詹姆斯计量经济学第三版偶数题4 8解答

Putting these results together
= σ β2ˆ1
1= (0.22 ×1× 0.8) + ((1 − 0.2)2 × 4 × 0.2)
n
0.162
1 21.25 n
4.12. (a) Write
n
n
n
∑ ∑ ∑ ESS= (Yˆi − Y )2= (β垐 ?0 + β1Xi − Y )2= [β1( Xi − X )]2
(a) Substituting Height = 70, 65, and 74 inches into the equation, the predicted weights are 176.39, 156.69, and 192.15 pounds.
(b) ∆ Weight =3.94 × ∆Height =3.94 ×1.5 =5.91.
(c) Rm − Rf= 7.3% − 3.5%= 3.8%. Thus, the predicted returns are Rˆ = Rf + β垐(Rm − Rf ) = 3.5% + β × 3.8%
Wal-Mart: 3.5% + 0.3×3.8% = 4.6% Kellogg: 3.5% + 0.5 × 3.8% = 5.4% Waste Management: 3.5% + 0.6 × 3.8% = 5.8% Verizon: 3.5% + 0.6 × 3.8% = 5.8% Microsoft: 3.5% +1.0 × 3.8% = 7.3% Best Buy: 3.5% +1.3× 3.8% = 8.4% Bank of America: 3.5% + 2.4 × 3.8% = 11.9%
− Y )2 (Yi − Y )2
=
1 n−1
∑n i =1
(
X
i
−
X
)(Yi
−Y
)
2
= n1−1 ∑in 1= ( X i − X )2 n1−1 ∑in 1 (Yi − Y )2
= = ssXXsYY 2 rX2Y
(b) This follows from part (a) because rXY = rYX.
(b) var( Xi ) = 0.2 × (1− 0.2) = 0.16 and µX = 0.2. Also
var[( Xi − µX )ui ] =E[( Xi − µX )ui ]2 = E[( Xi − µX )ui |Xi = 0]2 × Pr( Xi = 0) + E[( Xi − µX )ui |Xi = 1]2 × Pr( Xi = 1)
Yi = (β0 + 2) + β1Xi + (ui − 2).
The new regression error is (ui − 2) and the new intercept is (β0 + 2). All of the assumptions of Key Concept 4.3 hold for this regression model.
Chapter 5
Regression with a Single Regressor: Hypothesis Tests and Confidence Inter vals
5.2. (a) The estimated gender gap equals $2.12/hour.
(b) The null and alternative hypotheses are H0 : β1 = 0 vs. H1 : β1 ≠ 0. The t-statistic is
4.10.
(a) E(ui|X = 0) = 0 and E(ui|X = 1) = 0. (Xi, ui) are i.i.d. so that (Xi, Yi) are i.i.d. (because Yi is a function of Xi and ui). Xi is bounded and so has finite fourth moment; the fourth moment is non-zero because Pr(Xi = 0) and Pr(Xi = 1) are both non-zero so that Xi has finite, non-zero kurtosis. Following calculations like those exercise 2.13, ui also has nonzero finite fourth moment.
=i 1=i 1
=i 1
Fra Baidu bibliotek
∑ =
n
βˆ12 ( X i − X=)2
i =1
∑in=1( X i ∑n
i =1
−X (Xi
)(Yi −X
− )2
Y
)
2
.
This implies
= = R2 ∑= in 1(EY= Si S− Y )2 ∑in∑1= (in=X1(i X−iX−)X2 ∑)(inY1i
= t act
βˆ
1
−0 =
2.= 12
5.89,
SE
(
βˆ
)
1
0.36
and the p-value for the test is
p-value = 2Φ(−|tact |) = 2Φ (−5.89) = 2 × 0.0000 = 0.000 (to four decimal places)
The p-value is less than 0.01, so we can reject the null hypothesis that there is no gender gap at a 1% significance level. (c) The 95% confidence interval for the gender gap β1 is {2.12 ±1.96 × 0.36}, that is, 1.41 ≤ β1 ≤ 2.83. (d) The sample average wage of women is β=ˆ0 $12.52/hour. The sample average wage of men is β垐0 + = β 1 $12.52 + $2.1=2 $14.64/hour. (e) The binary variable regression model relating wages to gender can be written as either
(c)
∑ ∑ ∑ ∑ Because rXY = = ssXXsYY , rXY ssYX= = ssXX2Y (n 1−(n1)1−i= 1=n1)(iXn1= i(−X iX−)(XYi)−2 Y ) i= =n1 (iXn1i(−X iX−)(XYi)−2 Y )
βˆ1
4.14. Because β垐0= Y − β1X , Y= βˆ0 + β1X . The sample regression line is=y βˆ0 + β1x , so that the sample regression line passes through ( X ,Y ).
where the first equality follows because E[(Xi − µX)ui] = 0, and the second equality follows from the law of iterated expectations.
E[( Xi − µX )ui |Xi = 0]2 = 0.22 ×1, and E[( Xi − µX )ui |Xi = 1]2 = (1 − 0.2)2 × 4.
(b) Yes. Using the expression in (a) var (R − Rf ) − var (Rm − Rf ) = (β 2 −1) × var (Rm − Rf ) + var(u), which will be positive if var(u) > (1 − β 2 ) × var (Rm − Rf ).
SER = 10.2 × 0.4536 = 4.6267 kg.
4.4. (a) (R − Rf )= β (Rm − Rf ) + u, so that var (R − Rf ) = β 2 × var(Rm − Rf ) + var(u) + 2β × cov(u, Rm − Rf ). But cov(u, Rm − Rf ) = 0, thus var(R − Rf ) =β 2 × var(Rm − Rf ) + var(u). With β > 1, var(R − Rf) > var(Rm − Rf), follows because var(u) ≥ 0.
The coefficients are γˆ0 =−99.41× 0.4536 =−45.092 kg;
γˆ1
=3.94 ×
0.4536 2.54
=0.7036
kg
per
cm.
The R2 is unit free, so it remains at R2 = 0.81. The standard error of the regression is
Wage = β0 + β1Male + ui ,
or
Wage = γ 0 + γ1Female + vi .
In the first regression equation, Male equals 1 for men and 0 for women; β0 is the population mean of wages for women and β0 + β1 is the population mean of wages for men. In the second regression equation, Female equals 1 for women and 0 for men; γ 0 is the population mean of wages for men and γ 0 + γ1 is the population mean of wages for women. We have the following relationship for the coefficients in the two regression equations:
γ 0 = β0 + β1, γ0 + γ1 = β0.
Given the coefficient estimates βˆ 0 and βˆ1 , we have
γˆ0 = β垐0 + β 1 = 14.64, γ垐1 =β垐0 − γ 0 =−β 1 =−2.12.
Due to the relationship among coefficient estimates, for each individual observation, the OLS
(c) We have the following relations: 1 in = 2.54 cm and 1 lb = 0.4536 kg. Suppose the regression equation in centimeter-kilogram units is
W eigh=t γ垐0 + γ 1 Height .
4.6. Using E(ui|Xi ) = 0, we have E(Yi |Xi ) =E(β0 + β1Xi + ui |Xi ) =β0 + β1E( Xi |Xi ) + E(ui |X i ) =β0 + β1X i .
4.8. The only change is that the mean of βˆ0 is now β0 + 2. An easy way to see this is this is to write the regression model as
Chapter 4
Linear Regr ession with One Regressor
4.2. The sample size n = 200. The estimated regression equation is
W eight = (2.15) − 99.41 + (0.31) 3.94Height, R2 = 0.81, SER = 10.2.