homework1

计量经济学第一次作业

彭岩 116120910046

chapter 2 problem 9

1. β0 means when income equals 0, people’s consumption. It is less than 0 which means when income is 0, people will not consume anything but receive money from government. The average subsidy is 124.84.

2. when income equals 30000, according to the equation:
   

   cons=β0+β1inc
   
   we can know the consumption is 25465.16.
   

chapter 3 problem 1

1. I think the effect of sibs meet the expectation, for that the more children one family have, the less resources and care each children allocated. The sibs must increase at least 10.6 to decrease one year education.
2. The more education the mother get, the higher education the children will get. And in average, if the mother receive one more year education, the child will receive 0.13 year education more.The difference between A and B is simply 140 times the coefficient on sat, because hsperc is the same for both students. So A is predicted to have a score .00148(140) ≈ .207
higher.
3.

educA=14.452

educB=15.816

educA−educB=−1.364

Perhaps not surprisingly, a large ceteris paribus difference in SAT score – almost two and one-half standard deviations – is
needed to obtain a predicted difference in college GPA or a half a point.

chapter 3 problem 5

1. which is to prove
   E(β1+β2)=E(β1)+E(β2)
   , no, by definition, study+sleep+work+leisure=168. therefore, we must change at least one of the other categories so that the sum is still 168.
2. var(θ1)=var(β1+β2)
   =var(β1)+var(β2)−2Cov(β1,β2)
   from part 1, we can say, study as a perfect linear function of independent variables: study=168-sleep-work-leisure. this holds for every observation so MLR.3 violated.
3. so we cannot decline the assumption. now for example, β1 is interpreted as the change in GPA when study increases by one hour, where sleep, work, and μ are all held fixed. if we are holding sleep and work fixed but increase study by one hour, then we must be reducing leisure by one hour. The other slope parameters have a similar interpretation.

chapter 4 problem 1

1. H0:β3=0
   H1:β3>0
2. The salary will increase 0.012 percent. it won’t effect lot on salary.
3. t=(β3−0)/se(β3)=0.44<1.28

chapter 4 problem 2

1. when the residual error do not satisfy the normal distribution, then OLS don’t follow the t distribution. so according to the following three factors, i think when there is an important variable left or there is strong correlationship between two variables will affect the OLS t test outcome.

H0:β3=0

H1:β3>0

2. The proportionate effect on salary is .00024(50) = .012. To obtain the percentage effect, we multiply this by 100: 1.2%. Therefore, a 50 point ceteris paribus increase in ros is predicted to increase salary by only 1.2%. Practically speaking, this is a very small effect for such a large change in ros.
3. The 10% critical value for a one-tailed test, using df = ∞, is obtained from Table G.2 as 1.282. The t statistic on ros is .00024/.00054 ≈ .44, which is well below the critical value. Therefore, we fail to reject H0 at the 10% significance level.
4. Based on this sample, the estimated ros coefficient appears to be different from zero only because of sampling variation. On the other hand, including ros may not be causing any harm; it depends on how correlated it is with the other independent variables (although these are very significant even with ros in the equation).

chapter 4 problem 6

1.

1. joint hypotheses test:
   H0:β1=β2=β3=β4=0
   F=(R2u−R2r)/q/(1−R2u)/(n−k−1)
2. and F value is 11.37, which is a strong rejection of H0: from Table G.3c, the 1% critical value with 2 and 90 df is 4.85.
3. using log made the coefficients larger. the dkr and eps coefficients are reasonable however the netinc and salary is too small. We use the R-squared form of the F statistic. We are testing q = 3 restrictions and there are 88 – 5 = 83 df in the unrestricted model. The F statistic is (.829 – .820)/(1 – .829) ≈ 1.46. The 10% critical value (again using 90 denominator df in Table G.3a) is 2.15, so we fail to reject H0 at even the 10% level. In fact, the p-value is about .23.
4. for the whole model the R-square is a little small, so the effect of the model is not very good. If heteroskedasticity were present, Assumption MLR.5 would be violated, and the F statistic would not have an F distribution under the null hypothesis. Therefore, comparing the F statistic against the usual critical values, or obtaining the p-value from the F distribution, would not be especially meaningful.