(continued) (h) For Chris,

(0.975,993)se 30.191 1.962 1.5163 (27.22, 33.17)

EXPER t EXPER

We estimate with 95% confidence that the number of years of experience after which the marginal return to experience becomes negative is between 27.2 and 33.2 years.

EXERCISE 5.20

ADVERT is not optimal.

(b) ADVERT₀ 1.9 will be optimal if ₃ 2 1.9 ₄ 1. Thus the null and alternative

ADVERT could be optimal.

(c) ADVERT₀ 2.3 will be optimal if ₃ 2 2.3 ₄ 1 . Thus the null and alternative

ADVERT could be optimal.

Note that we have found that both 1.9 and 2.3 could be optimal values for advertising expenditure. A null hypothesis that used any value for ADVERT in between these two values ₀ would also not be rejected. This outcome illustrates why we never accept null hypotheses as the truth. The best we can do is to say there is insufficient evidence to conclude a null hypothesis is not true.

You might be surprised by the fact that 2.3 lies outside the 95% interval estimate for ADVERT ₀ found on page 195 of the text. To appreciate how the difference can arise, note that for part (c) we could also have set up the hypothesis

3

Exercise 5.20 (continued)

3 4

3 4

1 2.3 1 12.1512 2.3

2 2 ( 2.76796)

2.219 0.12872

se 1 2 b t b

b b

The p-value is 0.0297, and H is rejected. The different outcome arises because the delta method ₀ used to find se 1 b₃ 2b is a large sample approximation needed for nonlinear functions of ₄ the b’s, whereas se b₃ 4.6b involves getting the standard error for a linear function of the b’s, ₄ something we can do exactly without a large sample approximation.

EXERCISE 5.21

(a) The estimated equation is

19.9166 0.36923 1.3353 2.7548

(se) (1.2548) (0.01553) (0.1390) (0.3038)

TIME DEPART REDS TRAINS

Interpretations of each of the coefficients are:

1: 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.

2: 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).

3: Each red light increases traveling time by 1.34 minutes.

4: Each train increases traveling time by 2.75 minutes.

(b) The 95% confidence intervals for the coefficients are:

1: b t₁ (0.975,227)se( ) 19.9166 1.970 1.2548 (17.44, 22.39)b1 2: b₂ t(0.975,227)se( ) 0.36923 1.970 0.01553 (0.339, 0.400)b2 3: b₃ t(0.975,227)se( ) 1.3353 1.970 0.1390 (1.06,1.61)b3 4: b₄ t(0.975,227)se( ) 2.7548 1.970 0.3038 (2.16, 3.35)b4

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 H₀: ₃ 2 and H₁: ₃ 2. The critical value is t_(0.05,227) 1.652. hypothesis that each train delays Bill by 3 minutes.

Exercise 5.21 (continued)

(e) Delaying the departure time by 30 minutes, increases travel time by 30 ₂. Thus, the null hypothesis is H₀: 30 ₂ 10, or H₀: ₂ 1 3, and the alternative is H₁: ₂ 1 3. We reject H if ₀ t t_(0.05,227) 1.652, where the calculated t-value is

0.36923 0.33333

0.01553 2.31 t

Since 2.31 1.652 , we do not reject H . The data are consistent with the hypothesis ₀ that delaying departure time by 30 minutes increases travel time by at least 10 minutes.

(f) If we assume that ₂, ₃ and ₄ are all non-negative, then the minimum time it takes Bill to travel to work is ₁. Thus, the hypotheses are H₀: ₁ 20 and H₁: ₁ 20. We reject

H if 0 t t_(0.95,227) 1.652, where the calculated t-value is

19.9166 20

0.066 1.2548

t

Since 0.066 1.652 , we do not reject H . The data support the null hypothesis that the ₀ minimum travel time is less than or equal to 20 minutes. It was necessary to assume that

2, 3 and 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 ₁.

EXERCISE 5.22

The estimated equation is

19.9166 0.36923 1.3353 2.7548

(se) (1.2548) (0.01553) (0.1390) (0.3038)

TIME DEPART REDS TRAINS

(a) The delay from a train is ₄ and the delay from a red light is ₃. Thus, the null and alternative hypotheses are

H₀: 3 _{3 4} and H₁: 3 _{3 4}

The critical values for the t-test are t(0.975,227) 1.970 and t(0.975,227) 1.970. The rejection region is t 1.970 or t 1.970. The calculated value of the t-test statistic is

where the standard error is computed from

from a train is not equal to three times the delay from a red light.

(b) This test is similar to that in part (a), but it is a one-tail test rather than a two-tail test. The hypotheses are

H₀: ₄ 3 ₃ and H₁: ₄ 3 ₃

The rejection region for the t-test is t t_(0.05,227) 1.652, and the calculated t-value is

^{4 3} delay from a red light.

(c) The delay from 3 trains is 3 . The extra time gained by leaving 5 minutes earlier is ₄ 5 5 . Thus, the hypotheses are 2

H₀: 3 ₄ 5 5 ₂ and H₀: 3 ₄ 5 5 ₂

The rejection region for the t-test is t t_(0.95,227) 1.652, where the t-value is calculated as

^{4 2}

Exercise 5.22(c) (continued)

and the standard error is computed from

4 2 4 2 2 4

se(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 H at a 5% significance level. Alternatively, we do ₀ not reject H because the p-value = 0.0617, which is greater than 0.05. There is ₀ 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 45 . Thus, the null and alternative hypotheses are ₂

H₀: ₁ 45 ₂ 45 and H₁: ₁ 45 ₂ 45

The rejection region for the t-test is t t_(0.95,227) 1.652, where the t-value is calculated as

^{1 2}

1 2

45 45 19.9166 45 0.36923 45

se( 45 ) 1.1377 7.44

b b

t b b

and the standard error is computed from

2

1 2 1 2 1 2

se( 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 H at a 5% significance level. Alternatively, we do ₀ not reject H because the p-value = 1.000, which is greater than 0.05. There is insufficient ₀ evidence to conclude that Bill will not get to the University before 8:00AM.

EXERCISE 5.23

The within sample predictions, with age expressed in terms of years (not units of 10 years) are graphed in the following figure. They are also given in a table on page 176.

Figure xr5.23 Fitted line and observations

(a) To test the hypothesis that a quadratic function is adequate we test H₀: ₄ 0. The t-value is 2.972, with corresponding p-value 0.0035. We therefore reject H and conclude that the ₀ quadratic function is not adequate. For suitable values of ₂, ₃ and , the cubic function ₄ can decrease at an increasing rate, then go past a point of inflection after which it decreases at a decreasing rate, and then it can reach a minimum and increase. These are characteristics worth considering for a golfer. That is, the golfer improves at an increasing rate, then at a decreasing rate, and then declines in ability. These characteristics are displayed in Figure xr5.23.

(b) (i) Using the predictions in the table on page 176, we find the predicted score is lowest ( 6.29) at the age of 30. Thus, we predict that Lion was at the peak of his career at

Setting this derivative equal to zero and solving for age yields

Exercise 5.23(b)(i) (continued)

Replacing ₂, ₃, ₄ by their estimates b b b gives the two solutions ₂, ,_{3 4}

* 2 1

2 ( 20.2222) 4 ( 20.2222) 12 47.02386 2.74934

3.008

2 ( 20.2222) 4 ( 20.2222) 12 47.02386 2.74934

1.895

(ii) Lion’s game is improving at an increasing rate between the ages of 20 and 25, where the differences between the predictions are increasing.

(iii) Lion’s game is improving at a decreasing rate between the ages of 25 and 30, where the differences between the predictions are declining.

We can consider (ii) and (iii) mathematically in the following way. When Lion’s game is improving the first derivative will be negative. It can be verified that the estimated first derivative will be negative for values of AGE between 2 and 3. If Lion’s game is improving at an increasing rate, the second derivative will also be negative; it will be positive when Lion’s game is improving at a decreasing rate. Thus, to find the age at which Lion’s improvement changes from an increasing rate to a decreasing rate we find that AGE for which the second derivative is zero, namely

* 3 which is equivalent to 24.52 years.

(iv) At the age of 20, Lion’s predicted score is –4.4403. His predicted score then declines and rises again, reaching –4.1145 at age 36. Thus, our estimates suggest that, when he reaches the age of 36, Lion will play worse than he did at age 20.

(v) At the age of 40 Lion’s predicted score becomes positive implying that he can no longer score less than par.

(c) At the age of 70, the predicted score (relative to par) for Lion Forrest is 241.71. To break 100 it would need to be less than 28 ( 100 72) . Thus, he will not be able to break 100 when he is 70.

Exercise 5.23 (continued)

Predicted scores at different ages age predicted scores

20 4.4403

21 4.5621

22 4.7420

23 4.9633

24 5.2097

25 5.4646

26 5.7116

27 5.9341

28 6.1157

29 6.2398

30 6.2900

31 6.2497

32 6.1025

33 5.8319

34 5.4213

35 4.8544

36 4.1145

37 3.1852

38 2.0500

39 0.6923

40 0.9042

41 2.7561

42 4.8799

43 7.2921

44 10.0092

EXERCISE 5.24

(a) The coefficient estimates, standard errors, t-values and p-values are in the following table.

Dependent Variable: ln(PROD)

All estimates have elasticity interpretations. For example, a 1% increase in labor will lead to a 0.4328% increase in rice output. A 1% increase 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 H₀: ₂ 0.5 and H₁: ₂ 0.5. The 1% critical values are t(0.995,348) 2.59 and t(0.005,348) 2.59. Thus, the rejection region is t 2.59 or

2.59

t . The calculated value of the test statistic is

0.3617 0.5 2.16

0.064 t

Since 2.59 2.16 2.59 , we do not reject H . The data are compatible with the ₀ 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

b₄ t(0.975,348) se( ) 0.2095 1.967 0.03826 (0.134, 0.285)b4

This relatively narrow interval implies the fertilizer elasticity has been precisely measured.

(d) This hypothesis test is a test of H₀: ₃ 0.3 against H₁: ₃ 0.3. The rejection region is

(0.95,348) 1.649

t t . The calculated value of the test statistic is

0.433 0.3 1.99

0.067 t

We reject 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

0: 3 0.3

H against H₁: ₃ 0.3, leads to a rejection region of t 1.649. The calculated t-value is t 1.99. The null hypothesis is not rejected because 1.99 1.649 .

EXERCISE 5.25

(a) Taking logarithms yields the equation

ln( )Y _{1 2}ln( )K ₃ln( )L ₄ln( )E ₅ln( )M e

where ₁ ln( ) . This form of the production function is linear in the coefficients ₁, ₂,

3, ₄ and ₅, and hence is suitable for least squares estimation.

(b) Coefficient estimates and their standard errors are given in the following table.

Estimated coefficient

Standard error

2 0.05607 0.25927

3 0.22631 0.44269

4 0.04358 0.38989

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 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.

178

CHAPTER 6

Exercise Solutions

EXERCISE 6.1

(a) To compute R², 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

Solving this equation for SST yields

SST ˆ²_y (N 1) (13.45222)² 39 7057.5267 Thus,

² 979.830

1 1 0.8612

7057.5267 R SSE

SST

(b) The F-statistic for testing H₀: _{2 3} 0 is defined as

( ) ( 1) (7057.5267 979.830) / 2

114.75

( ) 979.830 / (40 3)

SST SSE K

F SSE N K

At 0.05, the critical value is F(0.95, 2, 37) 3.25. Since the calculated F is greater than the critical F, we reject H . There is evidence from the data to suggest that _{0 2} 0 and/or

3 0.

EXERCISE 6.2

The model from Exercise 6.1 is y _{1 2}x ₃z e . The SSE from estimating this model is 979.830. The model after augmenting with the squares and the cubes of predictions yˆ² and yˆ³ is y _{1 2}x ₃z ₁yˆ² ₂yˆ³ e. The SSE from estimating this model is 696.5375. To use the RESET, we set the null hypothesis H₀: _{1 2} 0. The F-value for testing this hypothesis is

( ) (979.830 696.5375) 2

7.1175

( ) 696.5373 (40 5)

R U

U

SSE SSE J

F SSE N K

The critical value for significance level 0.05 is F(0.95,2,35) 3.267 . Since the calculated F is greater than the critical F we reject H and conclude that the model is ₀ misspecified.

EXERCISE 6.3

(a) Let the total variation, unexplained variation and explained variation be denoted by SST, SSE and SSR, respectively. Then, we have

SSE eˆ_i² N K ˆ² (20 3) 2.5193 42.8281

Also,

² 1 SSE 0.9466

R SST

and hence the total variation is

₂ 42.8281

802.0243

1 1 0.9466

SST SSE R and the explained variation is

SSR SST SSE 802.0243 42.8281 759.1962

(b) A 95% confidence interval for 2 is

explained variation 1 759.1962 / 2

unexplained variation 42.8281 / 17 151 F K

N K

The critical value for a 5% level of significance is F(0.95,2,17) 3.59. Since 151 3.59 , we reject H0 and conclude that the hypothesis 2 = 3 = 0 is not compatible with the data.

Exercise 6.3 (continued)

(e) The t-statistic for testing H₀: 2 _{2 3} against the alternative H₁: 2 _{2 3} is

^{2 3}

2 3

2 se 2

b b

t b b

For a 5% significance level we reject H₀ if t t_(0.025,17) 2.11 or t t_(0.975,17) 2.11. The standard error is given by

2

2 3 2 3 2 3

se 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

2b₂ b₃ 2 0.69914 1.7769 0.37862

leading to a t-value of

0.37862

0.634 0.59675

t

Since 2.11 0.634 2.11, we do not reject H . There is no evidence to suggest that ₀

2 3

2 .

EXERCISE 6.4

(a) The value of the t statistic for the significance tests is calculated from:

se( ) 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 both appear in ² an equation, their coefficients are not significantly different from zero, with the exception of eqn (B), where EDUC is significant. In addition, the interaction term between ²

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 ₆ 0 . The F-test value for testing H₀: ₆ 0 against

F . 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 hypothesis and conclude that the interaction term,

EDUC EXPER is not significant in determining the wage.

The t-value for testing H₀: ₆ 0 against H₁: ₆ 0 is –1.058. At the 5% level, its absolute value is less than the critical value, t(0.975,993) 1.962. Thus, the t-test gives the same result. The two tests are equivalent because 1.120 1.058 and 3.851 1.962.

Exercise 6.4 (continued)

(c) The restrictions imposed on eqn (A) that give eqn (C) are ₄ 0, ₅ 0 and ₆ 0 . Thus, we test

H₀: ₄ 0, ₅ 0 and ₆ 0

H₁: At least one of or or is nonzero _{4 5 6} . The F-value is calculated from:

( ) (233.8317 222.4166) 3

F . Since the F-value is greater than the critical value (or the p-value is less than 0.05), we reject the null hypothesis and conclude at least one of ₄ or or is _{5 6} nonzero.

By performing this test, we are asking whether experience is relevant for determining the wage level. All three coefficients relate to variables that include EXPER. The test outcome suggests that experience is indeed a relevant variable.

(d) The restrictions imposed on eqn (B) that give eqn (D) are ₂ 0 and ₃ 0 . Thus, we test

H₀: ₂ 0, ₃ 0

H₁: At least one of or is nonzero _{2 3} . The F-value is calculated from:

( ) (280.5061 222.6674) 2

F . Since the F-value is greater than the critical value (or the p-value is less than 0.05), we reject the null hypothesis and conclude at least one of ₂ or is nonzero. ₃ By performing this test, we are asking whether education is relevant for determining the wage level. Both coefficients relate to variables that include EDUC. The test outcome suggests that education is indeed a relevant variable.

Exercise 6.4 (continued)

(e) The restrictions imposed on eqn (A) that give eqn (E) are ₃ 0 and ₆ 0 . Thus, we test

H₀: ₃ 0, ₆ 0

H₁: At least one of or is nonzero _{3 6} . The F-value is calculated from:

( ) (223.6716 222.4166) 2

F . Since the F-value is less than the critical value (or the p-value is greater than 0.05), we do not reject the null hypothesis. The assumption ₃ 0, ₆ 0 is compatible with the data.

By performing this test, we are asking whether it is sufficient to include education as a linear term or whether we should also include it as a quadratic and/or interaction term. The test outcome suggests that including it as a linear term is adequate.

(f) Eqn (E) is the preferred model. All its estimated coefficients are significantly different

Eqn (B) is favored by the AIC criterion. Eqn (E) is favored by the SC criterion.

EXERCISE 6.5

(a) Education and experience will have the same effects on ln(WAGE if ) _{2 4} and

3 5. The null and alternative hypotheses are:

^{0 2 4 3 5}

1 2 4 3 5

: and

: or or both

H H

(b) The restricted model assuming the null hypothesis is true is

ln(WAGE) ^{1 4}(EDUC EXPER) ⁵(EDUC² EXPER²) ⁶HRSWK e (c) The F-value is calculated from:

( ) (254.1726 222.6674) 2

70.32

( ) 222.6674 994

R U

U

SSE SSE J

F SSE N K

The corresponding p-value is 0.0000. Also, 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 (or the p-value is less than 0.05), we reject the null hypothesis and conclude that education and experience have different effects on ln(WAGE . )

EXERCISE 6.6

Consider, for example, the model

y _{1 2}x ₃z e

If we augment the model with the predictions ˆy the model becomes

y _{1 2}x ₃z y e ˆ

However, ˆy b b x b z is perfectly collinear with x and z . This perfect collinearity _{1 2 3} means that least-squares estimation of the augmented model will fail.

EXERCISE 6.7

(a) Least squares estimation of y _{1 2}x ₃w e gives b₃ 0.4979 , se( ) 0.1174b₃ and t 0.4979 0.1174 4.24. This result suggests that b is significantly different from ₃ zero and therefore w should be included in the model. Additionally, the RESET based on the equation y _{1 2}x e gives F-values of 17.98 and 8.72 which are much higher than the 5% critical values of F(0.95,1,32) 4.15 and F(0.95,2,31) 3.30, respectively. Thus, the model omitting w is inadequate.

(b) Let b₂ be the least squares estimator for ₂ in the model that omits w . The omitted-variable bias is given by

( )₂^* _{2 3}cov( , )

var( ) E b x w

x

Now, cov( , )x w 0 because r_xw 0. Thus, the omitted variable bias will be positive. This result is consistent with what we observe. The estimated coefficient for ₂ changes from

0.9985 to 4.1072 when w is omitted from the equation.

(c) The high correlation between x and w suggests the existence of collinearity. The observed outcomes that are likely to be a consequence of the collinearity are the sensitivity of the estimates to omitting w (the large omitted variable bias) and the insignificance of

b when both variables are included in the equation. 2

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

1 and ₃. With this in mind, we rewrite the restrictions as

^{3 4}

1 2 3 4

1 3.8

80 6 1.9 3.61

Substituting the first restriction into the second yields

₁ 80 6 ₂ 1.9(1 3.8 ) 3.61_{4 4}

Substituting this restriction and the first one ₃ 1 3.8 into the equation ₄ SALES _{1 2}PRICE ₃ADVERT ₄ADVERT² e yields

2 4 4 2

2

4 4

80 6 1.9(1 3.8 ) 3.61 1 3.8

SALES PRICE

ADVERT ADVERT e Rearranging this equation into a form suitable for estimation yields

2

2 4

78.1 6 3.61 3.8

SALES ADVERT PRICE ADVERT ADVERT e

EXERCISE 6.9

The results of the tests in parts (a) to (e) appear in the following table. Note that, in all cases, there is insufficient evidence to reject the null hypothesis at the 5% level of following table suggest a high degree of collinearity in the model.

Correlation with Variables estimated equation, with t values in parentheses,

2

ln 0.035 0.056ln 0.226 ln 0.044 ln 0.670ln

( ) 0.800 0.216 0.511 0.112 1.855

0.952

Y K L E M

t

R

The very small t-values for all variables except ln( )M , our inability to reject any of the null hypotheses in parts (a) through (e), and the high R , are indicative of high ² collinearity. Collectively, all the variables produce a model with a high level of explanation and a good predictive ability. Furthermore, our economic theory tells us that all the variables are important ones in a production function. However, we have not been able to estimate the effects of the individual explanatory variables with any reasonable degree of precision.

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 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

H0. The evidence from the data is consistent with the notion that if prices and income go up in the same proportion, demand will not change. This idea is consistent with economic theory.

The F-value can be calculated from restricted and unrestricted sums of squared errors as follows

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 t(0.975, 25) 2.060 for the unrestricted model and

(0.975, 26) 2.056

t 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 t_c 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

EXERCISE 6.11

(a) The estimated Cobb-Douglas production function with standard errors in parentheses is

ln 0.129 0.559ln 0.488ln 2 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

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

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