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BUEC 333 MIDTERM
Multiple Choice (1 point each)
1.) The significance level of a test is the probability that you:
a) reject the null when it is true
b) fail to reject the null when it is false
c) reject the null when it is false
d) fail to reject the null when it is true
e) none of the above
2.) Consider the standard univariate regression model: Yi  0  1 X i   i . OLS estimates ˆ1  2.54.
What is the sign of the covariance between Y and X in this case?
a) positive
b) negative
c) depends on the underlying variation in X
d) there is insufficient information to answer this question
e) none of the above
3.) Suppose that on any given trading day, a stock’s price can increase, decrease, or remain unchanged.
Likewise, on any given day, the weather can be sunny or cloudy. You have the following information
about the joint distribution of weather and stock price changes:
Cloudy Sunny
Decrease
1/10
3/20
Unchanged
1/5
3/10
Increase
1/10
It is sunny today. What is the probability of a price increase?
a) 3/20
b) 3/12
c) 3/10
d) 3/5
e) none of the above
4.) If we add an extra independent variable that is highly related to the dependent variable, R-squared and
adjusted R-squared will most likely
a) decrease and increase, respectively
b) increase and decrease, respectively
c) both decrease
d) both increase
e) none of the above
5.) In the linear regression model, the stochastic error term:
a) measures the difference between the dependent variable and its conditional expectation
b) measures the difference between the independent variable and its predicted value
c) is unbiased
d) a and c
e) none of the above
6.) Suppose [L(X), U(X)] is a 95% confidence interval for a population mean. Which of the following
is/are true?
a) Pr L X   X  U  X   0.90


b) Pr  L  X     U  X   0.95




c) Pr X  L X   Pr U  X   X  0.05
d) a and c
e) none of the above
7.) Suppose you draw a random sample of n observations, X1, X2, …, Xn, from a population with unknown
mean μ. Which of the following estimators of μ is/are unbiased?
a) the first observation you sample, X1
b)
X2
c) X 2  s 2 / n
d) b and c
e) a, b, and c
8.) If q is an unbiased estimator of Q, then:
a) Q is the mean of the sampling distribution of q
b) q is the mean of the sampling distribution of Q
c) Var[q] = Var[Q] / n where n = the sample size
d) q = Q
e) a and c
9.) The OLS estimator is said to be BUE when:
a.) Assumptions 1 through 3 are satisfied
b.) Assumptions 1 through 6 are satisfied
c.) Assumptions 1 through 3 are satisfied and errors are normally distributed
d.) Assumptions 1 through 6 are satisfied and errors are normally distributed
e.) errors are normally distributed
10.) The Gauss-Markov Theorem says that when the 6 classical assumptions are satisfied:
a) The least squares estimator is unbiased
b) The least squares estimator has the smallest variance of all estimators
c) The least squares estimator has a normal sampling distribution in large samples
d) The least squares estimator always has an approximately normal sampling distribution
e) None of the above
11.) Suppose you have the following information about the pdf of a random variable X, which takes one
of 4 possible values:
Value of X
1
2
3
4
pdf
0.05
0.15
0.25
Which of the following is/are true?
a) Pr(X ≤ 2) = 0.2
b) E(X) = 2.5
c) Pr(X = 4) = 3.3
d) all of the above
e) none of the above
12.) If two random variables X and Y are independent,
a) their joint distribution equals the product of their marginal distributions
b) the conditional distribution of X given Y equals the marginal distribution of X
c) their covariance is zero
d) a and c
e) a, b, and c
13.) In the linear regression model, adjusted R2 measures
a) the proportion of variation in Y explained by X
b) the proportion of variation in X explained by Y
c) the proportion of variation in Y explained by X, adjusted for the number of dependent
variables
d) the proportion of variation in X explained by Y, adjusted for the number of independent
variables
e) none of the above
14.) In order for our independent variables to be labelled “exogenous” which of the following must be
true:
a.) E(εi) = 0
b.) Cov(Xi,εi) = 0
c.) Cov(εi,εj) = 0
d.) Var(εi) = σ2
e.) none of the above
15.) The cost of unleaded gasoline is Vancouver follows an unknown distribution with a mean of $1.59
per litre and a standard deviation of $0.03. Thirty gas stations in the area are randomly chosen and their
sample average, X , is calculated. The distribution to use for the average cost of gasoline for the 30 gas
stations is
a) X
N (1.59,0.03)
b) X
N (1.59,0.09)
c) X
N (1.59,0.03 / 30)
d) X
N (1.59,0.09 / 30)
e) None of the above
Short Answer #1 (5 points – show your work!)
Consider the standard univariate regression model:
Yi  0  1 X i   i
Suppose you also know the following, 0  0.
Verbally, explain the steps necessary to derive the least squares estimator.
The least squares estimator seeks to minimize the sum of the squared residuals from the estimating
equation.
1.) Thus, we first have to define our residual as the difference between that which is observed and that
which is predicted by the regression. In this way, the residual is best thought of as a prediction error, that
is, something we would like to make as small as possible.
2.) Next, we need to define a minimization problem. Because our residuals will likely be both positive and
negative, simply considering their sum is unsatisfactory as these will tend to cancel one another out.
Additionally, minimizing the sum of residuals does not generally yield a unique answer. A better way
forward is to minimize the sum of the squared “prediction errors” which will definitely yield a unique
answer and which will penalize us for making big errors.
3.) We must take the derivatives of the sum of squared residuals with respect to the beta-hats and set them
equal to zero. These first order conditions establish the values of beta-hat for which the sum of squared
residuals “bottoms out” and is, thus, minimized.
4.) Finally, we must solve for the values of the beta-hats which are consistent with these first order
conditions, thus, yielding our least squares estimators.
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Short Answer #2 (10 points – show your work!)
Consider the standard univariate regression model:
Yi  0  1 X i   i
Show that the OLS estimator ˆ0 is an unbiased estimator of  0 . (Hint: make use of the fact that ˆ1 is an
unbiased estimator of 1 and the fact that Yi  0  1 X i   i  Y  0  1 X   )
We know that fundamentally the property of unbiasedness entails the expected value of an estimator
match up with what it is trying to estimate, or expressed a little more precisely, E (q)  Q, where q is an
estimator of a population parameter Q. First, substitute the second hint into the definition of the
estimated intercept term from the formula sheet provided:
ˆ0  Y  ˆ1 X
ˆ0  0  1 X    ˆ1 X
ˆ0  0  ( 1  ˆ1 ) X  
Now, we need to evaluate the expected value of the estimate of the intercept term generated by OLS:
E ( ˆ0 )  E (  0  ( 1  ˆ1 ) X   )
E ( ˆ )  E (  )  E[(   ˆ ) X ]  E ( )
0
0
1
1
E ( ˆ0 )   0  XE ( 1  ˆ1 )
E ( ˆ )    X [ E (  )  E ( ˆ )]
0
0
1
E ( ˆ0 )   0  X ( 1  1 )
E ( ˆ )  
0
0
1
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Short Answer #3 (10 points – show your work!)
Unable to find an internship for the summer, you decide to open an ice cream stand. After some
painstaking research, you deduce that on any given day the weather can be sunny, cloudy, or rainy.
Likewise, on any given day, the demand for ice cream can be low or high. You have the following
information about the joint distribution of weather and demand for ice cream:
Low demand High demand
Rainy
10/100
Cloudy
1/4
3/12
Sunny
5/100
1/5
a) What is the joint probability of sunny weather and high demand for ice cream?
b) If your revenue is $10 when demand is low and $100 when demand is high, what is your expected
revenue on any given day?
c) The weather is sunny today. What is the probability that demand is high?
d) Explain the difference between your answer to part a) and your answer to part c).
a) 1/5 or 20%.
b) For this we need to know the marginal probabilities of a high and low demand day, respectively. Of
course, we also need to “fill in the blank” for the joint probability of rainy weather and high demand.
This is given by 1 – 10/100 – 1/4 – 3/12 – 5/100 – 1/5 = 1 – 10/100 – 25/100 – 25/100 – 5/100 – 20/100,
or in other words P(R,H) = 15/100.
For the probability of a high demand day then, this would simply be 15/100 + 25/100 + 20/100 = 60/100.
For the probability of a low demand day then, this would simply be 1 – 60/100 = 40/100.
Thus, expected revenue is equal to 0.4*$10 + 0.6*$100 = $64.
c) Again, we need a marginal probability, in particular, that of a sunny day. This is calculated as
5/100 + 20/100 = 25/100. Thus, the probability of a high demand day given that it is sunny is
(1/5)/(25/100) = 100/125 = 0.8
d) The answer in part a) is a joint probability—that is, the chances of simultaneously observing sunny
weather and high demand. The answer in part c) is a conditional probability—that is, the chances of
observing high demand given information that it is a sunny day.
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Short Answer #4 (10 points – show your work!)
Suppose you are hired by the Human Resources & Skills Development Canada (HRSDC) to explain the
evolution of average weekly earnings (AWE) of workers over their lifetime. To answer this question, you
randomly sample 1000 college educated full-time workers age 25-65. Using these data, you estimate the
regression:
Yi  0  1 X i   i
Yi = average weekly earnings in Canadian dollars of worker i
Xi = age of worker i in years
You estimate the model in EViews and obtain the following:
ˆ0  696.7 ˆ1  9.6 R2  0.023
a) Explain what the coefficient values 696.7 and 9.6 mean.
b) The R2 of the regression is 0.023. Interpret this value. What are the units of measurement for the R2?
c) The average age in this sample is 41.6 years. What is the average value of AWE in the sample?
d) What is the regression’s predicted earnings for a 100-year old worker? Do you think this prediction is
reliable? Why or why not?
e) Suppose that instead of measuring age in years this variable is measured in months. What are the
regression estimates from this new month-based regression (report both coefficients and R2)?
a) 696.7 is the value of the intercept term. Loosely speaking, this is the predicted value of AWE given that
X is equal to zero (but this makes no sense—workers are not infants and the sample is actually bounded
by 25 years). Strictly speaking, it is a garbage collector ensuring that the sum of residuals (that is, the
effects of all sorts of unobserved variables) is equal to zero. Of more interest is the slope coefficient of 9.6
which tells us we expect AWE to increase by $9.6 for each additional year of age on the part of workers.
b) This is the proportion of the variation in AWE which is explained by variation in the age of workers.
Since it is expressed as a share, it inherently has not units of measurement.
c) Although not necessary for credit since I am sure a lot of students just plugged in the value for the
average age into estimated regression line, it is still nice to understand why this works:
1
1
1
Yi   (Yˆi  ei )   ( ˆ0  ˆ1 X  ei  ei )

n
n
n
1
Y   ( ˆ0  ˆ1 X )  ˆ0  ˆ1 X
n
Y
So, in this the sample mean of AWE is 696.7 + 9.6 * 41.6 = $1096.06.
d) This is nothing more than 696.7 + 9.6 * 100 = $1656.70. But we probably should not take this at face
value. First of all, this is for an age which is well outside of our sample of values for X which is bounded
by 65. Second of all, we would expect some decline to set in as workers get older and less productive. It is
unlikely then that this would be a very accurate prediction.
e) This does nothing to change the amount of relative variation of X and Y around their mean and just
serves to scale the slope coefficient on age downwards. In particular, the constant should remain affected
and stay at 696.7, the slope coefficient will now be equal to 0.8 and the R2 will remain unaffected.
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Useful Formulas:
E( X ) 

k

  x  
Var ( X )  E  X   X  
p i xi
i 1
2
k
i
X
2 pi
i 1
Pr( X  x, Y  y)
Pr(Y  y | X  x) 
Pr( X  x)
k
Pr( X  x)   Pr X  x, Y  yi 
i 1
k
m
E Y | X  x    yi PrY  yi | X  x 
E Y    E Y | X  xi  Pr X  xi 
i 1
k
i 1
Ea  bX  cY   a  bE( X )  cE (Y )
Var (Y | X  x)    yi  E Y | X  x  PrY  yi | X  x 
2
i 1
Cov( X , Y )   x j   X  yi  Y  PrX  x j , Y  yi 
k
m
i 1 j 1
Corr  X , Y    XY 
Cov X , Y 
Var  X Var Y 
E Y 2   Var (Y )  E (Y ) 2
1
n
n

i 1
xi
s2 
Var aX  bY   a 2Var ( X )  b 2Var (Y )  2abCov( X ,Y )
Cova  bX  cV ,Y   bCov( X ,Y )  cCov(V ,Y )
E XY   Cov( X ,Y )  E( X ) E(Y )
X 
Var a  bY   b 2Var (Y )
1 n
xi  x 2

n  1 i 1
t
X 
Z
s/ n
s XY 
For the simple linear regression model Yi   0  1 X i   i we have : ˆ1 
 X
i 1
 X Yi  Y 
 X
n
i 1
Yˆi  ˆ0  ˆ1 X 1i  ˆ2 X 2i    ˆk X ki
i
X
2
i

rXY  s XY / s X sY
1 n
 xi  x  yi  y 
n  1 i 1
n
X 
βˆ 0  Y  ˆ1 X