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计量经济代写-EMESTER 1
时间：2022-06-14
UNIVERSITY OF MELBOURNE
DEPARTMENT OF ECONOMICS
SEMESTER 1 ASSESSMENT, 2019
ECOM30001: Basic Econometrics
Time Allowed: TWO Hours
Reading Time: 15 minutes
This examination paper contributes 60 percent to the assessment in ECOM30001.
This examination consists of four (4) questions in total.
You must answer ALL four questions.
Each question is worth twenty (20) marks for a maximum of 80 marks.
The examination paper should be inserted in the back of your answer booklet at the end
of the examination.
The following items are authorized in the examination room:
-Casio FX-82 (any suffix) calculator
This exam has 19 pages.
The paper may not be removed from the exam room.
Answer all questions in the examination booklet(s) provided.
The examination contains a formula sheet starting on page 14.
The examination contains critical values for a number of distributions starting on page 17.
Page 2 of 19
Question 1
a) [5 marks] Consider the following econometric model:
yi = β0 + β1Xi + εi
What is meant by the term heteroskedasticity in the random error? What are the con-
sequences for the OLS estimator if you ignore heteroskedasticity in the random error εi?
Briefly outline how you would test for the presence of heteroskedasticity using White’s
test. Your answer should clearly state the null and alternative hypotheses, the test statis-
tic and its distribution.
b) [5 marks] Consider the following econometric model:
yi = β0 + β1X
∗
i + εi
Consider the following measurement equation relating the observed value for X to the
‘true’ value X∗:
Xi = X
∗
i + υi
with COV(X∗i , εi) = 0 and COV(υi, εi) = 0.
What is meant by the term classical measurement error? Clearly explain the implications
for the OLS estimator of β1 if the explanatory variable X
∗ is observed with (classical)
measurement error.
c) [5 marks] Consider the following econometric model:
yt = β0 + β1Xt + εt
What is meant by the term autocorrelation in the random error? Clearly explain the
consequences for the OLS estimator if you ignore autocorrelation in the random error εt.
Briefly outline how you would test for first order AR(1) autocorrelation in the random
error. Your answer should clearly state the null and alternative hypotheses, the test
statistic and its distribution.
Page 3 of 19
d) [5 marks] Consider the following regression:
∆ inft = β0 + β1 inft−1 + β2 ∆ inft−1 + β3 ∆ inft−2 + β4 ∆ inft−3 + β5 ∆ inft−4 + εt
where inf represents the quarterly inflation rate. This econometric model was estimated
using the method of Ordinary Least Squares (OLS) for the period 1948:Q1 to 2016:Q1
and the results are presented in Figure 1.
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Outline how you would test whether the series for the quarterly inflation rate was sta-
tionary or not. Your answer should clearly state the null and alternative hypotheses, the
test statistic and its distribution. Using the results in Figure 1, what is the value of the
Augmented Dickey-Fuller test statistic? At the 5% level of significance, explain whether
the sample evidence is consistent with the null hypothesis. Based upon the estimation
results presented in Figure 1, the p-value associated with the Augmented Dickey-Fuller
test is 0.0080.
Page 4 of 19
Question 2
Consider the following demand function for airline seats on routes in a large country:
lnpasseni = β0 + β1lnfarei + β2lndisti + β3lndist
2
i + εi (1)
where:
passeni = average (log) number of passengers per day on route i
farei = average (log) fare on route i, in dollars
lndisti = average (log) distance of route i, in miles
conceni = average market share of largest carrier on routei
and ln(X) denotes the natural logarithm of variable X.
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a) [3 marks] The estimation results from estimating model (1) by the method of Ordinary
Least Squares (OLS) are reported in Figure 2. At the 5% level of significance, test
the hypothesis that the price elasticity of demand for air travel is elastic, that is the
price elasticity of demand is less than -1. Your answer should clearly state the null and
alternative hypotheses, the distribution of the test statistic, and your decision.
b) [3 marks] Do you think that the condition COV(fare, ε|dist) = 0 is likely to be satisfied?
Clearly explain why or why not. Explain the consequences for the OLS estimator if this
condition is not satisfied.
c) [4 marks] Consider a variable concen that might be suitable as an instrumental variable
for ln (fare). This variable is a measure of market concentration, measured by the market
share of the largest carrier on the route. Clearly, explain the two conditions that must be
satisfied for the variable concen to be a valid instrumental variable. Do you think these
two conditions are likely to be satisfied? Why or why not?
Page 5 of 19
d) [2 marks] Consider the following first stage:
lnfarei = pi0 + pi1 conceni + pi2lndisti + pi3lndist
2
i + υi (2)
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The results from estimating this first stage model by OLS are reported in Figure 3. Based
upon these results, explain whether the variable concen is an adequate instrumental
variable for ln (fare).
Page 6 of 19
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Figure 4: Instrumental Variable Regression Results for Model (1)
e) [3 marks] The estimation results from estimating model (1) by the method of Instru-
mental Variables (IV) are reported in Figure 4. At the 5% level of significance, test
the hypothesis that the price elasticity of demand for air travel is elastic, that is the
price elasticity of demand is less than -1. Your answer should clearly state the null and
alternative hypotheses, the distribution of the test statistic, and your decision.
f) [2 marks] Compare and contrast the estimate of β1 obtained using OLS (Figure 2) with
that obtained using the method of IV (Figure 4). In your answer comment on both the
magnitude of the estimate and the statistical significance of the estimate of β1.
g) [3 marks] An alternative instrumental variable for ln (fare) might be the total cost of
fuel on route i. Clearly, explain the two conditions that must be satisfied for this variable
to be a valid instrumental variable. Do you think these two conditions are likely to be
satisfied? Why or why not.
Page 7 of 19
Question 3
Consider the following econometric model for the approval of applications for (household)
mortgage finance:
deny∗i = β0 + β1 pii + β2 hse inci + β3 lvratmedi + β4 lvrathighi
+ β5 ccredlowi + β6 ccredmedi + β7 pubreci + β8 denpmii + εi (3)
where εi|Xi ∼ N (0, σ2) and deny∗i is a latent variable that determines the choice to reject a
mortgage application. In addition:
pi = Ratio of total montly debt payments to total monthly income
hse inc = Ratio of monthly housing expensese to total montly income
lvratlow = 1 if ratio of loan to assessed value of the property is below 80%, 0 otherwise
lvratmed = 1 if ratio of loan to assessed value of the property is above 80%
but below 95%, 0 otherwise
lvrathigh = 1 if ratio of loan to assessed value of the property is above 95%, 0 otherwise
ccredlow = 1 if poor consumer credit score, 0
ccredmed = 1 if satisfactory consumer credit score, 0 otherwise
ccredhigh = 1 if excellent consumer credit score, 0 otherwise
pubrec = 1 if public record of bad credit, 0 otherwise
denpmi = 1 if denied mortgage insurance, 0 otherwise
Note that lvratlow, ccredhigh are the omitted categories.
a) [1 mark] Provide a brief interpretation of the latent variable deny∗i .
b) Suppose you have a dataset with observations on 2,303 individuals. The data contain
information on mortgage applications, augmented with additional data from the banks
and lending institutions that reviewed these applications. The data relate to household
mortgage applications during a particular calendar year. The data provide an indicator
variable for the approval of housing finance applications:
deny = 1 if mortgate application denied, 0 otherwise
The parameters of the model (3) were estimated as a Probit model and the results are
provided in Figure 5.
Note that the probability density function for a standard normal random variable (Z) is
given by:
φ(Z) =
1√
2pi
exp
(−Z2
2
)
Page 8 of 19
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Page 9 of 19
i) [4 marks] Let pˆi represent the predicted probability that a mortgage application
is denied, for each individual in the sample. Consider the following decision rule: if
pˆi ≥ 0.5, predict that d̂enyi = 1, otherwise d̂enyi = 0. Based on the information
in Table 1, calculate the percentage of outcomes that are successfully predicted?
Using Table 1, comment on the usefulness of the model in predicting denyi = 1 or
denyi = 0.
true predicted frequency
0 0 2,015
1 0 189
0 1 20
1 1 79
2,303
Table 1: Predicted Probability Threshold pˆi ≥ 0.5
ii) [5 marks] Calculate the marginal effect for the payment-income ratio (pi) for an
individual with a pi ratio of 0.30, a housing expense-income ratio of 0.25, an excellent
credit score, a loan-valuation ratio below 80%, no record of bad credit, and who
has never been denied mortgage insurance. Provide a clear interpretation for your
calculated marginal effect.
iii) [5 marks] Explain how you would compute the average marginal effect (AME)
for being denied mortgage insurance (denpmi). You do not need to calculate the
marginal effect using the estimates reported in Figure 5. Instead, your answer should
clearly explain how you would calculate the marginal effect.
c) [5 marks] An alternative to the Probit model is the Linear Probability Model (LPM).
Provide a brief outline of the Linear Probability Model. Explain the main advantages
and disadvantages of using this linear (OLS) specification, compared to an alternative
non-linear procedure, such as the Probit model.
Page 10 of 19
Question 4
Consider the following econometric model for firms operating in the chemical industry:
lnYit = β0 + β1 lnKit + β2 lnLit + β3 lnMit
+ β4 YEAR2t + β5 YEAR3t + εit (4)
where:
Yit = Annual sales for firm i in year t, measured in dollars
Kit = Value of the capital stock for firm i in year t
Lit = Employment for firm i in year t
Mit = Materials inputs for firm i in year t
YEARJt = 1 if YEAR=J, 0 otherwise for J=1,2,3
where lnX denotes the natural logarithm of variable X. Note that YEAR1 is the omitted
category. Economic theory provides the following restrictions on the population parameters:
β1 > 0, β2 > 0, β3 > 0
Suppose you have a dataset with observations on 12,552 firms, all observed for three years,
providing 37,656 total observations.
a) [2 marks] What is the interpretation of the population parameter β4?
b) [2 marks] Suppose you estimate the econometric model (4) by Ordinary Least Squares
(OLS). Do you think that the standard errors are valid? Clearly explain why or why not?
c) Consider the following alternative econometric model for firms operating in the chemical
industry:
lnYit = β0 + β1 lnKit + β2 lnLit + β3 lnMit
+ β4 YEAR2t + β5 YEAR3t + υi + εit (5)
where υi represents an unobserved time-invariant random variable.
i) [4 marks] Suppose you estimate this econometric model using the Random Effects
(RE) estimator. Clearly explain the assumption about the relationship between υi
and each of the inputs Kit, Lit, and Mit that is imposed when estimating the model
using the Random Effects (RE) estimator. Clearly explain, and provide an example,
whether you think that this is a realistic assumption.
Lagrange Multiplier Test - (Breusch-Pagan) for balanced panels
data: lnsales ~ lncapital + lnlabour + lnmaterials + year2 + year3
chisq = 8356.3, df = 1, p-value < 0.00000000000000022
alternative hypothesis: significant effects
Figure 6: LM Test for Random Effects Model (5)
ii) [3 marks] The Random Effects estimator (RE) nests the Pooled OLS model when
σ2υ = 0. Using Figure 6, test the hypothesis that the pooled OLS model is the
most appropriate model for the data. Your answer should clearly state the null and
alternative hypotheses, the distribution of the test statistic, and your decision.
Page 11 of 19
iii) [6 marks] Clearly outline the important differences between the Random Effects
(RE) estimator and the Fixed Effects (FE) estimator. Your answer should clearly
explain the variation in the data that is used to identify the parameters of interest.
d) [3 marks] Consider the following alternative Correlated Random Effects (CRE) model
for firms operating in the chemical industry:
lnYit = β0 + β1 lnKit + β2 lnLit + β3 lnMit
+ β4 YEAR2t + β5 YEAR3t+
+ β6 ln Ki + β7 ln Li + β8 ln M i + ηi + εit (6)
where:
ln Ki = mean value for the (log) capital stock for firm i over time
ln Li = mean value for the (log) labour input for firm i over time
ln M i = mean value for the (log) materials input for firm i over time
and ηi represents an unobserved time-invariant random variable.
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The estimation results for the CRE model (6) are presented in Figure 7. Using the results
in Figure 8 or Figure 9, test the hypothesis that the Random Effects (RE) estimator is
Page 12 of 19
the most appropriate model, at the 5% level of significance. Your answer should clearly
state the null and alternative hypotheses, the distribution of the test statistic, and your
decision.
Untitled
Linear hypothesis test
Hypothesis:
lncapital = 0
lnlabour = 0
lnmaterials = 0
Model 1: restricted model
Model 2: lnsales ~ lncapital + lnlabour + lnmaterials + year2 + year3 +
mlncapital + mlnlabour + mlnmaterials
Note: Coefficient covariance matrix supplied.
Res.Df Df Chisq Pr(>Chisq)
1 37650
2 37647 3 6155.9 < 0.00000000000000022 ***
‐‐‐
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Page 1
Figure 8: H0 : β1 = β2 = β3 = 0 in Model (6)
Untitled
Linear hypothesis test
Hypothesis:
mlncapital = 0
mlnlabour = 0
mlnmaterials = 0
Model 1: restricted model
Model 2: lnsales ~ lncapital + lnlabour + lnmaterials + year2 + year3 +
mlncapital + mlnlabour + mlnmaterials
Note: Coefficient covariance matrix supplied.
Res.Df Df Chisq Pr(>Chisq)
1 37650
2 37647 3 475.36 < 0.00000000000000022 ***
‐‐‐
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Page 1
Figure 9: H0 : β6 = β7 = β8 = 0 in Model (6)
Page 13 of 19
Some Useful Formulas
Variance of the Sum of Two Random Variables
VAR(aX + b Y ) = a2 VAR(X) + b2 VAR(Y ) + 2 a b COV(X, Y )
Sample Variance
V̂AR(X) =
∑N
i=1 (xi − x¯)2
N − 1
Sample Covariance
ĈOV(X, Y ) =
∑N
i=1 (xi − x¯) (yi − y¯)
N − 1
Multiple Linear Regression Model
yi = β0 + β1X1i + β2X2i + . . . βK XKi + εi
OLS Residuals
êi = yi − (b0 + b1X1i + b2X2i + . . . bK XKi)
Estimator of Error Variance
σ̂2 =
∑
ê2i
N −K − 1 =
RSS
N −K − 1
Sample t Statistic
t =
bk − βk
se(bk)
Sample F-statistic
F =
RSSR −RSSUR/M
RSSUR/(N −K − 1) =
(R2UR −R2R)/M
(1−R2UR)/N −K − 1
when the dependent variable in both the restricted and unrestricted model is the same. Here
M denotes the number of restrictions, N denotes the sample size, and K + 1 the number of
estimated parameters in the unrestricted model.
Sample F-statistic for Test of Overall Significance
F =
(TSS −RSS)/K
RSS/(N −K − 1) =
R2/K
(1−R2)/(N −K − 1)
R2
R2 =
∑
(ŷi − y¯)2∑
(yi − y¯)2 = 1−
RSS
TSS
Adjusted R¯2
R¯2 = 1− RSS/(N −K − 1)
TSS/(N − 1)
Page 14 of 19
Probit Model
Latent Variable Formulation
y∗i = β0 + β1X1i + β2X2i + . . . βK XKi + εi εi|Xi ∼ N (0, σ2ε)
Response Probability
Pr(Yi = 1) = Φ
(
β0
σε
+
β1
σε
X1i +
β2
σε
X2i + . . .
βK
σε
XKi
)
where Φ(·) is the cumulative distribution function for the standard normal distribution.
Marginal Effect if Xj is a continuous variable
∂Pr(Yi = 1)
∂Xij
= φ
(
β0
σε
+
β1
σε
X1i +
β2
σε
X2i + . . .
βK
σε
XKi
)
βj
σε
where φ(·) is the probability density function for the standard normal distribution.
Marginal Effect if Xj is an indicator (dummy) variable
∂Pr(Yi = 1)
∂Xij
≈ Φ
(
β0
σε
+
β1
σε
X1i +
β2
σε
X2i + . . .
βj
σε
+ . . .
βK
σε
XKi
)
Xij=1
− Φ
(
β0
σε
+
β1
σε
X1i +
β2
σε
X2i + . . .
βK
σε
XKi
)
Xij=0
White Test for Heteroskedasticity
Econometric Model of Interest
yi = β0 + β1X1i + β2X2i + . . . βK XKi + εi
Auxiliary Regression
eˆ2i = γ0 + γ1 Z1i + γ2 Z2i + . . . γK ZMi + υi
The test statistic is N R2 ∼ χ2(M) where N is the sample size, R2 is the R2 from this auxiliary
regression, and (M + 1) is the number of parameters in the auxiliary regression.
LM Test for First Order AR(1) Autocorrelation
Econometric Model of Interest
yt = β0 + β1X1t + β2X2t + . . . βK XKt + εt
Auxiliary Regression
eˆt = γ0 + γ1X1t + γ2X2t + . . . γK XKt + ρ eˆt−1 + υt
The test statistic is T R2 ∼ χ2(1) where T is the sample size, and R2 is the R2 from this
auxiliary regression.
Page 15 of 19
Dickey-Fuller Test
• Version I: (no constant, no trend):
∆yt = γ yt−1 + υt
• Version II: (constant, no trend):
∆yt = α + γ yt−1 + υt
• Version III: (constant, trend):
∆yt = α + δ t+ γ yt−1 + υt
Augmented Dickey-Fuller test
With intercept :
∆yt = α + γ yt−1 +
m∑
s=1
δ∆ yt−s + υt
Page 16 of 19
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Table 4: Critical Values for 5% Upper Tail Probabilities of the F Distribution
Page 6 of 9
Page 17 of 19
Right-Tail Critical Values for the t-distribution
Table 2: Critical Values for Upper Tail Probabilities of Student’s t distribution
Shaded Area = P (t ≥ tν,α) = α
For example,
P (t ≥ t3,0.05) = P (t ≥ 2.3534) = 0.05.
Note that ν denotes the degrees
of freedom of the distribution.
ν 0.1 0.05 0.025 0.01 0.005 ν 0.1 0.05 0.025 0.01 0.005
1 3.0777 6.3137 12.7062 31.8210 63.6559 28 1.3125 1.7011 2.0484 2.4671 2.7633
2 1.8856 2.9200 4.3027 6.9645 9.9250 29 1.3114 1.6991 2.0452 2.4620 2.7564
3 1.6377 2.3534 3.1824 4.5407 5.8408 30 1.3104 1.6973 2.0423 2.4573 2.7500
4 1.5332 2.1318 2.7765 3.7469 4.6041 31 1.3095 1.6955 2.0395 2.4528 2.7440
5 1.4759 2.0150 2.5706 3.3649 4.0321 32 1.3086 1.6939 2.0369 2.4487 2.7385
6 1.4398 1.9432 2.4469 3.1427 3.7074 33 1.3077 1.6924 2.0345 2.4448 2.7333
7 1.4149 1.8946 2.3646 2.9979 3.4995 34 1.3070 1.6909 2.0322 2.4411 2.7284
8 1.3968 1.8595 2.3060 2.8965 3.3554 35 1.3062 1.6896 2.0301 2.4377 2.7238
9 1.3830 1.8331 2.2622 2.8214 3.2498 36 1.3055 1.6883 2.0281 2.4345 2.7195
10 1.3722 1.8125 2.2281 2.7638 3.1693 37 1.3049 1.6871 2.0262 2.4314 2.7154
11 1.3634 1.7959 2.2010 2.7181 3.1058 38 1.3042 1.6860 2.0244 2.4286 2.7116
12 1.3562 1.7823 2.1788 2.6810 3.0545 39 1.3036 1.6849 2.0227 2.4258 2.7079
13 1.3502 1.7709 2.1604 2.6503 3.0123 40 1.3031 1.6839 2.0211 2.4233 2.7045
14 1.3450 1.7613 2.1448 2.6245 2.9768 45 1.3007 1.6794 2.0141 2.4121 2.6896
15 1.3406 1.7531 2.1315 2.6025 2.9467 50 1.2987 1.6759 2.0086 2.4033 2.6778
16 1.3368 1.7459 2.1199 2.5835 2.9208 60 1.2958 1.6706 2.0003 2.3901 2.6603
17 1.3334 1.7396 2.1098 2.5669 2.8982 70 1.2938 1.6669 1.9944 2.3808 2.6479
18 1.3304 1.7341 2.1009 2.5524 2.8784 80 1.2922 1.6641 1.9901 2.3739 2.6387
19 1.3277 1.7291 2.0930 2.5395 2.8609 90 1.2910 1.6620 1.9867 2.3685 2.6316
20 1.3253 1.7247 2.0860 2.5280 2.8453 100 1.2901 1.6602 1.9840 2.3642 2.6259
21 1.3232 1.7207 2.0796 2.5176 2.8314 120 1.2886 1.6576 1.9799 2.3578 2.6174
22 1.3212 1.7171 2.0739 2.5083 2.8188 140 1.2876 1.6558 1.9771 2.3533 2.6114
23 1.3195 1.7139 2.0687 2.4999 2.8073 160 1.2869 1.6544 1.9749 2.3499 2.6069
24 1.3178 1.7109 2.0639 2.4922 2.7970 180 1.2863 1.6534 1.9732 2.3472 2.6034
25 1.3163 1.7081 2.0595 2.4851 2.7874 200 1.2858 1.6525 1.9719 2.3451 2.6006
26 1.3150 1.7056 2.0555 2.4786 2.7787 ∞ 1.2816 1.6449 1.9600 2.3263 2.5758
27 1.3137 1.7033 2.0518 2.4727 2.7707
Page 3 of 9
Shaded Area = Pr(t > tυ,α) = α
For example:
Pr(t > t45,0.05) = Pr(t > 1.6794) = 0.05
Note: υ denotes the degrees of freedom
of the distribution
Critical Values for Upper Tail Probabilities of the (Student’s) t-Distribution
df Upper Tail Probability df Upper Tail Probability
υ 0.100 0.050 0.025 0.010 0.005 υ 0.100 0.050 0.025 0.010 0.005
1 3.0777 6.3137 1 .7062 31.8210 63.6559 28 1 3125 1.7011 2.0484 2.4671 2.7633
2 1.8856 2.9200 4.3027 6.9645 9.9250 29 1.3114 1.6991 2.0452 2.4620 2.7564
3 1.6377 2.3534 3.1824 4.5407 5.8408 30 1.3104 1.6973 2.0423 2.4573 2.7500
4 1.5332 2.1318 2.7765 3.7469 4.6041 31 1.3095 1.6955 2.0395 2.4528 2.7440
5 1.4759 2.0150 2.5706 3.3649 4.0321 32 1.3086 1.6939 2.0369 2.4487 2.7385
6 1.4 98 1.9432 2.4469 3.1427 3.7074 33 1 3077 1.6924 2.0345 2.4448 2.7333
7 1.4149 1.8946 2.3646 2.9979 3.4995 34 1.3070 1.6909 2.0322 2.4411 2.7284
8 1.3968 1.8595 2.3060 2.8965 3.3554 35 1.3062 1.6896 2.0301 2.4377 2.7238
9 1.3830 1.8331 2.2622 2.8214 3.2498 36 1.3055 1.6883 2.0281 2.4345 2.7195
10 1.3722 1.8125 2.2281 .7638 3.1693 37 .30 9 1.6871 2.0262 2.4314 2.7154
11 1.3634 1.7959 .2010 2.71 1 3.1058 38 3042 1.6860 2.0244 2.4286 2.7116
12 1.3562 1.7823 2.1788 2.6810 3.0545 39 1.3036 1.6849 2.0227 2.4258 2.7079
13 1.3502 1.7709 2.1604 2.6503 3.0123 40 1.3031 1.6839 2.0211 2.4233 2.7045
14 1.3450 1.7613 2.1448 2.6245 2.9768 45 1.3007 1.6794 2.0141 2.4121 2.6896
15 1. 406 1.7531 2.1315 2.6025 2.9467 0 .2987 1.6759 2.0086 2.4033 2.6778
16 1.3368 1.7459 .1199 2.5835 2.9208 60 .2958 1.6 06 2.0003 2.3901 2.6603
17 1.3334 1.7396 2.1098 2.5669 2.8982 70 1.2938 1.6669 1.9944 2.3808 2.6479
18 1.3304 1.7341 2.1009 2.5524 2.8784 80 1.2922 1.6641 1.9901 2.3739 2.6387
19 1.3277 1.7291 2.0930 2.5395 2.8609 90 1.2910 1.6620 1.9867 2.3685 2.6316
20 1.3253 1.7247 2.0860 2.5280 2.8453 100 1.2901 1.6602 1.9840 2.3642 2.6259
21 1.3232 1.7207 2.0796 2.5176 2.8314 120 1.2886 1.6576 1.9799 2.3578 2.6174
22 1.3212 1.7171 2.0739 2.5083 2.8188 140 1.2876 1.6558 1.9771 2.3533 2.6114
23 1.3195 1.7139 2.0687 2.4999 2.8073 160 1.2869 1.6544 1.9749 2.3499 2.6069
24 1.3178 1.7109 2.0639 2.4922 2.7970 180 1.2863 1.6534 1.9732 2.3472 2.6034
25 1.3163 1.7081 2.0595 2.4851 2.7874 200 1.2858 1.6525 1.9719 2.3451 2.6006
26 1.3150 1.7056 2.0555 2.4786 2.7787 ∞ 1.2816 1.6449 1.9600 2.3263 2.5758
27 1.3137 1.7033 2.0518 2.4727 2.7707
Page 18 of 19
Critical Values for Upper Tail Probabilities of the χ2 Distribution
df Upper Tail Probability (α)
υ 0.2500 0.1000 0.0500 0.0250 0.0100 0.0050
1 1.3233 2.7055 3.8415 5.0239 6.6349 7.8794
2 2.7726 4.6052 5.9915 7.3778 9.2103 10.5966
3 4.1083 6.2514 7.8147 9.3484 11.3449 12.8382
4 5.3853 7.7794 9.4877 11.1433 13.2767 14.8603
5 6.6257 9.2364 11.0705 12.8325 15.0863 16.7496
6 7.8408 10.6446 12.5916 14.4494 16.8119 18.5476
7 9.0371 12.0170 14.0671 16.0128 18.4753 20.2777
8 10.2189 13.3616 15.5073 17.5345 20.0902 21.9550
9 11.3888 14.6837 16.9190 19.0228 21.6660 23.5894
10 12.5489 15.9872 18.3070 20.4832 23.2093 25.1882
11 13.7007 17.2750 19.6751 21.9200 24.7250 26.7568
12 14.8454 18.5493 21.0261 23.3367 26.2170 28.2995
13 15.9839 19.8119 22.3620 24.7356 27.6882 29.8195
14 17.1169 21.0641 23.6848 26.1189 29.1412 31.3193
15 18.2451 22.3071 24.9958 27.4884 30.5779 32.8013
16 19.3689 23.5418 26.2962 28.8454 31.9999 34.2672
17 20.4887 24.7690 27.5871 30.1910 33.4087 35.7185
18 21.6049 25.9894 28.8693 31.5264 34.8053 37.1565
19 22.7178 27.2036 30.1435 32.8523 36.1909 38.5823
20 23.8277 28.4120 31.4104 34.1696 37.5662 39.9968
25 29.3389 34.3816 37.6525 40.6465 44.3141 46.9279
30 34.7997 40.2560 43.7730 46.9792 50.8922 53.6720
35 40.2228 46.0588 49.8018 53.2033 57.3421 60.2748
40 45.6160 51.8051 55.7585 59.3417 63.6907 66.7660
45 50.9849 57.5053 61.6562 65.4102 69.9568 73.1661
50 56.3336 63.1671 67.5048 71.4202 76.1539 79.4900
60 66.9815 74.3970 79.0819 83.2977 88.3794 91.9517
80 88.1303 96.5782 101.8795 106.6286 112.3288 116.3211
100 109.1412 118.4980 124.3421 129.5612 135.8067 140.1695
120 130.0546 140.2326 146.5674 152.2114 158.9502 163.6482
END OF EXAMINATION
Page 19 of 19
DEPARTMENT OF ECONOMICS
SEMESTER 1 ASSESSMENT, 2019
ECOM30001: Basic Econometrics
Time Allowed: TWO Hours
Reading Time: 15 minutes
This examination paper contributes 60 percent to the assessment in ECOM30001.
This examination consists of four (4) questions in total.
You must answer ALL four questions.
Each question is worth twenty (20) marks for a maximum of 80 marks.
The examination paper should be inserted in the back of your answer booklet at the end
of the examination.
The following items are authorized in the examination room:
-Casio FX-82 (any suffix) calculator
This exam has 19 pages.
The paper may not be removed from the exam room.
Answer all questions in the examination booklet(s) provided.
The examination contains a formula sheet starting on page 14.
The examination contains critical values for a number of distributions starting on page 17.
Page 2 of 19
Question 1
a) [5 marks] Consider the following econometric model:
yi = β0 + β1Xi + εi
What is meant by the term heteroskedasticity in the random error? What are the con-
sequences for the OLS estimator if you ignore heteroskedasticity in the random error εi?
Briefly outline how you would test for the presence of heteroskedasticity using White’s
test. Your answer should clearly state the null and alternative hypotheses, the test statis-
tic and its distribution.
b) [5 marks] Consider the following econometric model:
yi = β0 + β1X
∗
i + εi
Consider the following measurement equation relating the observed value for X to the
‘true’ value X∗:
Xi = X
∗
i + υi
with COV(X∗i , εi) = 0 and COV(υi, εi) = 0.
What is meant by the term classical measurement error? Clearly explain the implications
for the OLS estimator of β1 if the explanatory variable X
∗ is observed with (classical)
measurement error.
c) [5 marks] Consider the following econometric model:
yt = β0 + β1Xt + εt
What is meant by the term autocorrelation in the random error? Clearly explain the
consequences for the OLS estimator if you ignore autocorrelation in the random error εt.
Briefly outline how you would test for first order AR(1) autocorrelation in the random
error. Your answer should clearly state the null and alternative hypotheses, the test
statistic and its distribution.
Page 3 of 19
d) [5 marks] Consider the following regression:
∆ inft = β0 + β1 inft−1 + β2 ∆ inft−1 + β3 ∆ inft−2 + β4 ∆ inft−3 + β5 ∆ inft−4 + εt
where inf represents the quarterly inflation rate. This econometric model was estimated
using the method of Ordinary Least Squares (OLS) for the period 1948:Q1 to 2016:Q1
and the results are presented in Figure 1.
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Outline how you would test whether the series for the quarterly inflation rate was sta-
tionary or not. Your answer should clearly state the null and alternative hypotheses, the
test statistic and its distribution. Using the results in Figure 1, what is the value of the
Augmented Dickey-Fuller test statistic? At the 5% level of significance, explain whether
the sample evidence is consistent with the null hypothesis. Based upon the estimation
results presented in Figure 1, the p-value associated with the Augmented Dickey-Fuller
test is 0.0080.
Page 4 of 19
Question 2
Consider the following demand function for airline seats on routes in a large country:
lnpasseni = β0 + β1lnfarei + β2lndisti + β3lndist
2
i + εi (1)
where:
passeni = average (log) number of passengers per day on route i
farei = average (log) fare on route i, in dollars
lndisti = average (log) distance of route i, in miles
conceni = average market share of largest carrier on routei
and ln(X) denotes the natural logarithm of variable X.
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a) [3 marks] The estimation results from estimating model (1) by the method of Ordinary
Least Squares (OLS) are reported in Figure 2. At the 5% level of significance, test
the hypothesis that the price elasticity of demand for air travel is elastic, that is the
price elasticity of demand is less than -1. Your answer should clearly state the null and
alternative hypotheses, the distribution of the test statistic, and your decision.
b) [3 marks] Do you think that the condition COV(fare, ε|dist) = 0 is likely to be satisfied?
Clearly explain why or why not. Explain the consequences for the OLS estimator if this
condition is not satisfied.
c) [4 marks] Consider a variable concen that might be suitable as an instrumental variable
for ln (fare). This variable is a measure of market concentration, measured by the market
share of the largest carrier on the route. Clearly, explain the two conditions that must be
satisfied for the variable concen to be a valid instrumental variable. Do you think these
two conditions are likely to be satisfied? Why or why not?
Page 5 of 19
d) [2 marks] Consider the following first stage:
lnfarei = pi0 + pi1 conceni + pi2lndisti + pi3lndist
2
i + υi (2)
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The results from estimating this first stage model by OLS are reported in Figure 3. Based
upon these results, explain whether the variable concen is an adequate instrumental
variable for ln (fare).
Page 6 of 19
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e) [3 marks] The estimation results from estimating model (1) by the method of Instru-
mental Variables (IV) are reported in Figure 4. At the 5% level of significance, test
the hypothesis that the price elasticity of demand for air travel is elastic, that is the
price elasticity of demand is less than -1. Your answer should clearly state the null and
alternative hypotheses, the distribution of the test statistic, and your decision.
f) [2 marks] Compare and contrast the estimate of β1 obtained using OLS (Figure 2) with
that obtained using the method of IV (Figure 4). In your answer comment on both the
magnitude of the estimate and the statistical significance of the estimate of β1.
g) [3 marks] An alternative instrumental variable for ln (fare) might be the total cost of
fuel on route i. Clearly, explain the two conditions that must be satisfied for this variable
to be a valid instrumental variable. Do you think these two conditions are likely to be
satisfied? Why or why not.
Page 7 of 19
Question 3
Consider the following econometric model for the approval of applications for (household)
mortgage finance:
deny∗i = β0 + β1 pii + β2 hse inci + β3 lvratmedi + β4 lvrathighi
+ β5 ccredlowi + β6 ccredmedi + β7 pubreci + β8 denpmii + εi (3)
where εi|Xi ∼ N (0, σ2) and deny∗i is a latent variable that determines the choice to reject a
mortgage application. In addition:
pi = Ratio of total montly debt payments to total monthly income
hse inc = Ratio of monthly housing expensese to total montly income
lvratlow = 1 if ratio of loan to assessed value of the property is below 80%, 0 otherwise
lvratmed = 1 if ratio of loan to assessed value of the property is above 80%
but below 95%, 0 otherwise
lvrathigh = 1 if ratio of loan to assessed value of the property is above 95%, 0 otherwise
ccredlow = 1 if poor consumer credit score, 0
ccredmed = 1 if satisfactory consumer credit score, 0 otherwise
ccredhigh = 1 if excellent consumer credit score, 0 otherwise
pubrec = 1 if public record of bad credit, 0 otherwise
denpmi = 1 if denied mortgage insurance, 0 otherwise
Note that lvratlow, ccredhigh are the omitted categories.
a) [1 mark] Provide a brief interpretation of the latent variable deny∗i .
b) Suppose you have a dataset with observations on 2,303 individuals. The data contain
information on mortgage applications, augmented with additional data from the banks
and lending institutions that reviewed these applications. The data relate to household
mortgage applications during a particular calendar year. The data provide an indicator
variable for the approval of housing finance applications:
deny = 1 if mortgate application denied, 0 otherwise
The parameters of the model (3) were estimated as a Probit model and the results are
provided in Figure 5.
Note that the probability density function for a standard normal random variable (Z) is
given by:
φ(Z) =
1√
2pi
exp
(−Z2
2
)
Page 8 of 19
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Page 9 of 19
i) [4 marks] Let pˆi represent the predicted probability that a mortgage application
is denied, for each individual in the sample. Consider the following decision rule: if
pˆi ≥ 0.5, predict that d̂enyi = 1, otherwise d̂enyi = 0. Based on the information
in Table 1, calculate the percentage of outcomes that are successfully predicted?
Using Table 1, comment on the usefulness of the model in predicting denyi = 1 or
denyi = 0.
true predicted frequency
0 0 2,015
1 0 189
0 1 20
1 1 79
2,303
Table 1: Predicted Probability Threshold pˆi ≥ 0.5
ii) [5 marks] Calculate the marginal effect for the payment-income ratio (pi) for an
individual with a pi ratio of 0.30, a housing expense-income ratio of 0.25, an excellent
credit score, a loan-valuation ratio below 80%, no record of bad credit, and who
has never been denied mortgage insurance. Provide a clear interpretation for your
calculated marginal effect.
iii) [5 marks] Explain how you would compute the average marginal effect (AME)
for being denied mortgage insurance (denpmi). You do not need to calculate the
marginal effect using the estimates reported in Figure 5. Instead, your answer should
clearly explain how you would calculate the marginal effect.
c) [5 marks] An alternative to the Probit model is the Linear Probability Model (LPM).
Provide a brief outline of the Linear Probability Model. Explain the main advantages
and disadvantages of using this linear (OLS) specification, compared to an alternative
non-linear procedure, such as the Probit model.
Page 10 of 19
Question 4
Consider the following econometric model for firms operating in the chemical industry:
lnYit = β0 + β1 lnKit + β2 lnLit + β3 lnMit
+ β4 YEAR2t + β5 YEAR3t + εit (4)
where:
Yit = Annual sales for firm i in year t, measured in dollars
Kit = Value of the capital stock for firm i in year t
Lit = Employment for firm i in year t
Mit = Materials inputs for firm i in year t
YEARJt = 1 if YEAR=J, 0 otherwise for J=1,2,3
where lnX denotes the natural logarithm of variable X. Note that YEAR1 is the omitted
category. Economic theory provides the following restrictions on the population parameters:
β1 > 0, β2 > 0, β3 > 0
Suppose you have a dataset with observations on 12,552 firms, all observed for three years,
providing 37,656 total observations.
a) [2 marks] What is the interpretation of the population parameter β4?
b) [2 marks] Suppose you estimate the econometric model (4) by Ordinary Least Squares
(OLS). Do you think that the standard errors are valid? Clearly explain why or why not?
c) Consider the following alternative econometric model for firms operating in the chemical
industry:
lnYit = β0 + β1 lnKit + β2 lnLit + β3 lnMit
+ β4 YEAR2t + β5 YEAR3t + υi + εit (5)
where υi represents an unobserved time-invariant random variable.
i) [4 marks] Suppose you estimate this econometric model using the Random Effects
(RE) estimator. Clearly explain the assumption about the relationship between υi
and each of the inputs Kit, Lit, and Mit that is imposed when estimating the model
using the Random Effects (RE) estimator. Clearly explain, and provide an example,
whether you think that this is a realistic assumption.
Lagrange Multiplier Test - (Breusch-Pagan) for balanced panels
data: lnsales ~ lncapital + lnlabour + lnmaterials + year2 + year3
chisq = 8356.3, df = 1, p-value < 0.00000000000000022
alternative hypothesis: significant effects
Figure 6: LM Test for Random Effects Model (5)
ii) [3 marks] The Random Effects estimator (RE) nests the Pooled OLS model when
σ2υ = 0. Using Figure 6, test the hypothesis that the pooled OLS model is the
most appropriate model for the data. Your answer should clearly state the null and
alternative hypotheses, the distribution of the test statistic, and your decision.
Page 11 of 19
iii) [6 marks] Clearly outline the important differences between the Random Effects
(RE) estimator and the Fixed Effects (FE) estimator. Your answer should clearly
explain the variation in the data that is used to identify the parameters of interest.
d) [3 marks] Consider the following alternative Correlated Random Effects (CRE) model
for firms operating in the chemical industry:
lnYit = β0 + β1 lnKit + β2 lnLit + β3 lnMit
+ β4 YEAR2t + β5 YEAR3t+
+ β6 ln Ki + β7 ln Li + β8 ln M i + ηi + εit (6)
where:
ln Ki = mean value for the (log) capital stock for firm i over time
ln Li = mean value for the (log) labour input for firm i over time
ln M i = mean value for the (log) materials input for firm i over time
and ηi represents an unobserved time-invariant random variable.
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GGG
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GGG
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N42?,-./0-2=.9;
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N42?,-./0N2:4;H235
FCBQDD
GGG
,FCFBFJ0
R=54;S2:H.?5 IQTELE
7
D
FCMMIJ
UVW95:4V7
D
FCMMII
XY$ +
G
AZFCFL[
GG
AZFCFB[
GGG
AZFCFFB
Figure 7: CRE Results (with cluster-robust standard errors) for Model (6)
The estimation results for the CRE model (6) are presented in Figure 7. Using the results
in Figure 8 or Figure 9, test the hypothesis that the Random Effects (RE) estimator is
Page 12 of 19
the most appropriate model, at the 5% level of significance. Your answer should clearly
state the null and alternative hypotheses, the distribution of the test statistic, and your
decision.
Untitled
Linear hypothesis test
Hypothesis:
lncapital = 0
lnlabour = 0
lnmaterials = 0
Model 1: restricted model
Model 2: lnsales ~ lncapital + lnlabour + lnmaterials + year2 + year3 +
mlncapital + mlnlabour + mlnmaterials
Note: Coefficient covariance matrix supplied.
Res.Df Df Chisq Pr(>Chisq)
1 37650
2 37647 3 6155.9 < 0.00000000000000022 ***
‐‐‐
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Page 1
Figure 8: H0 : β1 = β2 = β3 = 0 in Model (6)
Untitled
Linear hypothesis test
Hypothesis:
mlncapital = 0
mlnlabour = 0
mlnmaterials = 0
Model 1: restricted model
Model 2: lnsales ~ lncapital + lnlabour + lnmaterials + year2 + year3 +
mlncapital + mlnlabour + mlnmaterials
Note: Coefficient covariance matrix supplied.
Res.Df Df Chisq Pr(>Chisq)
1 37650
2 37647 3 475.36 < 0.00000000000000022 ***
‐‐‐
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Page 1
Figure 9: H0 : β6 = β7 = β8 = 0 in Model (6)
Page 13 of 19
Some Useful Formulas
Variance of the Sum of Two Random Variables
VAR(aX + b Y ) = a2 VAR(X) + b2 VAR(Y ) + 2 a b COV(X, Y )
Sample Variance
V̂AR(X) =
∑N
i=1 (xi − x¯)2
N − 1
Sample Covariance
ĈOV(X, Y ) =
∑N
i=1 (xi − x¯) (yi − y¯)
N − 1
Multiple Linear Regression Model
yi = β0 + β1X1i + β2X2i + . . . βK XKi + εi
OLS Residuals
êi = yi − (b0 + b1X1i + b2X2i + . . . bK XKi)
Estimator of Error Variance
σ̂2 =
∑
ê2i
N −K − 1 =
RSS
N −K − 1
Sample t Statistic
t =
bk − βk
se(bk)
Sample F-statistic
F =
RSSR −RSSUR/M
RSSUR/(N −K − 1) =
(R2UR −R2R)/M
(1−R2UR)/N −K − 1
when the dependent variable in both the restricted and unrestricted model is the same. Here
M denotes the number of restrictions, N denotes the sample size, and K + 1 the number of
estimated parameters in the unrestricted model.
Sample F-statistic for Test of Overall Significance
F =
(TSS −RSS)/K
RSS/(N −K − 1) =
R2/K
(1−R2)/(N −K − 1)
R2
R2 =
∑
(ŷi − y¯)2∑
(yi − y¯)2 = 1−
RSS
TSS
Adjusted R¯2
R¯2 = 1− RSS/(N −K − 1)
TSS/(N − 1)
Page 14 of 19
Probit Model
Latent Variable Formulation
y∗i = β0 + β1X1i + β2X2i + . . . βK XKi + εi εi|Xi ∼ N (0, σ2ε)
Response Probability
Pr(Yi = 1) = Φ
(
β0
σε
+
β1
σε
X1i +
β2
σε
X2i + . . .
βK
σε
XKi
)
where Φ(·) is the cumulative distribution function for the standard normal distribution.
Marginal Effect if Xj is a continuous variable
∂Pr(Yi = 1)
∂Xij
= φ
(
β0
σε
+
β1
σε
X1i +
β2
σε
X2i + . . .
βK
σε
XKi
)
βj
σε
where φ(·) is the probability density function for the standard normal distribution.
Marginal Effect if Xj is an indicator (dummy) variable
∂Pr(Yi = 1)
∂Xij
≈ Φ
(
β0
σε
+
β1
σε
X1i +
β2
σε
X2i + . . .
βj
σε
+ . . .
βK
σε
XKi
)
Xij=1
− Φ
(
β0
σε
+
β1
σε
X1i +
β2
σε
X2i + . . .
βK
σε
XKi
)
Xij=0
White Test for Heteroskedasticity
Econometric Model of Interest
yi = β0 + β1X1i + β2X2i + . . . βK XKi + εi
Auxiliary Regression
eˆ2i = γ0 + γ1 Z1i + γ2 Z2i + . . . γK ZMi + υi
The test statistic is N R2 ∼ χ2(M) where N is the sample size, R2 is the R2 from this auxiliary
regression, and (M + 1) is the number of parameters in the auxiliary regression.
LM Test for First Order AR(1) Autocorrelation
Econometric Model of Interest
yt = β0 + β1X1t + β2X2t + . . . βK XKt + εt
Auxiliary Regression
eˆt = γ0 + γ1X1t + γ2X2t + . . . γK XKt + ρ eˆt−1 + υt
The test statistic is T R2 ∼ χ2(1) where T is the sample size, and R2 is the R2 from this
auxiliary regression.
Page 15 of 19
Dickey-Fuller Test
• Version I: (no constant, no trend):
∆yt = γ yt−1 + υt
• Version II: (constant, no trend):
∆yt = α + γ yt−1 + υt
• Version III: (constant, trend):
∆yt = α + δ t+ γ yt−1 + υt
Augmented Dickey-Fuller test
With intercept :
∆yt = α + γ yt−1 +
m∑
s=1
δ∆ yt−s + υt
Page 16 of 19
C
ri
ti
ca
l
V
al
u
es
fo
r
th
e
5%
U
p
p
er
T
ai
l
P
ro
b
ab
il
it
ie
s
of
th
e
F
D
is
tr
ib
u
ti
on
Table 4: Critical Values for 5% Upper Tail Probabilities of the F Distribution
Page 6 of 9
Page 17 of 19
Right-Tail Critical Values for the t-distribution
Table 2: Critical Values for Upper Tail Probabilities of Student’s t distribution
Shaded Area = P (t ≥ tν,α) = α
For example,
P (t ≥ t3,0.05) = P (t ≥ 2.3534) = 0.05.
Note that ν denotes the degrees
of freedom of the distribution.
ν 0.1 0.05 0.025 0.01 0.005 ν 0.1 0.05 0.025 0.01 0.005
1 3.0777 6.3137 12.7062 31.8210 63.6559 28 1.3125 1.7011 2.0484 2.4671 2.7633
2 1.8856 2.9200 4.3027 6.9645 9.9250 29 1.3114 1.6991 2.0452 2.4620 2.7564
3 1.6377 2.3534 3.1824 4.5407 5.8408 30 1.3104 1.6973 2.0423 2.4573 2.7500
4 1.5332 2.1318 2.7765 3.7469 4.6041 31 1.3095 1.6955 2.0395 2.4528 2.7440
5 1.4759 2.0150 2.5706 3.3649 4.0321 32 1.3086 1.6939 2.0369 2.4487 2.7385
6 1.4398 1.9432 2.4469 3.1427 3.7074 33 1.3077 1.6924 2.0345 2.4448 2.7333
7 1.4149 1.8946 2.3646 2.9979 3.4995 34 1.3070 1.6909 2.0322 2.4411 2.7284
8 1.3968 1.8595 2.3060 2.8965 3.3554 35 1.3062 1.6896 2.0301 2.4377 2.7238
9 1.3830 1.8331 2.2622 2.8214 3.2498 36 1.3055 1.6883 2.0281 2.4345 2.7195
10 1.3722 1.8125 2.2281 2.7638 3.1693 37 1.3049 1.6871 2.0262 2.4314 2.7154
11 1.3634 1.7959 2.2010 2.7181 3.1058 38 1.3042 1.6860 2.0244 2.4286 2.7116
12 1.3562 1.7823 2.1788 2.6810 3.0545 39 1.3036 1.6849 2.0227 2.4258 2.7079
13 1.3502 1.7709 2.1604 2.6503 3.0123 40 1.3031 1.6839 2.0211 2.4233 2.7045
14 1.3450 1.7613 2.1448 2.6245 2.9768 45 1.3007 1.6794 2.0141 2.4121 2.6896
15 1.3406 1.7531 2.1315 2.6025 2.9467 50 1.2987 1.6759 2.0086 2.4033 2.6778
16 1.3368 1.7459 2.1199 2.5835 2.9208 60 1.2958 1.6706 2.0003 2.3901 2.6603
17 1.3334 1.7396 2.1098 2.5669 2.8982 70 1.2938 1.6669 1.9944 2.3808 2.6479
18 1.3304 1.7341 2.1009 2.5524 2.8784 80 1.2922 1.6641 1.9901 2.3739 2.6387
19 1.3277 1.7291 2.0930 2.5395 2.8609 90 1.2910 1.6620 1.9867 2.3685 2.6316
20 1.3253 1.7247 2.0860 2.5280 2.8453 100 1.2901 1.6602 1.9840 2.3642 2.6259
21 1.3232 1.7207 2.0796 2.5176 2.8314 120 1.2886 1.6576 1.9799 2.3578 2.6174
22 1.3212 1.7171 2.0739 2.5083 2.8188 140 1.2876 1.6558 1.9771 2.3533 2.6114
23 1.3195 1.7139 2.0687 2.4999 2.8073 160 1.2869 1.6544 1.9749 2.3499 2.6069
24 1.3178 1.7109 2.0639 2.4922 2.7970 180 1.2863 1.6534 1.9732 2.3472 2.6034
25 1.3163 1.7081 2.0595 2.4851 2.7874 200 1.2858 1.6525 1.9719 2.3451 2.6006
26 1.3150 1.7056 2.0555 2.4786 2.7787 ∞ 1.2816 1.6449 1.9600 2.3263 2.5758
27 1.3137 1.7033 2.0518 2.4727 2.7707
Page 3 of 9
Shaded Area = Pr(t > tυ,α) = α
For example:
Pr(t > t45,0.05) = Pr(t > 1.6794) = 0.05
Note: υ denotes the degrees of freedom
of the distribution
Critical Values for Upper Tail Probabilities of the (Student’s) t-Distribution
df Upper Tail Probability df Upper Tail Probability
υ 0.100 0.050 0.025 0.010 0.005 υ 0.100 0.050 0.025 0.010 0.005
1 3.0777 6.3137 1 .7062 31.8210 63.6559 28 1 3125 1.7011 2.0484 2.4671 2.7633
2 1.8856 2.9200 4.3027 6.9645 9.9250 29 1.3114 1.6991 2.0452 2.4620 2.7564
3 1.6377 2.3534 3.1824 4.5407 5.8408 30 1.3104 1.6973 2.0423 2.4573 2.7500
4 1.5332 2.1318 2.7765 3.7469 4.6041 31 1.3095 1.6955 2.0395 2.4528 2.7440
5 1.4759 2.0150 2.5706 3.3649 4.0321 32 1.3086 1.6939 2.0369 2.4487 2.7385
6 1.4 98 1.9432 2.4469 3.1427 3.7074 33 1 3077 1.6924 2.0345 2.4448 2.7333
7 1.4149 1.8946 2.3646 2.9979 3.4995 34 1.3070 1.6909 2.0322 2.4411 2.7284
8 1.3968 1.8595 2.3060 2.8965 3.3554 35 1.3062 1.6896 2.0301 2.4377 2.7238
9 1.3830 1.8331 2.2622 2.8214 3.2498 36 1.3055 1.6883 2.0281 2.4345 2.7195
10 1.3722 1.8125 2.2281 .7638 3.1693 37 .30 9 1.6871 2.0262 2.4314 2.7154
11 1.3634 1.7959 .2010 2.71 1 3.1058 38 3042 1.6860 2.0244 2.4286 2.7116
12 1.3562 1.7823 2.1788 2.6810 3.0545 39 1.3036 1.6849 2.0227 2.4258 2.7079
13 1.3502 1.7709 2.1604 2.6503 3.0123 40 1.3031 1.6839 2.0211 2.4233 2.7045
14 1.3450 1.7613 2.1448 2.6245 2.9768 45 1.3007 1.6794 2.0141 2.4121 2.6896
15 1. 406 1.7531 2.1315 2.6025 2.9467 0 .2987 1.6759 2.0086 2.4033 2.6778
16 1.3368 1.7459 .1199 2.5835 2.9208 60 .2958 1.6 06 2.0003 2.3901 2.6603
17 1.3334 1.7396 2.1098 2.5669 2.8982 70 1.2938 1.6669 1.9944 2.3808 2.6479
18 1.3304 1.7341 2.1009 2.5524 2.8784 80 1.2922 1.6641 1.9901 2.3739 2.6387
19 1.3277 1.7291 2.0930 2.5395 2.8609 90 1.2910 1.6620 1.9867 2.3685 2.6316
20 1.3253 1.7247 2.0860 2.5280 2.8453 100 1.2901 1.6602 1.9840 2.3642 2.6259
21 1.3232 1.7207 2.0796 2.5176 2.8314 120 1.2886 1.6576 1.9799 2.3578 2.6174
22 1.3212 1.7171 2.0739 2.5083 2.8188 140 1.2876 1.6558 1.9771 2.3533 2.6114
23 1.3195 1.7139 2.0687 2.4999 2.8073 160 1.2869 1.6544 1.9749 2.3499 2.6069
24 1.3178 1.7109 2.0639 2.4922 2.7970 180 1.2863 1.6534 1.9732 2.3472 2.6034
25 1.3163 1.7081 2.0595 2.4851 2.7874 200 1.2858 1.6525 1.9719 2.3451 2.6006
26 1.3150 1.7056 2.0555 2.4786 2.7787 ∞ 1.2816 1.6449 1.9600 2.3263 2.5758
27 1.3137 1.7033 2.0518 2.4727 2.7707
Page 18 of 19
Critical Values for Upper Tail Probabilities of the χ2 Distribution
df Upper Tail Probability (α)
υ 0.2500 0.1000 0.0500 0.0250 0.0100 0.0050
1 1.3233 2.7055 3.8415 5.0239 6.6349 7.8794
2 2.7726 4.6052 5.9915 7.3778 9.2103 10.5966
3 4.1083 6.2514 7.8147 9.3484 11.3449 12.8382
4 5.3853 7.7794 9.4877 11.1433 13.2767 14.8603
5 6.6257 9.2364 11.0705 12.8325 15.0863 16.7496
6 7.8408 10.6446 12.5916 14.4494 16.8119 18.5476
7 9.0371 12.0170 14.0671 16.0128 18.4753 20.2777
8 10.2189 13.3616 15.5073 17.5345 20.0902 21.9550
9 11.3888 14.6837 16.9190 19.0228 21.6660 23.5894
10 12.5489 15.9872 18.3070 20.4832 23.2093 25.1882
11 13.7007 17.2750 19.6751 21.9200 24.7250 26.7568
12 14.8454 18.5493 21.0261 23.3367 26.2170 28.2995
13 15.9839 19.8119 22.3620 24.7356 27.6882 29.8195
14 17.1169 21.0641 23.6848 26.1189 29.1412 31.3193
15 18.2451 22.3071 24.9958 27.4884 30.5779 32.8013
16 19.3689 23.5418 26.2962 28.8454 31.9999 34.2672
17 20.4887 24.7690 27.5871 30.1910 33.4087 35.7185
18 21.6049 25.9894 28.8693 31.5264 34.8053 37.1565
19 22.7178 27.2036 30.1435 32.8523 36.1909 38.5823
20 23.8277 28.4120 31.4104 34.1696 37.5662 39.9968
25 29.3389 34.3816 37.6525 40.6465 44.3141 46.9279
30 34.7997 40.2560 43.7730 46.9792 50.8922 53.6720
35 40.2228 46.0588 49.8018 53.2033 57.3421 60.2748
40 45.6160 51.8051 55.7585 59.3417 63.6907 66.7660
45 50.9849 57.5053 61.6562 65.4102 69.9568 73.1661
50 56.3336 63.1671 67.5048 71.4202 76.1539 79.4900
60 66.9815 74.3970 79.0819 83.2977 88.3794 91.9517
80 88.1303 96.5782 101.8795 106.6286 112.3288 116.3211
100 109.1412 118.4980 124.3421 129.5612 135.8067 140.1695
120 130.0546 140.2326 146.5674 152.2114 158.9502 163.6482
END OF EXAMINATION
Page 19 of 19
