5, Exercise Solutions, Principles of Econometrics, 103 EXERCISE 5.10 (a) The estimated regression models and 95% confidence intervals fo3 β follow. (i) All houses: n PRICE =− 41948+ 90.970SQFT − 755.04AGE (se) (6990) (2.403) (140.89) − − = × ± − bt± = b se( ) 755.04 1.962 140.89 ( 1031.5, 478.6) 3 (0.975,10773 (ii) Town houses: PRICE = 9+0415 44 −.014SQFT 2621.6AGE (se) (11437) (5.688) (375.4) − − = × ± − b3± (0.975,67)3 2621.6 1.996 375.4 ( 3370.8, 1872.3) (iii) French style homes: n PRICE =− 293804+ 184.21SQFT− 64.713AGE (se) (32988) (10.43) (3473.323) − = × ± − bt± = b se( ) 64.7 1.986 3473.3 ( 6961.1, 6831.7) 3 (0.975,94)3 The value of additional square feet is highest for French style homes and lowest for town houses. Town houses depreciate more with age than do French style homes. The larger number of observations used in the regression for all houses has led to standard errors that are much lower than those for the special-category houses. In terms of the confidence intervals for , it also means the confidence interval for all houses is much narrower than 3 those for the special-category houses. The confidence interval for French style homes is very wide and includes both positive and negative ranges. Age does not seem to be an important determinant of the price of French style homes. For town houses it is more important than for the total population of houses. (b) We wish to test the hypothesis H0 3 =− 1000 against the alternative 0 3− 1000. The results for each of the three cases follow. (i) All houses: The critical values (0.975,1077)62 and t(0.025,1077).962 . We reject H 0 if the calculated t-value is such t ≥1.962 or t ≤ −1.962. The calculated value is t = =−55.04 ( 1000) 1.74 140.89 Since −1<<62 1.74 1.962 , we do not reject H . The data is consistent with the 0 hypothesis that having an older house reduces its price by $1000 per year for each year of its age. Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 104 Exercise 5.10(b) (continued) (b) (i)ow houseTs:h e critical values are (0.975,67).996 and t (0.025,67) 1.996 . We reject H i0 the calculated t-value is such that t ≥1.996 or t ≤− 1.996 . The calculated value is −−2−621.6 ( 1000) − t= = 4.32 375.4 Since t =−4<.32 1.996 , we reject H 0 The data is not consistent with the hypothesis that having an older house reduces its price by $1000 per year for each year of its age. (iii) French style homes: The critical values are t(0.975,94)986 and (0.025,94) 1.986 . We reject H if the calculated t-value is such that t ≥1.986 or t ≤− 1.986 . The 0 calculated value is −−6−4.713 ( 1000) t = = 0.27 3473.3 Since −1<.<86 0.27 1.986, we do not reject H . The data is consistent with the 0 hypothesis that having an older house reduces its price by $1000 per year for each year of its age. Chapter 5, Exercise Solutions, Principles of Econometrics,105 EXERCISE 5.11 (a) The estimated regression model is VOTE = 5+2.44 0.6488−GROWTH 0.1862INFLATION (se) (1.49) (0.1675) (0.4320) The hypothesis test results on the significance of the coefficients are: H 0 2>:=01 2 p-value = 0.0003 significant at 10% level H H:0<: β p-value = 0.335 not significant at 10% level 0 3 1 3 One-tail tests were used because more growth is considered favorable, and more inflation is considered not favorable, for re-election of the incumbent party. (b) The predicted percentage vote for the incumbent party when INFLATION = 4 and GROWTH =− 4 is VOTE =− +2.4×4 0.6488 ( 4) 0.1862 4 49.104 0 (c) Ignoring the error term, the incumbent party will get the majority of the vote when β+β GROWT+ Hβ INFLAT> ION 50 1 2 3 Assuming β1 1 and β3 3 , and given INFLATION = 4 and GROWTH =− 4 , the incumbent party will get the majority of the vote when 52.44−β4−2 0.186≥2 4 50 which is equivalent to β<−(−50 0.18× 62+21 4 52.44= 357) 4 0.42467 2 To ensure accuracy, we have included more decimal places in this calculation than are in the reported equation. teFirg H 0 2 0.42467 against the alternative H1 2 > 0.42467, the t-value is 0.64876−0.42467 t = = 1.338 0.16746 For a 5% significance level, the critical valut(0.95, 28)01. The rejection region is t ≥1.701. Thus, H 0 is not rejected. The incumbent party might still get elected if GROWTH =− 4% . Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 106 EXERCISE 5.12 (a) The estimated equation is n TIME =1 +9.9166 0.3+ 6923DE+ PART 1.3353REDS 2.7548TRAINS (se) (1.2548) (0.01553) (0.1390) (0.3038) Interpretations of each of the coefficients are: β1: The estimated time it takes Bill to get to work when he leaves Carnegie at 6:30AM and encounters no red lights and no trains is 19.92 minutes. β2: If Bill leaves later than 6:30AM, his traveling time increases by 3.7 minutes for every 10 minutes that his departure time is later than 6:30AM (assuming the number of red lights and trains are constant). β3: Each red light increases traveling time by 1.34 minutes. β : Each train increases traveling time by 2.75 minutes. 4 (b) The 95% confidence intervals for the coefficients are: β1 ×bt1±= (0.975,227)1 19.9166 1.970 1.2548 (17.44, 22.39) β2=×bt2±= (0.975,227)2 0.36923 1.970 0.01553 (0.339, 0.400) =3× bt3± =(0.975,227)3 1.3353 1.970 0.1390 (1.06,1.61) = × bt±± = b se( ) 2.7548 1.970 0.3038 (2.16,3.35) 4 4 (0.975,227)4 In the context of driving time, these intervals are relatively narrow ones. We have obtained precise estimates of each of the coefficients. (c) Thyepotheserse H 0 32≥ and H :1 32 < . The critical value is t(0.05,227)1.652 . We reject H 0hen the calculated t-value is less than −1.652 . This t-value is 1.3353−2 − t= = 4.78 0.1390 Since −4<.78 1.652, we reject H .0We conclude that the delay from each red light is less than 2 minutes. (d) Thyepothesaerse H :β3= and H :β 3 ≠ . The critical values are t =− 1.652 0 4 1 4 (0.05,227) and t =1.652 . We reject H when the calculated t-value is such that t <− 1.652 or (0.95,227) 0 t >1.652 . This t-value is 2.7548−3 t= = 0.807 0.3038 Since −1.65<2 0.807 1.652, we do not reject H 0 The data are consistent with the hypothesis that each train delays Bill by 3 minutes. Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 107 Exercise 5.12 (continued) (e) Delaying the departure time by 30 minutes, increases travel time by 30 β 2. Thus, the null hypothesis is H :30 β≥ 10 , or H 3:β 1 ≥ , and the alternative is H3:β1< . We 0 2 0 2 1 2 reject H 0f tt≤ (0.05,227) 1.652, where the calculated t-value is t = =36923−0.33333 2.31 0.01553 Since 2.31>− 1.652 , we do not reject H . The data are consistent with the hypothesis that 0 delaying departure time by 30 minutes increases travel time by at least 10 minutes. (f) If we assume that ββ,2 3 and β4 are all non-negative, then the minimum time it takes Bill to travel to work is β1 . Thus, the hypotheses are H00 1 ≤ and H1 12> . We reject H if tt≥ = 1.652 , where the calculated t-value is 0 (0.95,227) 19.9166− 20 − t= = 0.066 1.2548 Since −0<.066 1.652 , we do not reject H 0. The data support the null hypothesis that the minimum travel time is less than or equal to 20 minutes. It was necessary to assume that β2 3 and β 4 are all positive or zero, otherwise increasing one of the other variables will lower the travel time and the hypothesis would need to be framed in terms of more coefficients than β1. Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 108 EXERCISE 5.13 (a) The coefficient estimates, standard errors, t-values and p-values are in the following table. Dependent Variable: ln(PROD) Error Std. Coeff t-value p-value C -1.5468 0.2557 -6.0503 0.0000 ln(AREA) 0.3617 0.06405.6550.0000 ln(LABOR) 0.43280.0669 6.47180.0000 ln(FERT) 0.20950.03835.47500.0000 All estimates have elasticity interpretations. For example, a 1% increase in labor will lead to a 0.4328% increase in rice output. A 1% in crease in fertilizer will lead to a 0.2095% increase in rice output. All p-values are less than 0.0001 implying all estimates are significantly different from zero at conventional significance levels. (b) The null and alternative hypotheses are 5 H.0 20 = and5 H.1 20 ≠ . The 1% critical values are t(0.995,348).59 and t(0.005,348)2.59 . Thus, the rejection region is t ≥ 2.59 or t ≤− 2.59. The calculated value of the test statistic is − t = =.3617 −0.5 2.16 0.064 Since −2<.59 2.16 2.59, we do not reject H 0 The data are compatible with the hypothesis that the elasticity of production with respect to land is 0.5. (c) A 95% interval estimate of the elasticity of production with respect to fertilizer is given by bt± ×= ± × = b se( ) 0.2095 1.967 0.03826 (0.134, 0.285) 4 (0.975,348) 4 This relatively narrow interval implies the fertilizer elasticity has been precisely measured. (d) This hypothesis test is a test of3 0 30 ≤ agains3 H.1 30 > . The rejection region is tt≥ = 1.649 . The calculated value of the test statistic is (0.95,348) t = =.433−0.3 1.99 0.067 rejeet H because 1.99 >1.649. There is evidence to conclude that the elasticity of 0 production with respect to labor is greater than 0.3. Reversing the hypotheses and testing 3 H. :β0≥ against 3 H. :β0< , leads to a rejection region of t ≤− 1.649 . The 0 3 1 3 calculated t-value ist =1.99 . The null hypothesis is not rejected because 1.99 >− 1.649 . Chapter 5, Exercise Solutions, Principles of Econometri109 3e EXERCISE 5.14 (a) The predicted logarithm of rice productioAREA=1 , LABOR = 50, and FERT =100 is ln(PRO−D) =+×.+4×8 =0.3617 ln(1) 0.4328 ln(50) 0.2095 ln(100) 1.111318 Using the corrected predictor given on page 95 of the text, the corresponding prediction of rice production is = × PRσOD = exp (n(PROD) ) exp(ˆ2 2) exp(1.111318) exp(0.341419 2 ) = 3.22071 (b) The marginal principle says apply more fertilizer if ∂ PROD PRICE FERT > = 0.004 ∂FERT PRICE RICE Since ∂PROD =β × PROD , ∂FERT 4 FERT the above marginal principle can be stated as: Apply more fertilizer if PROD β4 FERT> 0.004 Substituting PROD = 3.22071 and FERT =100 , this inequality is the same as 100 β4 0.004 = 0.1242 3.22071 (c) The hypoth2 4 2 1 H. 0 4≤ and H 1 4> 0.1242 . The 5% critical value is t(0.95,348)49 , leading to a rejection regt ≥1.649. The calculated t-value is 0.2095−0.1242 t= = 2.23 0.0383 Since 2.23>1.649, we reject H . The farmer should apply more fertilizer. We chose 0 β>40.1242 as the alternative hypothesis because th e farmer would not want to incur the cost of more fertilizer unless there was strong evidence from the data that it is profitable to do so. Chapter 5, Exercise Solutions, Principles of Econo110rics, 3e Exercise 5.14 (continued) (d) The predicted logarithm of rice production for AREA=1, LABOR = 50, and FERT =1 is ln(PR−D) =+×.+4=8 0.3617 ln(1) 0.4328 ln(50) 0.2095 ln(10.146525 Nthitg ln(1)= 0, the estimated variance of the prediction error can be written as n n 2 n n 2 var( b=+ar(1)33n(5+)]1 varb( )b [ln(50] cov( , ) ˆ − = .06535875 15.303924 0.004473275 2 3.912023 ( 0.01360617) + 0.341419 = 0.143929 and n se( f )f= =var( 0.143929 0.37938 A 95% interval estimate for the logarithm of rice production at the specified values is n(PROD=)± (0.975,348) 0.146525 1.9668 0.37938 ( 0.59964, 0.89269) The corresponding 95% interval estimate for rice production is ( )p(−0=9964), exp(0.89269)(0.549, 2.442) Chapter 5, Exercise Solutions, Principles of Econometrics, 3e 111 EXERCISE 5.15 (a) Taking logarithms yields the equation eln(tt=ttt t1 2 ln( 3)β l4( +)β 5ln( +β) ln( + ) where β1 lαn( ). This form of the production function is linear in the coefficients β 1 β 2 β3, β 4nd β , 5nd hence is suitable for least squares estimation. (b) Coefficient estimates and their standard errors are given in the following table. Estimated Standard coefficient error β 2 0.05607 0.25927 β 3 0.22631 0.44269 0.04358 0.38989 β 4 β 5 0.66962 0.36106 (c) The estimated coefficients show the proportional change in output that results from proportional changes in K, L, E and M. All these estimated coefficients have positive signs, and lie between zero and one, as is requi red for profit maximization to be realistic. Furthermore, they sum to approximately one, indicating that the production function has constant returns to scale. However, from a statistical point of view, all the estimated coefficients are not significantly different from zero; the large standard errors suggest the estimates are not reliable.

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "I made $350 in just two days after posting my first study guide."

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.