Question

Consider the below model (Wage is per $100,000):

???? = ?0 + ?1??? + ?2??? + ?3?????? + ?4??? + ?5???2 + ?1??? ∗ ??? + ?

where P.edu is parents’ year of schooling.

a) Complete the below table if the number of the observations is fifteen, and α=0.05 with the following null and alternative hypotheses:

?0: ?1, ?2, ?3, ?4, ?5, ?6 = 0

?1: ?1, ?2, ?3, ?4, ?5, ?6 ≠ 0

Value ?̂ 1: 0.08 ??(?̂ 1): 0.035 ?̂ 2: 0.02 ??(?̂ 2): 0.008 ?̂ 3: 0.06 ??(?̂ 3): 0.03 ?̂ 4: 0.005 ??(?̂ 4): 0.0021 ?̂ 5: -0.003 ??(?̂ 5): 0.0013 ?̂ 6: 0.001 ??(?̂ 6): 0.00045

Find the T-value for each Beta.

b) Interpret the impact of each explanatory variables on the estimated wage?

c) If SST=2, SSRr =0.4 (excluding education squared and the interaction between education and experience vars), using the adjusted R-square parameter, what would be the SSR for the complete model in order to keep both variables (edu2 and Edu*Exp) in the model?

d) If SST=2, SSRr =0.4 (excluding education squared and the interaction between education and experience vars), using the F test, what would be the SSRur for the complete model if two added variables (edu2 and Edu*Exp) can jointly explain variation in wage?

e) If SST=2, SSRur=0.2, SSRr =0.4 (excluding education squared and the interaction between education and experience vars), SSRr =0.43 (excluding Gender and the interaction between education and experience vars), which variables should we keep in the model among them all?

f) What is the optimal educational level in the above model? Explain (Extra)

Answer 1

We want to test H0 :β2 + 2β4 + β6 = 0 against the alternative 1 2 4 6 H :β + 2β + β ≠ 0.
The value of the test statistic is
( )
2 4 6
2 4 6
2 8.011 2 1.944 0.567 7.367
se 2 0.233
t b b b
b b b
+ + − × −
= = =
+ +
At a 5% significance level, the critical t-value is c ±t where (0.975, 21) 2.080 ct = t = .
Since t > 2.080 we reject the null hypothesis and conclude that the marginal product of
yield to nitrogen is not zero when NITRO = 1 and PHOS = 1.
(ii) We want to test 0 2 4 6 H :β + 4β + β = 0 against the alternative 1 2 4 6 H :β + 4β + β ≠ 0 .
The value of the test statistic is
( )
2 4 6
2 4 6
4 8.011 4 1.944 0.567 1.660
se 4 0.040
t b b b
b b b
+ + − × −
= = = −
+ +
Since |t| < 2.080 (0.975, 21) = t , we do not reject the null hypothesis. A zero marginal yield
with respect to nitrogen is compatible with the data when NITRO = 1 and PHOS = 2.