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程序代写案例-ECON61001
时间：2022-01-09
University of Manchester
ECON61001: Econometric Methods
Final Exam
January 2021
Release date/time: 28/01/21, 14.00hrs GMT
Submission deadline: 30/01/21, 14.00hrs GMT
Instructions:
• You must answer all five questions in Section A and two out of the four ques-
tions in Section B. If you answer more questions than are required and do not
indicate which answers should be ignored, we will mark the requisite number of
answers in the order in which they appear in your answer submission: answers
beyond that number will not be considered.
• Your answers could be typed or hand-written (and scanned to a single pdf file
that can be submitted) or a combination of a typed answer with included images
of algebra or figures.
• Where relevant, questions include word limits. These are limits, not targets. Ex-
cellent answers can be shorter than the word limit. If you go beyond the word
limit the additional text will be ignored. Where a question includes a word limit
you HAVE to include a word count for your answer (excluding formulae). You
could use https://wordcounter.net to obtain word counts.
• Candidates are advised that the examiners attach considerable importance to
the clarity with which answers are expressed.
• You must correctly enter your registration number and the course code on
your answer.
c©The University of Manchester, 2021.
ECON61001
SECTION A
1. Suppose a researcher is interested in the following linear regression model
yi = x
′
iβ0 + ui, i = 1, 2, . . . N,
where xi = (1, x2,i, x3,i, x4,i)
′ and β0 = (β0,1, β0,2, β0,3, β0,4)
′ . Given the context, the
researcher is able to assume that {ui, x
′
i}
N
i=1 form a sequence of independently
and identically distributed random vectors with E[xix
′
i] = Q, a finite, positive
definite matrix of constants, E[ui|xi] = 0 and V ar[ui|xi] = σ
2
0 . Therefore, she
estimates the model via Ordinary Least Squares (OLS), obtaining the following
estimated equation
yˆi = 1.0213
(0.1416)
− 0.0020
(0.0005)
x2,i − 0.0208
(0.0080)
x3,i + 0.0095
(0.0088)
x4,i,
where the number in parenthesis is the conventional OLS standard error for the
coefficient in question. The OLS estimator of σ20 in this model is σˆ
2
N = 0.0081.
Given these results, the researcher concludes that β0,4 = 0 and so decides to
estimate the model via OLS with x4,i excluded, obtaining the following estimated
equation
yˆi = 1.1485 − 0.0020x2,i − 0.0211x3,i. (1)
Given that the sample size is N = 108 in both estimations, calculate the OLS
estimator of the error variance σ20 from the estimation in (1). Be sure to care-
fully explain your calculations. Hint: consider the F-statistic for testing β0,4 =
0. [8 marks]
2. Suppose it is desired to predict zt using zˆt = w
′
tγ where wt is a vector of ob-
servable variables and γ is a vector of constants that needs to be specified.
The choice of γ associated with the linear projection of zt on wt is γ0 where
E[(zt − w
′
tγ0)wt] = 0.
(a) What optimality property does zˆot = w
′
tγ0 possess? (Word limit: 50) [1mark]
(b) Now consider the regression model zt = w
′
tγ0 + vt. Is wt contempora-
neously exogenous or strictly exogenous for estimation of γ0 in this model?
Justify your answer. (Word limit: 150) [4 marks]
(c) Suppose the model in part (b) is dynamically complete. What optimal-
ity property does zˆot possess? Briefly justify your answer. (Word limit:
150) [3 marks]
Continued over
1
ECON61001
SECTION A continued
3.(a) Let A be n × n nonsingular symmetric matrix. Show that if A is positive definite
then A−1 is positive definite. Hint: AA−1 = In. [4 marks]
3.(b) Consider the classical linear regression model
y = Xβ0 + u, (2)
whereX is the T×k observable data matrix that is fixed in repeated samples with
rank(X) = k, and u is a T × 1 vector with E[u] = 0 and V ar[u] = σ20IT where σ
2
0
is an unknown positive finite constant. Let βˆT be the OLS estimator of β0 based
on (2) and βˆR,T be the Restricted Least Squares (RLS) estimator of β0 based on
(2) subject to the restrictions Rβ0 = r where R is a nr × k matrix of specified
constants with rank(R) = nr and r is a specified nr × 1 vector of constants.
Assuming the restrictions are correct, prove that βˆR,T is at least as efficient as
βˆT . Hint: you may quote the formula for the variance-covariance matrix of the
OLS and RLS estimators without proof; you may also take advantage of the
stated result in part (a). [4 marks]
4. Consider the linear regression model
y = Xβ0 + u,
where y and u are T × 1 vectors, X is T × k matrix, and β0 is the k × 1 vector
of unknown regression coefficients. Assume that X is fixed in repeated sam-
ples with rank(X) = k, and u ∼ N ( 0, σ20IT ) where σ
2
0 is an unknown positive
constant. Let θˆT denote the maximum likelihood estimator of the unknown pa-
rameter vector θ0 = (β
′
0, σ
2
0)
′. Derive the information matrix for this model. Hint:
you may state the form of the log likelihood function and score function for this
model without proof. [8 marks]
5. Consider the model
yi = x
′
iβ0 + ui, i = 1, 2, . . . , N,
where β0 is the k × 1 vector of unknown regression coefficients, {(x
′
i, ui)}
N
i=1
is a sequence of independently and identically distributed random vectors with
E[ui|xi] = 0, V ar[ui|xi] = σ
2
0, an unknown finite positive constant and E[xix
′
i] =
Q, a finite positive definite matrix of constants. Let σˆ2N be the OLS estimator of
σ20. Show that N
1/2(σˆ2N − σ
2
0)
d
→ N( 0, µ4 − σ
4
0) where µ4 = E[u
4
i ].
Hint: You may assume that: N−1
∑N
i=1 xix
′
i
p
→ Q; (ii) N−1/2
∑N
i=1 vi
d
→ N(0,Ω)
where vi = (u
2
i − σ
2
0, x
′
iui)
′, and Ω = V ar[vi] is a finite, positive definite
(k + 1) × (k + 1) matrix whose elements you must specify as needed to develop
your answer. [8 marks]
Continued over
2
ECON61001
SECTION B
6. Consider the regression model
yi = x
′
iβ0 + ui, i = 1, 2, . . . , N,
where β0 is the k × 1 vector of unknown regression coefficients, {(x
′
i, ui)}
N
i=1
is a sequence of independently and identically distributed random vectors with
E[ui|xi] = 0, V ar[ui|xi] = σ
2
0, an unknown finite positive constant and E[xix
′
i] =
Q, a finite positive definite matrix of constants. You may further assume that: (i)
N−1
∑N
i=1 xix
′
i
p
→ Q; (ii) N−1/2
∑N
i=1 xiui
d
→ N(0, σ20Q).
Let βˆR,N denote the RLS estimator based on the linear restrictionsRβ = r where
R is a nr × k matrix of pre-specified constants with rank equal to nr and r is a
nr × 1 vector of pre-specified constants, and let λˆN be the vector of Lagrange
Multipliers associated with this RLS estimation. Assuming Rβ0 = r, answer the
following questions.
(a) Show that N1/2(βˆR,N − β0)
d
→ N ( 0, VR ) where
VR = σ
2
0
(
Q−1 − Q−1R′(RQ−1R′)−1RQ−1
)
.
Hint: you may quote the formulae for βˆR,N and βˆN , the Ordinary Least
Squares estimator of β0, without proof. [10 marks]
(b) A colleague proposes testing H0 : Rβ0 = r versus H1 : Rβ0 6= r using the
decision rule of the form: reject H0 at the (approximate) 100α% significance
level if
λˆ′NMN λˆN > cnr(1− α)
where cnr(1− α) is the 100(1 − α)
th percentile of the χ2nr distribution. How-
ever, your colleague is unsure what the matrix MN should be in order that
this decision rule has the properties implied by the stated significance level.
Provide a suitable choice of MN , being sure to justify your choice carefully.
Hint: you may quote without proof: (i) the formulae for λˆN and βˆN ; (ii) that
both the OLS and RLS estimators of σ20 are consistent under the conditions
of the question. [20 marks]
Continued over
3
ECON61001
SECTION B continued
7.(a) Let {vt}
T
t=−3 be a weakly stationary time series process. Consider the following
statistic,
ρˆ4,T =
∑T
t=1 vtvt−4∑T
t=1 v
2
t
.
Let {εt}
∞
t=−∞ denote a sequence of independently and identically distributed
(i.i.d.) random variables with mean zero and variance σ2ε .
(i) Assume that vt = εt. Show that T
1/2ρˆ4,T
d
→ N(0, 1). [6 marks]
(ii) Assume that
vt = θ4vt−4 + εt,
where |θ4| < 1. What is the probability limit of ρˆ4,T as T → ∞? Be sure
to justify your answer carefully. Hint: vt has the following representation,
vt =
∑
∞
i=0 θ
i
4εt−4i. [9 marks]
7.(b) A researcher wishes to test the simple efficient-markets hypothesis in the foreign
exchange market. Let st = ln(St) and ft,n = ln[Ft,n], where St and Ft,n are the
levels of the spot exchange rate at time t and the n−period forward exchange
rate at time t. The simple efficient-markets hypothesis is that ft,n = E[st+n | It]
where It is the information set at time t which for the purposes of this question
can be taken to be It = {st, ft,n, st−1, ft−1,n, st−2, ft−2,n, . . .}. Using daily spot and
thirty-day forward exchange rate data for the US dollar UK pound exchange rate,
the researcher estimates the model,
yt+n = x
′
tβ0 + ut,n, (3)
where yt+n = st+n − ft,n, xt is the 3× 1 vector given by
x′t = ( 1, st − ft−n,n, st−1 − ft−1−n,n ) ,
n = 30 and ut,n is the error term. If the simple efficient markets hypothesis
holds in this foreign exchange market then E[ut,n | It] = 0 and the regression
coefficients in (3) satisfy a set of restrictions denoted here by g(β0) = 0 where
g( · ) is ng × 1 vector.
(i) What is g(β0)? Briefly justify your answer. (Word limit: 75) [4 marks]
Continued over
4
ECON61001
SECTION B continued
7.(b) continued
(ii) The researcher tests H0 : g(β0) = 0 versus H1 : g(β0) 6= 0 using the test
statistic
ST = Tg(βˆT )
′
(
G(βˆT ) Vˆβ G(βˆT )
′
)
−1
g(βˆT ), (4)
where G(βˆT ) = ∂g(β)/∂β
∣∣
β=βˆT
. Assuming T 1/2(βˆT − β0)
d
→ N(0, Vβ) and
Vˆβ
p
→ Vβ , write down a suitable decision rule for this test. If ST = 8.2 then
what is the outcome of the test? [3 marks]
(iii) Since the yt+n is a financial variable, the researcher is concerned that the
errors may exhibit autoregressive conditional heteroscedasticity and so has
calculated Vˆβ in (4) using White’s heteroscedasticity robust estimator. Given
this information, do you have any concerns about the test in part (ii)? If so
then explain your concerns and how you would modify the test to address
these concerns. (Word limit: 350) [8 marks]
Continued over
5
ECON61001
SECTION B continued
8. Consider the linear regression model
y1,i = γ0y2,i + z
′
1,iδ0 + u1,i = x
′
iβ0 + u1,i,
where x′i = (y2,i, z
′
1,i), β0 = (γ0, δ
′
0)
′ and assume that
y2,i = z
′
iη0 + u2,i, (5)
where y1,i and y2,i are observable random variables, zi = (z
′
1,i, z
′
2,i)
′ is a random
vector of observable variables, u1,i and u2,i are the error terms (unobservable
scalar random variables), γ0 is an unknown scalar parameter, and δ0, and η0 are
vectors of unknown parameters. Suppose there is a sample of N observations,
and let yˆ2,i denote the predicted value of y2,i based on Ordinary Least Squares
(OLS) estimation of (5). Define xˆi = (yˆ2,i, z
′
1,i)
′. Let X be the N × k matrix with ith
row x′i, Xˆ be the N × k matrix with i
th row xˆ′i, Z be the N × q matrix with i
th row
z′i and y1 be the N × 1 vector with i
th element y1,i. Consider the following three
estimators of β0:
• βˆ1 = (Xˆ
′Xˆ)−1Xˆ ′y1;
• βˆ2 = (Xˆ
′X)−1Xˆ ′y1;
• βˆ3 = {X
′Z(Z ′Z)−1Z ′X }
−1
X ′Z(Z ′Z)−1Z ′y1.
(a) Show that βˆ1 = βˆ2 = βˆ3. [15 marks]
(b) Let φˆ = (βˆ ′, θˆ)′ be the OLS estimator of φ0 = (β
′
0, θ0)
′ based on the model
y1,i = x
′
iβ0 + θuˆ2,i + “error”
where uˆ2,i is the i
th element of uˆ2, theN×1 vector of residual fromOLS estimation
of (5). Via an application of the Frisch-Waugh-Lovell Theorem or otherwise, show
that βˆ = βˆ1.
[15 marks]
Continued over
6
ECON61001
SECTION B continued
9.(a) Let {(yi, x
′
i)}
N
i=1 be a sequence of independently and identically distributed (i.i.d.)
random vectors. Suppose that yi is a dummy variable and so has a sample space
of {0, 1}. Consider the model
yi = x
′
iβ0 + ui.
(i) Assume that E[ui|xi] = 0. Derive the Generalized Least Squares (GLS)
estimator of β0 in this model? Hint: you may quote the generic formula for
the GLS estimator that is, βˆGLS = (X
′Σ−1X)−1X ′Σ−1y, but you must derive
Σ for this model. [6 marks]
(ii) Is your answer to part (i) a feasible or infeasible GLS estimator? If infeasible
then suggest a feasible GLS estimator. Do you foresee any potential prob-
lems in implementing your proposed Feasible GLS estimator? [4 marks]
(b) Let {Vi}
N
i=1 be a sequence of i.i.d. Bernoulli random variableswith P (Vi = 1) = θ0.
We assume here that θ0 ∈ (0, 1) and that our sample size is large enough for both
outcomes to occur.
(i) Derive the Wald, Likelihood Ratio and LagrangeMultiplier statistics for test-
ing H0 : θ0 = θ∗ against H1 : θ0 6= θ∗. Hint: you may quote the form of the
log likelihood function, the score equation and the formula for the maximum
likelihood estimator for this model without proof. [16 marks]
(ii) Given that N = 100 and the sample contains 55 outcomes that are one,
use your statistics in part (i) to test the hypothesis H0 : θ0 = 0.5 against
H1 : θ0 6= 0.5 at the 5% significance level. [4 marks]
END OF EXAMINATION
7
1 TABLE 1: PERCENTAGE POINTS FOR THE T DISTRIBUTION
1 Table 1: Percentage Points for the t distribution
Student’s t Distribution Function for Selected Probabilities
The table provides values of tα,v where Pr(T ≤ tα,v) = α and T ∼ tv
α 0.750 0.800 0.900 0.950 0.975 0.990 0.995 0.9975 0.999 0.9995
ν Values of tα,v
1 1.000 1.376 3.078 6.314 12.706 31.821 63.657
2 0.816 1.061 1.886 2.920 4.303 6.965 9.925
3 0.765 0.978 1.638 2.353 3.182 4.541 5.841
4 0.741 0.941 1.533 2.132 2.776 3.747 4.604
5 0.727 0.920 1.476 2.015 2.571 3.365 4.032 4.773
6 0.718 0.906 1.440 1.943 2.447 3.143 3.707 4.317 5.208
7 0.711 0.896 1.415 1.895 2.365 2.998 3.499 4.029 4.785 5.408
8 0.706 0.889 1.397 1.860 2.306 2.896 3.355 3.833 4.501 5.041
9 0.703 0.883 1.383 1.833 2.262 2.821 3.250 3.690 4.297 4.781
10 0.700 0.879 1.372 1.812 2.228 2.764 3.169 3.581 4.144 4.587
11 0.697 0.876 1.363 1.796 2.201 2.718 3.106 3.497 4.025 4.437
12 0.695 0.873 1.356 1.782 2.179 2.681 3.055 3.428 3.930 4.318
13 0.694 0.870 1.350 1.771 2.160 2.650 3.012 3.372 3.852 4.221
14 0.692 0.868 1.345 1.761 2.145 2.624 2.977 3.326 3.787 4.140
15 0.691 0.866 1.341 1.753 2.131 2.602 2.947 3.286 3.733 4.073
16 0.690 0.865 1.337 1.746 2.120 2.583 2.921 3.252 3.686 4.015
17 0.689 0.863 1.333 1.740 2.110 2.567 2.898 3.222 3.646 3.965
18 0.688 0.862 1.330 1.734 2.101 2.552 2.878 3.197 3.610 3.922
19 0.688 0.861 1.328 1.729 2.093 2.539 2.861 3.174 3.579 3.883
20 0.687 0.860 1.325 1.725 2.086 2.528 2.845 3.153 3.552 3.850
21 0.686 0.859 1.323 1.721 2.080 2.518 2.831 3.135 3.527 3.819
22 0.686 0.858 1.321 1.717 2.074 2.508 2.819 3.119 3.505 3.792
23 0.685 0.858 1.319 1.714 2.069 2.500 2.807 3.104 3.485 3.768
24 0.685 0.857 1.318 1.711 2.064 2.492 2.797 3.091 3.467 3.745
25 0.684 0.856 1.316 1.708 2.060 2.485 2.787 3.078 3.450 3.725
26 0.684 0.856 1.315 1.706 2.056 2.479 2.779 3.067 3.435 3.707
27 0.684 0.855 1.314 1.703 2.052 2.473 2.771 3.057 3.421 3.690
28 0.683 0.855 1.313 1.701 2.048 2.467 2.763 3.047 3.408 3.674
29 0.683 0.854 1.311 1.699 2.045 2.462 2.756 3.038 3.396 3.659
30 0.683 0.854 1.310 1.697 2.042 2.457 2.750 3.030 3.385 3.646
40 0.681 0.851 1.303 1.684 2.021 2.423 2.704 2.971 3.307 3.551
50 0.679 0.849 1.299 1.676 2.009 2.403 2.678 2.937 3.261 3.496
60 0.679 0.848 1.296 1.671 2.000 2.390 2.660 2.915 3.232 3.460
70 0.678 0.847 1.294 1.667 1.994 2.381 2.648 2.899 3.211 3.435
80 0.678 0.846 1.292 1.664 1.990 2.374 2.639 2.887 3.195 3.416
90 0.677 0.846 1.291 1.662 1.987 2.368 2.632 2.878 3.183 3.402
100 0.677 0.845 1.290 1.660 1.984 2.364 2.626 2.871 3.174 3.390
110 0.677 0.845 1.289 1.659 1.982 2.361 2.621 2.865 3.166 3.381
120 0.677 0.845 1.289 1.658 1.980 2.358 2.617 2.860 3.160 3.373
∞ 0.674 0.842 1.282 1.645 1.960 2.326 2.576 2.808 3.090 3.297
8
2 TABLE 2: PERCENTAGE POINTS FOR THE χ2 DISTRIBUTION
2 Table 2: Percentage Points for the χ2 distribution
The χ2 Distribution Function for Selected Probabilities
The table provides values of χ2α,v where Pr(χ
2 ≤ χ2α,v) = α and χ
2 ∼ χ2v
α 0.005 0.01 0.025 0.05 0.1 0.5 0.9 0.95 0.975 0.99 0.995
v Values of χ2α,v
1 0.000 0.000 0.001 0.004 0.016 0.455 2.706 3.841 5.024 6.635 7.879
2 0.010 0.020 0.051 0.103 0.211 1.386 4.605 5.991 7.378 9.210 10.60
3 0.072 0.115 0.216 0.352 0.584 2.366 6.251 7.815 9.348 11.34 12.84
4 0.207 0.297 0.484 0.711 1.064 3.357 7.779 9.488 11.14 13.28 14.86
5 0.412 0.554 0.831 1.145 1.610 4.351 9.236 11.07 12.83 15.09 16.75
6 0.676 0.872 1.237 1.635 2.204 5.348 10.64 12.59 14.45 16.81 18.55
7 0.989 1.239 1.690 2.167 2.833 6.346 12.02 14.07 16.01 18.48 20.28
8 1.344 1.646 2.180 2.733 3.490 7.344 13.36 15.51 17.53 20.09 21.95
9 1.735 2.088 2.700 3.325 4.168 8.343 14.68 16.92 19.02 21.67 23.59
10 2.156 2.558 3.247 3.940 4.865 9.342 15.99 18.31 20.48 23.21 25.19
11 2.603 3.053 3.816 4.575 5.578 10.34 17.28 19.68 21.92 24.72 26.76
12 3.074 3.571 4.404 5.226 6.304 11.34 18.55 21.03 23.34 26.22 28.30
13 3.565 4.107 5.009 5.892 7.042 12.34 19.81 22.36 24.74 27.69 29.82
14 4.075 4.660 5.629 6.571 7.790 13.34 21.06 23.68 26.12 29.14 31.32
15 4.601 5.229 6.262 7.261 8.547 14.34 22.31 25.00 27.49 30.58 32.80
16 5.142 5.812 6.908 7.962 9.312 15.34 23.54 26.30 28.85 32.00 34.27
17 5.697 6.408 7.564 8.672 10.09 16.34 24.77 27.59 30.19 33.41 35.72
18 6.265 7.015 8.231 9.390 10.86 17.34 25.99 28.87 31.53 34.81 37.16
19 6.844 7.633 8.907 10.12 11.65 18.34 27.20 30.14 32.85 36.19 38.58
20 7.434 8.260 9.591 10.85 12.44 19.34 28.41 31.41 34.17 37.57 40.00
21 8.034 8.897 10.28 11.59 13.24 20.34 29.62 32.67 35.48 38.93 41.40
22 8.643 9.542 10.98 12.34 14.04 21.34 30.81 33.92 36.78 40.29 42.80
23 9.260 10.20 11.69 13.09 14.85 22.34 32.01 35.17 38.08 41.64 44.18
24 9.886 10.86 12.40 13.85 15.66 23.34 33.20 36.42 39.36 42.98 45.56
25 10.52 11.52 13.12 14.61 16.47 24.34 34.38 37.65 40.65 44.31 46.93
26 11.16 12.20 13.84 15.38 17.29 25.34 35.56 38.89 41.92 45.64 48.29
27 11.81 12.88 14.57 16.15 18.11 26.34 36.74 40.11 43.19 46.96 49.64
28 12.46 13.56 15.31 16.93 18.94 27.34 37.92 41.34 44.46 48.28 50.99
29 13.12 14.26 16.05 17.71 19.77 28.34 39.09 42.56 45.72 49.59 52.34
30 13.79 14.95 16.79 18.49 20.60 29.34 40.26 43.77 46.98 50.89 53.67
35 17.19 18.51 20.57 22.47 24.80 34.34 46.06 49.80 53.20 57.34 60.27
40 20.71 22.16 24.43 26.51 29.05 39.34 51.81 55.76 59.34 63.69 66.77
45 24.31 25.90 28.37 30.61 33.35 44.34 57.51 61.66 65.41 69.96 73.17
50 27.99 29.71 32.36 34.76 37.69 49.33 63.17 67.50 71.42 76.15 79.49
60 35.53 37.48 40.48 43.19 46.46 59.33 74.40 79.08 83.30 88.30 91.95
70 43.28 45.44 48.76 51.74 55.33 69.33 85.53 90.53 95.02 100.4 104.2
80 51.17 53.54 57.15 60.39 64.28 79.33 96.58 101.9 106.6 112.3 116.3
90 59.20 61.75 65.65 69.13 73.29 89.33 107.6 113.1 118.1 124.1 128.3
100 67.33 70.06 74.22 77.93 82.36 99.33 118.5 124.3 129.6 135.8 140.2
150 109.1 112.7 118.0 122.7 128.3 149.3 172.6 179.6 185.8 193.2 198.4
200 152.2 156.4 162.7 168.3 174.8 199.3 226.0 234.0 241.1 249.4 255.3
9
3 TABLE 3: UPPER 5% PERCENTAGE POINTS FOR THE F DISTRIBUTION
3 Table 3: Upper 5% percentage points for the F
distribution
The F Distribution Function for α = 0.05
The table provides values of Fα,v1,v2 where Pr(F ≥ Fα,v1,v2) = 0.05 and F ∼ F (v1, v2)
v1 →
v2 ↓ 1 2 3 4 5 6 7 8 9 10 12 15
5 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 4.74 4.68 4.62
6 5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10 4.06 4.00 3.94
7 5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68 3.64 3.57 3.51
8 5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39 3.35 3.28 3.22
9 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.14 3.07 3.01
10 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 2.98 2.91 2.85
11 4.84 3.98 3.59 3.36 3.20 3.09 3.01 2.95 2.90 2.85 2.79 2.72
12 4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80 2.75 2.69 2.62
13 4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71 2.67 2.60 2.53
14 4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 2.60 2.53 2.46
15 4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 2.54 2.48 2.40
16 4.49 3.63 3.24 3.01 2.85 2.74 2.66 2.59 2.54 2.49 2.42 2.35
17 4.45 3.59 3.20 2.96 2.81 2.70 2.61 2.55 2.49 2.45 2.38 2.31
18 4.41 3.55 3.16 2.93 2.77 2.66 2.58 2.51 2.46 2.41 2.34 2.27
19 4.38 3.52 3.13 2.90 2.74 2.63 2.54 2.48 2.42 2.38 2.31 2.23
20 4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 2.35 2.28 2.20
21 4.32 3.47 3.07 2.84 2.68 2.57 2.49 2.42 2.37 2.32 2.25 2.18
22 4.30 3.44 3.05 2.82 2.66 2.55 2.46 2.40 2.34 2.30 2.23 2.15
23 4.28 3.42 3.03 2.80 2.64 2.53 2.44 2.37 2.32 2.27 2.20 2.13
24 4.26 3.40 3.01 2.78 2.62 2.51 2.42 2.36 2.30 2.25 2.18 2.11
25 4.24 3.39 2.99 2.76 2.60 2.49 2.40 2.34 2.28 2.24 2.16 2.09
30 4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.27 2.21 2.16 2.09 2.01
35 4.12 3.27 2.87 2.64 2.49 2.37 2.29 2.22 2.16 2.11 2.04 1.96
40 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.18 2.12 2.08 2.00 1.92
45 4.06 3.20 2.81 2.58 2.42 2.31 2.22 2.15 2.10 2.05 1.97 1.89
50 4.03 3.18 2.79 2.56 2.40 2.29 2.20 2.13 2.07 2.03 1.95 1.87
55 4.02 3.16 2.77 2.54 2.38 2.27 2.18 2.11 2.06 2.01 1.93 1.85
60 4.00 3.15 2.76 2.53 2.37 2.25 2.17 2.10 2.04 1.99 1.92 1.84
70 3.98 3.13 2.74 2.50 2.35 2.23 2.14 2.07 2.02 1.97 1.89 1.81
80 3.96 3.11 2.72 2.49 2.33 2.21 2.13 2.06 2.00 1.95 1.88 1.79
90 3.95 3.10 2.71 2.47 2.32 2.20 2.11 2.04 1.99 1.94 1.86 1.78
100 3.94 3.09 2.70 2.46 2.31 2.19 2.10 2.03 1.97 1.93 1.85 1.77
110 3.93 3.08 2.69 2.45 2.30 2.18 2.09 2.02 1.97 1.92 1.84 1.76
120 3.92 3.07 2.68 2.45 2.29 2.18 2.09 2.02 1.96 1.91 1.83 1.75
150 3.90 3.06 2.66 2.43 2.27 2.16 2.07 2.00 1.94 1.89 1.82 1.73
10
4 TABLE 4: UPPER 1% PERCENTAGE POINTS FOR THE F DISTRIBUTION
4 Table 4: Upper 1% percentage points for the F
distribution
The F Distribution Function for α = 0.01
The table provides values of Fα,v1,v2 where Pr(F ≥ Fα,v1,v2) = 0.01 and F ∼ F (v1, v2)
v1 →
v2 ↓ 1 2 3 4 5 6 7 8 9 10 12 15
5 16.3 13.3 12.1 11.4 11.0 10.7 10.5 10.3 10.2 10.1 9.89 9.72
6 13.7 10.9 9.78 9.15 8.75 8.47 8.26 8.10 7.98 7.87 7.72 7.56
7 12.2 9.55 8.45 7.85 7.46 7.19 6.99 6.84 6.72 6.62 6.47 6.31
8 11.3 8.65 7.59 7.01 6.63 6.37 6.18 6.03 5.91 5.81 5.67 5.52
9 10.6 8.02 6.99 6.42 6.06 5.80 5.61 5.47 5.35 5.26 5.11 4.96
10 10.0 7.56 6.55 5.99 5.64 5.39 5.20 5.06 4.94 4.85 4.71 4.56
11 9.65 7.21 6.22 5.67 5.32 5.07 4.89 4.74 4.63 4.54 4.40 4.25
12 9.33 6.93 5.95 5.41 5.06 4.82 4.64 4.50 4.39 4.30 4.16 4.01
13 9.07 6.70 5.74 5.21 4.86 4.62 4.44 4.30 4.19 4.10 3.96 3.82
14 8.86 6.51 5.56 5.04 4.69 4.46 4.28 4.14 4.03 3.94 3.80 3.66
15 8.68 6.36 5.42 4.89 4.56 4.32 4.14 4.00 3.89 3.80 3.67 3.52
16 8.53 6.23 5.29 4.77 4.44 4.20 4.03 3.89 3.78 3.69 3.55 3.41
17 8.40 6.11 5.18 4.67 4.34 4.10 3.93 3.79 3.68 3.59 3.46 3.31
18 8.29 6.01 5.09 4.58 4.25 4.01 3.84 3.71 3.60 3.51 3.37 3.23
19 8.18 5.93 5.01 4.50 4.17 3.94 3.77 3.63 3.52 3.43 3.30 3.15
20 8.10 5.85 4.94 4.43 4.10 3.87 3.70 3.56 3.46 3.37 3.23 3.09
21 8.02 5.78 4.87 4.37 4.04 3.81 3.64 3.51 3.40 3.31 3.17 3.03
22 7.95 5.72 4.82 4.31 3.99 3.76 3.59 3.45 3.35 3.26 3.12 2.98
23 7.88 5.66 4.76 4.26 3.94 3.71 3.54 3.41 3.30 3.21 3.07 2.93
24 7.82 5.61 4.72 4.22 3.90 3.67 3.50 3.36 3.26 3.17 3.03 2.89
25 7.77 5.57 4.68 4.18 3.85 3.63 3.46 3.32 3.22 3.13 2.99 2.85
30 7.56 5.39 4.51 4.02 3.70 3.47 3.30 3.17 3.07 2.98 2.84 2.70
35 7.42 5.27 4.40 3.91 3.59 3.37 3.20 3.07 2.96 2.88 2.74 2.60
40 7.31 5.18 4.31 3.83 3.51 3.29 3.12 2.99 2.89 2.80 2.66 2.52
45 7.23 5.11 4.25 3.77 3.45 3.23 3.07 2.94 2.83 2.74 2.61 2.46
50 7.17 5.06 4.20 3.72 3.41 3.19 3.02 2.89 2.78 2.70 2.56 2.42
55 7.12 5.01 4.16 3.68 3.37 3.15 2.98 2.85 2.75 2.66 2.53 2.38
60 7.08 4.98 4.13 3.65 3.34 3.12 2.95 2.82 2.72 2.63 2.50 2.35
70 7.01 4.92 4.07 3.60 3.29 3.07 2.91 2.78 2.67 2.59 2.45 2.31
80 6.96 4.88 4.04 3.56 3.26 3.04 2.87 2.74 2.64 2.55 2.42 2.27
90 6.93 4.85 4.01 3.53 3.23 3.01 2.84 2.72 2.61 2.52 2.39 2.24
100 6.90 4.82 3.98 3.51 3.21 2.99 2.82 2.69 2.59 2.50 2.37 2.22
110 6.87 4.80 3.96 3.49 3.19 2.97 2.81 2.68 2.57 2.49 2.35 2.21
120 6.85 4.79 3.95 3.48 3.17 2.96 2.79 2.66 2.56 2.47 2.34 2.19
150 6.81 4.75 3.91 3.45 3.14 2.92 2.76 2.63 2.53 2.44 2.31 2.16
ECON61001: Econometric Methods
Final Exam
January 2021
Release date/time: 28/01/21, 14.00hrs GMT
Submission deadline: 30/01/21, 14.00hrs GMT
Instructions:
• You must answer all five questions in Section A and two out of the four ques-
tions in Section B. If you answer more questions than are required and do not
indicate which answers should be ignored, we will mark the requisite number of
answers in the order in which they appear in your answer submission: answers
beyond that number will not be considered.
• Your answers could be typed or hand-written (and scanned to a single pdf file
that can be submitted) or a combination of a typed answer with included images
of algebra or figures.
• Where relevant, questions include word limits. These are limits, not targets. Ex-
cellent answers can be shorter than the word limit. If you go beyond the word
limit the additional text will be ignored. Where a question includes a word limit
you HAVE to include a word count for your answer (excluding formulae). You
could use https://wordcounter.net to obtain word counts.
• Candidates are advised that the examiners attach considerable importance to
the clarity with which answers are expressed.
• You must correctly enter your registration number and the course code on
your answer.
c©The University of Manchester, 2021.
ECON61001
SECTION A
1. Suppose a researcher is interested in the following linear regression model
yi = x
′
iβ0 + ui, i = 1, 2, . . . N,
where xi = (1, x2,i, x3,i, x4,i)
′ and β0 = (β0,1, β0,2, β0,3, β0,4)
′ . Given the context, the
researcher is able to assume that {ui, x
′
i}
N
i=1 form a sequence of independently
and identically distributed random vectors with E[xix
′
i] = Q, a finite, positive
definite matrix of constants, E[ui|xi] = 0 and V ar[ui|xi] = σ
2
0 . Therefore, she
estimates the model via Ordinary Least Squares (OLS), obtaining the following
estimated equation
yˆi = 1.0213
(0.1416)
− 0.0020
(0.0005)
x2,i − 0.0208
(0.0080)
x3,i + 0.0095
(0.0088)
x4,i,
where the number in parenthesis is the conventional OLS standard error for the
coefficient in question. The OLS estimator of σ20 in this model is σˆ
2
N = 0.0081.
Given these results, the researcher concludes that β0,4 = 0 and so decides to
estimate the model via OLS with x4,i excluded, obtaining the following estimated
equation
yˆi = 1.1485 − 0.0020x2,i − 0.0211x3,i. (1)
Given that the sample size is N = 108 in both estimations, calculate the OLS
estimator of the error variance σ20 from the estimation in (1). Be sure to care-
fully explain your calculations. Hint: consider the F-statistic for testing β0,4 =
0. [8 marks]
2. Suppose it is desired to predict zt using zˆt = w
′
tγ where wt is a vector of ob-
servable variables and γ is a vector of constants that needs to be specified.
The choice of γ associated with the linear projection of zt on wt is γ0 where
E[(zt − w
′
tγ0)wt] = 0.
(a) What optimality property does zˆot = w
′
tγ0 possess? (Word limit: 50) [1mark]
(b) Now consider the regression model zt = w
′
tγ0 + vt. Is wt contempora-
neously exogenous or strictly exogenous for estimation of γ0 in this model?
Justify your answer. (Word limit: 150) [4 marks]
(c) Suppose the model in part (b) is dynamically complete. What optimal-
ity property does zˆot possess? Briefly justify your answer. (Word limit:
150) [3 marks]
Continued over
1
ECON61001
SECTION A continued
3.(a) Let A be n × n nonsingular symmetric matrix. Show that if A is positive definite
then A−1 is positive definite. Hint: AA−1 = In. [4 marks]
3.(b) Consider the classical linear regression model
y = Xβ0 + u, (2)
whereX is the T×k observable data matrix that is fixed in repeated samples with
rank(X) = k, and u is a T × 1 vector with E[u] = 0 and V ar[u] = σ20IT where σ
2
0
is an unknown positive finite constant. Let βˆT be the OLS estimator of β0 based
on (2) and βˆR,T be the Restricted Least Squares (RLS) estimator of β0 based on
(2) subject to the restrictions Rβ0 = r where R is a nr × k matrix of specified
constants with rank(R) = nr and r is a specified nr × 1 vector of constants.
Assuming the restrictions are correct, prove that βˆR,T is at least as efficient as
βˆT . Hint: you may quote the formula for the variance-covariance matrix of the
OLS and RLS estimators without proof; you may also take advantage of the
stated result in part (a). [4 marks]
4. Consider the linear regression model
y = Xβ0 + u,
where y and u are T × 1 vectors, X is T × k matrix, and β0 is the k × 1 vector
of unknown regression coefficients. Assume that X is fixed in repeated sam-
ples with rank(X) = k, and u ∼ N ( 0, σ20IT ) where σ
2
0 is an unknown positive
constant. Let θˆT denote the maximum likelihood estimator of the unknown pa-
rameter vector θ0 = (β
′
0, σ
2
0)
′. Derive the information matrix for this model. Hint:
you may state the form of the log likelihood function and score function for this
model without proof. [8 marks]
5. Consider the model
yi = x
′
iβ0 + ui, i = 1, 2, . . . , N,
where β0 is the k × 1 vector of unknown regression coefficients, {(x
′
i, ui)}
N
i=1
is a sequence of independently and identically distributed random vectors with
E[ui|xi] = 0, V ar[ui|xi] = σ
2
0, an unknown finite positive constant and E[xix
′
i] =
Q, a finite positive definite matrix of constants. Let σˆ2N be the OLS estimator of
σ20. Show that N
1/2(σˆ2N − σ
2
0)
d
→ N( 0, µ4 − σ
4
0) where µ4 = E[u
4
i ].
Hint: You may assume that: N−1
∑N
i=1 xix
′
i
p
→ Q; (ii) N−1/2
∑N
i=1 vi
d
→ N(0,Ω)
where vi = (u
2
i − σ
2
0, x
′
iui)
′, and Ω = V ar[vi] is a finite, positive definite
(k + 1) × (k + 1) matrix whose elements you must specify as needed to develop
your answer. [8 marks]
Continued over
2
ECON61001
SECTION B
6. Consider the regression model
yi = x
′
iβ0 + ui, i = 1, 2, . . . , N,
where β0 is the k × 1 vector of unknown regression coefficients, {(x
′
i, ui)}
N
i=1
is a sequence of independently and identically distributed random vectors with
E[ui|xi] = 0, V ar[ui|xi] = σ
2
0, an unknown finite positive constant and E[xix
′
i] =
Q, a finite positive definite matrix of constants. You may further assume that: (i)
N−1
∑N
i=1 xix
′
i
p
→ Q; (ii) N−1/2
∑N
i=1 xiui
d
→ N(0, σ20Q).
Let βˆR,N denote the RLS estimator based on the linear restrictionsRβ = r where
R is a nr × k matrix of pre-specified constants with rank equal to nr and r is a
nr × 1 vector of pre-specified constants, and let λˆN be the vector of Lagrange
Multipliers associated with this RLS estimation. Assuming Rβ0 = r, answer the
following questions.
(a) Show that N1/2(βˆR,N − β0)
d
→ N ( 0, VR ) where
VR = σ
2
0
(
Q−1 − Q−1R′(RQ−1R′)−1RQ−1
)
.
Hint: you may quote the formulae for βˆR,N and βˆN , the Ordinary Least
Squares estimator of β0, without proof. [10 marks]
(b) A colleague proposes testing H0 : Rβ0 = r versus H1 : Rβ0 6= r using the
decision rule of the form: reject H0 at the (approximate) 100α% significance
level if
λˆ′NMN λˆN > cnr(1− α)
where cnr(1− α) is the 100(1 − α)
th percentile of the χ2nr distribution. How-
ever, your colleague is unsure what the matrix MN should be in order that
this decision rule has the properties implied by the stated significance level.
Provide a suitable choice of MN , being sure to justify your choice carefully.
Hint: you may quote without proof: (i) the formulae for λˆN and βˆN ; (ii) that
both the OLS and RLS estimators of σ20 are consistent under the conditions
of the question. [20 marks]
Continued over
3
ECON61001
SECTION B continued
7.(a) Let {vt}
T
t=−3 be a weakly stationary time series process. Consider the following
statistic,
ρˆ4,T =
∑T
t=1 vtvt−4∑T
t=1 v
2
t
.
Let {εt}
∞
t=−∞ denote a sequence of independently and identically distributed
(i.i.d.) random variables with mean zero and variance σ2ε .
(i) Assume that vt = εt. Show that T
1/2ρˆ4,T
d
→ N(0, 1). [6 marks]
(ii) Assume that
vt = θ4vt−4 + εt,
where |θ4| < 1. What is the probability limit of ρˆ4,T as T → ∞? Be sure
to justify your answer carefully. Hint: vt has the following representation,
vt =
∑
∞
i=0 θ
i
4εt−4i. [9 marks]
7.(b) A researcher wishes to test the simple efficient-markets hypothesis in the foreign
exchange market. Let st = ln(St) and ft,n = ln[Ft,n], where St and Ft,n are the
levels of the spot exchange rate at time t and the n−period forward exchange
rate at time t. The simple efficient-markets hypothesis is that ft,n = E[st+n | It]
where It is the information set at time t which for the purposes of this question
can be taken to be It = {st, ft,n, st−1, ft−1,n, st−2, ft−2,n, . . .}. Using daily spot and
thirty-day forward exchange rate data for the US dollar UK pound exchange rate,
the researcher estimates the model,
yt+n = x
′
tβ0 + ut,n, (3)
where yt+n = st+n − ft,n, xt is the 3× 1 vector given by
x′t = ( 1, st − ft−n,n, st−1 − ft−1−n,n ) ,
n = 30 and ut,n is the error term. If the simple efficient markets hypothesis
holds in this foreign exchange market then E[ut,n | It] = 0 and the regression
coefficients in (3) satisfy a set of restrictions denoted here by g(β0) = 0 where
g( · ) is ng × 1 vector.
(i) What is g(β0)? Briefly justify your answer. (Word limit: 75) [4 marks]
Continued over
4
ECON61001
SECTION B continued
7.(b) continued
(ii) The researcher tests H0 : g(β0) = 0 versus H1 : g(β0) 6= 0 using the test
statistic
ST = Tg(βˆT )
′
(
G(βˆT ) Vˆβ G(βˆT )
′
)
−1
g(βˆT ), (4)
where G(βˆT ) = ∂g(β)/∂β
∣∣
β=βˆT
. Assuming T 1/2(βˆT − β0)
d
→ N(0, Vβ) and
Vˆβ
p
→ Vβ , write down a suitable decision rule for this test. If ST = 8.2 then
what is the outcome of the test? [3 marks]
(iii) Since the yt+n is a financial variable, the researcher is concerned that the
errors may exhibit autoregressive conditional heteroscedasticity and so has
calculated Vˆβ in (4) using White’s heteroscedasticity robust estimator. Given
this information, do you have any concerns about the test in part (ii)? If so
then explain your concerns and how you would modify the test to address
these concerns. (Word limit: 350) [8 marks]
Continued over
5
ECON61001
SECTION B continued
8. Consider the linear regression model
y1,i = γ0y2,i + z
′
1,iδ0 + u1,i = x
′
iβ0 + u1,i,
where x′i = (y2,i, z
′
1,i), β0 = (γ0, δ
′
0)
′ and assume that
y2,i = z
′
iη0 + u2,i, (5)
where y1,i and y2,i are observable random variables, zi = (z
′
1,i, z
′
2,i)
′ is a random
vector of observable variables, u1,i and u2,i are the error terms (unobservable
scalar random variables), γ0 is an unknown scalar parameter, and δ0, and η0 are
vectors of unknown parameters. Suppose there is a sample of N observations,
and let yˆ2,i denote the predicted value of y2,i based on Ordinary Least Squares
(OLS) estimation of (5). Define xˆi = (yˆ2,i, z
′
1,i)
′. Let X be the N × k matrix with ith
row x′i, Xˆ be the N × k matrix with i
th row xˆ′i, Z be the N × q matrix with i
th row
z′i and y1 be the N × 1 vector with i
th element y1,i. Consider the following three
estimators of β0:
• βˆ1 = (Xˆ
′Xˆ)−1Xˆ ′y1;
• βˆ2 = (Xˆ
′X)−1Xˆ ′y1;
• βˆ3 = {X
′Z(Z ′Z)−1Z ′X }
−1
X ′Z(Z ′Z)−1Z ′y1.
(a) Show that βˆ1 = βˆ2 = βˆ3. [15 marks]
(b) Let φˆ = (βˆ ′, θˆ)′ be the OLS estimator of φ0 = (β
′
0, θ0)
′ based on the model
y1,i = x
′
iβ0 + θuˆ2,i + “error”
where uˆ2,i is the i
th element of uˆ2, theN×1 vector of residual fromOLS estimation
of (5). Via an application of the Frisch-Waugh-Lovell Theorem or otherwise, show
that βˆ = βˆ1.
[15 marks]
Continued over
6
ECON61001
SECTION B continued
9.(a) Let {(yi, x
′
i)}
N
i=1 be a sequence of independently and identically distributed (i.i.d.)
random vectors. Suppose that yi is a dummy variable and so has a sample space
of {0, 1}. Consider the model
yi = x
′
iβ0 + ui.
(i) Assume that E[ui|xi] = 0. Derive the Generalized Least Squares (GLS)
estimator of β0 in this model? Hint: you may quote the generic formula for
the GLS estimator that is, βˆGLS = (X
′Σ−1X)−1X ′Σ−1y, but you must derive
Σ for this model. [6 marks]
(ii) Is your answer to part (i) a feasible or infeasible GLS estimator? If infeasible
then suggest a feasible GLS estimator. Do you foresee any potential prob-
lems in implementing your proposed Feasible GLS estimator? [4 marks]
(b) Let {Vi}
N
i=1 be a sequence of i.i.d. Bernoulli random variableswith P (Vi = 1) = θ0.
We assume here that θ0 ∈ (0, 1) and that our sample size is large enough for both
outcomes to occur.
(i) Derive the Wald, Likelihood Ratio and LagrangeMultiplier statistics for test-
ing H0 : θ0 = θ∗ against H1 : θ0 6= θ∗. Hint: you may quote the form of the
log likelihood function, the score equation and the formula for the maximum
likelihood estimator for this model without proof. [16 marks]
(ii) Given that N = 100 and the sample contains 55 outcomes that are one,
use your statistics in part (i) to test the hypothesis H0 : θ0 = 0.5 against
H1 : θ0 6= 0.5 at the 5% significance level. [4 marks]
END OF EXAMINATION
7
1 TABLE 1: PERCENTAGE POINTS FOR THE T DISTRIBUTION
1 Table 1: Percentage Points for the t distribution
Student’s t Distribution Function for Selected Probabilities
The table provides values of tα,v where Pr(T ≤ tα,v) = α and T ∼ tv
α 0.750 0.800 0.900 0.950 0.975 0.990 0.995 0.9975 0.999 0.9995
ν Values of tα,v
1 1.000 1.376 3.078 6.314 12.706 31.821 63.657
2 0.816 1.061 1.886 2.920 4.303 6.965 9.925
3 0.765 0.978 1.638 2.353 3.182 4.541 5.841
4 0.741 0.941 1.533 2.132 2.776 3.747 4.604
5 0.727 0.920 1.476 2.015 2.571 3.365 4.032 4.773
6 0.718 0.906 1.440 1.943 2.447 3.143 3.707 4.317 5.208
7 0.711 0.896 1.415 1.895 2.365 2.998 3.499 4.029 4.785 5.408
8 0.706 0.889 1.397 1.860 2.306 2.896 3.355 3.833 4.501 5.041
9 0.703 0.883 1.383 1.833 2.262 2.821 3.250 3.690 4.297 4.781
10 0.700 0.879 1.372 1.812 2.228 2.764 3.169 3.581 4.144 4.587
11 0.697 0.876 1.363 1.796 2.201 2.718 3.106 3.497 4.025 4.437
12 0.695 0.873 1.356 1.782 2.179 2.681 3.055 3.428 3.930 4.318
13 0.694 0.870 1.350 1.771 2.160 2.650 3.012 3.372 3.852 4.221
14 0.692 0.868 1.345 1.761 2.145 2.624 2.977 3.326 3.787 4.140
15 0.691 0.866 1.341 1.753 2.131 2.602 2.947 3.286 3.733 4.073
16 0.690 0.865 1.337 1.746 2.120 2.583 2.921 3.252 3.686 4.015
17 0.689 0.863 1.333 1.740 2.110 2.567 2.898 3.222 3.646 3.965
18 0.688 0.862 1.330 1.734 2.101 2.552 2.878 3.197 3.610 3.922
19 0.688 0.861 1.328 1.729 2.093 2.539 2.861 3.174 3.579 3.883
20 0.687 0.860 1.325 1.725 2.086 2.528 2.845 3.153 3.552 3.850
21 0.686 0.859 1.323 1.721 2.080 2.518 2.831 3.135 3.527 3.819
22 0.686 0.858 1.321 1.717 2.074 2.508 2.819 3.119 3.505 3.792
23 0.685 0.858 1.319 1.714 2.069 2.500 2.807 3.104 3.485 3.768
24 0.685 0.857 1.318 1.711 2.064 2.492 2.797 3.091 3.467 3.745
25 0.684 0.856 1.316 1.708 2.060 2.485 2.787 3.078 3.450 3.725
26 0.684 0.856 1.315 1.706 2.056 2.479 2.779 3.067 3.435 3.707
27 0.684 0.855 1.314 1.703 2.052 2.473 2.771 3.057 3.421 3.690
28 0.683 0.855 1.313 1.701 2.048 2.467 2.763 3.047 3.408 3.674
29 0.683 0.854 1.311 1.699 2.045 2.462 2.756 3.038 3.396 3.659
30 0.683 0.854 1.310 1.697 2.042 2.457 2.750 3.030 3.385 3.646
40 0.681 0.851 1.303 1.684 2.021 2.423 2.704 2.971 3.307 3.551
50 0.679 0.849 1.299 1.676 2.009 2.403 2.678 2.937 3.261 3.496
60 0.679 0.848 1.296 1.671 2.000 2.390 2.660 2.915 3.232 3.460
70 0.678 0.847 1.294 1.667 1.994 2.381 2.648 2.899 3.211 3.435
80 0.678 0.846 1.292 1.664 1.990 2.374 2.639 2.887 3.195 3.416
90 0.677 0.846 1.291 1.662 1.987 2.368 2.632 2.878 3.183 3.402
100 0.677 0.845 1.290 1.660 1.984 2.364 2.626 2.871 3.174 3.390
110 0.677 0.845 1.289 1.659 1.982 2.361 2.621 2.865 3.166 3.381
120 0.677 0.845 1.289 1.658 1.980 2.358 2.617 2.860 3.160 3.373
∞ 0.674 0.842 1.282 1.645 1.960 2.326 2.576 2.808 3.090 3.297
8
2 TABLE 2: PERCENTAGE POINTS FOR THE χ2 DISTRIBUTION
2 Table 2: Percentage Points for the χ2 distribution
The χ2 Distribution Function for Selected Probabilities
The table provides values of χ2α,v where Pr(χ
2 ≤ χ2α,v) = α and χ
2 ∼ χ2v
α 0.005 0.01 0.025 0.05 0.1 0.5 0.9 0.95 0.975 0.99 0.995
v Values of χ2α,v
1 0.000 0.000 0.001 0.004 0.016 0.455 2.706 3.841 5.024 6.635 7.879
2 0.010 0.020 0.051 0.103 0.211 1.386 4.605 5.991 7.378 9.210 10.60
3 0.072 0.115 0.216 0.352 0.584 2.366 6.251 7.815 9.348 11.34 12.84
4 0.207 0.297 0.484 0.711 1.064 3.357 7.779 9.488 11.14 13.28 14.86
5 0.412 0.554 0.831 1.145 1.610 4.351 9.236 11.07 12.83 15.09 16.75
6 0.676 0.872 1.237 1.635 2.204 5.348 10.64 12.59 14.45 16.81 18.55
7 0.989 1.239 1.690 2.167 2.833 6.346 12.02 14.07 16.01 18.48 20.28
8 1.344 1.646 2.180 2.733 3.490 7.344 13.36 15.51 17.53 20.09 21.95
9 1.735 2.088 2.700 3.325 4.168 8.343 14.68 16.92 19.02 21.67 23.59
10 2.156 2.558 3.247 3.940 4.865 9.342 15.99 18.31 20.48 23.21 25.19
11 2.603 3.053 3.816 4.575 5.578 10.34 17.28 19.68 21.92 24.72 26.76
12 3.074 3.571 4.404 5.226 6.304 11.34 18.55 21.03 23.34 26.22 28.30
13 3.565 4.107 5.009 5.892 7.042 12.34 19.81 22.36 24.74 27.69 29.82
14 4.075 4.660 5.629 6.571 7.790 13.34 21.06 23.68 26.12 29.14 31.32
15 4.601 5.229 6.262 7.261 8.547 14.34 22.31 25.00 27.49 30.58 32.80
16 5.142 5.812 6.908 7.962 9.312 15.34 23.54 26.30 28.85 32.00 34.27
17 5.697 6.408 7.564 8.672 10.09 16.34 24.77 27.59 30.19 33.41 35.72
18 6.265 7.015 8.231 9.390 10.86 17.34 25.99 28.87 31.53 34.81 37.16
19 6.844 7.633 8.907 10.12 11.65 18.34 27.20 30.14 32.85 36.19 38.58
20 7.434 8.260 9.591 10.85 12.44 19.34 28.41 31.41 34.17 37.57 40.00
21 8.034 8.897 10.28 11.59 13.24 20.34 29.62 32.67 35.48 38.93 41.40
22 8.643 9.542 10.98 12.34 14.04 21.34 30.81 33.92 36.78 40.29 42.80
23 9.260 10.20 11.69 13.09 14.85 22.34 32.01 35.17 38.08 41.64 44.18
24 9.886 10.86 12.40 13.85 15.66 23.34 33.20 36.42 39.36 42.98 45.56
25 10.52 11.52 13.12 14.61 16.47 24.34 34.38 37.65 40.65 44.31 46.93
26 11.16 12.20 13.84 15.38 17.29 25.34 35.56 38.89 41.92 45.64 48.29
27 11.81 12.88 14.57 16.15 18.11 26.34 36.74 40.11 43.19 46.96 49.64
28 12.46 13.56 15.31 16.93 18.94 27.34 37.92 41.34 44.46 48.28 50.99
29 13.12 14.26 16.05 17.71 19.77 28.34 39.09 42.56 45.72 49.59 52.34
30 13.79 14.95 16.79 18.49 20.60 29.34 40.26 43.77 46.98 50.89 53.67
35 17.19 18.51 20.57 22.47 24.80 34.34 46.06 49.80 53.20 57.34 60.27
40 20.71 22.16 24.43 26.51 29.05 39.34 51.81 55.76 59.34 63.69 66.77
45 24.31 25.90 28.37 30.61 33.35 44.34 57.51 61.66 65.41 69.96 73.17
50 27.99 29.71 32.36 34.76 37.69 49.33 63.17 67.50 71.42 76.15 79.49
60 35.53 37.48 40.48 43.19 46.46 59.33 74.40 79.08 83.30 88.30 91.95
70 43.28 45.44 48.76 51.74 55.33 69.33 85.53 90.53 95.02 100.4 104.2
80 51.17 53.54 57.15 60.39 64.28 79.33 96.58 101.9 106.6 112.3 116.3
90 59.20 61.75 65.65 69.13 73.29 89.33 107.6 113.1 118.1 124.1 128.3
100 67.33 70.06 74.22 77.93 82.36 99.33 118.5 124.3 129.6 135.8 140.2
150 109.1 112.7 118.0 122.7 128.3 149.3 172.6 179.6 185.8 193.2 198.4
200 152.2 156.4 162.7 168.3 174.8 199.3 226.0 234.0 241.1 249.4 255.3
9
3 TABLE 3: UPPER 5% PERCENTAGE POINTS FOR THE F DISTRIBUTION
3 Table 3: Upper 5% percentage points for the F
distribution
The F Distribution Function for α = 0.05
The table provides values of Fα,v1,v2 where Pr(F ≥ Fα,v1,v2) = 0.05 and F ∼ F (v1, v2)
v1 →
v2 ↓ 1 2 3 4 5 6 7 8 9 10 12 15
5 6.61 5.79 5.41 5.19 5.05 4.95 4.88 4.82 4.77 4.74 4.68 4.62
6 5.99 5.14 4.76 4.53 4.39 4.28 4.21 4.15 4.10 4.06 4.00 3.94
7 5.59 4.74 4.35 4.12 3.97 3.87 3.79 3.73 3.68 3.64 3.57 3.51
8 5.32 4.46 4.07 3.84 3.69 3.58 3.50 3.44 3.39 3.35 3.28 3.22
9 5.12 4.26 3.86 3.63 3.48 3.37 3.29 3.23 3.18 3.14 3.07 3.01
10 4.96 4.10 3.71 3.48 3.33 3.22 3.14 3.07 3.02 2.98 2.91 2.85
11 4.84 3.98 3.59 3.36 3.20 3.09 3.01 2.95 2.90 2.85 2.79 2.72
12 4.75 3.89 3.49 3.26 3.11 3.00 2.91 2.85 2.80 2.75 2.69 2.62
13 4.67 3.81 3.41 3.18 3.03 2.92 2.83 2.77 2.71 2.67 2.60 2.53
14 4.60 3.74 3.34 3.11 2.96 2.85 2.76 2.70 2.65 2.60 2.53 2.46
15 4.54 3.68 3.29 3.06 2.90 2.79 2.71 2.64 2.59 2.54 2.48 2.40
16 4.49 3.63 3.24 3.01 2.85 2.74 2.66 2.59 2.54 2.49 2.42 2.35
17 4.45 3.59 3.20 2.96 2.81 2.70 2.61 2.55 2.49 2.45 2.38 2.31
18 4.41 3.55 3.16 2.93 2.77 2.66 2.58 2.51 2.46 2.41 2.34 2.27
19 4.38 3.52 3.13 2.90 2.74 2.63 2.54 2.48 2.42 2.38 2.31 2.23
20 4.35 3.49 3.10 2.87 2.71 2.60 2.51 2.45 2.39 2.35 2.28 2.20
21 4.32 3.47 3.07 2.84 2.68 2.57 2.49 2.42 2.37 2.32 2.25 2.18
22 4.30 3.44 3.05 2.82 2.66 2.55 2.46 2.40 2.34 2.30 2.23 2.15
23 4.28 3.42 3.03 2.80 2.64 2.53 2.44 2.37 2.32 2.27 2.20 2.13
24 4.26 3.40 3.01 2.78 2.62 2.51 2.42 2.36 2.30 2.25 2.18 2.11
25 4.24 3.39 2.99 2.76 2.60 2.49 2.40 2.34 2.28 2.24 2.16 2.09
30 4.17 3.32 2.92 2.69 2.53 2.42 2.33 2.27 2.21 2.16 2.09 2.01
35 4.12 3.27 2.87 2.64 2.49 2.37 2.29 2.22 2.16 2.11 2.04 1.96
40 4.08 3.23 2.84 2.61 2.45 2.34 2.25 2.18 2.12 2.08 2.00 1.92
45 4.06 3.20 2.81 2.58 2.42 2.31 2.22 2.15 2.10 2.05 1.97 1.89
50 4.03 3.18 2.79 2.56 2.40 2.29 2.20 2.13 2.07 2.03 1.95 1.87
55 4.02 3.16 2.77 2.54 2.38 2.27 2.18 2.11 2.06 2.01 1.93 1.85
60 4.00 3.15 2.76 2.53 2.37 2.25 2.17 2.10 2.04 1.99 1.92 1.84
70 3.98 3.13 2.74 2.50 2.35 2.23 2.14 2.07 2.02 1.97 1.89 1.81
80 3.96 3.11 2.72 2.49 2.33 2.21 2.13 2.06 2.00 1.95 1.88 1.79
90 3.95 3.10 2.71 2.47 2.32 2.20 2.11 2.04 1.99 1.94 1.86 1.78
100 3.94 3.09 2.70 2.46 2.31 2.19 2.10 2.03 1.97 1.93 1.85 1.77
110 3.93 3.08 2.69 2.45 2.30 2.18 2.09 2.02 1.97 1.92 1.84 1.76
120 3.92 3.07 2.68 2.45 2.29 2.18 2.09 2.02 1.96 1.91 1.83 1.75
150 3.90 3.06 2.66 2.43 2.27 2.16 2.07 2.00 1.94 1.89 1.82 1.73
10
4 TABLE 4: UPPER 1% PERCENTAGE POINTS FOR THE F DISTRIBUTION
4 Table 4: Upper 1% percentage points for the F
distribution
The F Distribution Function for α = 0.01
The table provides values of Fα,v1,v2 where Pr(F ≥ Fα,v1,v2) = 0.01 and F ∼ F (v1, v2)
v1 →
v2 ↓ 1 2 3 4 5 6 7 8 9 10 12 15
5 16.3 13.3 12.1 11.4 11.0 10.7 10.5 10.3 10.2 10.1 9.89 9.72
6 13.7 10.9 9.78 9.15 8.75 8.47 8.26 8.10 7.98 7.87 7.72 7.56
7 12.2 9.55 8.45 7.85 7.46 7.19 6.99 6.84 6.72 6.62 6.47 6.31
8 11.3 8.65 7.59 7.01 6.63 6.37 6.18 6.03 5.91 5.81 5.67 5.52
9 10.6 8.02 6.99 6.42 6.06 5.80 5.61 5.47 5.35 5.26 5.11 4.96
10 10.0 7.56 6.55 5.99 5.64 5.39 5.20 5.06 4.94 4.85 4.71 4.56
11 9.65 7.21 6.22 5.67 5.32 5.07 4.89 4.74 4.63 4.54 4.40 4.25
12 9.33 6.93 5.95 5.41 5.06 4.82 4.64 4.50 4.39 4.30 4.16 4.01
13 9.07 6.70 5.74 5.21 4.86 4.62 4.44 4.30 4.19 4.10 3.96 3.82
14 8.86 6.51 5.56 5.04 4.69 4.46 4.28 4.14 4.03 3.94 3.80 3.66
15 8.68 6.36 5.42 4.89 4.56 4.32 4.14 4.00 3.89 3.80 3.67 3.52
16 8.53 6.23 5.29 4.77 4.44 4.20 4.03 3.89 3.78 3.69 3.55 3.41
17 8.40 6.11 5.18 4.67 4.34 4.10 3.93 3.79 3.68 3.59 3.46 3.31
18 8.29 6.01 5.09 4.58 4.25 4.01 3.84 3.71 3.60 3.51 3.37 3.23
19 8.18 5.93 5.01 4.50 4.17 3.94 3.77 3.63 3.52 3.43 3.30 3.15
20 8.10 5.85 4.94 4.43 4.10 3.87 3.70 3.56 3.46 3.37 3.23 3.09
21 8.02 5.78 4.87 4.37 4.04 3.81 3.64 3.51 3.40 3.31 3.17 3.03
22 7.95 5.72 4.82 4.31 3.99 3.76 3.59 3.45 3.35 3.26 3.12 2.98
23 7.88 5.66 4.76 4.26 3.94 3.71 3.54 3.41 3.30 3.21 3.07 2.93
24 7.82 5.61 4.72 4.22 3.90 3.67 3.50 3.36 3.26 3.17 3.03 2.89
25 7.77 5.57 4.68 4.18 3.85 3.63 3.46 3.32 3.22 3.13 2.99 2.85
30 7.56 5.39 4.51 4.02 3.70 3.47 3.30 3.17 3.07 2.98 2.84 2.70
35 7.42 5.27 4.40 3.91 3.59 3.37 3.20 3.07 2.96 2.88 2.74 2.60
40 7.31 5.18 4.31 3.83 3.51 3.29 3.12 2.99 2.89 2.80 2.66 2.52
45 7.23 5.11 4.25 3.77 3.45 3.23 3.07 2.94 2.83 2.74 2.61 2.46
50 7.17 5.06 4.20 3.72 3.41 3.19 3.02 2.89 2.78 2.70 2.56 2.42
55 7.12 5.01 4.16 3.68 3.37 3.15 2.98 2.85 2.75 2.66 2.53 2.38
60 7.08 4.98 4.13 3.65 3.34 3.12 2.95 2.82 2.72 2.63 2.50 2.35
70 7.01 4.92 4.07 3.60 3.29 3.07 2.91 2.78 2.67 2.59 2.45 2.31
80 6.96 4.88 4.04 3.56 3.26 3.04 2.87 2.74 2.64 2.55 2.42 2.27
90 6.93 4.85 4.01 3.53 3.23 3.01 2.84 2.72 2.61 2.52 2.39 2.24
100 6.90 4.82 3.98 3.51 3.21 2.99 2.82 2.69 2.59 2.50 2.37 2.22
110 6.87 4.80 3.96 3.49 3.19 2.97 2.81 2.68 2.57 2.49 2.35 2.21
120 6.85 4.79 3.95 3.48 3.17 2.96 2.79 2.66 2.56 2.47 2.34 2.19
150 6.81 4.75 3.91 3.45 3.14 2.92 2.76 2.63 2.53 2.44 2.31 2.16
11
