CHAPTER 5. Exercise Solutions

CHAPTER 5 Exercise Solutions 91

Chapter 5, Exercise Solutions, Principles of Econometrics, e 9 EXERCISE 5.1 (a) y = 1, x =, x = x * * i x i 1 1 1 1 1 1 1 1 1 1 1 1 1 1 y * i (b) (c) yx = 1, x = 16, yx =, x = 1 * * * * * * i i i i i i ( * * * * * yx i i)( xi ) ( yx i i)( xi * xi * ) * * * * ( xi )( xi ) ( xixi) 1 1 b = = =.815 16 1 ( yx i )( ) ( )( ) * i * xi * yx i * i * xi * xi * * * * * ( xi )( xi ) ( xixi) 16 1 b = = =. 16 1 b1 = y bx bx = 1 (d) e ˆ = (.,.9875,.5,.75, 1.15,.5,.6,.15,.1875) (e) (f) (g) (h) eˆ i.875 σ ˆ =.696 N K = 9 = r ( xi x)( xi x) xixi = = = ( x x ) ( x x ) x x i i i i σˆ.696 se( b) = var( b) = = =.1999 ( x x ) (1 r ) 16 i ˆi.875 ( i ) 16, SSE = e = SST = y y = SSR 1.165 SSR = SST SSE = 1.165 R = = =.76 SST 16

Chapter 5, Exercise Solutions, Principles of Econometrics, e 9 EXERCISE 5. (a) A 95% confidence interval for is b ± t se( b ) =.815 ±.7.1999 = (., 1.17) (.975,6) (b) The null and alternative hypotheses are H : = 1, H : 1 1 The calculated t-value is b 1.815 1 t = = =.977 se( b ).1999 At a 5% significance level, we reject H if t > t(.975, 6) =.7. Since.977 <.7, we do not reject H.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 9 EXERCISE 5. b1.91 (a) (i) The t-statistic for b1 is.76 se( b 1) =.191 =..76 (ii) The standard error for b is se( b ) = =.18. 6.686 (iii) The estimate for is b =. ( 6.96) =.1. (iv) To compute R, we need SSE and SST. From the output, SSE = 5.75896. To find SST, we use the result SST σ ˆ y = =.6 N 1 which gives SST = 1518 (.6) = 6.86. Thus, SSE 5.759 R = 1 = 1 =.5 SST 6.86 (v) The estimated error standard deviation is σ= SSE 5.75896 ˆ.616 ( N K) = 1519 = (b) The value b =.76 implies that if ln( TOTEXP) increases by 1 unit the alcohol share will increase by.76. 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) =.76 TOTEXP. A 1% increase in total expenditure leads to a.76 increase in the alcohol share of expenditure. The value b =.1 suggests that if the age of the household head increases by 1 year the share of alcohol expenditure of that household decreases by.1. The value b =.1 suggests that if the household has one more child the share of the alcohol expenditure decreases by.1. (c) A 95% confidence interval for is b ± t se( b ) =.1 ± 1.96. = (.18,.1).975,1515) 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.18 and.1.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 95 Exercise 5. (Continued) (d) The null and alternative hypotheses are H: =, H1:. b The calculated t-value is t = =.75 se( b ) At a 5% significance level, we reject H if t > t(.975,1515) = 1.96. Since.75 > 1.96, we reject H and conclude that the number of children in the household influences the budget 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, e 96 EXERCISE 5. (a) The regression results are: ( ) WTRANS = + TOTEXP AGE NK R = ( ).15.1ln.1.1.7 se (.) (.71) (.) (.55) b = suggests that as ( ) (b) The value.1 ln TOTEXP increases by 1 unit the budget proportion for transport increases by.1. Alternatively, one can say that a 1% increase in total expenditure will increase the budget proportion for transportation by.. (See Section A..8 of Appendix A.) The positive sign of b is 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. The value b =.1 implies that as the age of the head of the household increases by 1 year the budget share for transport decreases by.1. The expected sign for b is not 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. The value b =.1 implies that an additional child decreases the budget share for transport by.1. 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 testing H : = against the alternative H1: where is the coefficient of AGE is.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.5. (d) (e) The proportion of variation in the budget proportion allocated to transport explained by this equation is.7. For a one-child household: WTRANS =.15 +.1ln( TOTEXP).1AGE.1NK =.15 +.1 ln(98.7).1 6.1 1 =.1 For a two-child household: WTRANS =.15 +.1ln( TOTEXP).1AGE.1NK =.15 +.1 ln(98.7).1 6.1 =.19

Chapter 5, Exercise Solutions, Principles of Econometrics, e 97 EXERCISE 5.5 (a) The estimated equation is VALUE = 8.67.18 CRIME.819 NITOX + 6.715 ROOMS.78 AGE (se) (5.659) (.65) (.167) (.9) (.11) 1.5DIST +.7ACCESS.16TAX 1.1768PTRATIO (.1) (.7) (.8) (.19) The estimated equation suggests that as the per capita crime rate increases by 1 unit the home value decreases by $18.. 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 $,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 $6,7. An increase in the proportion of owner-occupied units built prior to 19 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 $1,5 for every unit of weighted distance. The higher the tax rate per $1, 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 b ± t se( b ) =.18 ± 1.965.65 = (.55,.11). (.975,97) A 95% confidence interval for the coefficient of ACCESS is b ± t se( b ) =.7 ± 1.965.7 = (.1,.1) 7 (.975,97) 7 (c) We want to test H : room = 7 against H1: room 7. The value of the t statistic is brooms 7 6.715 7 t = = = 1.617 se( b ).9 rooms At α=.5, we reject H if the absolute calculated t is greater than 1.965. Since 1.617 < 1.965, we do not reject H. The data is consistent with the hypothesis that increasing the number of rooms by one increases the value of a house by $7. (d) We want to test H : 1 against H1: < 1. The value of the t statistic is ptratio 1.1768 + 1 t = = 1.68.19 At a significance level of α=.5, we reject H if the calculated t is less than the critical value t (.5,97) = 1.68. Since 1.68 > 1.68, we do not reject H. We cannot conclude that reducing the pupil-teacher ratio by 1 will increase the value of a house by more than $1,. ptratio

Chapter 5, Exercise Solutions, Principles of Econometrics, e 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..6658.7511.95 X.815.19995.68.66 X..59 1.58165.168 R-squared.76156 Mean dependent var 1. Adjusted R-squared.688 S.D. dependent var 1.11 S.E. of regression.7997 Akaike info criterion.651 Sum squared resid.875 Schwarz criterion.71788 Log likelihood -8.961 Hannan-Quinn criter. 1.7817

Chapter 5, Exercise Solutions, Principles of Econometrics, e 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.8798.179.816.19.191.881 Std Error.51.65.51.7.57.56 p-value....6878.78.16 ln(totexp) Estimate.177.56.99.7.1.59 Std Error.11.58.11.8.16.118 p-value....1.111.1 AGE Estimate.7..56..77.71 Std Error.55.9.55..6.58 p-value..15.6..167. NK Estimate.97.6.8.18.1.19 Std Error.8..8.61.9.88 p-value..1587.5658.15.191.1157 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 shares 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 of AGE and NK is only significant in the food and alcohol equations.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 1 Exercise 5.7 (continued) (b) The t-values and p-values for testing H: against H1: > are reported in the table below. Using a 5% level of significance, the critical value for each test is t (.95,96) = 1.68. t-value p-value decision WFOOD 1.8 1. Do not reject H WFUEL 9.569 1. Do not reject H WCLOTH 8.66. Reject H WALC.1. Reject H WTRANS.58.56 Reject H WOTHER.88.1 Reject H Those commodities which are regarded as necessities ( b < ) 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, e 11 EXERCISE 5.8 (a) The expected sign for is 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 to be positive; the purer the cocaine, the higher the price. The sign for will 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: PRICE = 9.867.6QUANT +.116QUAL.56TREND R =.597 (se) (8.58) (.1) (.) (1.861) ( t) (1.588) ( 5.89) (.5717) ( 1.6987) The estimated values for, and are.6,.116 and.56, respectively. They imply that as quantity (number of grams in one sale) increases by 1 unit, the price will go down by.6. Also, as the quality increases by 1 unit the price goes up by.116. As time increases by 1 year, the price decreases by.56. All the signs turn out according to our expectations, with 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.597. (d) For this hypothesis we test H : against H 1 : <. The calculated t-value is 5.89. We reject H if the calculated t is less than the critical t (.95,5) = 1.675. Since the calculated t is less than the critical t value, we reject 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 : against H 1 : >. The calculated t-value is.5717. At α=.5 we reject H 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 reject H. 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.56 b =. 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, e 1 EXERCISE 5.9 (a) (b) (c) The coefficients,, and 5 are expected to be positive, whereas 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 ( < ). Also, as the prices of lamb and pork increase, the consumption of beef should increase ( > and > ). If per capita disposable income increases, the consumption of beef should increase ( 5 > ). The least squares estimates and standard errors are given by ln( QB) =.676.866ln ( PB) +.1997ln ( PL) ( se) ( 1.6596)(.186) (.17) ( ) ( ) (.87) (.9).71ln.117 ln.769 + PP + IN R = The estimated coefficients are elasticities, implying that a 1% increase in the price of beef leads to an.8% decrease in the quantity of beef consumed. Likewise, 1% changes in the prices of lamb and pork will cause.% and.% 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. The estimated covariance matrix is given by.75.187.76.5.1.187...17.6 cov ( b1, b, b, b, b5) =.76..5.9.1661.5.17.9.17..1.6.1661..86 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 b1, b, b, b 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 (.975,1) =.179, the 95% confidence intervals are as follows. 1 5 lower limit 1.56 1.6.68.989.589 upper limit 8.888.86.66 1.7.7

Chapter 5, Exercise Solutions, Principles of Econometrics, e 1 EXERCISE 5.1 (a) The estimated regression models and 95% confidence intervals for follow. (i) All houses: PRICE = 198 + 9.97 SQFT 755. AGE (se) (699) (.) (1.89) b ± t se( b ) = 755. ± 1.96 1.89 = ( 11.5, 78.6) (.975,177) (ii) Town houses: PRICE = 915 +.1 SQFT 61.6 AGE (se) (117) (5.688) (75.) b ± t se( b ) = 61.6 ± 1.996 75. = ( 7.8, 187.) (.975,67) (iii) French style homes: PRICE = 98 + 18.1 SQFT 6.71 AGE (se) (988) (1.) (7.) b ± t se( b ) = 6.7 ± 1.986 7. = ( 6961.1, 681.7) (.975,9) 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 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 H : = 1 against the alternative H : 1. The results for each of the three cases follow. (i) All houses: The critical values are t (.975,177) = 1.96 and t (.5,177) = 1.96. We reject H if the calculated t-value is such that t 1.96 or t 1.96. The calculated value is 755. ( 1) t = = 1.7 1.89 Since 1.96 1.7 1.96 < <, we do not reject H. The data is consistent with the hypothesis that having an older house reduces its price by $1 per year for each year of its age.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 1 Exercise 5.1(b) (continued) (b) (ii) Town houses: The critical values are t (.975,67) = 1.996 and t (.5,67) = 1.996. We reject H if the calculated t-value is such that t 1.996 or t 1.996. The calculated value is 61.6 ( 1) t = =. 75. Since t =. < 1.996, we reject H. The data is not consistent with the hypothesis that having an older house reduces its price by $1 per year for each year of its age. (iii) French style homes: The critical values are t (.975,9) = 1.986 and t (.5,9) = 1.986. We reject H if the calculated t-value is such that t 1.986 or t 1.986. The calculated value is 6.71 ( 1) t = =.7 7. Since 1.986.7 1.986 < <, we do not reject H. The data is consistent with the hypothesis that having an older house reduces its price by $1 per year for each year of its age.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 15 EXERCISE 5.11 (a) The estimated regression model is VOTE = 5. +.688 GROWTH.186 INFLATION (se) (1.9) (.1675) (.) The hypothesis test results on the significance of the coefficients are: H : = H : > p-value =. significant at 1% level H 1 : = H : < p-value =.5 not significant at 1% level 1 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 = and GROWTH = is VOTE = 5. +.688 ( ).186 = 9.1 (c) Ignoring the error term, the incumbent party will get the majority of the vote when + + > 1 GROWTH INFLATION 5 Assuming 1 = b1 and = b, and given INFLATION = and GROWTH =, the incumbent party will get the majority of the vote when 5..186 5 which is equivalent to < ( 5.1861 + 5.57) =.67 To ensure accuracy, we have included more decimal places in this calculation than are in the reported equation. For testing H:.67 against the alternative H1: >.67, the t-value is.6876.67 t = = 1.8.1676 For a 5% significance level, the critical value is t (.95, 8) = 1.71. The rejection region is t 1.71. Thus, H is not rejected. The incumbent party might still get elected if GROWTH = %.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 16 EXERCISE 5.1 (a) The estimated equation is TIME = 19.9166 +.69 DEPART + 1.5 REDS +.758 TRAINS (se) (1.58) (.155) (.19) (.8) Interpretations of each of the coefficients are: 1 : The estimated time it takes Bill to get to work when he leaves Carnegie at 6:AM and encounters no red lights and no trains is 19.9 minutes. : If Bill leaves later than 6:AM, his traveling time increases by.7 minutes for every 1 minutes that his departure time is later than 6:AM (assuming the number of red lights and trains are constant). : Each red light increases traveling time by 1. minutes. : Each train increases traveling time by.75 minutes. (b) The 95% confidence intervals for the coefficients are: 1 : b1 t(.975,7) b1 ± se( ) = 19.9166 ± 1.97 1.58 = (17.,.9) : b t(.975,7) b ± se( ) =.69 ± 1.97.155 = (.9,.) : b t(.975,7) b ± se( ) = 1.5 ± 1.97.19 = (1.6,1.61) : b t(.975,7) b ± se( ) =.758 ± 1.97.8 = (.16,.5) In the context of driving time, these intervals are relatively narrow ones. We have obtained precise estimates of each of the coefficients. (c) The hypotheses are H : and H1: <. The critical value is t (.5,7) = 1.65. We reject H when the calculated t-value is less than 1.65. This t-value is 1.5 t = =.78.19 Since.78 < 1.65, we reject H. We conclude that the delay from each red light is less than minutes. (d) The hypotheses are H : = and H1:. The critical values are t (.5,7) = 1.65 and t (.95,7) = 1.65. We reject H when the calculated t-value is such that t < 1.65 or t > 1.65. This t-value is.758 t = =.87.8 Since 1.65.87 1.65 < <, we do not reject H. The data are consistent with the hypothesis that each train delays Bill by minutes.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 17 Exercise 5.1 (continued) (e) Delaying the departure time by minutes, increases travel time by. Thus, the null hypothesis is H: 1, or H : 1, and the alternative is H 1 : < 1. We reject H if t t(.5,7) = 1.65, where the calculated t-value is.69. t = =.1.155 Since.1 > 1.65, we do not reject H. The data are consistent with the hypothesis that delaying departure time by minutes increases travel time by at least 1 minutes. (f) If we assume that, and are all non-negative, then the minimum time it takes Bill to travel to work is 1. Thus, the hypotheses are H : 1 and H1: 1 >. We reject H if t t(.95,7) = 1.65, where the calculated t-value is 19.9166 t = =.66 1.58 Since.66 < 1.65, we do not reject H. The data support the null hypothesis that the minimum travel time is less than or equal to minutes. It was necessary to assume that, and 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 1.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 18 EXERCISE 5.1 (a) The coefficient estimates, standard errors, t-values and p-values are in the following table. Dependent Variable: ln(prod) Coeff Std. Error t-value p-value C -1.568.557-6.5. ln(area).617.6 5.655. ln(labor).8.669 6.718. ln(fert).95.8 5.75. All estimates have elasticity interpretations. For example, a 1% increase in labor will lead to a.8% increase in rice output. A 1% increase in fertilizer will lead to a.95% increase in rice output. All p-values are less than.1 implying all estimates are significantly different from zero at conventional significance levels. (b) The null and alternative hypotheses are H : =.5 and H1:.5. The 1% critical values are t (.995,8) =.59 and t (.5,8) =.59. Thus, the rejection region is t.59 or t.59. The calculated value of the test statistic is.617.5 t = =.16.6 Since.59 <.16 <.59, we do not reject H. The data are compatible with the hypothesis that the elasticity of production with respect to land is.5. (c) A 95% interval estimate of the elasticity of production with respect to fertilizer is given by b ± t se( b ) =.95 ± 1.967.86 = (.1,.85) (.975,8) This relatively narrow interval implies the fertilizer elasticity has been precisely measured. (d) This hypothesis test is a test of H :. against H1: >.. The rejection region is t t (.95,8) = 1.69. The calculated value of the test statistic is.. t = = 1.99.67 We reject H because 1.99 > 1.69. There is evidence to conclude that the elasticity of production with respect to labor is greater than.. Reversing the hypotheses and testing H :. against H 1 : <., leads to a rejection region of t 1.69. The calculated t-value is t = 1.99. The null hypothesis is not rejected because 1.99 > 1.69.

Chapter 5, Exercise Solutions, Principles of Econometrics, e 19 EXERCISE 5.1 (a) The predicted logarithm of rice production for AREA = 1, LABOR = 5, and FERT = 1 is ln( PROD ) = 1.568 +.617 ln(1) +.8 ln(5) +.95 ln(1) = 1.11118 Using the corrected predictor given on page 95 of the text, the corresponding prediction of rice production is ( ) ( ) ( ) ˆ PROD = exp ln( PROD) exp σ = exp(1.11118) exp.119 =.71 (b) The marginal principle says apply more fertilizer if PROD PRICEFERT > =. FERT PRICE RICE Since PROD PROD =, FERT FERT the above marginal principle can be stated as: Apply more fertilizer if PROD >. FERT Substituting PROD =.71 and FERT = 1, this inequality is the same as 1 >. =.1.71 (c) The hypotheses are H :.1 and H 1 : >.1. The 5% critical value is t =, leading to a rejection region of t 1.69. The calculated t-value is (.95,8) 1.69.95.1 t = =..8 Since. > 1.69, we reject H. The farmer should apply more fertilizer. We chose >.1 as the alternative hypothesis because the 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 Econometrics, e 11 Exercise 5.1 (continued) (d) The predicted logarithm of rice production for AREA = 1, LABOR = 5, and FERT = 1 is ln( PROD ) = 1.568 +.617 ln(1) +.8 ln(5) +.95 ln(1) =.1655 Noting that ln(1) =, the estimated variance of the prediction error can be written as and [ ] [ ] ˆ 1 1 var( f) = var( b ) + ln(5) var( b ) + ln(5) cov( b, b ) +σ =.655875 + 15.9.775 +.91 (.16617) +.119 =.199 se( f) = var( f) =.199 =.798 A 95% interval estimate for the logarithm of rice production at the specified values is ln( PROD) ± t se( f ) =.1655 ± 1.9668.798 = (.5996,.8969) (.975,8) The corresponding 95% interval estimate for rice production is ( exp(.5996), exp(.8969) ) = (.59,.)

Chapter 5, Exercise Solutions, Principles of Econometrics, e 111 EXERCISE 5.15 (a) (b) Taking logarithms yields the equation ln( Y ) = + ln( K ) + ln( L ) + ln( E ) + ln( M ) + e t 1 t t t 5 t t where 1 = ln( α ). This form of the production function is linear in the coefficients 1,,, and 5, and hence is suitable for least squares estimation. Coefficient estimates and their standard errors are given in the following table. Estimated coefficient Standard error.567.597.61.69.58.8989 5.6696.616 (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 required 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.