false

Textbook Notes
(368,439)

Canada
(161,878)

York University
(12,845)

Economics
(1,011)

ECON 3210
(6)

Razvan Sufana
(6)

Chapter 6

Unlock Document

Description

CHAPTER 6
Exercise Solutions
178 Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 179
EXERCISE 6.1
(a) compute R 2, we need SSE and SST. We are given SSE. We can find SST from the
equation
()i 2 SST
y 13.45222
N 1
Solving this equation for SST yields
2 2
SST ˆy (N 1) (13.45222) 39 7057.5267
Thus,
R 1 S 1 97.830 0.8612
SST 7057.5267
(b) The F-statistic for testin0 203 is defined as
(SST SSE) (K 1) (7057.5267 979.830) / 2
F SSE (N K) 979.830 / (40 3) 114.75
At 0.05 , the critical value i(0.95, 2,37)5 . Since the calculated F is greater than
the critical F, we rejec0 HThere is evidence from the data to suggest t20 and/or
0 .
3 Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 180
EXERCISE 6.2
The model from Exercise 6.1 is e z x 1 23 . The SSE from estimating this
model is 979.830. The model after augmen ting with the squares and the cubes of
2 3 2 3
predictions y yand e yisy z x1 2 31 2 ˆˆ . The SSE from estimating
this model is 696.5375. To use the RESET, we set the null hypothesis H 0 12 .
The F-value for testing this hypothesis is
(S)SER USSE J (979.830 96.5375) 2
F 7.1175
SSE U (N K ) 696.5373 (40 5)
The critical value for significance level 0.05 is F(0.95,2,35)67 . Since the
calculated F is greater than the criticalF we reject H 0 and conclude that the model is
misspecified. Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 181
EXERCISE 6.3
(a) Let the total variation, unexplained vari ation and explained variation be denoted by SST,
SSE and SSR, respectively. Then, we have
SS eiˆ N K (20 3) 2.5193 42.8281
Also,
2 SSE
R 1 SST 0.9466
and hence the total variation is
SSE 42.8281
SST 802.0243
1 2 1 0.9466
and the explained variation is
SSR SST SSE 80 2.0243 42.8281 759.1962
(b) A 95% confidence interval for is
2
2(0.975,172 0.69914 2.110 0.048526 (0.2343,1.1639)
A 95% confidence interval for 3s
bt b se( ) 1.7769 2.110 0.037120 (1.3704, 2.1834)
2 (0.975,173
(c) est H 0 2 1 against the alternative1H 2 < 1, we calculate
b2 2 0.69914 1
t 1.3658
se2 0.048526
At a 5% significance level, we reject0Hif tt 1.740 . Since 1.3658 1.740 ,
(0.05,17)
we fail to rejecH 0. There is insufficient evidence to conclu21 .
(d) tst H 0 23 against the alternativH 1 2 and/or 3 0 , we calculate
explained variationK 1 759.1962 / 2
F 151
unexplained variation K 42.8281/17
The critical value for a 5% level of significance i(0.95,2,17)59. Since 151 3.59 , we
reject 0 and conclude that the hypothesis 2 = 3 = 0 is not compatible with the data. Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 182
Exercise 6.3 (continued)
(e) The t-statistic for testH :2 against the alternatiH :2 is
0 2 3 1 2 3
t 2bb2 3
se b
2 3
For a 5% significance level we reject H0if tt (0.025,17)11 or tt (0.975,17)11.
The standard error is given by
se 2 3 2 b2ar( )3 var( ) 223 cov( , )
4 0.048526 0.03712 2 2 .031223
0.59675
The numerator of the t-statistic is
2bb2 3 2 0 .6991 1.7769 0.37862
leading to a t-value of
0.37862
t 0.634
0.59675
Since .1 0.634 2.11 , we do not rejectH 0. There is no evidence to suggest that
.
22 3 Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 183
EXERCISE 6.4
(a) The value of the t statistic for the significance tests is calculated from:
bk
t
se( k
We reject the null hypothesis H 00k if ttc 2 . The t-values for each of the
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 2 both appear in
an equation, their coefficients are not signi ficantly different from zero, with the exception
of eqn (B), where EDUC 2 is significant. In addition, the interaction term between
EXPER and EDUC is not significant in eqn (A).
Variable t-valuesa
Eqn (A) Eqn (B) Eqn (C) Eqn (D) Eqn (E)
C 3.97* 6.59* 8.38*23.82* 9.42*
1
EDUC 2 1.26 0.84 1.04 15.90*
2
EDUC 3 1.89 2.12*1.73
EXPER 4 4.58* 6.28* 5.17* 6.11*
2
EXPER 5 –5.38* –5.31* –4.90* –5.13*
EXPER*EDUC –1.06
6
HRSWK 7 8.34* 8.43* 9.87*10.11* 8.71*
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 0 . The F-test value for testing H :0 against
6 0 6
H 1 6 can be calculated from restricted and unr estricted sums of squared errors as
follows:
(S)SE SSE J (222.6674 22.4166) 1
F R U 1.120
SSEU (N K) 222.4166 993
corhesponding p-value is 0.290. The critical valu e at the 5% significance level is
F (0.95,1,993)51. 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 hypothsis and conclude that the interaction term,
EDUCEXPER is not significant in determining the wage.
The t-value for testingH 0 60 against H 1 6 is –1.058. At the 5% level, its
absolute value is less than the critical valut(0.975,993)62 . Thus, the t-test gives the
same result. The two tests are equivalent because 1.120 1.058 and 3.8511.962 . Chapter 6, Exercise Solutions, Principles of Econometrics, 4e 184
Exercise 6.4 (continued)
(c) The restrictions imposed on eqn (A) that give eqn (C) are4 50,nd60 . Thus,
we test
0d n a 0H, :0
0 4 5 6

More
Less
Related notes for ECON 3210

Join OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.