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Solutions Manual: Introductory Econometrics 1e

ANSWERS TO SELECTED QUESTIONS FROM EXERCISES SECTIONS OF THE TEXTBOOK (Chapters 2-8 & Chapter 10) (Note: By accessing and using this document you agree that you will not share or discuss this document with anyone else not enrolled in MAE256 in T1 2020 at Deakin)

Chapter 02: Basic mathematical tools Solutions to Problems 1 The table is extended with required calculatyions as follows, Observation

Xi

Yi

Xi2

Xi Yi

X

1

2

1

4

2

(2–2.6)2=.36

2

0

3

0

0

(0–2.6)2=6.76

3

–1

–2

1

2

(–1–2.6)2=12.96

4

5

4

25

20

(5–2.6)2=5.76

5

7

3

49

21

(7–2.6)2=19.36

5

X

i

 13

i 1

5

5

Y  9

X

i

2 i

 79

i1

i 1

5

XY i

i1

i

 45

i

X

5

 X i1

i

2

 X

2

=45.2

5

Note that X  (i)

5

 Xi i 1

5

 2.6 and Y 

As the tabole shows,

Y i 1

i

5

5

5

i1

i1

 1.8

 X i =13 and Yi = 9

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Solutions Manual: Introductory Econometrics 1e 2

5

(ii)

From the table,

(iii)

 X iYi

5

i 1

 5  2 2 X = 79 and  i   X i  =(13) =169 i1  i 1 

 5  5  =45 and   Xi   Yi  =13*9= 117  i 1  i 1  

5

(iv)

5

5



i 1



  X i Yi  =13+9 = 22 and  X i  Yi  =13+9=22. Actually, these are same  i 1

i 1

quantities. (v)

(vi)

5

5

5

i 1

i1

i1

  X i Yi  =  X i   Yi  13  9  4 5

5

i 1

i 1

 2 Xi = 2 X i  2*13  26 5

(vii)

 2 = 2*5=10 i1

(viii)

  X i  X  =0 5

i 1

4. If price = 15 and income = 200, quantity = 20 – 1.8(15) + .03(200) = –1, which is nonsense. This shows that linear demand functions generally cannot describe demand over a wide range of prices and income.

5. (i) The percentage point change is 6 – 4 = 2, or a two percentage point increase in the unemployment rate. (ii) The percentage change in the unemployment rate is 100[(6 – 4)/4] = 50%. i.e., unemployment increased by 50%.

6. The majority shareholder is referring to the percentage point increase in the stock return, while the CEO is referring to the change relative to the initial return of 15%. To be precise, the shareholder should specifically refer to a 3 percentage point increase.

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2

Solutions Manual: Introductory Econometrics 1e 9. (i) The relationship between yield and fertilizer is graphed (see over page).

y ield

122

121

120 0

50

100

f ertilizer

(ii) Compared with a linear function, the function

yield = 120 + .13

fertilizer

has a diminishing marginal effect, and the slope approaches zero as fertilizer gets large. The initial kilogram of fertilizer has the largest effect, and each additional kilogram has an marginal effect smaller than the previous kilogram. 10. (i) The value 20.5 is the intercept in the equation, so it literally means that if age = 0 then the BMI is 20.5. Of course, age = 0, measured in years would indicate the BMI of 20.5 as shown by that the intercept is the BMI of new born babies or precisely, babies less than a year old. The intercept by itself is not much of interest since body fat 1-of babies less than a year old is not usually a concern. Also, the intercept should ideally reflect a dataset on age and BMI of the adult population, so, by itself, 20.5 is not of much interest.

(ii) We use calculus to obtain the maximum BMI: dBMI  .2  .004 Age dAge

and

d 2 BMI  .004  0. dAge2

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3

Solutions Manual: Introductory Econometrics 1e Hence. The BMI function has a maximum. Letting, the first derivative equal to 0,

dBMI  .2  .004Age  0 dAge .2  50 Age  .004 Therefore, BMI is maximum at the age of 50 years.

(iii) The following graph shows the solution rounded to the nearest integer: BMI and AGE

26 25 24

BMI

23 22 21 20 19 0

20

40

60

80

100

120

AGE

(iv) It is not at all realistic to think that BMI and age will have a determinictic relationship. BMI is also scientifically measured in accordance with the height of a person. Besides there are many other factors that affect BMI of a person, such as a person’s general lifestyle, eating habits, health awareness, income. Multiple regression analysis allows for many observed factors to affect a variable like BMI, and also recognizes that there are unobserved factors that are important and that we can never directly account for.

Multiple Choice Questions: 1. 2. 3. 4. 5. 6.

d b d a c b © Cengage Learning 2017. All rights reserved.

4

Solutions Manual: Introductory Econometrics 1e 7. 8. 9. 10.

c c a a

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5

Solutions Manual: Introductory Econometrics 1e

Chapter 03: Fundamentals of statistics: a review Solutions to Problems 1. (i)

P(0< Z < 1) = .8413–.5=.3413

(ii)

P(–1< Z < 1) = .8413–.1587 = .6826

(iii)

P(Z > 2.55) = 1–.9946 =.0054

(iv)

P(Z < –1.60) = .0548

2. (i) P(X ≤ 6) = P[(X – 5)/2 ≤ (6 – 5)/2] = P(Z ≤ .5)  .692, where Z denotes a Normal (0,1) random variable. [We obtain P(Z ≤ .5) from Table A.1.] (ii) P(X > 4) = P[(X – 5)/2 > (4 – 5)/2] = P(Z > .5) = P(Z  .5)  .692. (iii) P(|X – 5| > 1) = P(X – 5 > 1) + P(X – 5 < –1) = P(X > 6) + P(X < 4)  (1 – .692) + (1 – .692) = .616, where we have used answers from parts (i) and (ii).

3. Let X denote family income. Then given the information we find the required probabilities as shown below, (i)

P ( X  30000)  P (Z 

30000  50000 )  P (Z   2)  .0228 10000

(ii)

P ( X  70000)  P (Z 

70000  50000 )  P (Z  2)  1  .9772  .0228 10000

4. Let X representds the marks obtained by the students and X* denote the lowest mark that will be awarded an A grade. Given that X ) we first find out the value of standard normal variable Z, such that the probability of Z exceeding this value is 10% or .10. That is, we need to find the value of Z that leaves out 10% area under the right tail of the Z distribution. From the appendix on areas under the standard normal distribution, we find that the relevant value of Z is 1.28 (approximately). Hence we get,

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6

Solutions Manual: Introductory Econometrics 1e

X * 70  1.28 6  X *  (1.28)  6  70  77.68 Hence, the lowest mark that will be awarded an A grade is 77.68 or 78 (apporximately).

7. Tossing three coins give the following sample space or the possible combinations of events, HHH, HHT, HTH, HTT, THH, THT,TTH, TTT Since P(H)= .5 and P(T) =.5 hence probability of each event, say of the event HTH is P(HTH) = .5*.5*.5=.125 Given the X represents the number of tails, we can construct the probability distribution of X, that takes the values of 0 (no tail), 1 (one tail), 2(two tails) and 3 (three tails). X Prob.

0 .125

1 .375

2 .375

3 .125

E(X) = 0*.125 + 1*.375 + 2*.375 + 3*.125 = 1.5 E(X2) = 02*.125 + 12*.375 + 22*.375 + 32*.125 = 3

Profit = (X2+X) – 5 E(Profit) = E(X2+X) –5 = E(X2)+E(X) – 5 = 3 + 1.5 –5 = –.5 hence there is a loss of 50 cents.

9. (i) P(male wins) = 40/60 = .667 apx

10 P (married & male) 10 60 10  60  *   .25 (ii) P(married/ male) = 40 60 40 40 P( male) 60 11. (i) This is just a special case of what we covered in the text, with n = 4: E( Y ) = µ and Var( Y ) = 2/4. (ii) E(W) = E(Y1)/8 + E(Y2)/8 + E(Y3)/4 + E(Y4)/2 = µ[(1/8) + (1/8) + (1/4) + (1/2)] = µ(1 + 1 + 2 + 4)/8 = µ, which shows that W is unbiased. Because the Yi are independent,

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Solutions Manual: Introductory Econometrics 1e Var(W) = Var(Y1)/64 + Var(Y2)/64 + Var(Y3)/16 + Var(Y4)/4 = 2[(1/64) + (1/64) + (4/64) + (16/64)] = 2(22/64) =  2(11/32). (iii) Because 11/32 > 8/32 = 1/4, Var(W) > Var( Y ) for any 2 > 0, so Y is preferred to W because each is unbiased.

12. (i) E(Wa) = a1E(Y1) + a2E(Y2) + + anE(Yn) = (a1 + a2 + + a2 + + an = 1 for unbiasedness.

(ii) Var(Wa) = a12 Var(Y1) + a22 Var(Y2) +

+ an)µ. Therefore, we must have a1

+ an2 Var(Yn) = (a12 + a22 +

+ an2 ) 2.

(iii) From the hint, when a1 + a2 + + an = 1 – the condition needed for unbiasedness of Wa 2 2 2 – we have 1/n  a1 + a2 + + an2 ) = + an . But then Var( Y ) =  2/n   2( a12 + a22 + Var(Wa).

Multiple Choice Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

a c b a c b d d a b c

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8

Solutions Manual: Introductory Econometrics 1e

Chapter 04: The simple regression model Solutions to Problems 1. (i) Income, age, and family background (such as number of siblings) are just a few possibilities. 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) A simple regression as shown will not be sufficient 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. However as per one of the basic assumptions of simple regression, we know that u is not to be correlated with the explanatory variable. Hence, if u is correlated with educ then E(u|educ)  0, and so SLR.4 fails. 2. The estimated regression models are shown in the following Figure.

It is clear that the estimates of the intercept,

1

is same for both the samples. However, the

slopes are different. The estimates of the slope  2 from the first and second samples are 0.75 and 0.95, respectively. The above figure shows that while the fitted line representing the second sample is steeper due to its higher slope estimate, both the lines have the same intercept (1.2).

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9

Solutions Manual: Introductory Econometrics 1e 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).

(iv) Since both the independent variable (cons) and the independent variable (income) are measured in natural logarithm, the slope coefficient represents the elasticity. The estimated slope of 1.839 indicates that as income increases by 1%, consumption increases by 1.839%.

6. (i) Yes. If living closer to an incinerator depresses housing prices, then being farther away 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)]. 8. (i) The scatter plot of the two variables is shown below:

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10

Solutions Manual: Introductory Econometrics 1e

10 9 8 7

y

6 5 4 3 2 1 0 0

1

2

3

4

5

6

7

8

9

10

x From the scatter plot, it is clear that on average, the relationship between the two variables, x and y, is positive.

(ii) The estimated regression model (SRF) using OLS is given below:

yˆ  1.2086  0.6871x The fitted regression line is plotted below:

(iii) First we calculate x  6.1 . Now when x  x  6.1Then yˆ  1.2086  0.6871(6.1)  5.4 . One could easily see and verify that this predicted value of y is exactly the value of y , that is, in this case

yˆ  y  5.4 . © Cengage Learning 2017. All rights reserved.

11

Solutions Manual: Introductory Econometrics 1e (iv) We put various values of x in the SRF yˆ  1.2086  0.6871x and calculate the predicted y’s. After

ˆ i and square the terms uˆi2 and obtain this we calculate uˆ i  yi  y

10

 uˆ

2

for 10 sets of observations.

i1

This is shown below,

yˆ  1.2087  0.6871x

uˆi  yi  yˆ i

uˆi2

2.583 3.270 5.331 4.644 4.644 6.706 6.018 6.706 7.393 6.706

–0.583 0.730 –2.331 1.356 –0.644 –1.706 2.982 0.294 0.607 –0.706

0.340 0.533 5.435 1.838 0.415 2.909 8.890 0.087 0.369 0.498 10

 uˆ

2

 21.313

i 1

10

Hence, the sum of squared residuals is

 uˆ

2

 21.313

i 1

(v) Given the values of SSR obtained in part (iv), we use the following formula for calculating R2

R2  1 

SSR SST

Let’s calculate, n

2 2 2 2 2 2 2 SST   ( y i  y )  (2  5.4)  (4  5.4)  (3  5.4)  (6  5.4)  (4  5.4)  (5  5.4) i 1

 (9  5.4)2  (7  5.4)2  (8  5.4)2  (6  5.4)2  44.4 Therefore, R 2  1

21.313  0.51998  0.52(apx ) 44.4

The value of R2 indicates that about 52 percent of the variations in the dependent variable y is explained by the variations in the independent variable x. The estimated model seems to be a reasonable fit of the data.

11

(i) We would want to randomly assign the number of hours in the preparation course so that

hours is independent of other factors that affect performance on the VCE exam. Then, we would collect information on VCE score for each student in the experiment, yielding a data set

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12

Solutions Manual: Introductory Econometrics 1e

{(vce i, hoursi) : i  1,..., n} , where n is the number of students we can afford to have in the study. We should try to get as much variation in hoursi as is feasible. (ii) Here are two factors: innate ability and general health during the days of the exam. If we think students with higher intelligence would require to spend less hours for preparing for the VCE exam, then ability and hours will be negatively correlated. Ruling out chronic health problems, health on the days of the exam should be roughly uncorrelated with hours spent in a preparation course. (iii) If preparation courses are effective, 1 should be positive: other factors equal, an increase in hours should increase vce. (iv) The intercept,  0 , has a useful interpretation in this example: because E(u) = 0,  0 is the average VCE score for students in the population with hours = 0. 12.

(i) As the number of finance and economics graduates working in a company increases by 1 person, the dividend per share increases by $22.8 dollars. When the number of business graduates hired by a company is zero, the dividend per share is $5.4.

(ii)

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13

Solutions Manual: Introductory Econometrics 1e

SSR 1399037.44   11282.56 . To calculate n 2 126  2 we use the expression for variance of ˆ . We know We first calculate ˆ  2

(iii)

n

x

i

x

2

i 1

1

Var ( ˆ1 ) 

ˆ n

 x

2

x 

2

i

i 1

n

Or,   xi  x   i 1

2



11282.56 n

 x

i

x



 (6.8)2

2

i 1

11282.56  244 46.24

2 Hence, ˆ  11282.56 and

n

x i 1

i

 x   244 2

(iv) A log-level model will show the percentage change in dividend per share due to the hiring of one additional business graduate by a company.

Multiple Choice Questions 1. 2. 3. 4. 5. 6. 7. 8.

b c d d a c a b

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14

Solutions Manual: Introductory Econometrics 1e

Chapter 05: Multiple regression analysis: estimation Solutions to Problems 2. (i) If adults trade off sleep for work, more work implies less sleep (other things equal), so (ii) The signs of

2

and

 3 are not obvious, at least to me.

 1 < 0.

One could argue that more

educated people like to get more out of life, and so, other things equal, they sleep less (

 2 < 0).

The

relationship between sleeping and age is more complicated than this model suggests, and economists are not in the best position to judge such things. (iii) Since totwrk is in minutes, we must convert five hours into minutes: totwrk = 5(60) = 300. Then sleep is predicted to fall by .148(300) = 44.4 minutes. For a week, 45 minutes less sleep is not an overwhelming change.

(iv) More education implies less predicted time sleeping, but the effect is quite small. If we assume the difference between college and high school is four y...