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Introduction to Econometrics – 3rd Edition – James H. Stock & Mark W. Watson | Complete Test Bank | All Chapters | Verified Questions & Answers | Economics & Statistics Exam Prep

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Designed for economics, econometrics, statistics, finance, business, and data analytics students, this complete test bank provides chapter-based practice questions and verified answers covering the major concepts presented in Introduction to Econometrics, 3rd Edition by James H. Stock and Mark W. Watson. The material supports understanding of statistical inference, regression analysis, hypothesis testing, multiple regression models, panel data, time series analysis, forecasting, instrumental variables, causal inference, and applied econometric methods. This resource is suitable for exam preparation, coursework review, quantitative analysis studies, and strengthening statistical reasoning and econometric problem-solving skills.

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Introduction to Econometrics – 3rd Edition
– James H. Stock & Mark W. Watson |
Complete Test Bank
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Trustedscholar

,Chapter 2
Review of Probability

2.1. (a) Probability distribution function for Y
Outcome (number of heads) Y=0 Y=1 Y=2
Probability 0.25 0.50 0.25

(b) Cumulative probability distribution function for Y

Outcome (number of heads) Y0 0Y1 1Y2 Y2
Probability 0 0.25 0.75 1.0

(c) Y = E(Y ) = (0  0.25) + (1 0.50) + (2  0.25) = 1.00 . F →
d
Fq, .
Tr


Using Key Concept 2.3: var(Y ) = E(Y 2 ) −[E(Y )]2 ,
and
(ui |Xi )
us


so that
var(Y ) = E(Y 2 ) −[E(Y )]2 = 1.50 − (1.00)2 = 0.50.

2.2. We know from Table 2.2 that Pr (Y = 0) = 022, Pr (Y = 1) = 078, Pr ( X = 0) = 030,
Pr( X = 1) = 070. So
(a) Y = E(Y ) = 0  Pr (Y = 0) + 1 Pr (Y = 1)
te


= 0  022 + 1 078 = 078,
X = E( X ) = 0  Pr ( X = 0) + 1 Pr ( X = 1)
= 0  030 + 1 070 = 070
ds


(b)  2 = E[( X −  )2 ]
X X

= (0 − 0.70)  Pr ( X = 0) + (1 − 0.70)2  Pr ( X = 1)
2


= (−070)2  030 + 0302  070 = 021,
 Y2 = E[(Y −  Y )2 ]
ch


= (0 − 0.78)2  Pr (Y = 0) + (1 − 0.78)2  Pr (Y = 1)
= (−078)2  022 + 0222  078 = 01716
ol
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YTREW
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, YTREW



(c)  XY = cov (X , Y ) = E[( X − X )(Y − Y )]
= (0 − 0.70)(0 − 0.78) Pr( X = 0, Y = 0)
+ (0 − 070)(1 − 078) Pr ( X = 0 Y = 1)
+ (1 − 070)(0 − 078) Pr ( X = 1 Y = 0)
+ (1 − 070)(1 − 078) Pr ( X = 1 Y = 1)
= (−070)  (−078)  015 + (−070)  022  015
+ 030  (−078)  007 + 030  022  063
= 0084,
 XY 0084
corr (X , Y ) = = = 04425
 XY 021 01716

2.3. For the two new random variables W = 3 + 6 X and V = 20 − 7Y , we have:
(a) E(V ) = E(20 − 7Y ) = 20 − 7E(Y ) = 20 − 7  078 = 1454,
E(W ) = E(3 + 6X ) = 3 + 6E( X ) = 3 + 6  070 = 72
Tr


(b)  2 = var(3 + 6X ) = 62   2 = 36  021 = 756,
W X

V = var(20 − 7Y ) = (−7)   Y2 = 49  01716 = 84084
2 2


(c) WV = cov(3 + 6X , 20 − 7Y ) = 6  (−7) cov(X , Y ) = −42  0084 = −3528
WV −3528
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corr (W , V ) = = = −04425
WV 756  84084

2.4. (a) E( X 3 ) = 03  (1− p) +13  p = p
(b) E( X k ) = 0k  (1− p) +1k  p = p
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(c) E( X ) = 0.3 , and var(X) = E(X2)−[E(X)]2 = 0.3 −0.09 = 0.21. Thus  = 0.21 = 0.46.
var( X ) = E( X 2 ) −[E( X )]2 = 0.3 − 0.09 = 0.21  = 0.21 = 0.46. To compute the skewness, use
the formula from exercise 2.21:
E( X − )3 = E( X 3 ) − 3[E( X 2 )][E( X )] + 2[E( X )]3
ds


= 0.3 − 3 0.32 + 2  0.33 = 0.084
Alternatively, E( X − )3 =[(1− 0.3)3  0.3] +[(0 − 0.3)3  0.7] = 0.084
Thus, skewness = E( X − )3/ 3 = 0.084/0.463 = 0.87.
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To compute the kurtosis, use the formula from exercise 2.21:
E( X − )4 = E( X 4 ) − 4[E( X )][E( X 3 )] + 6[E( X )]2 [E( X 2 )] − 3[E( X )]4
= 0.3 − 4  0.32 + 6  0.33 − 3 0.34 = 0.0777
Alternatively, E( X − )4 =[(1− 0.3)4  0.3] +[(0 − 0.3)4  0.7] = 0.0777
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Thus, kurtosis is E( X − )4/ 4 = 0.0777/0.464 =1.76
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, 4 Stock/Watson • Introduction to Econometrics, Third Edition


2.5. Let X denote temperature in F and Y denote temperature in C. Recall that Y = 0 when X = 32 and
Y =100 when X = 212; this implies Y = (100/180)  ( X − 32) or Y = −17.78 + (5/9)  X. Using Key
Concept 2.3, X = 70oF implies that  Y = −17.78 + (5/9)  70 = 21.11C, and X = 7oF implies
Y = (5/9)  7 = 3.89C.

2.6. The table shows that Pr ( X = 0, Y = 0) = 0037, Pr ( X = 0, Y = 1) = 0622,
Pr ( X = 1, Y = 0) = 0009, Pr ( X = 1, Y = 1) = 0332, Pr ( X = 0) = 0659, Pr ( X = 1) = 0341,
Pr(Y = 0) = 0046, Pr (Y = 1) = 0954.
(a) E(Y ) = Y = 0  Pr(Y = 0) + 1 Pr (Y = 1)
= 0  0046 +1 0954 = 0954
#(unemployed)
(b) Unemployment Rate =
#(labor force)
= Pr (Y = 0) = 1 − Pr(Y = 1) = 1 − E(Y ) = 1 − 0954 = 0.046
(c) Calculate the conditional probabilities first:
Pr ( X = 0, Y = 0) 0037
Tr


Pr (Y = 0| X = 0) = = = 0056,
Pr ( X = 0) 0659
Pr ( X = 0, Y = 1) 0622
Pr (Y = 1| X = 0) = = = 0944,
Pr ( X = 0) 0659
Pr ( X = 1, Y = 0) 0009
us


Pr (Y = 0| X = 1) = = = 0026,
Pr ( X = 1) 0341
Pr ( X = 1, Y = 1) 0332
Pr (Y = 1| X = 1) = = = 0974
Pr ( X = 1) 0341
The conditional expectations are
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E(Y|X = 1) = 0  Pr (Y = 0| X = 1) +1 Pr (Y = 1| X = 1)
= 0  0026 + 1 0974 = 0974,
E(Y|X = 0) = 0  Pr (Y = 0| X = 0) + 1 Pr (Y = 1|X = 0)
ds


= 0  0056 +1 0944 = 0944
(d) Use the solution to part (b),
Unemployment rate for college graduates = 1 − E(Y|X = 1) = 1 − 0.974 = 0.026
Unemployment rate for non-college graduates = 1 − E(Y|X = 0) = 1 − 0.944 = 0.056
(e) The probability that a randomly selected worker who is reported being unemployed is a
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college graduate is
Pr ( X = 1, Y = 0) 0009
Pr ( X = 1|Y = 0) = = = 0196
Pr (Y = 0) 0046

The probability that this worker is a non-college graduate is
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Pr ( X = 0|Y = 0) = 1 − Pr ( X = 1|Y = 0) = 1 − 0196 = 0804
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©2011 Pearson Education, Inc. Publishing as Addison Wesley


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