TEST BANK for Introduction to Econometrics 2nd Second Edition by M. W. Watson and J. H. Stock. by welldoneassistant

Chapter 1 Economic Questions and Data 1.1 Multiple Choice 1) Analyzing the behavior of unemployment rates across U.S. states in March of 2006 is an example of using A) time series data. B) panel data. C) cross-sectional data. D) experimental data. Answer: C 2) Studying inflation in the United States from 1970 to 2006 is an example of using A) randomized controlled experiments. B) time series data. C) panel data. D) cross-sectional data. Answer: B 3) Analyzing the effect of minimum wage changes on teenage employment across the 48 contiguous U.S. states from 1980 to 2004 is an example of using A) time series data. B) panel data. C) having a treatment group vs. a control group, since only teenagers receive minimum wages. D) cross-sectional data. Answer: B 4) Panel data A) is also called longitudinal data. B) is the same as time series data. C) studies a group of people at a point in time. D) typically uses control and treatment groups. Answer: A 5) Econometrics can be defined as follows with the exception of A) the science of testing economic theory. B) fitting mathematical economic models to real-world data. C) a set of tools used for forecasting future values of economic variables. D) measuring the height of economists. Answer: D 6) To provide quantitative answers to policy questions A) it is typically sufficient to use common sense. B) you should interview the policy makers involved. C) you should examine empirical evidence. D) is typically impossible since policy questions are not quantifiable. Answer: C 7) An example of a randomized controlled experiment is when A) households receive a tax rebate in one year but not the other. B) one U.S. state increases minimum wages and an adjacent state does not, and employment differences are observed. C) random variables are controlled for by holding constant other factors. D) some 5 th graders in a specific elementary school are allowed to use computers at school while others are not, and their end-of-year performance is compared holding constant other factors. Answer: D Stock/Watson 2e -- CVC2 8/23/06 -- Page 1

8) Ideal randomized controlled experiments in economics are A) often performed in practice. B) often used by the Federal Reserve to study the effects of monetary policy. C) useful because they give a definition of a causal effect. D) sometimes used by universities to determine who graduates in four years rather than five. Answer: C 9) Most economic data are obtained A) through randomized controlled experiments. B) by calibration methods. C) through textbook examples typically involving ten observation points. D) by observing real-world behavior. Answer: D 10) One of the primary advantages of using econometrics over typical results from economic theory, is that A) it potentially provides you with quantitative answers for a policy problem rather than simply suggesting the direction (positive/negative) of the response. B) teaching you how to use statistical packages C) learning how to invert a 4 by 4 matrix. D) all of the above.

Answer: A 11) In a randomized controlled experiment A) there is a control group and a treatment group. B) you control for the effect that random numbers are not truly randomly generated C) you control for random answers D) the control group receives treatment on even days only. Answer: A 12) The reason why economists do not use experimental data more frequently is for all of the following reasons except that real-world experiments A) cannot be executed in economics. B) with humans are difficult to administer. C) are often unethical. D) have flaws relative to ideal randomized controlled experiments. Answer: A 13) The most frequently used experimental or observational data in econometrics are of the following type: A) cross-sectional data. B) randomly generated data. C) time series data. D) panel data. Answer: A

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14) In the graph below, the vertical axis represents average real GDP growth for 65 countries over the period 1960-1995, and the horizontal axis shows the average trade share within these countries.

This is an example of A) cross-sectional data. B) experimental data. C) a time series. D) longitudinal data. Answer: A

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15) The accompanying graph

Is an example of A) cross-sectional data. B) experimental data. C) a time series. D) longitudinal data. Answer: A

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16) The accompanying graph

is an example of A) experimental data. B) cross-sectional data. C) a time series. D) longitudinal data. Answer: C

1.2 Essays 1) Give at least three examples from economics where each of the following type of data can be used: cross-sectional data, time series data, and panel data. Answer: Answers will vary by student. At this level of economics, students most likely have heard of the following use of cross-sectional data: earnings functions, growth equations, the effect of class size reduction on student performance (in this chapter), demand functions (in this chapter: cigarette consumption); time series: the Phillips curve (in this chapter), consumption functions, Okunʹs law; panel data: various U.S. state panel studies on road fatalities (in this book), unemployment rate and unemployment benefits variations, growth regressions (across states and countries), and crime and abortion (Freakonomics).

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Chapter 2 Review of Probability 2.1 Multiple Choice 1) The probability of an outcome A) is the number of times that the outcome occurs in the long run. B) equals M × N, where M is the number of occurrences and N is the population size. C) is the proportion of times that the outcome occurs in the long run. D) equals the sample mean divided by the sample standard deviation. Answer: C 2) The probability of an event A or B (Pr(A or B)) to occur equals A) Pr(A) × Pr(B). B) Pr(A) + Pr(B) if A and B are mutually exclusive. Pr(A) C) . Pr(B) D) Pr(A) + Pr(B) even if A and B are not mutually exclusive. Answer: B 3) The cumulative probability distribution shows the probability A) that a random variable is less than or equal to a particular value. B) of two or more events occurring at once. C) of all possible events occurring. D) that a random variable takes on a particular value given that another event has happened. Answer: A 4) The expected value of a discrete random variable A) is the outcome that is most likely to occur. B) can be found by determining the 50% value in the c.d.f. C) equals the population median. D) is computed as a weighted average of the possible outcome of that random variable, where the weights are the probabilities of that outcome. Answer: D 5) Let Y be a random variable. Then var(Y) equals A) E[Y - μY)2 ]. B) E (Y - μY) . C) E (Y -μ )2 . Y D) E (Y - μY) . Answer: C

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6) The skewness of the distribution of a random variable Y is defined as follows: E (Y3 - μY) A) 2 σY B) E (Y - μ )3 Y 3 E Y3 - μ Y C)

3 σY E (Y - μ )3 Y 3 σY

Answer: D 7) The skewness is most likely positive for one of the following distributions: A) The grade distribution at your college or university. B) The U.S. income distribution. C) SAT scores in English. D) The height of 18 year old females in the U.S. Answer: B 8) The kurtosis of a distribution is defined as follows: E Y-μ 4 Y A) 4 σY 4 E Y4 - μ Y B)

2 σY skewness var(Y)

D) E[(Y - μY)4 ) Answer: A 9) For a normal distribution, the skewness and kurtosis measures are as follows: A) 1.96 and 4 B) 0 and 0 C) 0 and 3 D) 1 and 2 Answer: C

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10) The conditional distribution of Y given X = x, Pr(Y = y X=x), is Pr(Y = y) A) . Pr(X = x) l

∑ Pr(X = xi, Y = y).

i=1 Pr(X = x, Y = y) C) Pr(Y = y) D)

Pr(X = x, Y = y) . Pr(X = x)

Answer: D 11) The conditional expectation of Y given X, E(Y X = x), is calculated as follows: k A) ∑ Yi Pr(X = x i Y= y) i=1 B) E E(Y X)] k C) ∑ y i Pr(Y = y i X= x) i=1 l D) ∑ E(Y X= x i) Pr(X = x i) i=1 Answer: C 12) Two random variables X and Y are independently distributed if all of the following conditions hold, with the exception of A) Pr(Y = y X = x) = Pr(Y = y). B) knowing the value of one of the variables provides no information about the other. C) if the conditional distribution of Y given X equals the marginal distribution of Y. D) E(Y) = E[E(Y X)]. Answer: D 13) The correlation between X and Y A) cannot be negative since variances are always positive. B) is the covariance squared. C) can be calculated by dividing the covariance between X and Y by the product of the two standard deviations. cov(X, Y) . D) is given by corr(X, Y) = var(X) var(Y) Answer: C 14) Two variables are uncorrelated in all of the cases below, with the exception of A) being independent. B) having a zero covariance. C) σXY ≤

2 2 σ X σ Y.

D) E(Y X) = 0. Answer: C

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15) var(aX + bY) = 2 2 A) a2 σ X + b2 σ Y . 2 2 B) a2 σ X + 2abσXY + b2 σ Y . C) σXY + μXμY. 2 2 D) a σ X + b σ Y . Answer: B 16) To standardize a variable you A) subtract its mean and divide by its standard deviation. B) integrate the area below two points under the normal distribution. C) add and subtract 1.96 times the standard deviation to the variable. D) divide it by its standard deviation, as long as its mean is 1. Answer: A 17) Assume that Y is normally distributed N(μ, σ2 ). Moving from the mean (μ) 1.96 standard deviations to the left and 1.96 standard deviations to the right, then the area under the normal p.d.f. is A) 0.67 B) 0.05 C) 0.95 D) 0.33 Answer: C ci – μ , you need 18) Assume that Y is normally distributed N(μ, σ2 ). To find Pr(c1 ≤ Y ≤ c2 ), where c1 < c2 and d i = σ to calculate Pr(d 1 ≤ Z ≤ d 2 ) = A) Φ(d 2 ) - Φ(d 1 ) B) Φ(1.96) - Φ(1.96) C) Φ(d 2 ) - (1 - Φ(d 1 )) D) 1 - (Φ(d 2 ) - Φ(d 1 )) Answer: A 19) If variables with a multivariate normal distribution have covariances that equal zero, then A) the correlation will most often be zero, but does not have to be. B) the variables are independent. C) you should use the χ2 distribution to calculate probabilities. D) the marginal distribution of each of the variables is no longer normal. Answer: B 20) The Student t distribution is A) the distribution of the sum of m squared independent standard normal random variables. B) the distribution of a random variable with a chi-squared distribution with m degrees of freedom, divided by m. C) always well approximated by the standard normal distribution. D) the distribution of the ratio of a standard normal random variable, divided by the square root of an independently distributed chi-squared random variable with m degrees of freedom divided by m. Answer: D

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21) When there are ∞ degrees of freedom, the t∞ distribution A) can no longer be calculated. B) equals the standard normal distribution. C) has a bell shape similar to that of the normal distribution, but with “fatter” tails. 2 D) equals the χ ∞ distribution. Answer: B 22) The sample average is a random variable and A) is a single number and as a result cannot have a distribution. B) has a probability distribution called its sampling distribution. C) has a probability distribution called the standard normal distribution. D) has a probability distribution that is the same as for the Y1 ,..., Yn i.i.d. variables. Answer: B 23) To infer the political tendencies of the students at your college/university, you sample 150 of them. Only one of the following is a simple random sample: You A) make sure that the proportion of minorities are the same in your sample as in the entire student body. B) call every fiftieth person in the student directory at 9 a.m. If the person does not answer the phone, you pick the next name listed, and so on. C) go to the main dining hall on campus and interview students randomly there. D) have your statistical package generate 150 random numbers in the range from 1 to the total number of students in your academic institution, and then choose the corresponding names in the student telephone directory. Answer: D 2 24) The variance of Y, σ Y , is given by the following formula: 2 A) σ Y . B)

σY . n 2 σY

2 σY D)

Answer: C

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25) The mean of the sample average Y, E(Y), is 1 A) μY . n B) μY. C)

μY n

σY for n > 30. μY

Answer: B 26) In econometrics, we typically do not rely on exact or finite sample distributions because A) we have approximately an infinite number of observations (think of re -sampling). B) variables typically are normally distributed. C) the covariances of Yi, Yj are typically not zero. D) asymptotic distributions can be counted on to provide good approximations to the exact sampling distribution (given the number of observations available in most cases). Answer: D 27) Consistency for the sample average Y can be defined as follows, with the exception of A) Y converges in probability to μY. B) Y has the smallest variance of all estimators. p C) Y μY . D) the probability of Y being in the range μY ± c becomes arbitrarily close to one as n increases for any constant c > 0. Answer: B 28) The central limit theorem states that A) the sampling distribution of B) Y

Y-μY σY

is approximately normal.

μY.

C) the probability that Y is in the range μY ± c becomes arbitrarily close to one as n increases for any constant c > 0. D) the t distribution converges to the F distribution for approximately n > 30. Answer: A 29) The central limit theorem A) states conditions under which a variable involving the sum of Y1 ,..., Yn i.i.d. variables becomes the standard normal distribution. B) postulates that the sample mean Y is a consistent estimator of the population mean μY. C) only holds in the presence of the law of large numbers. D) states conditions under which a variable involving the sum of Y1 ,..., Yn i.i.d. variables becomes the Student t distribution. Answer: A

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30) The covariance inequality states that 2 A) 0 ≤ σ ≤ 1. XY B) σ

2 2 2 ≤σ σ . XY X Y

C) σ

2 2 2 -σ ≤σ . XY X Y

D) σ

2 ≤ XY

2 X

2 Y

Answer: B n

31)

∑ (axi + byi + c)= i=1 n n A) a ∑ x i + b ∑ y i + n× c i=1 i=1 n n B) a ∑ x i + b ∑ y i + c i=1 i=1 C) ax + by + n×c n n D) a ∑ x i + b ∑ y i i=1 i=1 Answer: A n

32)

∑ (axi+b) i=1 A) n×a×x+ n×b B) n(a+b) C) D) Answer: A

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33) Assume that you assign the following subjective probabilities for your final grade in your econometrics course (the standard GPA scale of 4 = A to 0 = F applies): Probability 0.20 0.50 0.20 0.08

Grade A B C D F

0.02

The expected value is: A) 3.0 B) 3.5 C) 2.78 D) 3.25 Answer: C 34) The mean and variance of a Bernoille random variable are given as A) cannot be calculated B) np and np(1-p) C) p and p(1-p) D) p and (1- p) Answer: D 35) Consider the following linear transformation of a random variable y =

x-μx σx

where μx is the mean of x and σx

is the standard deviation. Then the expected value and the standard deviation of Y are given as A) 0 and 1 B) 1 and 1 C) Cannot be computed because Y is not a linear function of X D)

μ σx

and σx

Answer: A

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2.2 Essays and Longer Questions 1) Think of the situation of rolling two dice and let M denote the sum of the number of dots on the two dice. (So M is a number between 1 and 12.) (a) In a table, list all of the possible outcomes for the random variable M together with its probability distribution and cumulative probability distribution. Sketch both distributions. (b) Calculate the expected value and the standard deviation for M. (c) Looking at the sketch of the probability distribution, you notice that it resembles a normal distribution. Should you be able to use the standard normal distribution to calculate probabilities of events? Why or why not? Answer: (a) 2 3 4 5 6 7 8 9 10 11 12 Outcome (sum of dots) Probability 0.028 0.056 0.083 0.111 0.139 0.167 0.139 0.111 0.083 0.056 0.028 distribution Cumulative 0.028 0.083 0.167 0.278 0.417 0.583 0.722 0.833 0.912 0.972 1.000 probability distribution

(b) 7.0; 2.42. (c) You cannot use the normal distribution (without continuity correction) to calculate probabilities of events, since the probability of any event equals zero.

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2) What is the probability of the following outcomes? (a) Pr(M = 7) (b) Pr(M = 2 or M = 10) (c) Pr(M = 4 or M ≠ 4) (d) Pr(M = 6 and M = 9) (e) Pr(M < 8) (f) Pr(M = 6 or M > 10) Answer: (a) 0.167 or

1 6 = ; 36 6

(b) 0.111 or

1 4 = ; 39 9

2 8 = . 36 9

3) Probabilities and relative frequencies are related in that the probability of an outcome is the proportion of the time that the outcome occurs in the long run. Hence concepts of joint, marginal, and conditional probability distributions stem from related concepts of frequency distributions. You are interested in investigating the relationship between the age of heads of households and weekly earnings of households. The accompanying data gives the number of occurrences grouped by age and income. You collect data from 1,744 individuals and think of these individuals as a population that you want to describe, rather than a sample from which you want to infer behavior of a larger population. After sorting the data, you generate the accompanying table: Joint Absolute Frequencies of Age and Income, 1,744 Households

Household Income Y1 $0-under $200 Y2 $200-under $ 400

Age of head of household X1 X2 X3 X4 X5 16-under 20 20-under 25 25-under 45 45-under 65 65 and > 80 76 130 86 24 13

346

140

Y3 $400-under $600

251

101

Y4 $600-under $800

110

Y5 $800 and >

108

The median of the income group of $800 and above is $1,050. (a) Calculate the joint relative frequencies and the marginal relative frequencies. Interpret one of each of these. Sketch the cumulative income distribution. (b) Calculate the conditional relative income frequencies for the two age categories 16 -under 20, and 45-under 65. Calculate the mean household income for both age categories. (c) If household income and age of head of household were independently distributed, what would you expect these two conditional relative income distributions to look like? Are they similar here? (d) Your textbook has given you a primary definition of independence that does not involve conditional relative frequency distributions. What is that definition? Do you think that age and income are independent here, using this definition?

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Answer: (a) The joint relative frequencies and marginal relative frequencies are given in the accompanying table. 5.2 percent of the individuals are between the age of 20 and 24, and make between $200 and under $400. 21.6 percent of the individuals earn between $400 and under $600. Joint Relative and Marginal Frequencies of Age and Income, 1,744 Households Age of head of household X1 X2 X3 X4 Household Income 16-under 20 20-under 25 25-under 45 45-under 65 Y1 $0-under $200 0.046 0.044 0.075 0.049

X5 65 and > Total 0.014 0.227

Y2 $200-under $400 Y3 $400-under $600

0.007

0.052

0.198

0.080

0.005

0.342

0.000

0.011

0.144

0.058

0.003

0.216

Y4 $600-under $800

0.001

0.006

0.063

0.032

0.001

0.102

Y5 $800 and >

0.001

0.062

0.048

0.001

0.112

(b) The mean household income for the 16-under 20 age category is roughly $144. It is approximately $489 for the 45-under 65 age category. Conditional Relative Frequencies of Income and Age 16-under 20, and 45-under 65, 1,744 Households Age of head of household X1 X4 Household Income 16-under 20 Y1 $0-under $200 0.842

45-under 65 0.185

Y2 $200-under $400

0.300

0.137

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Y3 $400-under $600

0.000

0.217

Y4 $600-under $800

0.001

0.118

Y5 $800 and >

0.001

0.180

(c) They would have to be identical, which they clearly are not. (d) Pr(Y = y, X = x) = Pr(Y = y) Pr(X = x). We can check this by multiplying two marginal probabilities to see if this results in the joint probability. For example, Pr(Y = Y3 ) = 0.216 and Pr(X = X3 ) = 0.542, resulting in a product of 0.117, which does not equal the joint probability of 0.144. Given that we are looking at the data as a population, not a sample, we do not have to test how “close” 0.117 is to 0.144. 4) Math and verbal SAT scores are each distributed normally with N (500,10000). (a) What fraction of students scores above 750? Above 600? Between 420 and 530? Below 480? Above 530? (b) If the math and verbal scores were independently distributed, which is not the case, then what would be the distribution of the overall SAT score? Find its mean and variance. (c) Next, assume that the correlation coefficient between the math and verbal scores is 0.75. Find the mean and variance of the resulting distribution. (d) Finally, assume that you had chosen 25 students at random who had taken the SAT exam. Derive the distribution for their average math SAT score. What is the probability that this average is above 530? Why is this so much smaller than your answer in (a)? Answer: (a) Pr(Y>750) = 0.0062; Pr(Y>600) = 0.1587; Pr(420<Y<530) = 0.4061; Pr(Y<480) = 0.4270; Pr(Y>530) = 0.3821. (b) The distribution would be N(1000, 2000), using equations (2.29) and (2.31) in the textbook. Note that the standard deviation is now roughly 141 rather than 200. (c) Given the correlation coefficient, the distribution is now N(1000, 35000) , which has a standard deviation of approximately 187. (d) The distribution for the average math SAT score is N(500, 400). Pr(Y > 530) = 0.0668. This probability is smaller because the sample mean has a smaller standard deviation (20 rather than 100). 5) The following problem is frequently encountered in the case of a rare disease, say AIDS, when determining the probability of actually having the disease after testing positively for HIV. (This is often known as the accuracy of the test given that you have the disease.) Let us set up the problem as follows: Y = 0 if you tested negative using the ELISA test for HIV, Y = 1 if you tested positive; X = 1 if you have HIV, X = 0 if you do not have HIV. Assume that 0.1 percent of the population has HIV and that the accuracy of the test is 0.95 in both cases of (i) testing positive when you have HIV, and (ii) testing negative when you do not have HIV. (The actual ELISA test is actually 99.7 percent accurate when you have HIV, and 98.5 percent accurate when you do not have HIV.) (a) Assuming arbitrarily a population of 10,000,000 people, use the accompanying table to first enter the column totals. Test Positive (Y=1) HIV (X=1) No HIV (X=0) Total

Test Negative (Y=0)

Total

10,000,000

(b) Use the conditional probabilities to fill in the joint absolute frequencies. (c) Fill in the marginal absolute frequencies for testing positive and negative. Determine the conditional probability of having HIV when you have tested positive. Explain this surprising result. (d) The previous problem is an application of Bayes’ theorem, which converts Pr( Y = y X = x) into Pr(X = x Y = y). Can you think of other examples where Pr( Y = y X = x) ≠ Pr(X = x Y = y)?

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Answer: (a) Test Positive (Y=1)

Test Negative (Y=0)

Total 10,000 9,990,000 10,000,000

Test Positive (Y=1) 9,500 499,500

Test Negative (Y=0) 500 9,490,500

Total 10,000 9,990,000 10,000,000

Test Positive (Y=1) 9,500 499,500 509,000

Test Negative (Y=0) 500 9,490,500 9,491,000

Total 10,000 9,990000 10,000,000

HIV (X=1) No HIV (X=0) Total (b) HIV (X=1) No HIV (X=0) Total (c) HIV (X=1) No HIV (X=0) Total

Pr(X=1 Y=1) = 0.0187. Although the test is quite accurate, there are very few people who have HIV (10,000), and many who do not have HIV (9,999,000). A small percentage of that large number (499,500/9,990,000) is large when compared to the higher percentage of the smaller number (9,500/10,000). d. Answers will vary by student. Perhaps a nice illustration is the probability to be a male given that you play on the college/university men’s varsity team, versus the probability to play on the college/university men’s varsity team given that you are a male student.

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6) You have read about the so-called catch-up theory by economic historians, whereby nations that are further behind in per capita income grow faster subsequently. If this is true systematically, then eventually laggards will reach the leader. To put the theory to the test, you collect data on relative (to the United States) per capita income for two years, 1960 and 1990, for 24 OECD countries. You think of these countries as a population you want to describe, rather than a sample from which you want to infer behavior of a larger population. The relevant data for this question is as follows:

0.023 0.014 …. 0.041 0.033 0.625

0.770 1.000 …. 0.200 0.130 13.220

Y × X1

2 X1

2 X2

1.030 1.000

0.018 0.00053 0.593 1.0609 0.014 0.00020 1.000 1.0000 …. …. …. …. …. 0.450 0.008 0.00168 0.040 0.2025 0.230 0.004 0.00109 0.017 0.0529 17.800 0.294 0.01877 8.529 13.9164

where X1 and X2 are per capita income relative to the United States in 1960 and 1990 respectively, and Y is the average annual growth rate in X over the 1960-1990 period. Numbers in the last row represent sums of the columns above. (a) Calculate the variance and standard deviation of X1 and X2 . For a catch-up effect to be present, what relationship must the two standard deviations show? Is this the case here? (b) Calculate the correlation between Y and . What sign must the correlation coefficient have for there to be evidence of a catch-up effect? Explain. Answer: (a) The variances of X1 and X2 are 0.0520 and 0.0298 respectively, with standard deviations of 0.2279 and 0.1726. For the catch-up effect to be present, the standard deviation would have to shrink over time. This is the case here. (b) The correlation coefficient is –0.88. It has to be negative for there to be evidence of a catch -up effect. If countries that were relatively ahead in the initial period and in terms of per capita income grow by relatively less over time, then eventually the laggards will catch -up. 7) Following Alfred Nobel’s will, there are five Nobel Prizes awarded each year. These are for outstanding achievements in Chemistry, Physics, Physiology or Medicine, Literature, and Peace. In 1968, the Bank of Sweden added a prize in Economic Sciences in memory of Alfred Nobel. You think of the data as describing a population, rather than a sample from which you want to infer behavior of a larger population. The accompanying table lists the joint probability distribution between recipients in economics and the other five prizes, and the citizenship of the recipients, based on the 1969-2001 period. Joint Distribution of Nobel Prize Winners in Economics and Non -Economics Disciplines, and Citizenship, 1969-2001

Economics Nobel Prize (X = 0) Physics, Chemistry, Medicine, Literature, and Peace Nobel Prize (X = 1) Total

U.S. Citizen (Y = 0) 0.118

Non= U.S. Citizen (Y = 1) 0.049

Total

0.345

0.488

0.833

0.463

0.537

1.00

(a) Compute E(Y) and interpret the resulting number. (b) Calculate and interpret E(Y X=1) and E(Y X=0). Stock/Watson 2e -- CVC2 8/23/06 -- Page 19

0.167

(c) A randomly selected Nobel Prize winner reports that he is a non-U.S. citizen. What is the probability that this genius has won the Economics Nobel Prize? A Nobel Prize in the other five disciplines? (d) Show what the joint distribution would look like if the two categories were independent. Answer: (a) E(Y) = 0.53.7 . 53.7 percent of Nobel Prize winners were non-U.S. citizens. (b) E(Y X=1) = 0.586 . 58.6 percent of Nobel Prize winners in non-economics disciplines were non-U.S. citizens. E(Y X=0) = 0.293 . 29.3 percent of the Economics Nobel Prize winners were non -U.S. citizens. (c) There is a 9.1 percent chance that he has won the Economics Nobel Prize, and a 90.9 percent chance that he has won a Nobel Prize in one of the other five disciplines. (d) Joint Distribution of Nobel Prize Winners in Economics and Non -Economics Disciplines, and Citizenship, 1969-2001, under assumption of independence

Economics Nobel Prize (X = 0) Physics, Chemistry, Medicine, Literature, and Peace Nobel Prize (X = 1) Total

U.S. Citizen (Y = 0) 0.077

Non= U.S. Citizen (Y = 1) 0.090

Total

0.386

0.447

0.833

0.463

0.537

1.00

0.167

8) A few years ago the news magazine The Economist listed some of the stranger explanations used in the past to predict presidential election outcomes. These included whether or not the hemlines of women’s skirts went up or down, stock market performances, baseball World Series wins by an American League team, etc. Thinking about this problem more seriously, you decide to analyze whether or not the presidential candidate for a certain party did better if his party controlled the house. Accordingly you collect data for the last 34 presidential elections. You think of this data as comprising a population which you want to describe, rather than a sample from which you want to infer behavior of a larger population. You generate the accompanying table: Joint Distribution of Presidential Party Affiliation and Party Control of House of Representatives, 1860 -1996

Democratic President (X = 0) Republican President (X = 1) Total

Democratic Control Republican Control of House (Y = 0) of House (Y = 1) 0.412 0.030

Total 0.441

0.176

0.382

0.559

0.588

0.412

1.00

(a) Interpret one of the joint probabilities and one of the marginal probabilities. (b) Compute E(X). How does this differ from E(X Y = 0 )? Explain. (c) If you picked one of the Republican presidents at random, what is the probability that during his term the Democrats had control of the House? (d) What would the joint distribution look like under independence? Check your results by calculating the two conditional distributions and compare these to the marginal distribution.

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Answer: (a) 38.2 percent of the presidents were Republicans and were in the White House while Republicans controlled the House of Representatives. 44.1 percent of all presidents were Democrats. (b) E(X)= 0.559. E(X Y = 0) = 0.701. E(X) gives you the unconditional expected value, while E(X Y = 0) is the conditional expected value. (c) E(X) = 0.559 . 55.9 percent of the presidents were Republicans. E(X Y = 0) = 0.299 . 29.9 percent of those presidents who were in office while Democrats had control of the House of Representatives were Republicans. The second conditions on those periods during which Democrats had control of the House of Representatives, and ignores the other periods. (d) Joint Distribution of Presidential Party Affiliation and Party Control of House of Representatives, 1860-1996, under the Assumption of Independence Democratic Control Republican Control of House (Y = 0) of House (Y = 1) 0.259 0.182

Democratic President (X = 0) Republican President (X = 1) Total

Total 0.441

0.329

0.230

0.559

0.588

0.412

1.00

Pr(X = 0 Y = 0) =

0.259 = 0.440 (there is a small rounding error). 0.588

Pr(Y = 1 X = 1) =

0.230 = 0.411 (there is a small rounding error). 0.559

9) The expectations augmented Phillips curve postulates △p = π – f (u – u), where △p is the actual inflation rate, π is the expected inflation rate, and u is the unemployment rate, with ʺ–ʺ indicating equilibrium (the NAIRU – Non-Accelerating Inflation Rate of Unemployment). Under the assumption of static expectations (π = △p –1), i.e., that you expect this period’s inflation rate to hold for the next period (ʺthe sun shines today, it will shine tomorrowʺ), then the prediction is that inflation will accelerate if the unemployment rate is below its equilibrium level. The accompanying table below displays information on accelerating annual inflation and unemployment rate differences from the equilibrium rate (cyclical unemployment), where the latter is approximated by a five-year moving average. You think of this data as a population which you want to describe, rather than a sample from which you want to infer behavior of a larger population. The data is collected from United States quarterly data for the period 1964:1 to 1995:4. Joint Distribution of Accelerating Inflation and Cyclical Unemployment, 1964:1-1995:4 (u – u) ≥ 0 (Y = 1) 0.383

Total

△p– △p –1 > 0

(u – u) > 0 (Y = 0) 0.156

(X = 0) △p– △p –1 ≤ 0

0.297

0.164

0.461

(X = 1) Total

0.453

0.547

1.00

0.539

(a) Compute E(Y) and E(X), and interpret both numbers. (b) Calculate E(Y X= 1) and E(Y X= 0). If there was independence between cyclical unemployment and acceleration in the inflation rate, what would you expect the relationship between the two expected values to Stock/Watson 2e -- CVC2 8/23/06 -- Page 21

be? Given that the two means are different, is this sufficient to assume that the two variables are independent? (c) What is the probability of inflation to increase if there is positive cyclical unemployment? Negative cyclical unemployment? (d) You randomly select one of the 59 quarters when there was positive cyclical unemployment (( u – u) > 0). What is the probability there was decelerating inflation during that quarter? Answer: (a) E(Y) = 0.547 . 54.7 percent of the quarters saw cyclical unemployment. E(Y) = 0.461 . 46.1 percent of the quarters saw decreasing inflation rates. (b) E(Y X = 1) = 0.356; E(Y X = 0 ) = 0.711. You would expect the two conditional expectations to be the same. In general, independence in means does not imply statistical independence, although the reverse is true. (c) There is a 34.4 percent probability of inflation to increase if there is positive cyclical unemployment. There is a 70 percent probability of inflation to increase if there is negative cyclical unemployment. (d) There is a 65.6 percent probability of inflation to decelerate when there is positive cyclical unemployment.

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10) The accompanying table shows the joint distribution between the change of the unemployment rate in an election year and the share of the candidate of the incumbent party since 1928. You think of this data as a population which you want to describe, rather than a sample from which you want to infer behavior of a larger population. Joint Distribution of Unemployment Rate Change and Incumbent Party’s Vote Share in Total Vote Cast for the Two Major -Party Candidates, 1928-2000

△u > 0 (X = 0) △u ≤ 0 (X = 1) Total

(Incumbent- 50%) > 0 (Incumbent- 50%) ≤ 0 (Y = 0) (Y = 1) 0.053 0.211 0.579 0.157 0.632 0.368

Total 0.264 0.736 1.00

(a) Compute and interpret E(Y) and E(X). (b) Calculate E(Y X = 1) and E(Y X = 0). Did you expect these to be very different? (c) What is the probability that the unemployment rate decreases in an election year? (d) Conditional on the unemployment rate decreasing, what is the probability that an incumbent will lose the election? (e) What would the joint distribution look like under independence? Answer: (a) E(Y) = 0.368; E(X) = 0.736. The probability of an incumbent to have less than 50% of the share of votes cast for the two major-party candidates is 0.368. The probability of observing falling unemployment rates during the election year is 73.6 percent. (b) E(Y X = 1) = 0.213; E(Y X = 0) = 0.799 . A student who believes that incumbents will attempt to manipulate the economy to win elections will answer affirmatively here. (c) Pr(X = 1) = 0.736. (d) Pr(Y = 1 X = 1) = 0.213. (e) Joint Distribution of Unemployment Rate Change and Incumbent Party’s Vote Share in Total Vote Cast for the Two Major -Party Candidates, 1928-2000 under Assumption of Statistical Independence

△u > 0 (X = 0) △u ≤ 0 (X = 1) Total

(Incumbent- 50%) > 0 (Incumbent- 50%) > 0 (Y = 0) (Y = 1) 0.167 0.097 0.465 0.271 0.632 0.368

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Total 0.264 0.736 1.00

11) The table accompanying lists the joint distribution of unemployment in the United States in 2001 by demographic characteristics (race and gender). Joint Distribution of Unemployment by Demographic Characteristics, United States, 2001

Age 16-19 (X = 0) Age 20 and above (X = 1) Total

White (Y = 0) 0.13

Black and Other (Y = 1) 0.05

Total

0.60

0.22

0.82

0.73

0.27

1.00

0.18

(a) What is the percentage of unemployed white teenagers? (b) Calculate the conditional distribution for the categories ʺwhiteʺ and ʺblack and other.ʺ (c) Given your answer in the previous question, how do you reconcile this fact with the probability to be 60% of finding an unemployed adult white person, and only 22% for the category ʺblack and other.ʺ Answer: (a) Pr(Y = 0, X = 0) = 0.13. (b) Conditional Distribution of Unemployment by Demographic Characteristics, United States, 2001

Age 16-19 (X = 0) Age 20 and above (X = 1) Total

White (Y = 0) 0.18

Black and Other (Y = 1) 0.19

0.82

0.81

1.00

(c) The original table showed the joint probability distribution, while the table in (b) presented the conditional probability distribution. 12) From the Stock and Watson (http://www.pearsonhighered.com/stock_watson ) website the chapter 8 CPS data set (ch8_cps.xls) into a spreadsheet program such as Excel. For the exercise, use the first 500 observations only. Using data for average hourly earnings only (ahe), describe the earnings distribution. Use summary statistics, such as the mean, meadian, variance, and skewness. Produce a frequency distribution (“histogram”) using reasonable earnings class sizes. Answer: ahe Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum

19.79 0.51 16.83 19.23 11.49 131.98 0.23 0.96 58.44 2.14 Stock/Watson 2e -- CVC2 8/23/06 -- Page 24

Maximum Sum Count

60.58 9897.45 500.0

The mean is $19.79. The median ($16.83) is lower than the average, suggesting that the mean is being pulled up by individuals with fairly high average hourly earnings. This is confirmed by the skewness measure, which is positive, and therefore suggests a distribution with a long tail to the right. The variance is $2 131.96, while the standard deviation is $11.49. To generate the frequency distribution in Excel, you first have to settle on the number of class intervals. Once you have decided on these, then the minimum and maximum in the data suggests the class width. In Excel, you then define “bins” (the upper limits of the class intervals). Sturges’s formula can be used to suggest the number of class intervals (1+3.31log(n) ), which would suggest about 9 intervals here. Instead I settled for 8 intervals with a class width of $8 — minimum wages in California are currently $8 and approximately the same in other U.S. states. The table produces the absolute frequencies, and relative frequencies can be calculated in a straightforward way. bins 8 16 24 32 40 48 56 66 More

Frequency 50 187 115 68 38 33 8 1 0

rel. freq. 0.1 0.374 0.23 0.136 0.076 0.066 0.016 0.002

Substitution of the relative frequencies into the histogram table then produces the following graph (after eliminating the gaps between the bars).

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2.3 Mathematical and Graphical Problems 1) Think of an example involving five possible quantitative outcomes of a discrete random variable and attach a probability to each one of these outcomes. Display the outcomes, probability distribution, and cumulative probability distribution in a table. Sketch both the probability distribution and the cumulative probability distribution. Answer: Answers will vary by student. The generated table should be similar to Table 2.1 in the text, and figures should resemble Figures 2.1 and 2.2 in the text. 2) The height of male students at your college/university is normally distributed with a mean of 70 inches and a standard deviation of 3.5 inches. If you had a list of telephone numbers for male students for the purpose of conducting a survey, what would be the probability of randomly calling one of these students whose height is (a) taller than 6ʹ0ʺ? (b) between 5ʹ3ʺ and 6ʹ5ʺ? (c) shorter than 5ʹ7ʺ, the mean height of female students? (d) shorter than 5ʹ0ʺ? (e) taller than Shaq O’Neal, the center of the Miami Heat, who is 7ʹ1ʺ tall? Compare this to the probability of a woman being pregnant for 10 months (300 days), where days of pregnancy is normally distributed with a mean of 266 days and a standard deviation of 16 days. Answer: (a) Pr(Z > 0.5714) = 0.2839; (b) Pr( –2 < Z < 2) = 0.9545 or approximately 0.95; (c) Pr(Z < -0.8571) = 0.1957; (d) Pr(Z < -2.8571) = 0.0021; (e) Pr(Z > 4.2857) = 0.000009 (the text does not show values above 2.99 standard deviations, Pr(Z >2.99 = 0.0014) and Pr(Z > 2.1250) = 0.0168. 3) Calculate the following probabilities using the standard normal distribution. Sketch the probability distribution in each case, shading in the area of the calculated probability. (a) Pr(Z < 0.0) (b) Pr(Z ≤ 1.0) (c) Pr(Z > 1.96) (d) Pr(Z < –2.0) (e) Pr(Z > 1.645) (f) Pr(Z > –1.645) (g) Pr(–1.96 < Z < 1.96) (h.) Pr(Z < 2.576 or Z > 2.576) (i.) Pr(Z > z) = 0.10; find z. (j.) Pr(Z < –z or Z > z) = 0.05; find z. Answer: (a) 0.5000; (b) 0.8413; (c) 0.0250; (d) 0.0228; (e) 0.0500; (f) 0.9500; (g) 0.0500; (h) 0.0100; (i) 1.2816; (j) 1.96.

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4) Using the fact that the standardized variable Z is a linear transformation of the normally distributed random variable Y, derive the expected value and variance of Z. Answer: Z =

Y - μY

μY σY

σY +

μY σY

μY 1 1 Y = a + bY, with a = and b = . Given (2.29) and (2.30) in the text, E(Z) = σY σY σY

1 μ = 0, and σZ = σY Y

1 2 σZ

2 = 1. Z

5) Show in a scatterplot what the relationship between two variables X and Y would look like if there was (a) a strong negative correlation. (b) a strong positive correlation. (c) no correlation. Answer: (a)

(b)

6) What would the correlation coefficient be if all observations for the two variables were on a curve described by Y = X2 ? Answer: The correlation coefficient would be zero in this case, since the relationship is non -linear. 7) Find the following probabilities: 2 (a) Y is distributed χ 4 . Find Pr(Y > 9.49). (b) Y is distributed t∞. Find Pr(Y > –0.5). (c) Y is distributed F4,∞. Find Pr(Y < 3.32). (d) Y is distributed N(500, 10000). Find Pr(Y > 696 or Y < 304). Answer: (a) 0.05. (b) 0.6915. (c) 0.99. (d) 0.05.

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8) In considering the purchase of a certain stock, you attach the following probabilities to possible changes in the stock price over the next year. Stock Price Change During Next Twelve Months (%) +15 +5 0 –5 –15

Probability 0.2 0.3 0.4 0.05 0.05

What is the expected value, the variance, and the standard deviation? Which is the most likely outcome? Sketch the cumulative distribution function. 2 Answer: E(Y) = 3.5; σ Y = 8.49; σY = 2.91; most likely: 0.

9) You consider visiting Montreal during the break between terms in January. You go to the relevant Web site of the official tourist office to figure out the type of clothes you should take on the trip. The site lists that the average high during January is –7° C, with a standard deviation of 4° C. Unfortunately you are more familiar with Fahrenheit than with Celsius, but find that the two are related by the following linear function: 5 C= (F – 32). 9 Find the mean and standard deviation for the January temperature in Montreal in Fahrenheit. Answer: Using equations (2.29) and (2.30) from the textbook, the result is 19.4 and 7.2.

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10) Two random variables are independently distributed if their joint distribution is the product of their marginal distributions. It is intuitively easier to understand that two random variables are independently distributed if all conditional distributions of Y given X are equal. Derive one of the two conditions from the other. Answer: If all conditional distributions of Y given X are equal, then Pr(Y = y X = 1) = Pr(Y = y X = 2) = ... = Pr(Y = y X = l). But if all conditional distributions are equal, then they must also equal the marginal distribution, i.e., Pr(Y = y X = x) = Pr(Y - y). Given the definition of the conditional distribution of Y given X = x, you then get Pr(Y = y X = x) =

Pr(Y = y, X = x) = Pr(Y = y), Pr(X = x)

which gives you the condition Pr(Y = y, X = x) = Pr(Y = y) Pr(X = x). 11) There are frequently situations where you have information on the conditional distribution of Y given X, but Pr(X = x, Y = y) , derive a are interested in the conditional distribution of X given Y. Recalling Pr(Y = y X = x) = Pr(X = x) relationship between Pr(X = x Y = y) and Pr(Y = y X = x). This is called Bayes’ theorem. Answer: Given Pr(Y = y X = x) =

Pr(X = x Y = y) , Pr(X = x)

Pr(Y = y X = x) × Pr(X = x) = Pr(X = x, Y = y); Pr(X = x Y = y) similarly Pr(X = x Y = y) = and Pr(Y = y) Pr(X = x Y = y) × Pr(Y = y) = Pr(X = x, Y = y). Equating the two and solving for Pr(X = x Y = y) then results in Pr(Y = y X = x) × Pr(X = x) . Pr(X = x Y = y) = Pr(Y = y) 12) You are at a college of roughly 1,000 students and obtain data from the entire freshman class (250 students) on height and weight during orientation. You consider this to be a population that you want to describe, rather than a sample from which you want to infer general relationships in a larger population. Weight ( Y) is measured in pounds and height (X) is measured in inches. You calculate the following sums: n

∑ y i = 94,228.8, ∑ x i = 1,248.9, ∑ xiyi = 7,625.9 i=1

i=1

(small letters refer to deviations from means as in z i = Zi – Z). (a) Given your general knowledge about human height and weight of a given age, what can you say about the shape of the two distributions? (b) What is the correlation coefficient between height and weight here? Answer: (a) Both distributions are bound to be normal. (b) 0.703.

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13) Use the definition for the conditional distribution of Y given X = x and the marginal distribution of X to derive the formula for Pr(X = x, Y = y). This is called the multiplication rule. Use it to derive the probability for drawing two aces randomly from a deck of cards (no joker), where you do not replace the card after the first draw. Next, generalizing the multiplication rule and assuming independence, find the probability of having four girls in a family with four children. 3 1 4 1 4 . Answer: × = 0.0045; 0.0625 or = 2 16 52 51 14) The systolic blood pressure of females in their 20s is normally distributed with a mean of 120 with a standard deviation of 9. What is the probability of finding a female with a blood pressure of less than 100? More than 135? Between 105 and 123? You visit the women’s soccer team on campus, and find that the average blood pressure of the 25 members is 114. Is it likely that this group of women came from the same population? Answer: Pr(Y<100) = 0.0131; Pr(Y>135) = 0.0478; Pr(105<Y<123) = 0.6784; Pr(Y< 114) = Pr(Z < -3.33) = 0.0004. (The smallest z-value listed in the table in the textbook is –2.99, which generates a probability value of 0.0014.) This unlikely that this group of women came from the same population. 15) Show that the correlation coefficient between Y and X is unaffected if you use a linear transformation in both variables. That is, show that corr(X,Y) = corr(X*, Y*), where X* = a + bX and Y* = c + dY, and where a, b, c, and d are arbitrary non–zero constants. Answer: corr(X*, Y*) =

cov(X*, Y*) = var(X*) var(Y*)

bd cov(X, Y) corr(X, Y). 2 b var(X) d 2 var(Y)

16) The textbook formula for the variance of the discrete random variable Y is given as k 2 σ Y = ∑ (y i – μY)2 p i. i=1 Another commonly used formulation is k 2 2 2 σ Y = ∑ y i pi – μ Y . i=1 Prove that the two formulas are the same. k k k 2 2 2 2 2 Answer: σ Y = ∑ (y i - μY)2 p i = ∑ (y i + μ Y - 2μYy i) p i = ∑ ( y i p i + μ Y p i - 2μYy ip i). i=1 i=1 i=1 Moving the summation sign through results in k k k k k 2 2 2 σ Y = ∑ y i p i+ μ Y ∑ p i - 2 μY ∑ y i p i. But ∑ p i = 1 and μY ∑ y ip i , giving you the second i=1 i=1 i=1 i=1 i=1 expression after simplification.

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17) The Economic Report of the President gives the following age distribution of the United States population for the year 2000: United States Population By Age Group, 2000 Outcome (age category Percentage

Under 5 5-15

16-19

20-24

25-44

45-64

0.06

0.07

0.30

0.22

0.16

65 and over 0.13

Imagine that every person was assigned a unique number between 1 and 275,372,000 (the total population in 2000). If you generated a random number, what would be the probability that you had drawn someone older than 65 or under 16? Treating the percentages as probabilities, write down the cumulative probability distribution. What is the probability of drawing someone who is 24 years or younger? Answer: Pr(Y < 16 or Y > 65) = 0.35; Outcome (age category Cumulative probability distribution

Under 5 5-15

16-19

20-24

25-44

45-64

0.06

0.28

0.35

0.65

0.87

0.22

65 and over 1.00

Pr(Y ≤ 24) = 0.35. 18) The accompanying table gives the outcomes and probability distribution of the number of times a student checks her e-mail daily: Probability of Checking E-Mail Outcome (number of email checks) Probability distribution

0.05

0.15

0.30

0.25

0.15

0.08

0.02

Sketch the probability distribution. Next, calculate the c.d.f. for the above table. What is the probability of her checking her e-mail between 1 and 3 times a day? Of checking it more than 3 times a day? Answer: Outcome (number of email checks) Cumulative probability distribution

0.05

0.20

0.50

0.75

0.90

0.98

1.00

Pr(1 ≤ Y ≤ 3) 0.70 ; Pr(Y > 0.25).

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19) The accompanying table lists the outcomes and the cumulative probability distribution for a student renting videos during the week while on campus. Video Rentals per Week during Semester Outcome (number of weekly 0 video rentals) Probability distribution 0.05

0.55

0.25

0.05

0.07

0.02

0.01

Sketch the probability distribution. Next, calculate the cumulative probability distribution for the above table. What is the probability of the student renting between 2 and 4 a week? Of less than 3 a week? Answer: The cumulative probability distribution is given below. The probability of renting between two and four videos a week is 0.37. The probability of renting less than three a week is 0.85. Outcome (number of weekly video rentals) Cumulative probability distribution

0.05

0.60

0.85

0.90

0.97

0.99

1.00

20) The textbook mentioned that the mean of Y, E(Y) is called the first moment of Y, and that the expected value of the square of Y, E(Y2 ) is called the second moment of Y, and so on. These are also referred to as moments about the origin. A related concept is moments about the mean, which are defined as E[(Y – μY)r]. What do you call the second moment about the mean? What do you think the third moment, referred to as ʺskewness,ʺ measures? Do you believe that it would be positive or negative for an earnings distribution? What measure of the third moment around the mean do you get for a normal distribution? Answer: The second moment about the mean is the variance. Skewness measures the departure from symmetry. For the typical earnings distribution, it will be positive. For the normal distribution, it will be zero. 21) Explain why the two probabilities are identical for the standard normal distribution: Pr(–1.96 ≤ X ≤ 1.96) and Pr(–1.96 < X < 1.96). Answer: For a continuous distribution, the probability of a point is zero.

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22) SAT scores in Mathematics are normally distributed with a mean of 500 and a standard deviation of 100. The 1 Y-μY 2 ) - ( 1 2 σY formula for the normal distribution is f(Y)= e Use the scatter plot option in a standard 2 2πσ Y

spreadsheet program, such as Excel, to plot the Mathematics SAT distribution using this formula. Start by entering 300 as the first SAT score in the first column (the lowest score you can get in the mathematics section as long as you fill in your name correctly), and then increment the scores by 10 until you reach 800. In the second column, use the formula for the normal distribution and calculate f(Y). Then use the scatter plot option, where you eventually remove markers and substitute these with the solid line option. Answer:

23) Use a standard spreadsheet program, such as Excel, to find the following probabilities from various distributions analyzed in the current chapter: a. If Y is distributed N (1,4), find Pr(Y≤3) b. If Y is distributed N (3,9), find Pr(Y>0) c. If Y is distributed N (50,25), find Pr(40≤Y≤52) d. If Y is distributed N (5,2), find Pr(6≤Y≤8) Answer: The answers here are given together with the relevant Excel commands. a. =NORMDIST(3,1,2,TRUE) = 0.8413 b. =1-NORMDIST(0,3,3,TRUE) = 0.8413 c. =NORMDIST(52,50,5,TRUE)-NORMDIST(40,50,5,TRUE) = 0.6326 d. =NORMDIST(8,5,SQRT(2),TRUE)-NORMDIST(6,5,SQRT(2),TRUE) = 0.2229

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24) Looking at a large CPS data set with over 60,000 observations for the United States and the year 2004, you find that the average number of years of education is approximately 13.6. However, a surprising large number of individuals (approximately 800) have quite a low value for this variable, namely 6 years or less. You decide to drop these observations, since none of your relatives or friends have that few years of education. In addition, you are concerned that if these individuals cannot report the years of education correctly, then the observations on other variables, such as average hourly earnings, can also not be trusted. As a matter of fact you have found several of these to be below minimum wages in your state. Discuss if dropping the observations is reasonable. Answer: While it is always a good idea to check the data carefully before conducting a quantitative analysis, you should never drop data before carefully thinking about the problem at hand. While it is not plausible to find many individuals in the U.S. who were raised here with that few years of education, there will be immigrants in the survey. Average years of education can be quite low in other countries. For example, Brazil’s average years of schooling is less than 6 years. The point of the exercise is to think hard whether or not observations are outliers generated by faulty data entry or if there is a reason for observing values which may appear strange at first. 25) Use a standard spreadsheet program, such as Excel, to find the following probabilities from various distributions analyzed in the current chapter: 2

If Y is distributed χ 4 , find Pr( Y ≤ 7.78)

If Y is distributed χ 10 , find Pr( Y > 18.31)

c. d.

If Y is distributed F10,∞, find Pr( Y > 1.83) If Y is distributed t15, find Pr( Y > 1.75)

e. f. g. h.

If Y is distributed t90, find Pr( -1.99 ≤Y ≤ 1.99) If Y is distributed N(0,1), find Pr( -1.99 ≤Y ≤ 1.99) If Y is distributed F7,4, find Pr( Y > 4.12) If Y is distributed F7,120, , find Pr( Y > 2.79)

Answer: The answers here are given together with the relevant Excel commands. a. =1-CHIDIST(7.78,4) = 0.90 b. =CHIDIST(18.31,10) = 0.05 c. =FDIST(1.83,10,1000000) = 0.05 d. =TDIST(1.75,15,1) = 0.05 e. =1-TDIST(1.99,90,2) = 0.95 f. =NORMDIST(1.99,0,1,1)-NORMDIST(-1.99,0,1,1) = 0.953 g. =FDIST(4.12,7,4) = 0.10 h. =FDIST(2.79,7,120) = 0.01

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Chapter 3 Review of Statistics 3.1 Multiple Choice 1) An estimator is A) an estimate. B) a formula that gives an efficient guess of the true population value. C) a random variable. D) a nonrandom number. Answer: C 2) An estimate is A) efficient if it has the smallest variance possible. B) a nonrandom number. C) unbiased if its expected value equals the population value. D) another word for estimator. Answer: B ^

3) An estimator μY of the population value μY is unbiased if ^

A) μY = μ . Y B) Y has the smallest variance of all estimators. p C) Y μY . ^

D) E(μY) = μY. Answer: D ^

4) An estimator μY of the population value μY is consistent if ^

A) μY p μ . Y B) its mean square error is the smallest possible. C) Y is normally distributed. p D) Y 0. Answer: A

5) An estimator μY of the population value μY is more efficient when compared to another estimator μY, if ^

A) E(μY) > E(μY). B) it has a smaller variance. C) its c.d.f. is flatter than that of the other estimator. ^

D) both estimators are unbiased, and var(μY) < var(μY). Answer: D 6) With i.i.d. sampling each of the following is true except A) E(Y) = μY. 2 B) var(Y) = σ Y /n. C) E(Y) < E(Y). D) Y is a random variable. Answer: C

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7) The standard error of Y, SE(Y) = σY is given by the following formula: A)

n 1 (Yi – Y)2 . ∑ n i=1 2 SY

C) SY. D)

SY . n

Answer: D 8) The critical value of a two-sided t-test computed from a large sample A) is 1.64 if the significance level of the test is 5%. B) cannot be calculated unless you know the degrees of freedom. C) is 1.96 if the significance level of the test is 5%. D) is the same as the p-value. Answer: C 9) A type I error is A) always the same as (1-type II) error. B) the error you make when rejecting the null hypothesis when it is true. C) the error you make when rejecting the alternative hypothesis when it is true. D) always 5%. Answer: B 10) A type II error A) is typically smaller than the type I error. B) is the error you make when choosing type II or type I. C) is the error you make when not rejecting the null hypothesis when it is false. D) cannot be calculated when the alternative hypothesis contains an ʺ=ʺ. Answer: C 11) The size of the test A) is the probability of committing a type I error. B) is the same as the sample size. C) is always equal to (1-the power of test). D) can be greater than 1 in extreme examples. Answer: A 12) The power of the test is A) dependent on whether you calculate a t or a t2 statistic. B) one minus the probability of committing a type I error. C) a subjective view taken by the econometrician dependent on the situation. D) one minus the probability of committing a type II error. Answer: D

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13) When you are testing a hypothesis against a two-sided alternative, then the alternative is written as A) E(Y) > μY,0. B) E(Y) = μY,0. C) Y ≠ μY,0. D) E(Y) ≠ μY,0. Answer: D 14) A scatterplot A) shows how Y and X are related when their relationship is scattered all over the place. B) relates the covariance of X and Y to the correlation coefficient. C) is a plot of n observations on Xi and Yi, where each observation is represented by the point (Xi, Yi). D) shows n observations of Y over time. Answer: C 15) The following types of statistical inference are used throughout econometrics, with the exception of A) confidence intervals. B) hypothesis testing. C) calibration. D) estimation. Answer: C 16) Among all unbiased estimators that are weighted averages of Y1 ,..., Yn Y, is A) the only consistent estimator of μY. B) the most efficient estimator of μY. C) a number which, by definition, cannot have a variance. D) the most unbiased estimator of μY. Answer: B 17) To derive the least squares estimator μY, you find the estimator m which minimizes n

∑ (Yi – m)2 .

i=1 n

∑ (Yi – m) . i=1 n 2 C) ∑ m Y i . i=1 n D) ∑ (Yi – m) . i=1 B)

Answer: A 18) If the null hypothesis states H0 : E(Y) = μY,0, then a two-sided alternative hypothesis is A) H1 : E(Y) ≠ μY,0. B) H1 : E(Y) ≈ μY,0. C) H1 : μY < μY,0. D) H1 : E(Y) > μY,0. Answer: A Stock/Watson 2e -- CVC2 8/23/06 -- Page 39

19) The p-value is defined as follows: A) p = 0.05. B) PrH0 [ Y – μY,0 > Y act– μY,0 ]. C) Pr(z > 1.96). D) PrH0 [ Y – μY,0 < Y act– μY,0 ].. Answer: B 20) A large p-value implies A) rejection of the null hypothesis. B) a large t-statistic. C) a large Yact. D) that the observed value Yact is consistent with the null hypothesis. Answer: D 21) The formula for the sample variance is n 2 1 (Yi – Y). A) S Y = n–1 ∑ i=1 n 2 1 (Yi – Y)2 . B) S Y = ∑ n–1 i=1 n 2 1 (Yi – μ )2 . C) S Y = ∑ Y n–1 i=1 n–1 2 1 (Yi – Y)2 . D) S Y = n–1 ∑ i=1 Answer: B 22) Degrees of freedom A) in the context of the sample variance formula means that estimating the mean uses up some of the information in the data. B) is something that certain undergraduate majors at your university/college other than economics seem to have an ∞ amount of. C) are (n-2) when replacing the population mean by the sample mean. 2 2 D) ensure that S Y = σ Y . Answer: A

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23) The t-statistic is defined as follows: A) t =

Y – μY,0 2 σY

n B) t =

C) t =

Y – μY,0 SE(Y)

(Y – μY,0)2 SE(Y)

D) 1.96. Answer: A 24) The power of the test A) is the probability that the test actually incorrectly rejects the null hypothesis when the null is true. B) depends on whether you use Y or Y2 for the t-statistic. C) is one minus the size of the test. D) is the probability that the test correctly rejects the null when the alternative is true. Answer: D 25) The sample covariance can be calculated in any of the following ways, with the exception of: n 1 (Xi – X)(Yi – Y). A) ∑ n–1 i=1 n 1 XiYi – n XY. B) ∑ n–1 n–1 i=1 n 1 (Xi – μX)(Yi – μ ). ∑ Y n i=1 D) rXYSYSY, where rXY is the correlation coefficient. C)

Answer: C 26) When the sample size n is large, the 90% confidence interval for μY is A) Y ± 1.96SE(Y). B) Y ± 1.64SE(Y). C) Y ± 1.64σY. D) Y ± 1.96. Answer: B

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27) The standard error for the difference in means if two random variables M and W , when the two population variances are different, is 2 2 S M+ S W A)

SM SW . + nM n W 2 SM

2 SW

1 ( ). + 2 nM nW 2 SM

nM + n W

2 SW . nW

Answer: D 28) The t-statistic has the following distribution: A) standard normal distribution for n < 15 B) Student t distribution with n–1 degrees of freedom regardless of the distribution of the Y. C) Student t distribution with n–1 degrees of freedom if the Y is normally distributed. D) a standard normal distribution if the sample standard deviation goes to zero. Answer: C 29) The following statement about the sample correlation coefficient is true. A) –1 ≤ rXY ≤ 1. p 2 B) r XY corr(Xi, Yi). C) rXY < 1.

D) rXY =

2 S XY 2 2 SXSY

Answer: A 30) The correlation coefficient A) lies between zero and one. B) is a measure of linear association. C) is close to one if X causes Y. D) takes on a high value if you have a strong nonlinear relationship. Answer: B

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31) When testing for differences of means, the t-statistic t =

Ym-Yw SE(Ym-Yw )

, where SE(Ym-Yw )=

2 sm nm

2 sw nw

has

A) a student t distribution if the population distribution of Y is not normal B) a student t distribution if the population distribution of Y is normal C) a normal distribution even in small samples D) cannot be computed unless nw=nm Answer: B 32) When testing for differences of means, you can base statistical inference on the A) Student t distribution in general B) normal distribution regardless of sample size C) Student t distribution if the underlying population distribution of Y is normal, the two groups have the same variances, and you use the pooled standard error formula D) Chi-squared distribution with (nw + nm - 2) degrees of freedom Answer: C 33) Assume that you have 125 observations on the height ( H) and weight (W) of your peers in college. Let sHW = 68, sH = 3.5, sW = 29. The sample correlation coefficient is A) 1.22 B) 0.50 C) 0.67 D) Cannot be computed since males and females have not been separated out. Answer: C 34) You have collected data on the average weekly amount of studying time ( T) and grades (G) from the peers at your college. Changing the measurement from minutes into hours has the following effect on the correlation coefficient: A) decreases the rTG by dividing the original correlation coefficient by 60 B) results in a higher rTG C) cannot be computed since some students study less than an hour per week D) does not change the rTG Answer: A, D 35) A low correlation coefficient implies that A) the line always has a flat slope B) in the scatterplot, the points fall quite far away from the line C) the two variables are unrelated D) you should use a tighter scale of the vertical and horizontal axis to bring the observations closer to the line Answer: B

3.2 Essays and Longer Questions 1) Think of at least nine examples, three of each, that display a positive, negative, or no correlation between two economic variables. In each of the positive and negative examples, indicate whether or not you expect the correlation to be strong or weak. Answer: Answers will vary by student. Students frequently bring up the following correlations. Positive correlations: earnings and education (hopefully strong), consumption and personal disposable income (strong), per capita income and investment-output ratio or saving rate (strong); negative correlation: Okun’s Law (strong), income velocity and interest rates (strong), the Phillips curve (strong); no correlation: productivity growth and initial level of per capita income for all countries of the world (beta-convergence regressions), consumption and the (real) interest rate, employment and real wages.

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2) Adult males are taller, on average, than adult females. Visiting two recent American Youth Soccer Organization (AYSO) under 12 year old (U12) soccer matches on a Saturday, you do not observe an obvious difference in the height of boys and girls of that age. You suggest to your little sister that she collect data on height and gender of children in 4th to 6th grade as part of her science project. The accompanying table shows her findings. Height of Young Boys and Girls, Grades 4-6, in inches

YBoys

Boys SBoys

nBoys

57.8

3.9

YGirls

Girls SGirls

nGirls

58.4

4.2

(a) Let your null hypothesis be that there is no difference in the height of females and males at this age level. Specify the alternative hypothesis. (b) Find the difference in height and the standard error of the difference. (c) Generate a 95% confidence interval for the difference in height. (d) Calculate the t-statistic for comparing the two means. Is the difference statistically significant at the 1% level? Which critical value did you use? Why would this number be smaller if you had assumed a one -sided alternative hypothesis? What is the intuition behind this? Answer: (a) H0 : μBoys - μGirls = 0 vs. H1 : μBoys - μGirls ≠ 0 (b) YBoys - YGirls = -0.6, SE(YBoys - YGirls) =

3.92 4.22 + = 0.77. 55 57

(c) -0.6 ± 1.96 × 0.77 = (-2.11, 0.91). (d) t = -0.78, so t < 2.58, which is the critical value at the 1% level. Hence you cannot reject the null hypothesis. The critical value for the one-sided hypothesis would have been 2.33. Assuming a one-sided hypothesis implies that you have some information about the problem at hand, and, as a result, can be more easily convinced than if you had no prior expectation.

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3) Math SAT scores (Y) are normally distributed with a mean of 500 and a standard deviation of 100. An evening school advertises that it can improve students’ scores by roughly a third of a standard deviation, or 30 points, if they attend a course which runs over several weeks. (A similar claim is made for attending a verbal SAT course.) The statistician for a consumer protection agency suspects that the courses are not effective. She views the situation as follows: H0 : μY = 500 vs. H1 : μY = 530. (a) Sketch the two distributions under the null hypothesis and the alternative hypothesis. (b) The consumer protection agency wants to evaluate this claim by sending 50 students to attend classes. One of the students becomes sick during the course and drops out. What is the distribution of the average score of the remaining 49 students under the null, and under the alternative hypothesis? (c) Assume that after graduating from the course, the 49 participants take the SAT test and score an average of 520. Is this convincing evidence that the school has fallen short of its claim? What is the p-value for such a score under the null hypothesis? (d) What would be the critical value under the null hypothesis if the size of your test were 5%? (e) Given this critical value, what is the power of the test? What options does the statistician have for increasing the power in this situation? Answer: (a)

(b) Y of the 49 participants is normally distributed, with a mean of 500 and a standard deviation of 14.286 under the null hypothesis. Under the alternative hypothesis, it is normally distributed with a mean of 530 and a standard deviation of 14.286. (c) It is possible that the consumer protection agency had chosen a group of 49 students whose average score would have been 490 without attending the course. The crucial question is how likely it is that 49 students, chosen randomly from a population with a mean of 500 and a standard deviation of 100, will score an average of 520. The p-value for this score is 0.081, meaning that if the agency rejected the null hypothesis based on this evidence, it would make a mistake, on average, roughly 1 out of 12 times. Hence the average score of 520 would allow rejection of the null hypothesis that the school has had no effect on the SAT score of students at the 10% level. (d) The critical value would be 523. (e) Pr(Y < 523 H1 is true) = 0.312. Hence the power of the test is 0.688. She could increase the power by decreasing the size of the test. Alternatively, she could try to convince the agency to hire more test subjects, i.e., she could increase the sample size. 4) Your packaging company fills various types of flour into bags. Recently there have been complaints from one chain of stores: a customer returned one opened 5 pound bag which weighed significantly less than the label indicated. You view the weight of the bag as a random variable which is normally distributed with a mean of 5 pounds, and, after studying the machine specifications, a standard deviation of 0.05 pounds. (a) You take a sample of 20 bags and weigh them. Sketch below what the average pattern of individual weights might look like. Let the horizontal axis indicate the sampled bag number (1, 2, …, 20). On the vertical axis, mark the expected value of the weight under the null hypothesis, and two (≈ 1.96) standard deviations above and below the expected value. Draw a line through the graph for E(Y) + 2 σY, E(Y), and E(Y) – 2σY. How many of the bags in a sample of 20 will you expect to weigh either less than 4.9 pounds or more than 5.1 pounds? (b) You sample 25 bags of flour and calculate the average weight. What is the distribution of the average weight of these 25 bags? Repeating the same exercise 20 times, sketch what the distribution of the average weights would look like in a graph similar to the one you drew in (b), where you have adjusted the standard Stock/Watson 2e -- CVC2 8/23/06 -- Page 45

error of Y accordingly. (c) For each of the twenty observations in (c) a 95% confidence interval is constructed. Draw these confidence intervals, using the same graph as in (c). How many of these 20 confidence intervals would you expect to weigh 5 pounds under the null hypothesis? Answer: (a) On average, there should be one bag in every sample of 20 which weighs less than 4.9 pounds or more than 5.1 pounds.

(b) The average weight of 25 bags will be normally distributed, with a mean of 5 pounds and a standard deviation of 0.01 pounds. (Same graph as in (a), but with the following lower and upper bounds.)

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18.2 Essays and Longer Questions 1) Write an essay on the difference between the OLS estimator and the GLS estimator. Answer: Answers will vary by student, but some of the following points should be made. The multiple regression model is Yi = β0 + β1 X1i + β0 X2i + ... + βkXki + ui, i = 1, …, n which, in matrix form, can be written as Y = Xβ + U. The OLS estimator is derived by minimizing the ^ squared prediction mistakes and results in the following formula: β = (X′X)-1 X′Y. There are two GLS ^

estimators. The infeasible GLS estimator is β GLS = (X′Ω-1 X)-1 (X′Ω-1 Y). Since Ω is typically unknown, the estimator cannot be calculated, and hence its name. However, a feasible GLS estimator can be calculated if Ω is a known function of a number of parameters which can be estimated. Once ^

these parameters have been estimated, they can then be used to calculate Ω, the estimator of Ω. The ^ ^ ^ feasible GLS estimator is defined as β GLS= (X′Ω -1 )-1 (X′Ω -1 Y). There are extended least squares assumptions. · ·

E(ui Xi) = 0 (ui has conditional mean zero); (Xi,Yi), i = 1, …, n are independently and identically distributed (i.i.d.) draws from their

joint

distribution; Xi and ui have nonzero finite fourth moments; · ·

X has full column rank (there is no perfect multicollinearity);

2 var(ui Xi) = σ u (homoskedasticity);

the conditional distribution of ui given Xi is normal (normal errors),

2 These assumptions imply E(U X) = 0 n and E(UU′ X) = σ u In, the Gauss-Markov conditions for multiple regression. If these hold, then OLS is BLUE. If assumptions 5 and 6 do not hold, but assumptions 1 to 4 still hold, then OLS is consistent and asymptotically normally distributed. Small sample statistics can be derived for the case where the errors are i.i.d. and normally distributed, conditional on X. The GLS assumptions are 1.

E(U X) = 0 n;

2. 3.

E(UU′ X) = Ω(X), where Ω(X) is n×n matrix-valued that can depend on X; Xi and ui have nonzero finite fourth moments;

X has full column rank (there is no perfect multicollinearity).

The major differences between the two sets of assumptions relevant to the estimators themselves are that (i) GLS allows for homoskedastic errors to be serially correlated (dropping assumption 2 of OLS list), and (ii) there is the possibility that the errors are heteroskedastic (adding assumption 2 to GLS list). For 2 the case of independent sampling, replacing E(UU′ X) =Ω(X) with E(UU′ X) = σ u In turns the GLS estimator into the OLS estimator. In the case of the infeasible GLS estimator, the model can be transformed in such a way that the Gauss-Markov assumptions apply to the transformed model, if the four GLS assumptions hold. In that case, GLS is BLUE and therefore more efficient than the OLS estimator. This is of little practical value Stock/Watson 2e -- CVC2 8/23/06 -- Page 421

since the estimator typically cannot be computed. The result also holds if an estimator of Ω exists. However, for the feasible GLS estimator to be consistent, the first GLS assumption must apply, which is much stronger than the first OLS assumption, particularly in time series applications. It is therefore possible for the OLS estimator to be consistent while the GLS estimator is not consistent. 2) Give several economic examples of how to test various joint linear hypotheses using matrix notation. Include specifications of Rβ = r where you test for (i) all coefficients other than the constant being zero, (ii) a subset of coefficients being zero, and (iii) equality of coefficients. Talk about the possible distributions involved in finding critical values for your hypotheses. Answer: Answers will vary by student. Many restrictions involve the equality of coefficients across different types of entities in cross-sections (“stability”). Using earnings functions, students may suggest testing for the presence of regional effects, as in the textbook example at the end of Chapter 5 (exercises). The textbook tested jointly for the presence of interaction effects in the student achievement example at the end of Chapter 6. Students may want to test for the equality of returns to education and on-the-job training. The panel chapter allowed for the presence of fixed effects, the presence of which can be tested for. Testing for constant returns to scale in production functions is also frequently mentioned. Consider the multiple regression model with k regressors plus the constant. Let R be of order q × (k+ 1), where q are the number of restrictions. Then to test (i) for all coefficients other than the constant to be zero, H0 : β1 = 0, β2 = 0,. . ., βk = 0 vs. H1 : βj ≠ 0, at least one j, j=1, ..., n, you have R = [0 k×1 Ik ] and r = 0 k×1 . In large samples, the test will produce the overall regression F-statistic, which has a Fk, ∞ distribution. In case (ii), reorder the variables so that the regressors with non-zero coefficients appear first, followed by the regressors with coefficients that are hypothesized to be zero. This leads to the following formulation Yi = β0 + β1 X1i + β2 X2i + ... + βk-qXk-q,i + βk-q+1Xk-q+1,i + βk-q+2 Xk-q+2,i + ... + βkXki + ui, i = 1, …, n. R = [0 q× (k-q+1) Iq ] and r = 0 q×1 . In large samples, the test will produce an F-statistic, which has an Fq,∞ distribution. In (iii), assume that the task at hand is to test the equality of two coefficients, say H0 : β1 = β1 vs. H1 : β1 ≠ β2 , as in section 5.8 of the textbook. Then R = [0 1 -1 0 … 0], r = 0 and q = 1. This is a single restriction, and the F-statistic is the square of the corresponding t-statistic. Hence critical values can be found either from Fq,∞ or from the standard normal table, after taking the square root. 3) Define the GLS estimator and discuss its properties when Ω is known. Why is this estimator sometimes called infeasible GLS? What happens when Ω is unknown? What would the Ω matrix look like for the case of 2 independent sampling with heteroskedastic errors, where var( ui Xi) = ch(Xi) = σ2 X 1i ? Since the inverse of the error variance-covariance matrix is needed to compute the GLS estimator, find Ω -1 . The textbook shows that ~ ~ ~ ~ ~ ~ the original model Y = Xβ + U will be transformed into Y = X β + U, where Y = FY, X = FX, and U = FU, and F′F = Ω-1 . Find F in the above case, and describe what effect the transformation has on the original data. ^ Answer: β GLS= (X′Ω -1 X)-1 (X′Ω -1 Y). The key point for the GLS estimator with Ω known is that Ω is used to create a transformed regression model such that the resulting error term satisfies the Gauss-Markov conditions. In that case, GLS is BLUE. However, since Ω is typically unknown, the estimator cannot be calculated, and is therefore sometimes referred to as infeasible GLS. If Ω is unknown, then a feasible GLS estimator can be calculated if Ω is a known function of a number of parameters which can be estimated. ^

Once the parameters have been estimated, they can then be used to calculate Ω, which is the estimator of Ω. The feasible GLS estimator is then

Stock/Watson 2e -- CVC2 8/23/06 -- Page 422

β GLS= (X′Ω-1 X)-1 (X′Ω-1 Y). In the above example of heteroskedasticity, 2 X 11 0

E(UU′ X) = Ω(X) = σ2 ,

2 X 12 N

R O

2 N X 1n

2 X 11

0 1 Ω-1 (X) = σ2 O

2 X 12 O

1 0 X11

1 N 0 X12

,F= R O 1 N 2 X 1n

N 0

R O 1 N X1n

The transformation in effect scales all variables by X1 . 4) Consider the multiple regression model from Chapter 5, where k = 2 and the assumptions of the multiple regression model hold. (a) Show what the X matrix and the β vector would look like in this case. (b) Having collected data for 104 countries of the world from the Penn World Tables, you want to estimate the effect of the population growth rate (X1i) and the saving rate (X2i) (average investment share of GDP from 1980 to 1990) on GDP per worker (relative to the U.S.) in 1990. What are your expected signs for the regression coefficient? What is the order of the (X′X) here? ^

(c) You are asked to find the OLS estimator for the intercept and slope in this model using the formula β = ( X′X)-1 X′Y. Since you are more comfortable in inverting a 2×2 matrix (the inverse of a 2×2 matrix is, 1 a b -1 = d -b ) ad bc c d -c a you decide to write the multiple regression model in deviations from mean form. Show what the X matrix, the ( X′X) matrix, and the X′Y matrix would look like now. (Hint: use small letters to indicate deviations from mean, i.e., zi = Zi - Z and note that ^

Yi = β0 + β1 X1i + β2 X2i + ui Y = β0 + β1 X1 + β2 X2 . Subtracting the second equation from the first, you get

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y i = β1 x 1i + β2 x 2i + ui) (d) Show that the slope for the population growth rate is given by n

∑ yix1i ∑ x 2i - ∑ yix2i ∑ x1ix2i

β1 =

i=1

i=1 n

∑ x 1i ∑ x 2i - ( ∑ x1ix2i )2 i=1

i=1

(e) The various sums needed to calculate the OLS estimates are given below: n

∑ y i = 8.3103; ∑ x 1i = .0122; ∑ x 2i = 0.6422 i=1

i=1

∑ yix1i = -0.2304; ∑ yix2i = 1.5676; ∑ x1ix2i = -0.0520 Find the numerical values for the effect of population growth and the saving rate on per capita income and interpret these. (f) Indicate how you would find the intercept in the above case. Is this coefficient of interest in the interpretation of the determinants of per capita income? If not, then why estimate it? Answer: (a) 1 X 11 X21 β0 X 12 X22 X= 1 , and β = β1 ... ... ... β2 1 X 1n X2n (b) You would expect the population growth rate to have a negative coefficient, and the saving rate to have a positive coefficient. The order of X′X is 3×3. n n n 2 y ix 1i x 1ix 2i x ∑ ∑ ∑ 1i x 11 x21 i=1 i=1 i=1 x 12 x22 (c) X = , X′X = n , X′X . = n n 2 ... ... x x y x 1i 2i X i 2i ∑ ∑ 2i ∑ x 1n x2n i=1 i=1 i=1 (d)

-1 n 2 x x x 1i 2i ∑ 1i ∑ i=1 i=1 = n n 2 ∑ x1ix2i ∑ x 2i i=1 i=1 n

n - ∑ x 1ix 2i i=1 i=1 1 n n 2 n n n 2 2 - ∑ x 1ix 2i ∑ x 1i 2 x x ) ( 1i 2i x x ∑ 1i ∑ 1i ∑ i=1 i=1 i=1 i=1 i=1 2

∑ x 2i

∑ yix1i Post multiplying this expression with

i=1 n

∑ yix2i

results in the two least squares estimators

i=1 Stock/Watson 2e -- CVC2 8/23/06 -- Page 424

β1 ^

β2

∑ yix1i ∑ x 2i - ∑ yix2i ∑ x1ix2i i=1

i=1

i=1 n

∑ x 1i ∑ x 2i - ( ∑ x1ix2i )2 i=1

i=1

n n 2 y x i 1i x ∑ 1i ∑ ∑ x1ix2i i=1 i=1 i=1 i=1 n n n 2 2 ∑ x 1i ∑ x 2i - ( ∑ x1ix2i )2 i=1 i=1 i=1 n

∑ yix2i

, and hence gives the formula for β1 .

(e) ^

-0.2304×0.6422-(1.5676×(-0.0520)) 0.0122×0.6422-(-0.0520)2

= 1.5676×0.0122- ((-0.2304)× (-0.0520)2

β1 β2

0.0122×0.6422- (-0.0520)2

= -12.953 . 1.393

A reduction of the population growth rate by one percent increases the per capita income relative to the United States by roughly 0.13. An increase in the saving rate by ten percent increases per capita income relative to the United States by roughly 0.14. (f) The first order condition for the OLS estimator in the case of k = 2 is n n n ^ ^ ^ ^ ^ ^ Y X nβ β i β 1i = + + ∑ 0 1 ∑ 2 ∑ X2i , which, after dividing by n, results in β1 = Y - β1 X1 - β2 X2 . The i=1 i=1 i=1 intercept is only of interest if there are observations close to the origin, which is not the case here. If it is set to zero, then the regression is forced through the origin, instead being allowed to choose a level. 5) In Chapter 10 of your textbook, panel data estimation was introduced. Panel data consist of observations on the same n entities at two or more time periods T. For two variables, you have (Xit, Yit), i = 1,..., n and t = 1,..., T where n could be the U.S. states. The example in Chapter 10 used annual data from 1982 to 1988 for the fatality rate and beer taxes. Estimation by OLS, in essence, involved “stacking” the data. (a) What would the variance-covariance matrix of the errors look like in this case if you allowed for homoskedasticity-only standard errors? What is its order? Use an example of a linear regression with one regressor of 4 U.S. states and 3 time periods. (b) Does it make sense that errors in New Hampshire, say, are uncorrelated with errors in Massachusetts during the same time period (“contemporaneously”)? Give examples why this correlation might not be zero. (c) If this correlation was known, could you find an estimator which was more efficient than OLS? 2 Answer: (a) Under the extended least least squares assumptions, E(UU′ X) = σ u In. In the above example of 4 U.S. states and 3 time periods, the identity matrix will be of order 12 ×12, or (nT) × (nT) in general. Specifically

. Stock/Watson 2e -- CVC2 8/23/06 -- Page 425

(b) It is reasonable to assume that a shock to an adjacent state would have an effect on its neighboring state, particularly when the shock affects the larger of the two such as the case in Massachusetts. Other examples may be Texas and Arkansas, Michigan and Indiana, California and Arizona, New York and New Jersey, etc. A negative oil price shock, which affects the demand for automobiles produced in Michigan, will have repercussions for suppliers located not only in Michigan, but also elsewhere. (c) In case of a known variance-covariance matrix of the error terms, the GLS estimator ^ β GLS = (X′Ω-1 X)-1 (X′Ω-1 Y) could be used. The variance-covariance matrix would be of the form

(There is a subtle issue here for the case of a feasible GLS estimator, where the variances and covariances have to be estimated. It can be shown, in that case, that the GLS estimator does not exist unless n ≤ T, which is not the case for most panels. It is easier to see that the variance-covariance matrix is singular for n>T if the data is stacked by time period.)

18.3 Mathematical and Graphical Problems 1) Your textbook derives the OLS estimator as ^ β = (X′X)-1 X′Y.

Show that the estimator does not exist if there are fewer observations than the number of explanatory variables, including the constant. What is the rank of X′X in this case? Answer: In order for a matrix to be invertible, it must have full rank. Since X′X is of order (k + 1) × (k + 1), then in order to invert X′X , it must have rank (k+1). In the case of a product such as X′X, the rank is less than or equal to the rank of X′ or X, whichever is smaller. X is of order n × (k + 1), and assuming that there is no perfect multicollinearity, will have either rank n or rank (k+1), whichever is the smaller of the two. Hence if there are fewer observations than the number of explanatory variables (including the constant), then the rank of X will be n(< k+1), and the rank of X′X is also n( < k +1). Hence X′X does not have full rank, and therefore cannot be inverted. The OLS estimator does not exist as a result.

Stock/Watson 2e -- CVC2 8/23/06 -- Page 426