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

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CHAPTER 6

Exercise Solutions

EXERCISE 6.

(a) To compute R^2 , we need SSE and SST. We are given SSE. We can find SST from the equation

( )^2 ˆ 13. 1 1

i y

y y (^) SST N N

σ = = = − −

∑

Solving this equation for SST yields ˆ (^2) ( 1) (13.45222) 2 39 7057. SST = σ × y N − = × =

Thus,

2 1 1 979.830 0.

SSE

R

SST

(b) The F -statistic for testing H (^) 0 : β 2 = β 3 = 0 is defined as

( ) ( 1) (7057.5267 979.830) / 2

( ) 979.830 /(40 3)

SST SSE K

F

SSE N K

At α = 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 (^) 0. There is evidence from the data to suggest that (^) β 2 ≠ 0 and/or β 3 ≠ 0.

EXERCISE 6.

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

SSE = (^) ∑ e ˆ i^2^ = (^) ( N − K )× σˆ 2 = (20 − 3) × 2.5193 =42.

Also,

R^2 1^ SSE 0. SST

and hence the total variation is

2

SST SSE

R

and the explained variation is SSR = SST − SSE = 802.0243 − 42.8281 =759.

(b) A 95% confidence interval for β 2 is

b 2 (^) ± t (0.975,17) (^) se( b 2 ) = 0.69914 ± 2.110 × 0.048526 =(0.2343, 1.1639)

A 95% confidence interval for β 3 is

b 2 (^) ± t (0.975,17) (^) se( b 3 ) = 1.7769 ± 2.110 × 0.037120 =(1.3704, 2.1834)

(c) To test H 0 : β 2 ≥ 1 against the alternative H 1 : β 2 < 1, we calculate

( )

2 2 2

se (^) 0.

b t b

− β − = = = −

At a 5% significance level, we reject H 0 if t < t (0.05,17) = −1.740. Since −1.3658 > −1.740 , we fail to reject H 0. There is insufficient evidence to conclude β 2 < 1.

(d) To test H (^) 0 : β 2 = β 3 = 0 against the alternative H (^) 1 : β 2 ≠ 0 and/or β 3 ≠ 0 , we calculate

( ) ( )

explained variation (^1) 759.1962 / 2 151 unexplained variation 42.8281/

K

F

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 H 0 and conclude that the hypothesis β 2 = β 3 = 0 is not compatible with the data.

Exercise 6.3 (continued)

(e) The t -statistic for testing H 0 (^) : 2β 2 = β 3 against the alternative H 1 (^) : 2β 2 ≠ β 3 is

( ) ( )

2 3 2 3

se 2

b b t b b

For a 5% significance level we reject H (^) 0 if t < t (0.025,17) = −2.11 or t > t (0.975,17) = 2.11. The standard error is given by

( ) n^ n^ n

( )

2 se 2 2 3 2 var( 2 ) var( 3 ) 2 2 cov( 2 , 3 )

4 0.048526 0.03712 2 2 0.

b − b = × b + b − × × b b

= × + − × × −

=

The numerator of the t -statistic is 2 b 2 (^) − b 3 = 2 × 0.69914 − 1.7769 = −0.

leading to a t -value of 0.37862 (^) 0.

t = − = −

Since −2.11 < −0.634 < 2.11, we do not reject H (^) 0. There is no evidence to suggest that 2 β 2 ≠ β 3.

Exercise 6.4 (continued)

(c) For testing H 0 (^) : β − β 1 2 + β 3 = 4 against the alternative H 1 (^) : β 1 − β 2 + β 3 ≠ 4 , we use the

statistic

1 2 3 1 2 3

se( )

b b b t b b b

Now, ( b 1 (^) − b 2 (^) + b 3 ) − 4 = 2 − 3 − 1 − 4 = − 6

and

n

n n n n n n

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

4

t

Since − 2 < −1.5 < 2 , we fail to reject H 0 and conclude that there is insufficient sample evidence to suggest that β 1 − β 2 + β 3 = 4 is incorrect.

EXERCISE 6.

Consider, for example, the model

yi = β + β 1 2 xi + β 3 zi + ei

If we augment the model with the predictions y ˆ i the model becomes

yi = β + β 1 2 xi + β 3 zi + γ y ˆ i + ei

However, y ˆ i = b 1 (^) + b x 2 i + b z 3 i is perfectly collinear with xi and wi. This perfect collinearity means that least-squares estimation of the augmented model will fail.

EXERCISE 6.

(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 H (^) 0 : β 2 + β 3 = 1 is 6.1048 with

a p -value of 0.014. This p -value is less than the significance level of 0.05. We reject H 0 and conclude that constant returns to scale do not exist.

(f) The F -value and the p -value for testing H (^) 0 : β 4 + β + β 5 6 = 1 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 H (^) 0 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.

Exercise 6.7 (continued)

(g) To test H (^) 0 : β 2 + β 3 = 1 , the value of the t statistic is

2 3 2 3

se( ) 0.

b b t b b

where the standard error is calculated from

n

n n n

2 3 2 3

2 3 2 3

se( ) var( )

var( ) var( ) 2cov( , )

0.002851 0.002796 2( 0.002753)

b b b b

b b b b

We reject H (^) 0 because 2.48 > t (0.975,261)= 1.969. Note t^2 = (2.48)^2 = 6.15 ≈ F = 6.10. The difference between t^2 and F is due to rounding error.

To test H (^) 0 : β 4 + β + β 5 6 = 1 , the value of the t -statistic is

4 5 6 4 5 6

se( ) 0.

b b b t b b b

where

se( b 4 (^) + b 5 (^) + b 6 (^) ) = nvar( b 4 (^) + b 5 (^) + b 6 ) = 0.002938 =0.

with nvar( 4 5 6 ) nvar( 4 ) nvar( 5 ) (^) var(n 6 ) (^) 2cov(n 4 , 5 ) (^) 2cov(n 4 , 6 ) (^) 2cov(n 5 , 6 )

0.001919 0.001303 0.010068 2 0. 2 0.002159 2 0.

b + b + b = b + b + b + b b + b b + b b

= + + − × − × − × =

We reject H (^) 0 because −8.69 < t (0.025,261)= −1.969. Note that t^2 = −( 8.69) 2 = 75.52which is approximately equal to F = 75.43.

EXERCISE 6.

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 significance.

Part H 0 F -value df F (^) c (5%) p -value

(a) (^) β 2 = 0 0.047 (1,20) 4.35 0. (b) (^) β 2 = β 3 = 0 0.150 (2,20) 3.49 0. (c) β 2 = β 4 = 0 0.127 (2,20) 3.49 0. (d) β 2 = β 3 = β 4 = 0 0.181 (3,20) 3.10 0. (e) (^) β 2 + β 3 + β 4 + β 5 = 1 0.001 (1,20) 4.35 0.

(f) The auxiliary R^2 s and the explanatory-variable correlations that are exhibited in the following table suggest a high degree of collinearity in the model.

Correlation with Variables

Variable (^) Auxiliary R^2 ln( L ) ln( E ) ln( M )

ln( K ) 0.969 0.947 0.984 0. ln( L ) 0.973 0.972 0. ln( E ) 0.987 0. ln( M ) 0.

To examine the effect of collinearity on the reliability of estimation, we examine the estimated equation, with t values in parentheses, n ( ) (^) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) 2

ln 0.035 0.056ln 0.226ln 0.044ln 0.670ln ( ) 0.800 0.216 0.511 0.112 1.

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^2 , are indicative of high 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.

(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

Unrestricted (^) −3.243 −1.020 −0.583 0.210 0. (3.743) (0.239) (0.560) (0.080) (0.416) Restricted (^) −4.798 −1.299 0.187 0.167 0. (3.714) (0.166) (0.284) (0.077) (0.427)

(b) The high auxiliary R^2^ s and sample correlations between the explanatory variables that 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 correlation.

Sample Correlation With Variable Auxiliary R^2 ln( PL ) ln( PR ) ln( I ) ln( PB ) 0.955 0.967 0.774 0. ln( PL ) 0.955 0.809 0. ln( PR ) 0.694 0. ln( I ) 0.

(c) We use the F -test to test the restriction H 0 (^) : β 2 + β + β 3 4 + β 5 = 0 against the alternative

hypothesis H 1 (^) : β 2 + β + β 3 4 + β 5 ≠ 0. The value of the test statistic is^ F^ = 2.50, with a^ p - value of 0.127. The critical value is F (0.95,1,25) (^) = 4.24. Since 2.50 < 4.24, we do not reject H (^) 0. 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 ( ) (^) (0.098901 0.08992) 1

( ) 0.08992 25

R U U

SSE SSE J

F

SSE N K

EXERCISE 6.

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

n ( ) (^) ( ) ( )

( ) ( ) ( )

ln 0.129 0.559ln 0.488ln 2 0. (se) 0.546 0.816 0.

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

n ( ) (^) ( ) ( )

( ) ( ) ( )

ln 0.020 0.398ln 0.602ln (se) 0.053 0.559 0.

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.

EXERCISE 6.

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. 2 terms 0.3581 (2,23) 3.422 0. Linear 1 term 8.8377 (1,24) 4.260 0. 2 terms 4.7618 (2,23) 3.422 0.

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.

EXERCISE 6.

(a) The estimated model is

n

( ) ( ) ( ) ( ) ( ) ( )

(se) 4.1583 0.1801 0. ( ) 1.954 12.182 2.

HW HE HA R

t

An increase of one year of a husband’s education leads to a $2.19 increase in wages. Also, older husbands earn 20 cents more on average per year of age, other things equal.

(b) A RESET test with one term yields F = 9.528with p -value = 0.0021, and with two terms F = 4.788and p -value = 0.0086. Both p -values are smaller than a significance level of 0.05, leading us to conclude that the linear model suggested in part (a) is not adequate.

(c) The estimated equation is:

n

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(se) 17.5436 1.1228 0.0458 0.7329 0. ( ) 2.597 1.298 3.298 3.943 3.

HW HE HE HA HA R

t

Wages are now quadratic functions of age and education. The effects of changes in education and in age on wages are given by the partial derivatives n 1.4580 0.

HW

HE

HE

n 2.8895 0.

HW

HA

HA

The first of these two derivatives suggests that the wage rate declines with education up to an education level of HE min (^) = 1.458 0.30522 = 4.8 years, and then increases at an increasing rate. A negative value of ∂ HW ∂ HE for low values of HE is not realistic. Only 7 of the 753 observations have education levels less than 4.8, so the estimated relationship might not be reliable in this region. The derivative with respect to age suggests the wage rate increases with age, but at a decreasing rate, reaching a maximum at the age HA max (^) = 2.8895 0.06022 = 48 years.

(d) A RESET test with one term yields F = 0.326with p -value = 0.568, and with two terms F = 0.882and p -value = 0.414. Both p -values are much larger than a significance level of 0.05. Thus, there is no evidence from the RESET test to suggest the model in part (c) is inadequate.

Exercise 6.14 (continued)

(e) The estimated model is:

n

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(se) 17.0160 1.0914 0.0444 0. ( ) 2.178 2.023 3.800 3.

HW HE HE HA

t

( ) ( ) ( ) ( )

− 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 b 6 , the coefficient associated with CIT , is 0.0000. This suggests that b 6 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

n 1.4580 0.

HW

HE

HE

n 2.8895 0.

HW

HA

HA

and from part (f) n HW (^) 2.2076 0.3376 HE HE

n HW (^) 2.6213 0.0556 HA HA

Evaluating these expressions for HE = 6 , HE = 15 , HA = 35 and HA = 50 leads to the following results.

∂ HW ∂ HE ∂ HW ∂ HA

HE = 6 HE = 15 HA = 35 HA = 50

Part (c) (^) 0.356 3.076 0.781 −0. Part (e) −0.182 2.855 0.678 −0.

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 HE = 6 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 ( CIT HE , )=0. and corr( CIT HA , ) = 0.0676.