Income Determination
To estimate the relation between income, education and experience, we run the
regression and get the equation,
log(income) = 7.307423 + 0.159746edu – 0.002961expr
standard error (0.046202) (0.159746)
(0.001266)
Hypothesis Test
1. H0: β1=0,
H1:β1>0
t-statistics = 51.38789, degree of freedom: 5778-2-1=5775 (almost normal
distribution)
we can get the p-value is 0.0000, so we can reject H0 and think β1 is larger than
0.
2. H0: β1=0.2, H1: β1≠0.2
t-statistics = (0.159746-0.2)/0.003109=-12.9275716, degree of freedom: 5775,
we can get the p-value is 1.045E-37, so we can reject the null hypothesis and
think β1 is not equal to 0.2; that is, with each additional year of schooling, the
increase of income is not 20%.
3. The 95% confidence interval forβ1:
Standard error ofβ1 is 0.003109, degree of freedom is 5775, cα /2=1.96. Thus the
confidence
interval
is[0.159746-1.96*0.003109,0.159746+1.96*0.003109]=
[0.15365236,0.16583946]
Testing Cobb-Douglas Production Function
According to Cobb-Douglas function (Q=AKα1Lα2) and the competitive economy
condition (w/p=α2Q/L), we can get the implication that V/L=w/α2. To test the
soundness of the implication, we regress log(V/L) on log(w) to obtain the equation,
log(V/L) =β0 +β1log(w)+u
To examine the soundness, we need to test whetherβ1 equals 1.
Thus we have the hypothesis test,
H0: β1=1, H1: β1≠1
1. Dairy industry
log(V/L) = -4.005473 + 0.718909log(w); obs=16
standard error (0.381983) (0.055186)
t-statistics: (0.718909-1)/ 0.055186 = -5.09352
degree of freedom: 16-1-1=14
critical value: tα/2 =2.14479, given α=95%
p-value: 0.00016
Since |t-statistics|=5.09352>2.14479, and p-value=0.00016<5%, we can reject
the null hypothesis and thinkβ1 does not equal to 1 and the implication is not
sound.
Considering the implication is not sound, we can infer that either the
Cobb-Douglas function fits dairy industry poorly or the industry is not
competitive.
2. Chemical industry
log(V/L) = -4.673098 + 0.829709log(w); obs=16
standard error (0.510452) (0.072281)
t-statistics: (0.829709-1)/0.072281=-2.355958
degree of freedom: 16-1-1=14
critical value: tα/2 = 2.14479, given α=95%
p-value: 0.033580
Under the significance level of 95%, since |t-statistics|=2.355958>2.14479 and
p-value=0.033580<5%, we can reject the null hypothesis and thinkβ1 does not
equal to 1.
Considering the implication is not sound, we can infer that either the
Cobb-Douglas function fits dairy industry poorly or the industry is not
competitive.
The Influence of Parents
As a further investigation of the factors influence wage, this time we add parents’
education into the regression to examine the influence of parents. The model we use
is that,
Log(wage)= β0 +β1 edu+β2 expr+β3 mothedu+β4 fathedu
The equation we get is,
Log(wage)=0.173620+0.115893edu+0.034192expr+0.008330mothedu+0.020931fath
edu; obs=1230
We can see that both mother’s education and father’s education positively affect
children’s salary. However, father’s education has a more significant influence.
However, since this equation is based on the particular sample, to see whether in the
population father’s education has the same influence as mother’s education we
make the hypothesis test.
H0: β3=β4, H1: β3≠β4
The restricted equation is,
Log(wage)= β0 +β1 edu+β2 expr+β3 (mothedu+fathedu).
After running the regression, the equation is,
Log(wage)= 0.150042 + 0.115197edu + 0.033956expr+0.016128 (mothedu+ fathedu)
SSRu=350.5707
SSRR=350.8486
The number of restriction: j=1
(𝑆𝑆𝑅 −𝑆𝑆𝑅 )/1
𝑈
F-statistics= 𝑆𝑆𝑅 𝑅/(𝑛−𝑘−1)
=
𝑈
(350.8486−350.5707)/1
350.5707/(1230−4−1)
= 0.9711~F1,1225
Critic value: 3.849061, given α=95%
P-value: 0.324601
Since |F-statistics|=0.9711<3.849061 and the P-value 0.324601>5%, we cannot
reject the null hypothesis. As a result, although in this sample we see that mother’s
education has a less impact. However, in the population we cannot make this
inference.