Chapter2-面板计量习题答案

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Chapter 2 essay(题目+答案)A level AP IB 经济学 Economics 历年Essay真题完美打印

Chapter 2 essay(题目+答案)A level AP IB 经济学 Economics 历年Essay真题完美打印

Chapter 2 Essay 题型分类1. Explain the way in which economists measure how much the supply of a good changes as its price changes. Explain two factors that influence the result. (m18-22)➢Definition and formula of PES (1 mark)➢elastic supply and inelastic supply (1 mark)➢explanation of two factors that influence the price elasticity of supply. (Up to 3 marks per factor explained)2.(a) Explain the factors that can affect the price elasticity of demand for a product. [8](b) Discuss the extent to which knowledge of a product’s cross-elasticity of demand is likely to be important to a firm supplying that product. s18-21➢Analysis of a positive XED. (Up to 4 marks)➢Analysis of a negative XED. (Up to 4 marks)➢Evaluation (4 marks):➢The extent to which XED is likely to be important to a firm will depend on many factors, such as how close a substitute or a complement is to the firm’s product. XED is likely to be more important when the relationship between two products is a close one.If two goods are unrelated, the XED =0 and it is not important.3.(a) Explain the factors that can affect the value of the price elasticity of supply for a product. [8]➢Definition (2 marks)➢Different values (2 marks)➢At least two factors (4 marks)(b) Discuss whether income elasticity of demand is the most useful measure of elasticity to a firm. w18-21➢Comment on whether it is positive, which will be the case with normal goods, or whether it is negative, which will be the case with inferior goods.YED is important to the decisions of a firm, enabling it to estimate the effect on the demand for its products of a change in incomes. For example, if incomes are rising in an economy, a firm would be expected to increase the production of normal goods and decrease the production of inferior goods.➢Candidates need to consider that the extent to which YED is likely to be important toa firm will depend on a number of factors, such as the proportion of income that isspend on a particular product. The demand for some products will not be very sensitive to a change in income because they are not very expensive, whereas the demand for other more expensive products will be much more sensitive to a change in income.(Up to 4 marks)➢Candidates could then discuss other elasticities of demand, such as PED or XED. One of these done very well could gain all 4 marks, or two done reasonably well. (Up to 4 marks)➢For evaluative comment on whether YED is likely to be the most important measure of elasticity to a firm. (EV: up to 4 marks)4.(a) Explain the significance of cross elasticity of demand values that are negative, positive and zero. [8](b) Discuss the extent to which the concepts of price elasticity of demand, income elasticity of demand and price elasticity of supply would be helpful to an organisation responsible for the growth of tourism to a holiday resort. w18-225.(a) Explain the factors that are likely to make the supply of a product relatively price inelastic. [8](b) Discuss the ways in which businesses might attempt to increase the price elasticity of supply of their products. Assess whether these attempts are likely to be successful. [12] m17-226.(a) Explain the factors that determine whether the price elasticity of supply for a good is likely to be relatively elastic or relatively inelastic. [8](b) Discuss how governments might attempt to increase the elasticity of supply of an agricultural product. Consider whether they are likely to be successful. [12] w17-217.(a) Explain how economists use the concept of income elasticity of demand to distinguish between different types of goods. [8](b) Discuss whether an understanding of price elasticity of demand is of more benefit to a producer of agricultural goods than an understanding of income elasticity of demand. [12] w17-228.(a) Explain the factors that determine whether the price elasticity of demand for a good is likely to be relatively inelastic. [8](b) Discuss if it is more useful for governments to have knowledge of the price elasticity of demand or the income elasticity of demand for certain products. [12] w17-239.(a) Explain two factors that are likely to make the supply of a product relatively price-inelastic. [8](b) Discuss the policies that governments might use to increase the price elasticity of supply of essential goods, and assess the likely effectiveness of such policies.m16-2210.(a) Explain how economists use the concept of elasticity to distinguish between normal and inferior goods and between substitutes and complements. [8](b) Discuss how knowledge of the differences between these types of goods would help government policy makers and entrepreneurs to make decisions. s16-2311.✓the impact of a rise in the price of a substitute using a supply and demand diagram(s).(Up to 4 marks)The diagram should be correctly labelled (1 mark) and show an increase in demand for a good as the price of its substitute rises (1 mark). It should also show a rise in the equilibrium price of the good (1 mark) and a rise in the equilibrium quantity (1 mark).✓the cross elasticity of demand in terms of what it measures (1 mark)✓an accurate formula (1 mark)✓explaining why substitutes have a positive value for cross elasticity. For stating that substitutes have a positive cross elasticity (1 mark) and explaining why this occurs (1 mark)✓explaining how habit-forming demerit goods and goods with lots of substitutes are each likely to respond to price rises (Up to 4 marks)✓explaining how government policy makers could use the information (Up to 4 marks)✓For evaluative comment on the extent to which this information would be useful. A concluding comment is essential for full marks. (Up to 4 marks)Habit-forming demerit goods are likely to be price inelastic and goods with lots of substitutes are likely to be price elastic. Government policy makers will use this information to assess which products to tax and also the extent to which the consumption of demerit goods will fall if the goods are taxed.12.(a) Explain any two factors that cause the supply of a product to be price-inelastic. [8](b) Discuss the policies that a government might adopt to increase the price elasticity of supply of agricultural goods in an economy and consider which policy is likely to be most effective. [12] w16-22✓income elasticity of demand:• What it measures 1 mark• accurate formula 1 mark✓Income elasticity for inferior goods: 2• For recognition that the coefficient is negative 1 mark• For an explanation of why a negative coefficient arises when incomes change✓Income elasticity for necessary goods:• For recognition that the coefficient is positive (accept zero) 1 mark• For an explanation linking the extent of necessity to the size of the coefficient.Analysis:✓Explains any methods that might be used to change the price elasticity of demand fora product with due reference to any difficulties that might arise with each methodexplained.Up to 4 marks for each method explained✓Evaluation:Compares the likely success of the methods explained (3 marks) and reaches a conclusion on which is most likely to be successful (1 mark)14. Explain how economists use the concept of elasticity to distinguish between substitute goods and complementary goods. s17-22(b) Discuss which measure of the different types of elasticity of demand is most useful for a business when setting the price for its product. s17-2215. Use the concept of income elasticity of demand to explain how a rise in incomes would affect the demand for an inferior good and for a necessary good. [8](b) Discuss the range of policies that are available to businesses to increase sales when incomes are falling. Consider which is most likely to be successful. s17-2316. Explain the meaning of ‘price elasticity of demand’ and, using examples, outline the factors that would cause the demand for a good to be relatively price-elastic. [8](b) Discuss why entrepreneurs might want to change the price elasticity of demand for their products, and consider the extent to which this is achievable. s16-22。

高分子材料成型加工唐颂超第三版第2-0章课后习题答案(仅供参考)

高分子材料成型加工唐颂超第三版第2-0章课后习题答案(仅供参考)

高分子材料成型加工Chapter2-10 课后习题答案(仅供参考)Chapter2 高分子材料学1. 分别区分“通用塑料”和“工程塑料” 、“热塑性塑料”和“热固性塑料” ,并请各举2、3 例。

答:通用塑料:一般指产量大、用途广、成型性好、价廉的塑料。

通用塑料有PE、PP、PVC 、PS 等工程塑料是指拉伸强度大于50MPa 冲击强度大于6kJ/m2 ,长期耐热温度超过100℃,刚性好、蠕变小、自润滑、电绝缘、耐腐蚀等可代替金属用作结构件的塑料。

工程塑料有PA、PET 、PBT、POM 等。

热塑性塑料:加热时变软以至流动,冷却变硬。

这种过程是可逆的、可以反复进行。

如聚乙烯、聚丙烯、聚氯乙烯、聚苯乙烯、聚甲醛、聚砜、聚苯醚好和氯化聚醚等都是热塑性塑料。

热固性塑料:第一次加热时可以软化流动,加热到一定温度,产生化学反应一交链固化而变硬,这种变化是不可逆的。

此后,再次加热时,已不能再变软流动了。

正是借助这种特性进行成型加工,利用第一次加热时的塑化流动在压力下充满型腔,进而固化成为确定形状和尺寸的制品。

这种材料称为热固性塑料。

酚醛、脲醛、三聚氰胺甲醛、不饱和聚酯、有机硅等塑料都是热固性塑料。

2. 什么是聚合物的结晶和取向?它们有何不同?研究结晶和取向对高分子材料加工有何实际意义?聚合物的结晶:高聚物发生的分子链在三维空间形成局部区域的、高度有序的排列的过程。

聚合物的取向:高聚物的分子链沿某特定方向作优势的平行排列的过程。

包括分子链、链段和结晶高聚物的晶片、晶带沿特定方向择优排列。

不同之处:(1)高分子的结晶属于高分子的一个物理特性,不是所有的高聚物都会结晶,而所有的高聚物都可以在合适的条件下发生取向。

(2)结晶是某些局部区域内分子链在三维空间的规整排列,而取向一般是在一定程度上的一维或二维有序,是在外力作用下整个分子链沿特定方向发生较为规整排列。

(3)结晶是在分子链内部和分子链之间的相互作用下发生的,外部作用也可以对结晶产生一定的影响;取向一般是在外力作用和环境中发生的,没有外力的作用,取向一般不会内部产生。

计量经济学第二章练习题及参考解答

计量经济学第二章练习题及参考解答

第二章练习题及参考解答2.1 为研究中国的货币供应量(以货币与准货币M2表示)与国内生产总值(GDP)的相互依存关系,分析表中1990年—2007年中国货币供应量(M2)和国内生产总值(GDP )的有关数据:表2.9 1990年—2007年中国货币供应量和国内生产总值(单位:亿元)资料来源:中国统计年鉴2008,中国统计出版社对货币供应量与国内生产总值作相关分析,并说明相关分析结果的经济意义。

练习题2.1 参考解答:计算中国货币供应量(以货币与准货币M2表示)与国内生产总值(GDP)的相关系数为:计算方法: 2222()()i i i iXY i i i i n X Y X Y r n X X n Y Y -=--∑∑∑∑∑∑∑或 ,22()()()()ii X Y iiX X Y Y r X X Y Y --=--∑∑∑计算结果:M2GDPM2 1 0.996426148646 GDP0.9964261486461经济意义: 这说明中国货币供应量与国内生产总值(GDP)的线性相关系数为0.996426,线性相关程度相当高。

2.2 为研究美国软饮料公司的广告费用X 与销售数量Y 的关系,分析七种主要品牌软饮料公司的有关数据表2.10 美国软饮料公司广告费用与销售数量 资料来源:(美) Anderson D R 等. 商务与经济统计.机械工业出版社.1998. 405绘制美国软饮料公司广告费用与销售数量的相关图, 并计算相关系数,分析其相关程度。

能否在此基础上建立回归模型作回归分析?练习题2.2参考解答美国软饮料公司的广告费用X 与销售数量Y 的散点图为年份货币供应量M2国内生产总值GDP1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 200715293.4 19349.9 25402.2 34879.8 46923.5 60750.5 76094.9 90995.3 104498.5 119897.9 134610.4 158301.9 185007.0 221222.8 254107.0 298755.7 345603.6 40342.218718.3 21826.2 26937.3 35260.0 48108.5 59810.5 70142.5 78060.8 83024.3 88479.2 98000.5 108068.2 119095.7 135174.0 159586.7 184088.6 213131.7 251483.2品牌名称广告费用X(百万美销售数量Y(百万箱)Coca-Cola Classic 131.3 1929.2 Pepsi-Cola 92.4 1384.6 Diet-Coke60.4 811.4 Sprite 55.7 541.5 Dr.Pepper 40.2 546.9 Moutain Dew 29.0 535.6 7-Up11.6219.5说明美国软饮料公司的广告费用X 与销售数量Y 正线性相关。

计量经济学软件应用教程讲义-chapter2 (更新)

计量经济学软件应用教程讲义-chapter2 (更新)
y and x: linearity. A one-unit change in x has the same effect on y, regardless of the initial value of x.
This is unrealistic for many economic applications. E.g., in the wage-education example, we might want to allow for increasing returns: the next year of education has a larger effect on wages than did the previous year.
yield b1 fertilizer
9
DERIVING THE ORDINARY LEAST SQUARES ESTIMATES
10
Estimate the population parameters from a sample
Basic idea of regression is to estimate the population parameters from a sample
Q
bˆ0 , bˆ1
F.O.C.
Q
b0
bˆ0 , bˆ1
b1
0 0
2 2
n
i 1 n
i 1
yi xi
yi
bˆ0 bˆ1xi bˆ0 bˆ1xi
0
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15
Derivation of OLS
Using the properties of summation, we can rewrite the first condition as follows

计量经济学第二版课后习题1-14章中文版答案汇总

计量经济学第二版课后习题1-14章中文版答案汇总

第四章习题 1.(1)22ˆ=TSR estScore T =520.4-5.82×22=392.36 (2)ΔTestScore=-5.82×(23-19)=-23.28即平均测试成绩所减少的分数回归预测值为23.28。

(3)core est S T =βˆ0 +βˆ1×CS =520.4-5.82×1.4=395.85 (4)SER 2=∑=-n i u n 1ˆ21i 2=11.5 ∴SSR=∑=ni u1ˆi2=SER 2×(n-2)=11.5×(100-2)=12960.5R 2=T SS ESS =1-T SSSSR =0.08∴TSS=SSR ÷(1-R 2)=12960.5÷(1-0.08)=14087.5=21)(Y ∑=-ni iY∴s Y 2=1-n 121)(Y ∑=-ni iY =14087.5÷(100-1)≈140.30∴s Y ≈11.93 2. (1)①70ˆ=Height eight W =-99.41+3.94×70=176.39 ②65ˆ=Height eight W =-99.41+3.94×65=156.69 ③74ˆ=Height eight W=-99.41+3.94×74=192.15(2)ΔWeight=3.94×1.5=5.91(3)1inch=2.54cm,1lb=0.4536kg①eight Wˆ(kg)=-99.41×0.4536+54.24536.0×94.3Height(cm)=-45.092+0.7036×Height(cm)②R 2无量纲,与计量单位无关,所以仍为0.81③SER=10.2×0.4536=4.6267kg 3.(1)①系数696.7为回归截距,决定回归线的总体水平②系数9.6为回归系数,体现年龄对周收入的影响程度,每增加1岁周收入平均增加$9.6 (2)SER=624.1美元,其度量单位为美元。

chapter2 课后答案

chapter2 课后答案

4 × ( 25 ) + 1× ( 35 ) + 25 × ( −5 ) ( 3 − 1)( 4 + 1 + 25) 4350 = 0.0009mm 60

= +0.0025mm
ww
L0 = L ± 3σ L = 1000.0020 ± 0.0027 mm
第二测回 pi 7 αi 34º55′40〞 34º55′30〞 34º55′20〞 1 34º55′0〞 pi 3 2 1 1
da
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0.0055 ⎞ ⎛ = ⎜10 + ⎟ mm = 10.00055mm 10 ⎠ ⎝

x9 = 10.0005mm
σ 9 = 0.12μ m
k (10, 0.05 ) = 2.43

解:按测量顺序,用表格记下测得数据。 (1) 求算术平均值
υi = xi − x (数据见上表)
1 2 3 4 5 6 7 8 ∑
150.0015 150.0017 150.0016 150.0014 150.0013 150.0015 150.0016 150.0014 1200.0120mm
n
0 +0.2 +0.1 -0.2 -0.1 0
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σ=
查 t 分布表得
例4 在万能测长仪上测量某校对量具。 重复测量 8 次, 测得值 (单位 mm) 为 150.0015, 150.0017,150.0016,150.0014,150.0013,150.0015,150.0016,150.0014。试分别以 99.73% 和 95%的概率确定测量结果 解:列表计算如下:

数字通信第五版课后题答案chap2

数字通信第五版课后题答案chap2

E
Y2
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σY2 = E Y 2 − [E (Y )]2 = 2nσ4 + 4σ2s2
Problem 2.9 :
The
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=
−j
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=
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−nσ4 (n + 2) −s2(n + 4)σ2 − ns2σ2
−s4
(1 − j2vσ2)n/2+2 + (1 − j2vσ2)n/2+3 + (1 − j2vσ2)n/2+4
ej vs2 /(1−j 2vσ2 )
Hence, and
i=1
4
P (B2) = P (Ai, B2) = 0.08 + 0.03 + 0.12 + 0.04 = 0.27
i=1
4
P (B3) = P (Ai, B3) = 0.13 + 0.09 + 0.14 + 0.06 = 0.42
i=1
Problem 2.2 :
The relationship holds for n = 2 (2-1-34) : p(x1, x2) = p(x2|x1)p(x1) Suppose it holds for n = k, i.e : p(x1, x2, ..., xk) = p(xk|xk−1, ..., x1)p(xk−1|xk−2, ..., x1) ...p(x1) Then for n = k + 1 :

伍德里奇计量经济学第六版答案Chapter 2

伍德里奇计量经济学第六版答案Chapter 2

CHAPTER 2TEACHING NOTESThis is the chapter where I expect students to follow most, if not all, of the algebraic derivations. In class I like to derive at least the unbiasedness of the OLS slope coefficient, and usually I derive the variance. At a minimum, I talk about the factors affecting the variance. To simplify the notation, after I emphasize the assumptions in the population model, and assume random sampling, I just condition on the values of the explanatory variables in the sample. Technically, this is justified by random sampling because, for example, E(u i|x1,x2,…,x n) = E(u i|x i) by independent sampling. I find that students are able to focus on the key assumption SLR.4 and subsequently take my word about how conditioning on the independent variables in the sample is harmless. (If you prefer, the appendix to Chapter 3 does the conditioning argument carefully.) Because statistical inference is no more difficult in multiple regression than in simple regression, I postpone inference until Chapter 4. (This reduces redundancy and allows you to focus on the interpretive differences between simple and multiple regression.)You might notice how, compared with most other texts, I use relatively few assumptions to derive the unbiasedness of the OLS slope estimator, followed by the formula for its variance. This is because I do not introduce redundant or unnecessary assumptions. For example, once SLR.4 is assumed, nothing further about the relationship between u and x is needed to obtain the unbiasedness of OLS under random sampling.Incidentally, one of the uncomfortable facts about finite-sample analysis is that there is a difference between an estimator that is unbiased conditional on the outcome of the covariates and one that is unconditionally unbiased. If the distribution of the x i is such that they can all equal the same value with positive probability – as is the case with discreteness in the distribution –then the unconditional expectation does not really exist. Or, if it is made to exist then the estimator is not unbiased. I do not try to explain these subtleties in an introductory course, but I have had instructors ask me about the difference.SOLUTIONS TO PROBLEMS2.1 (i) Income, age, and family background (such as number of siblings) are just a fewpossibilities. It seems that each of these could be correlated with years of education. (Income and education are probably positively correlated; age and education may be negatively correlated because women in more recent cohorts have, on average, more education; and number of siblings and education are probably negatively correlated.)(ii) Not if the factors we listed in part (i) are correlated with educ . Because we would like to hold these factors fixed, they are part of the error term. But if u is correlated with educ then E(u|educ ) ≠ 0, and so SLR.4 fails.2.2 In the equation y = β0 + β1x + u , add and subtract α0 from the right hand side to get y = (α0 + β0) + β1x + (u - α0). Call the new error e = u - α0, so that E(e ) = 0. The new intercept is α0 + β0, but the slope is still β1.2.3 (i) Let y i = GPA i , x i = ACT i , and n = 8. Then x = 25.875, y =3.2125, 1ni =∑(x i – x )(y i – y ) =5.8125, and 1ni =∑(x i – x )2 = 56.875. From equation (2.9), we obtain the slope as 1ˆβ= 5.8125/56.875 ≈ .1022, rounded to four places after the decimal. From (2.17), 0ˆβ = y – 1ˆβx ≈ 3.2125 – (.1022)25.875 ≈ .5681. So we can writeGPA = .5681 + .1022 ACTn = 8.The intercept does not have a useful interpretation because ACT is not close to zero for the population of interest. If ACT is 5 points higher, GPA increases by .1022(5) = .511.(ii) The fitted values and residuals — rounded to four decimal places — are given along with the observation number i and GPA in the following table:You can verify that the residuals, as reported in the table, sum to -.0002, which is pretty close to zero given the inherent rounding error.(iii) When ACT = 20, GPA = .5681 + .1022(20) ≈ 2.61.(iv) The sum of squared residuals,21ˆni i u=∑, is about .4347 (rounded to four decimal places), and the total sum of squares,1ni =∑(y i – y )2, is about 1.0288. So the R -squared from the regressionisR 2 = 1 – SSR/SST ≈ 1 – (.4347/1.0288) ≈ .577.Therefore, about 57.7% of the variation in GPA is explained by ACT in this small sample of students.2.4 (i) When cigs = 0, predicted birth weight is 119.77 ounces. When cigs = 20, bwght = 109.49. This is about an 8.6% drop.(ii) Not necessarily. There are many other factors that can affect birth weight, particularly overall health of the mother and quality of prenatal care. These could be correlated withcigarette smoking during birth. Also, something such as caffeine consumption can affect birth weight, and might also be correlated with cigarette smoking.(iii) If we want a predicted bwght of 125, then cigs = (125 – 119.77)/( –.524) ≈–10.18, or about –10 cigarettes! This is nonsense, of course, and it shows what happens when we are trying to predict something as complicated as birth weight with only a single explanatory variable. The largest predicted birth weight is necessarily 119.77. Yet almost 700 of the births in the sample had a birth weight higher than 119.77.(iv) 1,176 out of 1,388 women did not smoke while pregnant, or about 84.7%. Because we are using only cigs to explain birth weight, we have only one predicted birth weight at cigs = 0. The predicted birth weight is necessarily roughly in the middle of the observed birth weights at cigs = 0, and so we will under predict high birth rates.2.5 (i) The intercept implies that when inc = 0, cons is predicted to be negative $124.84. This, of course, cannot be true, and reflects that fact that this consumption function might be a poor predictor of consumption at very low-income levels. On the other hand, on an annual basis, $124.84 is not so far from zero.(ii) Just plug 30,000 into the equation: cons = –124.84 + .853(30,000) = 25,465.16 dollars.(iii) The MPC and the APC are shown in the following graph. Even though the intercept is negative, the smallest APC in the sample is positive. The graph starts at an annual income level of $1,000 (in 1970 dollars).increases housing prices.(ii) If the city chose to locate the incinerator in an area away from more expensive neighborhoods, then log(dist) is positively correlated with housing quality. This would violate SLR.4, and OLS estimation is biased.(iii) Size of the house, number of bathrooms, size of the lot, age of the home, and quality of the neighborhood (including school quality), are just a handful of factors. As mentioned in part (ii), these could certainly be correlated with dist [and log(dist )].2.7 (i) When we condition on incbecomes a constant. So E(u |inc⋅e |inc) = ⋅E(e |inc⋅0 because E(e |inc ) = E(e ) = 0.(ii) Again, when we condition on incbecomes a constant. So Var(u |inc⋅e |inc2Var(e |inc ) = 2e σinc because Var(e |inc ) = 2e σ.(iii) Families with low incomes do not have much discretion about spending; typically, a low-income family must spend on food, clothing, housing, and other necessities. Higher income people have more discretion, and some might choose more consumption while others more saving. This discretion suggests wider variability in saving among higher income families.2.8 (i) From equation (2.66),1β = 1n i i i x y =⎛⎫ ⎪⎝⎭∑ / 21n i i x =⎛⎫ ⎪⎝⎭∑.Plugging in y i = β0 + β1x i + u i gives1β = 011()n i i i i x x u ββ=⎛⎫++ ⎪⎝⎭∑/ 21n i i x =⎛⎫ ⎪⎝⎭∑.After standard algebra, the numerator can be written as201111in n ni i i i i i x x x u ββ===++∑∑∑.Putting this over the denominator shows we can write 1β as1β = β01n i i x =⎛⎫ ⎪⎝⎭∑/ 21n i i x =⎛⎫ ⎪⎝⎭∑ + β1 + 1n i i i x u =⎛⎫ ⎪⎝⎭∑/ 21n i i x =⎛⎫⎪⎝⎭∑.Conditional on the x i , we haveE(1β) = β01n i i x =⎛⎫ ⎪⎝⎭∑/ 21n i i x =⎛⎫⎪⎝⎭∑ + β1。

计量经济学第2章习题参考答案

计量经济学第2章习题参考答案
ˆt = a ˆ0 + a ˆ1 + yt 却是两个完全不同的回归方程。③回归分析对资料的要求是:被解释变 和x
量 y 是随机变量, 解释变量 x 是非随机变量, 相关分析对资料的要求是两个变量都是随机变 量。 2. 答: 相关关系是指两个以上的变量的样本观测值序列之间表现出来的随机数学关系, 用相关 系数来衡量。 因果关系是指两个或两个以上变量在行为机制上的依赖性, 作为结果的变量是由作为原因的 变量所决定的, 原因变量的变化引起结果变量的变化。 因果关系有单向因果关系和互为因果 关系之分。 具有因果关系的变量之间一定具有数学上的相关关系。 而具有相关关系的变量之间并不一定 具有因果关系。 3. 答:主要区别:①描述的对象不同。总体回归模型描述总体中变量 y 与 x 的相互关系,而样 本回归模型描述所观测的样本中变量 y 与 x 的相互关系。 ②建立模型的不同。 总体回归模型 是依据总体全部观测资料建立的, 样本回归模型是依据样本观测资料建立的。 ③模型性质不 同。总体回归模型不是随机模型,样本回归模型是随机模型,它随着样本的改变而改变。 主要联系:样本回归模型是总体回归模型的一个估计式,之所以建立样本回归模型,目的是 用来估计总体回归模型。
1 n ∑ ui = 0 ,因为 n i =1
前者是条件期望,即针对给定的 X i 的随机干扰的期望,而后者是无条件的平均值,即针对 所有 X i 的随机干扰取平均值。
二、单项选择题 1. A 2. D 3. A 4. B 5. C 6. B 7. D 8. B 9. D 10. C 11. D 12. D 13. C
14. D 15. D 16. A 17. B
三、多项选择题 1. ACD 2. ABE 3. AC 4. BE 5. BEFH 6. DG, ABCG, G, EF 7. ABDE 8. ADE 9. ACDE

《计量经济学教程(第二版)》习题解答课后习题答案

《计量经济学教程(第二版)》习题解答课后习题答案

《计量经济学(第二版)》习题解答第一章1.1 计量经济学的研究任务是什么?计量经济模型研究的经济关系有哪两个基本特征? 答:(1)利用计量经济模型定量分析经济变量之间的随机因果关系。

(2)随机关系、因果关系。

1.2 试述计量经济学与经济学和统计学的关系。

答:(1)计量经济学与经济学:经济学为计量经济研究提供理论依据,计量经济学是对经济理论的具体应用,同时可以实证和发展经济理论。

(2)统计数据是建立和评价计量经济模型的事实依据,计量经济研究是对统计数据资源的深层开发和利用。

1.3 试分别举出三个时间序列数据和横截面数据。

1.4 试解释单方程模型和联立方程模型的概念,并举例说明两者之间的联系与区别。

1.5 试结合一个具体经济问题说明计量经济研究的步骤。

1.6 计量经济模型主要有哪些用途?试举例说明。

1.7 下列设定的计量经济模型是否合理,为什么?(1)ε++=∑=31i iiGDP b a GDPε++=3bGDP a GDP其中,GDP i (i =1,2,3)是第i 产业的国内生产总值。

答:第1个方程是一个统计定义方程,不是随机方程;第2个方程是一个相关关系,而不是因果关系,因为不能用分量来解释总量的变化。

(2)ε++=21bS a S其中,S 1、S 2分别为农村居民和城镇居民年末储蓄存款余额。

答:是一个相关关系,而不是因果关系。

(3)ε+++=t t t L b I b a Y 21其中,Y 、I 、L 分别是建筑业产值、建筑业固定资产投资和职工人数。

答:解释变量I 不合理,根据生产函数要求,资本变量应该是总资本,而固定资产投资只能反映当年的新增资本。

(4)ε++=t t bP a Y其中,Y 、P 分别是居民耐用消费品支出和耐用消费品物价指数。

答:模型设定中缺失了对居民耐用消费品支出有重要影响的其他解释变量。

按照所设定的模型,实际上假定这些其他变量的影响是一个常量,居民耐用消费品支出主要取决于耐用消费品价格的变化;所以,模型的经济意义不合理,估计参数时可能会夸大价格因素的影响。

计量经济学课后题答案

计量经济学课后题答案

第十三章面板数据模型一简单题1、简述面板数据模型的优点和局限性它能综合利用样本信息,同时反映应变量在截面和时序两个方向上的变化规律及特征。

由于面板数据模型在经济定量分析中,起着只用截面或只用时序数据模型不可替代的独特优点,而具有很高的应用价值。

总之:1.增加了样本容量;2. 可多层面分析经济问题局限性:模型设定错误与数据手机不慎引起较大的偏差;研究截面或者平行数据时,由于样本非随机性造成观测值的偏差,从而导致模型选择上的偏差。

2、你是如何理解面板数据的?在经济领域中,同时具有截面与时序特征的数据很多。

如统计年鉴中提供的各地区或各国的若干系列的年度(季度或月度)经济总量数据;在企业投资分析中,要用到多个企业若干指标的月度或季度时间序列数据;在城镇居民消费分析中,要用到不同省市反映居民消费和收入的年度时序数据。

我们将上述的企业、或地区等统称为个体,从行的方向看,是由若干个体在某个时期构成的截面观察值(截面样本),从列的方向看,是各时间序列。

这种具有三维(截面、时期、变量)信息的数据结构称为面板。

这是“面板”数据的由来,面板数据也称为时序截面数据或混合数据(Pooled Data)。

3、简述建立面板数据模型的过程。

(1)建立面板数据对象,即建立工作文件;(2)面板时序变量平稳性检验;(3)协整检验;(4)模型识别;(5)建立模型;(6)结论。

二填空题1、GDP界面变量是一维变量,面板变量为三维变量。

2、面板数据模型是无斜率系数非齐性、而截距齐性的模型。

3、面板数据模型识别包括效应模型识别和具体模型识别。

4、建立面板数据模型之前,要对面板变量进行平稳性检验和协整检验。

第十二章向量自回归(VAR)模型和向量误差修正(VEC)模型一简答题1、VAR模型的特点VAR模型不以经济理论为指导,它根据样本数据统计特征建模。

VAR模型对参数不施加零约束(如t检验),故称其为无约束VAR模型。

VAR模型的解释变量中不含t期变量,所有与线性联利方程组模型有关的问题均不存在。

Introduction to Management Science 5th Edition, 课后习题答案 Chapter 2

Introduction to Management Science 5th Edition, 课后习题答案 Chapter 2

CHAPTER 2LINEAR PROGRAMMING: BASIC CONCEPTSSOLUTION TO SOLVED PROBLEMS2.S1Back Savers Production ProblemBack Savers is a company that produces backpacks primarily for students. They are considering offering some combination of two different models—the Collegiate and the Mini. Both are made out of the same rip-resistant nylon fabric. Back Savers has a long-term contract with a supplier of the nylon and receives a 5000 square-foot shipment of the material each week. EachCollegiate requires 3 square feet while each Mini requires 2 square feet. The sales forecasts indicate that at most 1000 Collegiates and 1200 Minis can be sold per week. Each Collegiate requires 45 minutes of labor to produce and generates a unit profit of $32. Each Mini requires40 minutes of labor and generates a unit profit of $24. Back Savers has 35 laborers that eachprovides 40 hours of labor per week. Management wishes to know what quantity of each type of backpack to produce per week.a. Formulate and solve a linear programming model for this problem on a spreadsheet.To build a spreadsheet model for this problem, start by entering the data. The data for thisproblem are the unit profit of each type of backpack, the resource requirements (square feet of nylon and labor hours required), the availability of each resource, 5400 square feet of nylon and(35 laborers)(40 hours/laborer) = 1400 labor hours, and the sales forecast for each type ofbackpack (1000 Collegiates and 1200 Minis). In order to keep the units consistent in row 8(hours), the labor required for each backpack (in cells C8 and D8) are converted from minutes to hours (0.75 hours = 45 minutes, 0.667 hours = 40 minutes). The range names UnitProfit(C4:D4), Available (G7:G8), and SalesForecast (C13:D13) are added for these data.The decision to be made in this problem is how many of each type of backpack to make.Therefore, we add two changing cells with range name UnitsProduced (C11:D11). The values in CallsPlaced will eventually be determined by the Solver. For now, arbitrary values of 10 and 10 are entered.The goal is to produce backpacks so as to achieve the highest total profit. Thus, the objective cell should calculate the total profit, where the objective will be to maximize this objective cell. In this case, the total profit will beTotal Profit = ($32)(# of Collegiates) + ($24)(# of Minis)orTotal Cost = SUMPRODUCT(UnitProfit, UnitsProduced).This formula is entered into cell G11 and given a range name of TotalProfit. With 10 Collegiates and 10 Minis produced, the total profit would be ($32)(10) + ($24)(10) = $560.The first set of constraints in this problem involve the limited available resources (nylon and labor hours). Given the number of units produced (UnitsProduced in C11:D11), we calculate the total resources required. For nylon, this will be =SUMPRODUCT(C7:D7, UnitsProduced) in cell E7. By using a range name or an absolute reference for the units produced, this formula can be copied into cell E8 to calculate the labor hours required. The total resources used(TotalResources in E7:E8) must be <= Available (in cells G7:G8), as indicated by the <= in F7:F8.The final constraint is that it does not make sense to produce more backpacks than can be sold (as predicted by the sales forecast). Therefore UnitsProduced (C11:D11) should be less-than-or-equal-to the SalesForecast (C13:D13), as indicated by the <= in C12:D125678E Total Required=SUMPRODUCT(C7:D7,UnitsProduced)=SUMPRODUCT(C8:D8,UnitsProduced)The Solver information and solved spreadsheet are shown below.Thus, they should produce 1000 Collegiates and 975 Minis to achieve the maximum total profit of $55,400.b. Formulate this same model algebraically.To build an algebraic model for this problem, start by defining the decision variables. In this case, the two decisions are how many Collegiates to produce and how many Minis to produce. These variables are defined below: Let C = Number of Collegiates to produce,M = Number of Minis to produce.Next determine the goal of the problem. In this case, the goal is to produce the number of each type of backpack to achieve the highest possible total profit. Each Collegiate yields a unit profit of $32 while each Mini yields a unit profit of $24. The objective function is thereforeSolver ParametersSet Objective Cell: TotalProfit To: MaxBy Changing Variable Cells: UnitsProducedSubject to the Constraints: TotalRequired <= Available UnitsProduced <= SalesForecastSolver Options:Make Variables Nonnegative Solving Method: Simplex LP5678E Total Required=SUMPRODUCT(C7:D7,UnitsProduced)=SUMPRODUCT(C8:D8,UnitsProduced)Maximize Total Profit = $32C + $24M.The first set of constraints in this problem involve the limited resources (nylon and labor hours).Given the number of backpacks produced, C and M, and the required nylon and labor hours for each, the total resources used can be calculated. These total resources used need to be less than or equal to the amount available. Since the labor available is in units of hours, the labor required for each backpack needs to be in units of hours (3/4 hour and 2/3 hour) rather than minutes (45 minutes and 40 minutes). These constraints are as follows:Nylon: 3C + 2M≤ 5400 square feet,Labor Hours: (3/4)C + (2/3)M≤ 1400 hours.The final constraint is that they should not produce more of each backpack than the salesforecast. Therefore,Sales Forecast: C≤ 1000M≤ 1200.After adding nonnegativity constraints, the complete algebraic formulation is given below: Let C = Number of Collegiates to produce,M = Number of Minis to produce.Maximize Total Profit = $32C + $24M,subject toNylon: 3C + 2M≤ 5400 square feet,Labor Hours: (3/4)C + (2/3)M≤ 1400 hours,Sales Forecast: C≤ 1000M≤ 1200.and C≥ 0, M≥ 0.c. Use the graphical method by hand to solve this model.Start by plotting a graph with Collegiates (C) on the horizontal axis and Minis (M) on the vertical axis, as shown below.Next, the four constraint boundary lines (where the left-hand-side of the constraint exactly equals the right-hand-side) need to be plotted. The easiest way to do this is by determining where these lines intercepts the two axes. For the Nylon constraint boundary line (3C + 2M = 5400), setting M = 0 yields a C -intercept of 1800 while setting C = 0 yields an M -intercept of 2700. For the Labor constraint boundary line ((3/4)C + (2/3)M = 1400), setting M = 0 yields a C -intercept of 1866.67 while setting C = 0 yields an M -intercept of 2100. The sales forecast constraints are a horizontal line at M = 1200 and a vertical line at C = 1000. These constraint boundary lines are plotted below.100015005002000500100015002000MiniCollegiate 25003000A feasible solution must be below and/or to the left of all four of these constraints while being above the Collegiate axis (since C ≥ 0) and to the right of the Mini axis (since M ≥ 0). This yields the feasible region shown below.100015005002000500100015002000Mini Collegiate 25003000(3/4)C + (2/3)100015005002000500100015002000Mini Collegiate25003000To find the optimal solution, an objective function line is plotted by setting the objective function equal to a value. For example, the objective function line when the value of the objective function is $48,000 is plotted as a dashed line below.$32C + $24M = $48,000All objective function lines will be parallel to this one. To find the feasible solution that maximizes profit, slide this line out as far as possible while still touching the feasible region. This occurs when the profit is $55,400, and the objective function line intersect the feasible region at the single point with (C, M) = (1000, 975) as shown below.$32C + $24M = $55,400Therefore, the optimal solution is to produce 1000 Collegiates and 975 Minis, yielding a total profit of $55,400.2.S2Conducting a Marketing SurveyThe marketing group for a cell phone manufacturer plans to conduct a telephone survey todetermine consumer attitudes toward a new cell phone that is currently under development. In order to have a sufficient sample size to conduct the analysis, they need to contact at least 100 young males (under age 40), 150 older males (over age 40), 120 young females (under age 40), and 200 older females (over age 40). It costs $1 to make a daytime phone call and $1.50 to make an evening phone call (due to higher labor costs). This cost is incurred whether or not anyone answers the phone. The table below shows the likelihood of a given customer type answering each phone call. Assume the survey is conducted with whoever first answers the phone. Also, because of limited evening staffing, at most one-third of phone calls placed can be evening phone calls. How should the marketing group conduct the telephone survey so as to meet the sample size requirements at the lowest possible cost?Who Answers? Daytime Calls Evening CallsYoung Male 10% 20%Older Male 15% 30%Young Female 20% 20%Older Female 35% 25%No Answer 20% 5%a.Formulate and solve a linear programming model for this problem on a spreadsheet.To build a spreadsheet model for this problem, start by entering the data. The data for thisproblem are the cost of each type of phone call, the percentages of each customer type answering each type of phone call, and the total number of each customer type needed for the survey.The decision to be made in this problem is how many of each type of phone call to make.Therefore, we add two changing cells with range name CallsPlaced (C13:D13). The values in CallsPlaced will eventually be determined by the Solver. For now, arbitrary values of 10 and 5 are entered.The goal of the marketing group is to conduct the survey at the lowest possible cost. Thus, the objective cell should calculate the total cost, where the objective will be to minimize this objective cell. In this case, the total cost will beTotal Cost = ($1)(# of daytime calls) + ($1.50)(# of evening calls)orTotal Cost = SUMPRODUCT(UnitCost, CallsPlaced).This formula is entered into cell G13 and given a range name of TotalCost. With 10 daytime phone calls and 5 evening calls, the total cost would be ($1)(10) + ($1.50)(5) = $17.50.The first set of constraints in this problem involve the minimum responses required from each customer group. Given the number of calls placed (CallsPlaced in C13:D13), we calculate the total responses by each customer type. For young males, this will be =SUMPRODUCT(C7:D7, CallsPlaced). By using a range name or an absolute reference for the calls placed, this formula can be copied into cells E8-E10 to calculate the number of older males, young females, and older females reached. The total responses of each customer type (Total Responses in E7:E10) must be >= ResponsesNeeded (in cells G7:G10), as indicated by the >= in F7:F10.The final constraint is that at most one third of the total calls placed can be evening calls. In other words:Evening Calls <= (1/3)(Total Calls Placed)The two sides of this constraint (i.e., evening calls and 1/3 of total calls placed) are calculated in cells C15 and E15. Enter <= in D15 to show that C15 <= E15.5678910ETotal Responses=SUMPRODUCT(C7:D7,CallsPlaced)=SUMPRODUCT(C8:D8,CallsPlaced)=SUMPRODUCT(C9:D9,CallsPlaced)=SUMPRODUCT(C10:D10,CallsPlaced)The Solver information and solved spreadsheet are shown below.Thus, the marketing group should place 500 daytime calls and 250 evening calls at a total cost of $875.Solver ParametersSet Objective Cell: TotalCost To: MinBy Changing Variable Cells: CallsPlacedSubject to the Constraints: EveningCalls <= E15TotalResponses >= ResponsesNeeded Solver Options:Make Variables Nonnegative Solving Method: Simplex LP5678910ETotal Responses=SUMPRODUCT(C7:D7,CallsPlaced)=SUMPRODUCT(C8:D8,CallsPlaced)=SUMPRODUCT(C9:D9,CallsPlaced)=SUMPRODUCT(C10:D10,CallsPlaced)b. Formulate this same model algebraically.To build an algebraic model for this problem, start by defining the decision variables. In this case, the two decisions are how many daytime calls and how many evening calls to place. These variables are defined below:Let D = Number of daytime calls to placeE = Number of evening calls to place.Next determine the goal of the problem. In this case, the goal is to conduct the marketing survey at the lowest possible cost. Each daytime call costs $1 while each evening call costs $1.50. The objective function is thereforeMinimize Total Cost = $1D + $1.50E.The first set of constraints in this problem involve the minimum responses required from each customer group. Given the number of calls place, D and E, and the percentage of calls answered by each customer group, the total responses for each customer group is calculated. These total responses need to be greater than or equal to the minimum responses required. These constraints are as follows:Young Males: (10%)D + (20%)E≥ 100Older Males: (15%)D + (30%)E≥ 150Young Females: (20%)D + (20%)E≥ 120Older Females: (35%)D + (25%)E ≥ 200.The final constraint is that at most one third of the total calls placed can be evening calls. In other words:Evening Calls <= (1/3)(Total Calls Placed)Substituting E for Evening Calls, and D + E for Total Calls Placed yields the followingconstraint:E≤ (1/3)(D + E).After adding nonnegativity constraints, the complete algebraic formulation is given below: Let D = Number of daytime calls to placeE = Number of evening calls to place.Minimize Total Cost = $1D + $1.50E.subject toYoung Males: (10%)D + (20%)E≥ 100Older Males: (15%)D + (30%)E≥ 150Young Females: (20%)D + (20%)E≥ 120Older Females: (35%)D + (25%)E ≥ 200Evening Call Ratio: E≤ (1/3)(D + E)and D≥ 0, E≥ 0.。

计量课后习题第二章

计量课后习题第二章

2.1题:(1)Dependent Variable: YMethod: Least SquaresDate: 05/24/15 Time: 14:13Sample: 2001 2022Included observations: 22Variable Coefficient Std. Error t-Statistic Prob.C 56.64794 1.960820 28.88992 0.0000X1 0.128360 0.027242 4.711834 0.0001R-squared 0.526082 Mean dependent var 62.50000 Adjusted R-squared 0.502386 S.D. dependent var 10.08889 S.E. of regression 7.116881 Akaike info criterion 6.849324 Sum squared resid 1013.000 Schwarz criterion 6.948510 Log likelihood -73.34257 Hannan-Quinn criter. 6.872689 F-statistic 22.20138 Durbin-Watson stat 0.629074 Prob(F-statistic) 0.000134Y=56.64794+0.128360x1Dependent Variable: YMethod: Least SquaresDate: 05/24/15 Time: 15:30Sample: 2001 2022Included observations: 22Variable Coefficient Std. Error t-Statistic Prob.C 38.79424 3.532079 10.98340 0.0000X2 0.331971 0.046656 7.115308 0.0000R-squared 0.716825 Mean dependent var 62.50000 Adjusted R-squared 0.702666 S.D. dependent var 10.08889 S.E. of regression 5.501306 Akaike info criterion 6.334356 Sum squared resid 605.2873 Schwarz criterion 6.433542 Log likelihood -67.67792 Hannan-Quinn criter. 6.357721 F-statistic 50.62761 Durbin-Watson stat 1.846406 Prob(F-statistic) 0.000001Y=38.79424+0.331971x2Dependent Variable: YMethod: Least SquaresDate: 05/24/15 Time: 15:30Sample: 2001 2022Included observations: 22Variable Coefficient Std. Error t-Statistic Prob.C 31.79956 6.536434 4.864971 0.0001X3 0.387276 0.080260 4.825285 0.0001R-squared 0.537929 Mean dependent var 62.50000Adjusted R-squared 0.514825 S.D. dependent var 10.08889S.E. of regression 7.027364 Akaike info criterion 6.824009Sum squared resid 987.6770 Schwarz criterion 6.923194Log likelihood -73.06409 Hannan-Quinn criter. 6.847374F-statistic 23.28338 Durbin-Watson stat 0.952555Prob(F-statistic) 0.000103Y=31.79956+0.387276X3Dependent Variable: YMethod: Least SquaresDate: 05/24/15 Time: 15:27Sample: 2001 2022Included observations: 22Variable Coefficient Std. Error t-Statistic Prob.C 32.99309 3.138595 10.51206 0.0000X1 0.071619 0.014755 4.853871 0.0001X2 0.168727 0.039956 4.222811 0.0005X3 0.179042 0.048869 3.663731 0.0018R-squared 0.906549 Mean dependent var 62.50000Adjusted R-squared 0.890974 S.D. dependent var 10.08889S.E. of regression 3.331262 Akaike info criterion 5.407545Sum squared resid 199.7515 Schwarz criterion 5.605916Log likelihood -55.48299 Hannan-Quinn criter. 5.454275F-statistic 58.20479 Durbin-Watson stat 1.616536Prob(F-statistic) 0.000000Y=32.99309+0.071619x1+0.168727x2+0.179042x3(2)拟合优度的度量:人均寿命与人均GDP的R^2=0.526082,说明所建模型整体上对样本数据拟合还行,即解释变量“人均GDP”对被解释变量“人均寿命”的小部分差异作出了解释。

经济统计计量模型Chapter-2

经济统计计量模型Chapter-2

Chapter 21. A dependent variable is also known as a(n) _____.a. explanatory variableb. control variablec. predictor variabled. response variableAnswer: dDifficulty: EasyBloom’s: KnowledgeA-Head: Definition of the Simple Regression ModelBUSPROG:Feedback: A dependent variable is known as a response variable.2. If a change in variable x causes a change in variable y, variable x is called the _____.a. dependent variableb. explained variablec. explanatory variabled. response variableAnswer: cDifficulty: EasyBloom’s: ComprehensionA-Head: Definition of the Simple Regression ModelBUSPROG:Feedback: If a change in variable x causes a change in variable y, variable x is called the independentvariable or the explanatory variable.3. In the equation y = + x + u, is the _____.a. dependent variableb. independent variablec. slope parameterd. intercept parameterAnswer: dDifficulty: EasyBloom’s: KnowledgeA-Head: Definition of the Simple Regression ModelBUSPROG:Feedback: In the equation y = + x + u, is the intercept parameter.4. In the equation y = + x + u, what is the estimated value of?a.b.c.d.Answer: aDifficulty: EasyBloom’s: KnowledgeA-Head: Deriving the Ordinary Least Squares EstimatesBUSPROG:Feedback: The estimated value of is .5. In the equation c = + i + u, c denotes consumption and i denotes income. What is the residual forthe 5th observation if =$500 and =$475?a. $975b. $300c. $25d. $50Answer: cDifficulty: EasyBloom’s: KnowledgeA-Head: Deriving the Ordinary Least Squares EstimatesBUSPROG:Feedback: The formula for calculating the residual for the i th observation is . In this case, theresidual is =$500 -$475= $25.6. What does the equation denote if the regression equation is y = β0+ β1x1 + u?a. The explained sum of squaresb. The total sum of squaresc. The sample regression functiond. The population regression functionAnswer: cDifficulty: EasyBloom’s: KnowledgeA-Head: Deriving the Ordinary Least Squares EstimatesBUSPROG:Feedback: The equation denotes the sample regression function of the given regressionmodel.7. Consider the following regression model: y = β0+ β1x1 + u. Which of the following is a property ofOrdinary Least Square (OLS) estimates of this model and their associated statistics?a. The sum, and therefore the sample average of the OLS residuals, is positive.b. The sum of the OLS residuals is negative.c. The sample covariance between the regressors and the OLS residuals is positive.d. The point (, ) always lies on the OLS regression line.Answer: dDifficulty: EasyBloom’s: KnowledgeA-Head: Properties of OLS on Any Sample of DataBUSPROG:Feedback: An important property of the OLS estimates is that the point (, ) always lies on the OLS regression line. In other words, if , the predicted value of .8. The explained sum of squares for the regression function, , is defined as _____.a.b.c.d.Answer: bDifficulty: EasyBloom’s: KnowledgeA-Head: Properties of OLS on Any Sample of DataBUSPROG:Feedback: The explained sum of squares is defined as9. If the total sum of squares (SST) in a regression equation is 81, and the residual sum of squares (SSR) is25, what is the explained sum of squares (SSE)?a. 64b. 56c. 32d. 18Answer: bDifficulty: ModerateBloom’s: ApplicationA-Head: Properties of OLS on Any Sample of DataBUSPROG: AnalyticFeedback: Total sum of squares (SST) is given by the sum of explained sum of squares (SSE) and residual sum of squares (SSR). Therefore, in this case, SSE=81-25=56.10. If the residual sum of squares (SSR) in a regression analysis is 66 and the total sum of squares (SST) isequal to 90, what is the value of the coefficient of determination?a. 0.73b. 0.55c. 0.27d. 1.2Answer: cDifficulty: ModerateBloom’s: ApplicationA-Head: Properties of OLS on Any Sample of DataBUSPROG: AnalyticFeedback: The formula for calculating the coefficient of determination is . In this case,11. Which of the following is a nonlinear regression model?a. y = β0 + β1x1/2 + ub. log y = β0 + β1log x +uc. y = 1 / (β0 + β1x) + ud. y = β0 + β1x + uAnswer: cDifficulty: ModerateBloom’s: ComprehensionA-Head: Properties of OLS on Any Sample of DataBUSPROG:Feedback: A regression model is nonlinear if the equation is nonlinear in the parameters. In this case, y=1 / (β0 + β1x) + u is nonlinear as it is nonlinear in its parameters.12. Which of the following is assumed for establishing the unbiasedness of Ordinary Least Square (OLS)estimates?a. The error term has an expected value of 1 given any value of the explanatory variable.b. The regression equation is linear in the explained and explanatory variables.c. The sample outcomes on the explanatory variable are all the same value.d. The error term has the same variance given any value of the explanatory variable.Answer: dDifficulty: EasyA-Head: Expected Values and Variances of the OLS EstimatorsBUSPROG:Feedback: The error u has the same variance given any value of the explanatory variable.13. The error term in a regression equation is said to exhibit homoskedasticty if _____.a. it has zero conditional meanb. it has the same variance for all values of the explanatory variable.c. it has the same value for all values of the explanatory variabled. if the error term has a value of one given any value of the explanatory variable.Answer: bDifficulty: EasyBloom’s: KnowledgeA-Head: Expected Values and Variances of the OLS EstimatorsBUSPROG:Feedback: The error term in a regression equation is said to exhibit homoskedasticty if it has the samevariance for all values of the explanatory variable.14. In the regression of y on x, the error term exhibits heteroskedasticity if _____.a. it has a constant varianceb. Var(y|x) is a function of xc. x is a function of yd. y is a function of xAnswer: bDifficulty: EasyBloom’s: KnowledgeA-Head: Expected Values and Variances of the OLS EstimatorsBUSPROG:Feedback: Heteroskedasticity is present whenever Var(y|x) is a function of x because Var(u|x) = Var(y|x).15. What is the estimated value of the slope parameter when the regression equation, y = β0+ β1x1 +u passes through the origin?a.b.)c.d.Answer: cDifficulty: EasyA-Head: Regression through the Origin and Regression on a ConstantBUSPROG:Feedback: The estimated value of the slope parameter when the regression equation passes through theorigin is .16. A natural measure of the association between two random variables is the correlation coefficient.Answer: TrueDifficulty: EasyBloom’s: KnowledgeA-Head: Definition of the Simple Regression ModelBUSPROG:Feedback: A natural measure of the association between two random variables is the correlationcoefficient.17. The sample covariance between the regressors and the Ordinary Least Square (OLS) residuals isalways positive.Answer: FalseDifficulty: EasyBloom’s: KnowledgeA-Head: Properties of OLS on Any Sample of DataBUSPROG:Feedback: The sample covariance between the regressors and the Ordinary Least Square (OLS) residualsis zero.18. is the ratio of the explained variation compared to the total variation.Answer: TrueDifficulty: EasyBloom’s: KnowledgeA-Head: Properties of OLS on Any Sample of DataBUSPROG:Feedback: The sample covariance between the regressors and the Ordinary Least Square (OLS) residualsis zero.19. There are n-1 degrees of freedom in Ordinary Least Square residuals.Answer: FalseDifficulty: EasyA-Head: Expected Values and Variances of the OLS EstimatorsBUSPROG:Feedback: There are n-2 degrees of freedom in Ordinary Least Square residuals.20. The variance of the slope estimator increases as the error variance decreases.Answer: FalseDifficulty: EasyBloom’s: KnowledgeA-Head: Expected Values and Variances of the OLS EstimatorsBUSPROG:Feedback: The variance of the slope estimator increases as the error variance increases.。

计量经济学练习册答案(第二版)

计量经济学练习册答案(第二版)

计量经济学练习册参考答案数量经济学教研室第一章导论一、单项选择题1、C2、B3、A4、A5、B6、A7、D8、C9、B 10、 B 11、C 12、A 13、D 14、C 15、A 16、C 17、D 18、C 19、A 20、 A 21、D二、多选题1、ABCD2、ABD3、ABCD4、ABC5、ABCD三、判断题4、1.√2. ×3.√4.×5.√6.√7.×8.√9.× 10.×四、简答题1、计量经济学与经济理论、统计学、数学的联系主要体现在计量经济学对经济理论、统计学、数学的应用方面,分别如下:1)计量经济学对经济理论的利用主要体现在以下几个方面(1)计量经济模型的选择和确定(2)对经济模型的修改和调整(3)对计量经济分析结果的解读和应用2)计量经济学对统计学的应用(1)数据的收集、处理、(2)参数估计(3)参数估计值、模型和预测结果的可靠性的判断3)计量经济学对数学的应用(1)关于函数性质、特征等方面的知识(2)对函数进行对数变换、求导以及级数展开(3)参数估计(4)计量经济理论和方法的研究2、模型的检验主要包括:经济意义检验、统计检验、计量经济学检验、模型的预测检验。

①在经济意义检验中,需要检验模型是否符合经济意义,检验求得的参数估计值的符号、大小、参数之间的关系是否与根据人们的经验和经济理论所拟订的期望值相符合;②在统计检验中,需要检验模型参数估计值的可靠性,即检验模型的统计学性质,有拟合优度检验、变量显著检验、方程显著性检验等;③在计量经济学检验中,需要检验模型的计量经济学性质,包括随机扰动项的序列相关检验、异方差性检验、解释变量的多重共线性检验等;④模型的预测检验,主要检验模型参数估计量的稳定性以及对样本容量变化时的灵敏度,以确定所建立的模型是否可以用于样本观测值以外的范围。

3、怎样理解计量经济学与经济统计学的关系?二者的联系:第一,计量经济分析所采用的数据的收集与处理、参数的估计等,需要使用统计学的方法和技术来完成;第二,参数估计值、模型的预测结果的可靠性,需要使用统计方法加以分析、判断。

chapter2习题答案

chapter2习题答案

chapter2习题答案⼀.名词解释1.假根是Rhizopus(根霉属)等低等真菌匍匐菌丝与固体基质接触处分化出来的根状结构,具有固着和吸取养料等功能。

2.假菌丝当酵母菌进⾏⼀连串的芽殖后,如果长⼤的⼦细胞与母细胞不⽴即分离,其间仅以狭⼩的⾯积相连,则这种藕节状的细胞串就称为假菌丝3.⽓⽣菌丝伸展到空间的菌丝体,颜⾊较深、直径较粗的分枝菌丝,其成熟后分化成孢⼦丝4.⼦囊果能产⽣有性孢⼦的、结构复杂的⼦实体称为⼦囊果5.⽣活史⼜称⽣命周期,指上⼀代⽣物经⼀系列⽣长、发育阶段⽽产⽣下⼀代个体的全部过程为⽣活史6、异宗配合:指⽑霉在形成接合孢⼦时,凡是由不同性的菌丝体上形成的性器官结合⽽产⽣有性孢⼦的则称异宗配合。

7、同宗配合:指⽑霉在形成接合孢⼦时,凡是由同⼀个菌丝体上形成的配⼦囊结合⽽产⽣有性孢⼦的则称同宗配合。

8、锁状联合:指担⼦菌亚门的次⽣菌丝的菌丝尖端⽣长⽅式。

9Saccharomyces cerevisiae 酿酒酵母⼆.填空1、真菌的⽆性繁殖⽅式有裂殖、芽殖、⽆性孢⼦繁殖和菌丝体断裂。

2、酵母菌的⽆性繁殖⽅式主要有裂殖、芽殖。

3、真菌的有性孢⼦的种类有:卵孢⼦、接合孢⼦、⼦囊孢⼦和担孢⼦四种;真菌的⽆性孢⼦的种类有:游动孢⼦、⼦囊孢⼦、分⽣孢⼦、节孢⼦和厚垣孢⼦五种。

4、根霉的形态特征是具有假根和匍匐功丝且菌丝⽆隔;曲霉的形态特征是具顶囊和⾜细胞,菌丝有隔;青霉的形态特征是具扫帚状的分⽣孢⼦梗。

5.粘菌可分为__细胞粘菌___和_原质团粘菌__两个门.6.真核微⽣物细胞质核糖体类型为80S,原核微⽣物核糖体类型为_70S___.7.⼦囊果有闭囊果,_⼦囊壳_____ 和_____⼦囊盘________三种类型。

8. 同的酵母菌产⽣的有性孢⼦有⼦囊孢⼦和_担孢⼦9. 类酵母的营养体只能以⼆倍体形式存在,酿酒酵母的营养体能以单倍体和⼆倍体形式存在,⼋孢酵母的营养体能以单倍体形式存在。

10.根霉属于接合菌门,⾹菇属于担⼦菌门。

计量经济学: 面板数据模型习题与答案

计量经济学: 面板数据模型习题与答案

一、单选题1、一个因素具有3个不同的属性类型,则需要引入几个虚拟变量?()A.5B.2C.3D.4正确答案:B2、虚拟变量的设置方式包括()A.ABC都是B.乘法方式C.加法方式D.混合方式正确答案:A3、对于有限分布滞后模型Yt=α+β0Xt+β1Xt-1+β2Xt-2+…+βsXt-s+ut在一定条件下,参数βi可近似用一个关于i的多项式表示(i=0,1,2,…,k),其中多项式的阶数m必须满足()A.m<kB.m=kC.m>kD.m≥k正确答案:A4、在某个结构方程过度识别的条件下,不适用的估计方法是()A.间接最小二乘法B.二阶段最小二乘法C.工具变量法D.有限信息极大似然估计法正确答案:D5、对Logistic模型和Probit模型的参数估计通常采用极大似然估计法,用这种方法估计模型参数时,对样本抽取方法的要求是()A.随机抽样B.遵循抽样的基本原则C.独立同分布D.与线性回归模型一样正确答案:A二、多选题1、先决变量包括哪几种变量?()A.内生滞后变量B.外生滞后变量C.内生变量D.外生变量正确答案:A、B、D2、结构式方程中的解释变量可以是()A.内生滞后变量B.外生变量C.外生滞后变量D.虚拟变量正确答案:A、B、C、D3、结构式模型中,需要进行识别的方程是A.行为方程B.平衡方程C.技术方程D.制度方程正确答案:A、C、D4、以下选项中属于非随机方程的有()A.制度方程B.行为方程C.平衡方程D.经验方程正确答案:C、D三、判断题1、1990年至2016年全国31个省份的GDP数据是属于面板数据。

()正确答案:√2、元旦前,学院要统计学生回家与否的情况,获得的记录数据中变量X(学生回家与否)是二元选择变量。

()正确答案:√3、通过虚拟变量将属性因素引入计量经济模型,引入虚拟变量的个数仅与样本容量大小有关。

()正确答案:×4、在联立方程模型中,取值独立于模型的变量称为外生变量或前定变量。

计量地理第二章课后题答案

计量地理第二章课后题答案

徐建华版计量地理学其次章答案点击这里1. 地理数据有哪几种类型,各种类型地理数据之间的区分和联系是什么? (1)2. 各种类型的地理数据的测度方法分别是什么? (1)3. 地理数据的根本特征有哪些? (2)4. 地理数据采集的来源渠道有哪些? (2)5. 数学方法和地理信息系统在地理数据处理中各自发挥什么样的作用? (2)6. 对表2.4.1 中的分组数据,分别计算其平均值、中位数和众数。

(3)7. 查阅2011年的中国经济统计年鉴,以各省〔直辖市、自治区〕的 (3)8.某一地区各个亚区的GDP 数据如下表所示。

(7)9.假如我们在作罗伦次曲线时,不是把某要素各组分的数据由大到小排序, (13)11. 依据第10 题中的数据,计算锡尔系数L 指标和T 指标。

(21)1.地理数据有哪几种类型,各种类型地理数据之间的区分和联系是什么?答:地理数据就是用必需的测度方式描述和衡量地理对象的有关量化指标。

按类型可分为:1〕空间数据:点数据,线数据,面数据;2〕属性数据:数量标记数据,品质标记数据地理数据之间的区分及联系:数据包括空间数据和属性数据,空间数据的表达可以接受栅格和矢量两种形式。

空间数据表现了地理空间实体的位置、大小、形态、方向以及几何拓扑关系。

属性数据表现了空间实体的空间属性以外的其他属性特征,属性数据主要是对空间数据的说明。

如一个城市点,它的属性数据有人口,GDP,绿化率等等描述指标。

它们有密切的关系,两者相互结合才能将一个地理试题表达清楚。

2. 各种类型的地理数据的测度方法分别是什么?地理数据主要包括空间数据和属性数据:空间数据——对于空间数据的表达,可以将其归纳为点、线、面三种几何实体以及描述它们之间空间联系的拓扑关系;属性数据——对于属性数据的表达,须要从数量标记数据和品质标记数据两方面进展描述。

其测度方法主要有:(1) 数量标记数据①间隔尺度〔Interval Scale〕数据: 以有量纲的数据形式表示测度对象在某种单位〔量纲〕下的确定量。

第2章习题参考答案

第2章习题参考答案

习题:1、 如图所示,一平行板电容器充电后,A、B两极板上电荷的面密度分别为σ和-σ。

设P为两板间任一点,略去边缘效应,求: (1) A板上的电荷在P点产生的电场强度EA; (2) B板上的电荷在P点产生的电场强度EB; (3) A、B两板上的电荷在P点产生的电场强度E;(4) 若把B板拿走,A板上电荷分布如何A板上的电荷在P点产生的电场强度为多少解:略去边缘效应,两极板上的电荷是均匀分布的电荷,两极板间的电场是均匀电场。

由对称性和高斯定理可得(1)A板上的电荷在P点产生的电场强度 e E A2εσ=(A 板法线方向上的单位矢量,指向B 板);(2) B 板上的电荷在P点产生的电场强度 e E B2εσ=(3) A、B 两板上的电荷在P点产生的电场强度e E E E B Aεσ=+=(4) B板拿走后,A板上电荷将均匀分布在两个表面上,面电荷密度减小为一半。

在P 点产生的场强为两个表面上电荷产生场强的叠加,e E A2εσ=※2、证明:对于两个无限大的平行平面带电导体板来说,(1) 相向的两面上,电荷的面密度总是大小相等而符号相反;(2) 相背的两面上,电荷的面密度总是大小相等而符号相同。

(3) 若左导体板带电+3微库/米2,右导体板带电+7微库/米2,求四个表面上1 2 3 4的电荷。

解:由对称性可知,在每个面上,电荷必定都是均匀分布的,在两板间和两板外的电场必定都是均匀电场,电场强度的方向都与板面垂直。

(1) 作柱形高斯面如图所示,由高斯定理得32320)(10σσσσε-=∴+==⋅⎰SS d E S(2) 根据无限大带电平面均匀电荷产生电场强度的公式和电场强度的叠加原理,导体内任一点P 的电场强度为4143210040302010)(21)(2)(2)(22σσσσσσεεσεσεσεσ==---=-+-+-+=e e e e e E P (3) 应用前述结果及电荷守恒定律SQ S Q )()(4322114132σσσσσσσσ+=+==-= 解得:()()2214122132/521/221m C S Q Q m C S Q Q μσσμσσ=+==-=-=-=由此可知,当Q 1=Q 2时,相向的两面上无电荷分布,相背的两面上电荷等量同号;当Q 1=-Q 2时,相背的两面上无电荷分布,相向的两面上电荷等量异号。

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0 0 0 0 since .`i Ã0T / Q D .`i Ã0T /. I N T P / D .`i Ã0T / .`i Ã0T / D 0. 2.11 Using the Grunfeld (1958) data on investment given as Grunfeld.fil on the Wiley web site this gives the SAS program and output for the estimation of a one-way error component regression. The estimators are OLS, WITHIN, BETWEEN, WALHUS, AMEMIYA, SWAR, NERLOVE and IMLE. The results are reported in Table 2.1.
where there are T ones for the i th individual. This can be rewritten as 2 .` w D ¦¼ ÃT / where `i is the i th column of I N , i.e. `i is a vector that i has one in the i th position and zero elsewhere. ÃT is a vector of ones of dimension T . 0 0 0 (b) .`i Ã0T / P D .`i Ã0T /. I N ÃT Ã0T = T / D .`i Ã0T /. Therefore, in (2.39) " # 2 ¦¼ 1 1 0 1 2 0 0 w D ¦¼ .`i ÃT / P C Q D .`0 Ã0T / 2 2 2 i ¦¹ ¦1 ¦1
VAR_WT=SSQ(WT_RES)/(NT-N-NCOL(X)); /* Equation 2.24 */ V_C_WT=VAR_WT*INV(X‘*Q*X); STD_WT=SQRT(VECDIAG(V_C_WT)); T_WT=WT_BETA/STD_WT; LOOK1=WT_BETA||STD_WT||T_WT; CTITLE={‘PARAMETER’ ‘STANDARD ERROR’ ‘T-STATISTICS’}; RTITLE={‘FIRMV’ ‘CAPITALV’}; PRINT ‘RESULTS OF WITHIN’,, LOOK1(|COLNAME=CTITLE ROWNAME=RTITLE FORMAT=8.5|);
2
Econometric analysis of panel data D .Z 0 Q Z C Z 0 P Z / D . Z 0. Q C P / Z / since Q C P D I N T . (b) GLS on (2.28) yields 2
1 1
. Z 0 Qy C Z 0 P y /
1
X 0 Q Q Qy D . X 0 Q X /
1
e X 0 Qy D þ
(b) For the general linear model y D X þ C u with E .uu 0 / D , a necessary and sufficient condition for OLS to be equivalent to GLS is given by NX D 0; where P NX D I PX and PX D X . X 0 X / 1 X 0 , see Baltagi X0 1 P (1989b). For (2.6), this condition cZ 0
1
Z/
1
Z0
1
y
1 is given by (2.19). where 2 , for i D j and zero otherwise. 2.10 (a) From (2.2) and (2.38), E .u i ;T C S u jt / D ¦¼ The only correlation over time occurs because of the presence of the same individual across the panel. The ¹it ’s are not correlated for different time periods. In vector form 2 .0; : : : ; 0; : : : ; 1; : : : ; 1; : : : ; 0; : : : ; 0/0 w D E .u i ;T C S u / D ¦¼
2 The One-way Error Component Regression Model
NT is idempotent and is therefore its 2.2 (a) Q D I N T P where P D I N J own generalized inverse. The variance–covariance matrix of the disturbance e ¹ D Q ¹ in (2.6) is
0
2P ¦1
3 5 3 5
10
QZ PZ
@
10
0
2Q ¦¹
13 A7 5 1 A
1
0
2P ¦1
Qy Py
@
0
1
2 D4
2 Q =¦¹
0
2 P =¦1
3 5
0
1
2 2 b ŽGLS D [. Z 0 Q Z /=¦¹ C . Z 0 P Z /=¦1 ]
2 2 [. Z 0 Q Z /=¦¹ C . Z 0 P y /=¦1 ]
2 E .e ¹e ¹ 0 / D E . Q ¹¹ 0 Q / D ¦¹ Q 2 . OLS on (2.6) yields with generalized inverse Q =¦¹
b D .X0 Q Q X / þ
1
X 0 Q Qy D . X 0 Q X /
1
X 0 Qy
e given in (2.7). Also, GLS on (2.6) using the generalized inverse which is þ yields b D .X0 Q Q Q X / þ
/* Equation 2.27 */
LOOK1=BW_BETA||STD_BW||T_BW; CTITLE={‘PARAMETER’ ‘STANDARD ERROR’ ‘T-STATISTICS’}; RTITLE={‘INTERCEPT’ ‘FIRMV’ ‘CAPITALV’}; PRINT ‘RESULTS OF BETWEEN’,, LOOK1(|COLNAME=CTITLE ROWNAME=RTITLE FORMAT=8.5|);
*--------- WALLACE & HUSSAIN ESTIMATOR ----------* *--------OF VARIANCE COMPONENTS ---------*; WH_V_V=(OLS_RES‘*Q*OLS_RES)/(NT-N); /* Equation 2.22 with U=OLS residuals */ WH_V_1=(OLS_RES‘*P*OLS_RES)/N; /* Equation 2.21 with U=OLS residuals */
The one-way error component regression model
DATA GRUNFELD; INFILE ‘A:\GRUNFELD.FIL’; INPUT FIRM YEAR INVEST STOCK CAPITAL;
3
PROC IML; USE GRUNFELD; READ ALL INTO TEMP; N=10; T=20; NT=N*T; One=Repeat(1,NT,1); Y=Temp[,3]; X=Temp[,4:5]; Z=One||X; K=NCOL(X); /* I, J, P and Q matrices */ l_t=J(T,1,1); JT=(l_t*l_t‘); Z_U=I(N)@l_t; P=I(N)@JT/T; Q=I(NT)-P; JNT=Repeat(JT,N,N); JNT_BAR=JNT/NT; *--------- OLS ESTIMATORS ------------*; OLS_BETA=INV(Z‘*Z)*Z‘*Y; OLS_RES=Y-Z*OLS_BETA; VAR_REG=SSQ(OLS_RES)/(NT-NCOL(Z)); VAR_COV=VAR_REG*INV(Z‘*Z); STD_OLS=SQRT(VECDIAG(VAR_COV)); T_OLS=OLS_BETA/STD_OLS; LOOK1=OLS_BETA||STD_OLS||T_OLS; CTITLE={‘PARAMETER’ ‘STANDARD ERROR’ ‘T-STATISTICS’}; RTITLE={‘INTERCEPT’ ‘FIRMV’ ‘CAPITALV’}; PRINT ‘RESULTS OF OLS’,, LOOK1(|COLNAME=CTITLE ROWNAME=RTITLE FORMAT=8.5|); *--------- BETWEEN ESTIMATOR ------------------*; BW_BETA=INV(Z‘*P*Z)*Z‘*P*Y; BW_RES=P*Y-P*Z*BW_BETA; VAR_BW=SSQ(BW_RES)/(N-NCOL(Z)); V_C_BW=VAR_BW*INV(Z‘*P*Z); STD_BW=SQRT(VECDIAG(V_C_BW)); T_BW=BW_BETA/STD_BW;
2 N . X 0 Q /. Q =¦¹ / PQ X D 0
using the fact that Q is idempotent, the left-hand side can be written as 2 which is clearly 0, since P NQ X =¦¹ NQ X is the orthogonal projection . X 0 Q/ P of Q X . 2.5 E .u 0 Pu / D E ftr.uu 0 P /g D trf E .uu 0 / P g D tr. P /. From (2.18), P D 2 P since P Q D 0. Hence, from (2.21), E .b 2 / D E .u 0 Pu /=tr. P / D ¦1 ¦1 2 2 0 2 Q / where the tr.¦1 P /=tr. P / D ¦1 . Similarly, E .u Qu / D tr. Q / D tr.¦¹ 2 last equality follows from (2.18) and the fact that Q D ¦¹ Q since P Q D 2 / D E .u 0 Qu /=tr. Q / D tr.¦ 2 Q /=tr. Q / D ¦ 2 . 0. Hence, from (2.22), E .b ¦¹ ¹ ¹ 2.7 (a) OLS on (2.28) yields Ä Â Ã½ 1  à QZ Qy b ŽOLS D . Z 0 Q ; Z 0 P / . Z 0 Q; Z 0 P / PZ Py
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