Practice Problems for Midterm 1

Practice Problems for Midterm 1 Multiple Choice Questions

Chapter 2

1)

The probability of an outcome a. b. c. d.

is the the number number of times that the the outcome outcome occurs occurs in in the long long run. run. equals M × N , where M is is the number of occurrences and N is is the population size. is the the proportio proportion n of times times that the the outcome outcome occurs occurs in in the long run. run. equals the sample sample mean divided divided by by the sample sample standard standard deviation deviation..

Answer: c )

The pr probability of of an ev event A or B !"r! A or B)) to occur equals a. "r! A A) × "r! B B). b. "r! A A) # "r! B) if A and B are mutually e$clusive. "r! A) c. . "r! B ) d. "r! A A) # "r! B) even if A and B are not mutually e$clusive. Answer: b

)%

Thee cum Th cumul ulat ativ ivee pro proba babi bili lity ty dist distri ribut butio ion n show showss the the prob probabi abili lity ty e. that a random variable variable is less than or equal to to a particul particular ar value. value. f. of two two or or more more even events ts occurri occurring ng at at once. once. g. of all all poss possibl iblee event eventss occur occurri ring. ng. h. that a random variab variable le ta&es ta&es on a particul particular ar value given given that that another another event event has happened.

)'

Answer: a Thee e$p Th e$pec ecte ted d val value ue of a dis discr cret etee rand random om vari variab able le a. b. c. d.

is the the outcome outcome that that is is most most li&ely li&ely to occur occur.. can be found by determining the (* value in the c.d.f. equal equalss the pop popul ulat atio ion n media median. n. is computed computed as a weighted weighted average average of the the possible possible outcome outcome of of that random variable, where the weights are the probabilities of that outcome. 1

Answer: d et Y be be a random variable. Then var!Y ) equals

+)

E-!Y − µ Y )   . b. E-/ !Y − µ Y ) / .  c. E-!Y − µ Y )  . d. E-!Y − µ Y ) . a.

Answer: c The conditional distribution of Y given given X 0 0 x, "r!Y = y / X = x ) , is

1)

a.

"r!Y = y ) . "r! X = x ) l

b.

∑ "r! X = x ,Y = y) . i

i =1

c. d.

"r! X = x, Y = y ) . "r!Y = y ) "r! X = x, Y = y ) . "r! X = x )

Answer: d The conditional e$pectation of Y given given X , E !Y / X

11)

= x ) , is calculated as follows:

k

a.

∑ y "r! X = x / Y = y) . i

i

i =1

b. E- E !Y / X ) . k

c.

∑ y "r!Y = y / X = x) . i

i

i =1 l

d.

∑ E !Y / X = x ) "r! X = x ) . i

i

i =1

Answer: c +)

Two random variables X and and Y are are independently distributed if all of the following conditions hold, with the e$ception of



Answer: d et Y be be a random variable. Then var!Y ) equals

+)

E-!Y − µ Y )   . b. E-/ !Y − µ Y ) / .  c. E-!Y − µ Y )  . d. E-!Y − µ Y ) . a.

Answer: c The conditional distribution of Y given given X 0 0 x, "r!Y = y / X = x ) , is

1)

a.

"r!Y = y ) . "r! X = x ) l

b.

∑ "r! X = x ,Y = y) . i

i =1

c. d.

"r! X = x, Y = y ) . "r!Y = y ) "r! X = x, Y = y ) . "r! X = x )

Answer: d The conditional e$pectation of Y given given X , E !Y / X

11)

= x ) , is calculated as follows:

k

a.

∑ y "r! X = x / Y = y) . i

i

i =1

b. E- E !Y / X ) . k

c.

∑ y "r!Y = y / X = x) . i

i

i =1 l

d.

∑ E !Y / X = x ) "r! X = x ) . i

i

i =1

Answer: c +)

Two random variables X and and Y are are independently distributed if all of the following conditions hold, with the e$ception of



a. b. c. d.

"r!Y = y / X = x ) = "r "r!Y = y ) . &nowing the value of one of the variables provides no information about the other. if the the cond conditi itional onal distr distribu ibutio tion n of Y given given X equals equals the marginal distribution of Y . E !Y ) = E- E !Y / X ) .

Answer: d

9)

The correlation between X and and Y a. cannot be negativ negativee since since varian variances ces are are always always positive. positive. b. is the covariance squared. c. can be calcu calculat lated ed by dividi dividing ng the covar covarian iance ce betwee between n X and and Y by by the product of the two standard deviations. cov! X , Y ) d. is giv given by by corr! X , Y ) = . var! X ) va var!Y ) Answer: c

10)

Two Two variables are uncorrelated in all of the cases below, with the e$ception of a. bein being g ind indep epen ende dent nt.. b. having a zero covariance. c. / σ XY /≤ σ X σ Y  . d. E !Y / X ) =  . Answer: c

11) 11)

var!aX

+ bY ) =

  a. a σ X

+ bσ Y  . + abσ XY + bσ Y  .

  b. a σ X c. σ XY XY + µ X µ Y .  d. aσ X

+ bσ Y  .

Answer: b

12)

To standardize a variable you a. b. c. d.

subtract subtract its mean mean and divide by its its standard standard deviation. deviation. integrate the area below two points under the normal distribution. add and subtract subtract 1.+ times the standard standard deviation deviation to the variable. variable. divide it by its its standard standard deviation, deviation, as long as its its mean is 1.

%

Answer: a

13)

 Assume that Y is is normally distributed N ! µ , σ ) . To find "r!c1 ≤ Y c − µ and d i = i , you need to calculate "r! d1 ≤ Z ≤ d  ) = σ

≤ c ) , where c1 < c

a. Φ ! d  ) − Φ ! d 1 ) . b. Φ !1.+) − Φ ! −1.+) . c. Φ ! d  ) − !1 − Φ ! d 1 )) . )) . d. 1 − !Φ ! d  ) − Φ ! d 1 )) Answer: a The 2tudent t distribution distribution is a. the distri distributio bution n of the the sum of m squared independent standard normal random variables. b. the distribution of a random variable with a chi3squared distribution distribution with m degrees of freedom, divided by m. c. always always well appro$imated appro$imated by the standard standard normal distrib distribution ution.. d. the distribution of the ratio ratio of a standard standard normal random variable, variable, divided by the the square root of an independently distributed chi3squared random variable with m degrees of freedom divided by m. Answer: d 14)

5hen there are a. b. c. d.

∞ degrees of freedom, the t ∞ distribution

can no longe longerr be calcul calculate ated. d. equals the standard normal distribution. has a bell shape similar to to that of the normal distribution, distribution, but with with 6fatter7 tails. tails.  equal equalss the the χ ∞ distribution.

Answer: b 18) 18)

Thee sam Th sampl plee aver averag agee is is a rand random om vari variab able le and and a. .b .c .d

is a single single number and as a result result cannot cannot have have a distri distributio bution. n. has a probabil probability ity distri distributio bution n called called its sampli sampling ng distribut distribution. ion. has a probabil probability ity distri distributio bution n called called the standard standard normal normal distr distributi ibution. on. has a probabi probability lity distr distributi ibution on that is the the same same as for for the the Y1 ,..., Y n i.i.d. variables.

Answer: b '

1+)

To infer the political tendencies of the students at your college9university, you sample 1( of them. nly one of the following is a simple random sample: ;ou a. ma&e 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 + a.m.
)

The variance of Y , σ Y , is given by the following formula: 

a. σ Y  . b. c. d.

σ Y

n σ Y 

n σ Y 

n

. . .

Answer: c 1)

The mean of the sample average Y , E !Y ) , is a.

1

n

µ Y .

b. µ Y . µ Y c. .

n

d.

σ Y for n = %. µ Y

Answer: b

(

)

we have appro$imately an infinite number of observations !thin& of re3sampling). variables typically are normally distributed. the covariances of Yi , Y j are typically not zero. asymptotic distributions can be counted on to provide good appro$imations to the e$act sampling distribution.

Answer: d %) The central limit theorem states that a. the distribution for

Y − µ Y σ Y

becomes arbitrarily well appro$imated by the standard

normal distribution. p

b. Y → µ . Y c. the probability that Y is in the range µ Y ± c becomes arbitrarily close to one as n increases for any constant c >  . d. the t distribution converges to the  distribution for appro$imately n = %. Answer: a ')

The covariance inequality states that  a.  ≤ σ XY

≤ σ X σ Y  . − σ X ≤ σ Y  .

 b. σ XY  c. σ XY  XY

d. σ

≤ 1.

σ X  ≤ . σ Y

Answer: b Chapter 3

1)

An estimator is a. b. c. d.

an estimate. a formula that gives an efficient guess of the true population value. a random variable. a nonrandom number. 

Answer: c )

An estimate is a. b. c. d.

)%

efficient if it has the smallest variance possible. a nonrandom number. unbiased if its e$pected value equals the population value. another word for estimator.

Answer: b >Y of the population value µ Y is consistent if An estimator µ p

a. µ> → µ . Y Y b. its mean square error is the smallest possible. c. Y is normally distributed. p

d.

Y →  .

Answer: a

)'

>Y of the population value µ Y is more efficient when compared to another An estimator µ % estimator µ Y , if > Y ) = E ! µ % a. E ! µ Y ). b. it has a smaller variance. c. its c.d.f. is flatter than that of the other estimator. > Y ) ? var! µ % d. both estimators are unbiased, and var! µ Y ).

Answer: d > Y is given by the following formula: () The standard error of Y , !E !Y ) = σ i.

n

1

∑ !Y − Y ) n i



.

i =1

"Y  @. . n &. "Y . "Y l. . n Answer: d 4

4)

5hen you are testing a hypothesis against a two3sided alternative, then the alternative is written as a. E !Y ) > µ Y , . b. E !Y ) = µ Y , . c. Y ≠ µ Y , . d. E !Y ) ≠ µ Y , . Answer: d

8)

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 X i and Y i , where each observation is represented by the point ! X i , Y i ). d. shows n observations of Y over time. Answer: c

+) The following types of statistical inference are used throughout econometrics, with the e$ception of a. b. c. d.

confidence intervals. hypothesis testing. calibration. estimation.

Answer: c 1)

Among all unbiased estimators that are weighted averages of Y1 ,..., Y n , Y is a. the only consistent estimator of µ Y . b. the most efficient estimator of µ Y . c. a number which, by definition, cannot have a variance. d. the most unbiased estimator of µ Y . Answer: b

8

11)

To derive the least squares estimator µ Y , you find the estimator m which minimizes n

e.

∑ !Y − m)



i

.

i =1

n

f. /

∑ !Y − m) / . i

i =1

n

g.

∑ mY . 

i

i =1 n

h.

∑ !Y − m)

.

i

i =1

Answer: a 1)

< µ Y , . h. #1 : E !Y ) > µ Y , . g. # 1 : µY

Answer: a 1')

A large p3value implies e. f. g. h.

re@ection of the null hypothesis. a large t 3statistic. a large Y act . that the observed value Y act is consistent with the null hypothesis.

Answer: d 1()

The formula for the sample variance is a. "

 Y

b. "

 Y

= =

1

n

∑ !Y − Y )

n − 1 i =1 1

.

i

n

∑ !Y − Y ) n −1



i

.

i =1

c. "

 Y

=

1

n

∑ !Y − µ ) n −1 i

Y



.

i =1

+

d. "

 Y

=

1

n −1

∑ !Y − Y ) n −1 i



.

i =1

Answer: b 1)

egrees of freedom a. in the conte$t 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 ma@ors at your university9college other than economics seem to have an ∞ amount of. c. are !n3) when replacing the population mean by the sample mean.   d. ensure that "Y = σ Y . Answer: a

14)

The t 3statistic is defined as follows:

a.

t =

Y − µ Y , σ Y 

.

n Y − µ Y , . !E !Y ) !Y − µ Y , )  .f t = . !E !Y ) .e t =

.g 1.+. Answer: b 18)

The power of the test e. is the probability that the test actually incorrectly re@ects the null hypothesis when the null is true. f. depends on whether you use Y or Y  for the t 3statistic. g. is one minus the size of the test. h. is the probability that the test correctly re@ects the null when the alternative is true. Answer: d

1+)

The sample covariance can be calculated in any of the following ways, with the e$ception of:

1

a. b.

n

1

∑ ! X − X )!Y − Y ) .

n − 1 i =1

i

i

n X Y − XY . ∑ n −1 n −1 n

1

i i

i =1

1

n

∑

! X − µ X )!Y i − µ Y ) . n i =1 i d. rXY "Y "Y , where r XY is the correlation coefficient. c.

Answer: c

)

5hen the sample size n is large, the +* confidence interval for µ Y is

± 1.+!E !Y ) . Y ± 1.'!E !Y ) . Y ± 1.'σ Y . Y ± 1.+ .

a. Y b. c. d. Answer: b

1)

The standard error for the difference in means if two random variables M and $ , when the two population variances are different, is

" M + "$  a. . n M + n$ " M "$ + . b. n M n$ c. d.

 1 " M !  n M

"$  + ). n$

 " M "$  + . n M n$

Answer: d ) The following statement about the sample correlation coefficient is true. 11

a. B1 ≤ r XY ≤ 1.

c.

p

→ corr ! X i , Yi ) . / r XY /< 1 .

 b. r XY

 " XY d. r XY =   . " X "Y

Answer: a

%)

The correlation coefficient a. b. c. d.

lies between zero and one. is a measure of linear association. is close to one if X causes Y . ta&es on a high value if you have a strong nonlinear relationship.

Answer: b

Chapter 4

1)

> , is zero, then 5hen the estimated slope coefficient in the simple regression model, β 1 a. b. c. d.

% 0 Y .  ? % ? 1. % 0 . % = !!!%9&!! ).

Answer: c )

Ceteros&edasticity means that

a' b' c' d'

homogeneity cannot be assumed automatically for the model. the variance of the error term is not constant. the observed units have different preferences. agents are not all rational.

Answer: b %)

5ith heteros&edastic errors, the weighted least squares estimator is DEF. ;ou should 1

use 2 with heteros&edasticity3robust standard errors because a. b. c. e.

this method is simpler. the e$act form of the conditional variance is rarely &nown. the Gauss3Har&ov theorem holds. your spreadsheet program does not have a command for weighted least squares.

Answer: b ')

5hich of the following statements is correctI

a' b' c' d'

&!! 0 E!! # !!% E!! 0 !!% # &!! E!! = &!! % 0 1 B ! E!! 9&!! )

Answer: a ()

Dinary variables a. b. c. d.

are generally used to control for outliers in your sample. can ta&e on more than two values. e$clude certain individuals from your sample. can ta&e on only two values.

Answer: d )

5hen estimating a demand function for a good where quantity demanded is a linear function of the price, you should a. b. c. d.

not include an intercept because the price of the good is never zero. use a one3sided alternative hypothesis to chec& the influence of price on quantity. use a two3sided alternative hypothesis to chec& the influence of price on quantity. re@ect the idea that price determines demand unless the coefficient is at least 1.+.

Answer: b 4)

The reason why estimators have a sampling distribution is that a. b. c. d.

economics is not a precise science. individuals respond differently to incentives. in real life you typically get to sample many times. the values of the e$planatory variable and the error term differ across samples.

1%

Answer: d

8)

The 2 estimator is derived by a. connecting the Y i corresponding to the lowest X i observation with the Y i corresponding to the highest X i observation. b. ma&ing sure that the standard error of the regression equals the standard error of the slope estimator. c. minimizing the sum of absolute residuals. d. minimizing the sum of squared residuals. Answer: d

+)

1)

The sample average of the 2 residuals is a. b. c. d.

some positive number since 2 uses squares. zero. unobservable since the population regression function is un&nown. dependent on whether the e$planatory variable is mostly positive or negative.

Answer: b

11)

The t 3statistic is calculated by dividing a. b. c. d.

the 2 estimator by its standard error. the slope by the standard deviation of the e$planatory variable. the estimator minus its hypothesized value by the standard error of the estimator. the slope by 1.+.

Answer: c 1'

1)

The slope estimator, β 1, has a smaller standard error, other things equal, if a. b. c. d.

there is more variation in the e$planatory variable, X . there is a large variance of the error term, (. the sample size is smaller. the intercept, β , is small.

Answer: a 1%)

The regression % is a measure of a. b. c. d.

whether or not X causes Y . the goodness of fit of your regression line. whether or not E!! = &!! . the square of the determinant of %.

Answer: b 1')

!Jequires Appendi$) The sample regression line estimated by 2 a. b. c. d.

will always have a slope smaller than the intercept. is e$actly the same as the population regression line. cannot have a slope of zero. will always run through the point ! X , Y ).

Answer: d

1()

The confidence interval for the sample regression function slope a. can be used to conduct a test about a hypothesized population regression function slope. b. can be used to compare the value of the slope relative to that of the intercept. c. adds and subtracts 1.+ from the slope. d. allows you to ma&e statements about the economic importance of your estimate. Answer: a

1)

c. re@ect the assumption that the error terms are homos&edastic. d. conclude that most of the actual values are very close to the regression line. Answer: a

14)

Ender the least squares assumptions !zero conditional mean for the error term, X i and Y i being i.i.d., and X i and (i having finite fourth moments), the 2 estimator for the slope and intercept a. b. c. d.

has an e$act normal distribution for n = 1(. is DEF. has a normal distribution even in small samples. is unbiased.

Answer: d

18)

To obtain the slope estimator using the least squares principle, you divide the a. b. c. d.

sample variance of X by the sample variance of Y . sample covariance of X and Y by the sample variance of Y . sample covariance of X and Y by the sample variance of X . sample variance of X by the sample covariance of X and Y .

Answer: c 1+)

To decide whether or not the slope coefficient is large or small, a. b. c. d.

you should analyze the economic importance of a given increase in X . the slope coefficient must be larger than one. the slope coefficient must be statistically significant. you should change the scale of the X variable if the coefficient appears to be too small.

Answer: a )

F!(i / X i) 0  says that a. b. c. d.

dividing the error by the e$planatory variable results in a zero !on average). the sample regression function residuals are unrelated to the e$planatory variable. the sample mean of the Ks is much larger than the sample mean of the errors. the conditional distribution of the error given the e$planatory variable has a zero 1

mean. Answer: d

1)

the population regression function. the sample regression function. e$ogenous variation. the right3hand variable or regressor.

Answer: a ) Hultiplying the dependent variable by 1 and the e$planatory variable by 1, leaves the a. b. c. d.

2 estimate of the slope the same. 2 estimate of the intercept the same. regression % the same. heteros&edasticity3robust standard errors of the 2 estimators the same.

Answer: c Analytical Questions

Chapter 2

1)

Thin& of the situation of rolling two dice and let M denote the sum of the number of dots on the two dice. !2o M is a number between 1 and 1.)

!a)

utcome !sum of dots) "robability distribution Lumulativ e probability



%

'

(



. 8 . 8

.(  .8 %

.8 % .1 4

.111 .1% + .4 .'1 8 4

14

4

8

+

1

11

1

.1 4 .(8 %

.1% + .4 

.111

.8 % .+1 

.(  .+4 

.8

.8% %

1.

distribution

Probability and Cumulative Probability Distribution of Number of Dots 2

3

4

5

6

7

8

9

10

11

12

0.18 0.16 0.14 y t 0.12 i l i b 0.1 a b 0.08 o r 0.06 P 0.04 0.02 0

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 2

3

4

5

6

7

8

9

10

11

12

Number of Dots

Probability

!b)

Cumulative Probability

Lalculate the e$pected value and the standard deviation for M . Answer: 4.M .'.

!c)

oo&ing at the s&etch of the probability distribution, you notice that it resembles a normal distribution. 2hould you be able to use the standard normal distribution to calculate probabilities of eventsI 5hy or why notI Answer: ;ou cannot use the normal distribution !without continuity correction) to calculate probabilities of events, since the probability of any event equals zero.

!d)

5hat is the probability of the following outcomesI !i) !ii) !iii) !iv) !v) !vi)

"r! M 0 4) "r! M 0  or M 0 1) "r! M 0 ' or M ≠ ') "r! M 0  and M 0 +) "r! M ? 8) "r! M 0  or M = 1)

18

Answer: !i) .14 or

% 8

=

1

M !ii) .111 or

 

' %+

=

1 +

M !iii) 1M !iv) M !v) .(8%M

= . % + "robabilities 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. Cence concepts of @oint, marginal, and conditional probability distributions stem from related concepts of frequency distributions. !vi) . or

)



;ou are interested in investigating the relationship between the age of heads of households and wee&ly earnings of households. The accompanying data gives the number of occurrences grouped by age and income. ;ou collect data from 1,4'' individuals and thin& 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 ouseholds Age of head of household

X 1

X %

X 

X '

X (

ousehold Income

13under  3under ( (3under '( '(3under ( ( and =

Y 1 Y  Y % Y ' Y (

N3under N

8

4

1%

8

'

N3under N'

1%

+

%'

1'

8

N'3under N



1+

(1

11



N3under N8

1

11

11

((

1

N8 and =

1

1

18

8'



The median of the income group of N8 and above is N1,(.

1+

!a)

Lalculate the @oint relative frequencies and the marginal relative frequencies.
X 1

X %

X 

X '

X (

ousehold Income

13under  3under ( (3under '( '(3under ( ( and = $otal

Y 1 Y  Y % Y ' Y (


.'

.''

.4(

.'+

.1'

.4


.4

.(

.1+8

.8

.(

.%'


.

.11

.1''

.(8

.%

.1


.1

.

.%

.%

.1

.1

N8 and =

.1

.1

.

.'8

.1

.11



Cumulative Income Distribution 1 0.9 0.8 0.7 t n e c r e P

0.6 0.5 0.4 0.3 0.2 0.1 0 0!"200

200!" 400

400!"600

600!"800

800 an# $

Income Class

Cumulative Income Distribution

!b)

Lalculate the conditional relative income frequencies for the two age categories 13under , and '(3under (. Lalculate the mean household income for both age categories. Answer: The mean household income for the 13under  age category is roughly N1''.
X 1

X '

ousehold Income

13under 

'(3under (

Y 1 Y  Y % Y ' Y (


.8'

.18(


.1%4

.%


.

.14


.1

.118

N8 and =

.1

.18 1

!c)

!d)

;our te$tboo& has given you a primary definition of independence that does not involve conditional relative frequency distributions. 5hat is that definitionI o you thin& that age and income are independent here, using this definitionI Answer: "r!Y = y, X = x) = "r!Y = y) "r! X = x) . 5e can chec& this by multiplying two marginal probabilities to see if this results in the @oint probability. Oor e$ample, "r!Y = Y % ) = .1 and "r! X = X % ) = .(' , resulting in a product of .114, which does not equal the @oint probability of .1''. Given that we are loo&ing at the data as a population, not a sample, we do not have to test how 6close7 .114 is to .1''.

%)

Hath and verbal 2AT scores are each distributed normally with N !(,1) .

!a)

5hat fraction of students scores above 4(I Above I Detween ' and (%I Delow '8I Above (%I Answer: "r!;=4() 0 .M "r!;=) 0 .1(84M "r!'?;?(%) 0 .'1M "r!;?'8) 0 .'4M "r!;=(%) 0 .%81.

!b)

!c)

Pe$t, assume that the correlation coefficient between the math and verbal scores is .4(. Oind the mean and variance of the resulting distribution. Answer: Given the correlation coefficient, the distribution is now N !1,%() , which has a standard deviation of appro$imately 184.

!d)

Oinally, assume that you had chosen ( students at random who had ta&en the 2AT e$am. erive the distribution for their average math 2AT score. 5hat is the probability that this average is above (%I 5hy is this so much smaller than your answer in !a)I



Answer: The distribution for the average math 2AT score is N !(,') . "r!Y > (%) 0 .8. This probability is smaller because the sample mean has a smaller standard deviation ! rather than 1). ;ou have read about the so3called catch3up theory by economic historians, whereby nations that are further behind in per capita income grow faster subsequently.
Y .% .1' Q. .'1 .%% .(

X 1

X 

Y × X 1

Y 

X 1

X 

.44 1. Q. . .1% 1%.

1.% 1. Q. .'( .% 14.8

.18 .1' Q. .8 .' .+'

.(% . Q. .18 .1+ .1844

.(+% 1. Q. .' .14 8.(+

1.+ 1. Q. .( .(+ 1%.+1'

where X 1 and X  are per capita income relative to the Enited 2tates in 1+ and 1++ respectively, and Y is the average annual growth rate in X over the 1+31++ period. Pumbers in the last row represent sums of the columns above. !a)

Lalculate the variance and standard deviation of X 1 and X  . Oor a catch3up effect to be present, what relationship must the two standard deviations showI
!b)

Lalculate the correlation between Y and X 1 . 5hat sign must the correlation coefficient have for there to be evidence of a catch3up effectI F$plain. Answer: The correlation coefficient is B.88.
%

')

Oollowing Alfred PobelRs will, there are five Pobel "rizes awarded each year. These are for outstanding achievements in Lhemistry, "hysics, "hysiology or Hedicine, iterature, and "eace.
Fconomics Pobel "rize ! X =  ) "hysics, Lhemistry, Hedicine, iterature, and "eace Pobel "rize ! X = 1 ) Total

!a)

E.2. Litizen ! Y =  ) .118

Pon3E.2. Litizen ! Y = 1 ) .'+

Total .14

.%'(

.'88

.8%%

.'%

.(%4

1.

Lompute E !Y ) and interpret the resulting number. Answer: E !Y ) = .(%4 . (%.4 percent of Pobel "rize winners were non3E.2. citizens.

!b)

Lalculate and interpret E !Y / X = 1) and E !Y / X = ) . Answer: E !Y / X = 1) = .(8 . (8. percent of Pobel "rize winners in non3economics disciplines were non3E.2. citizens. E !Y / X = ) = .+% . +.% percent of the Fconomics Pobel "rize winners were non3E.2. citizens.

!c)

A randomly selected Pobel "rize winner reports that he is a non3E.2. citizen. 5hat is the probability that this genius has won the Fconomics Pobel "rizeI A Pobel "rize in the other five disciplinesI Answer: There is a +.1 percent chance that he has won the Fconomics Pobel "rize, and a +.+ percent chance that he has won a Pobel "rize in one of the other five disciplines.

!d) .

2how what the @oint distribution would loo& li&e if the two categories were independent.

Answer: '

Soint istribution of Pobel "rize 5inners in Fconomics and Pon3Fconomics isciplines, and Litizenship, 1++31, under assumption of independence

Fconomics Pobel "rize ! X =  ) "hysics, Lhemistry, Hedicine, iterature, and "eace Pobel "rize ! X = 1 ) Total

E.2. Litizen ! Y =  ) .44

Pon= E.2. Litizen ! Y = 1 ) .+

.14

.%8

.''4

.8%%

.'%

.(%4

1.

Total

A few years ago the news magazine &* Economi"t listed some of the stranger e$planations used in the past to predict presidential election outcomes. These included whether or not the hemlines of womenRs s&irts went up or down, stoc& mar&et performances, baseball 5orld 2eries wins by an American eague team, etc. Thin&ing 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 %' presidential elections. ;ou thin& 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. ;ou generate the accompanying table:

4)

Soint istribution of "residential "arty Affiliation and "arty Lontrol of Couse of Jepresentatives, 1831++

emocratic "resident ! X =  ) Jepublican "resident ! X = 1 ) Total

!a)

emocratic Lontrol of Couse ! Y =  ) .'1

Jepublican Lontrol of Couse ! Y = 1 ) .%

Total .''1

.14

.%8

.((+

.(88

.'1

1.

Answer: %8. percent of the presidents were Jepublicans and were in the 5hite Couse while Jepublicans controlled the Couse of Jepresentatives. ''.1 percent of all presidents were emocrats.

(

!b)

Lompute E ! X ) . Cow does this differ from E ! X / Y = ) I F$plain.. Answer: E ! X ) 0 .((+. E ! X / Y = ) 0 .41. E ! X ) gives you the unconditional e$pected value, while E ! X / Y = ) is the conditional e$pected value.

!c)

!d)

5hat would the @oint distribution loo& li&e under independenceI Lhec& your results by calculating the two conditional distributions and compare these to the marginal distribution. Answer:

Soint istribution of "residential "arty Affiliation and "arty Lontrol of Couse of Jepresentatives, 1831++, under the Assumption of
emocratic "resident ! X =  ) Jepublican "resident ! X = 1 ) Total

8)

emocratic Lontrol of Couse ! Y =  ) .(+

Jepublican Lontrol of Couse ! Y = 1 ) .18

.''1

.%+

.%

.((+

.(88

.'1

1.

"r! X

=  / Y = ) =

"r!Y

= 1/ X = 1) =

.(+

.(88 .% .((+

Total

= .'' !there is a small rounding error).

= .'11 !there is a small rounding error).

The e$pectations augmented "hillips curve postulates

∆ p = π − + !( − ( ) , where ∆ p is the actual inflation rate, π is the e$pected inflation rate, and ( is the unemployment rate, with 6B7 indicating equilibrium !the PA
), i'*' that you e$pect this periodRs inflation rate to hold for the ne$t period !6 the sun shines today, it will shine tomorrow7), then the prediction is that inflation will accelerate if the unemployment rate is below its equilibrium level. The accompanying table below displays information on accelerating annual inflation and unemployment rate differences from the equilibrium rate !cyclical unemployment), where the latter is appro$imated by a five 0 year moving average. ;ou thin& 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 Enited 2tates quarterly data for the period 1+':1 to 1++(:'. Soint istribution of Accelerating
∆ p − ∆ p−1 >  ! X =  ) ∆ p − ∆ p−1 ≤  ! X = 1 ) Total !a)

!( − ( ) >  ! Y =  ) .1(

!( − ( ) ≥  ! Y = 1 ) .%8%

.(%+

.+4

.1'

.'1

.'(%

.('4

1.

Total

Lompute E !Y ) and E ! X ) , and interpret both numbers. Answer: E !Y ) = .('4 . ('.4 percent of the quarters saw cyclical unemployment. E ! X ) = .'1 . '.1 percent of the quarters saw decreasing inflation rates.

!b)

Lalculate E !Y / X = 1) and E !Y / X = ) .
!c)

5hat is the probability of inflation to increase if there is positive cyclical unemploymentI Pegative cyclical unemploymentI Answer: There is a %'.' percent probability of inflation to increase if there is positive cyclical unemployment. There is a 4 percent probability of inflation to increase if there is negative cyclical unemployment.

4

!d)

;ou randomly select one of the (+ quarters when there was positive cyclical unemployment ! !( − ( ) >  ) . 5hat is the probability there was decelerating inflation during that quarterI Answer: There is a (. percent probability of inflation to decelerate when there is positive cyclical unemployment. +) The accompanying table shows the @oint distribution between the change of the unemployment rate in an election year and the share of the candidate of the incumbent party since 1+8. ;ou thin& 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. Soint istribution of Enemployment Jate Lhange and
∆( >  ! X =  ) ∆( ≤  ! X = 1 ) Total !a)

! ,nc(mb*nt − (*) >  ! Y =  ) .(% .(4+ .%

! ,nc(mb*nt − (*) ≤  ! Y = 1 ) .11 .1(4 .%8

Total .' .4% 1.

Lompute and interpret E !Y ) and E ! X ) . Answer: E !Y ) = .%8 M E ! X ) = .4% . The probability of an incumbent to have less than (* of the share of votes cast for the two ma@or3party candidates is .%8. The probability of observing falling unemployment rates during the election year is 4%. percent.

!b)

Lalculate E !Y / X = 1) and E !Y / X = ) . id you e$pect these to be very differentI Answer: E !Y / X = 1) = .1% M E !Y / X = ) = .4++ . A student who believes that incumbents will attempt to manipulate the economy to win elections will answer affirmatively here.

!c)

5hat is the probability that the unemployment rate decreases in an election yearI Answer: "r! X = 1) = .4%.

!d)

Londitional on the unemployment rate decreasing, what is the probability that an incumbent will lose the electionI Answer: "r!Y

= 1/ X = 1) = .1% . 8

!e)

5hat would the @oint distribution loo& li&e under independenceI Answer: Soint istribution of Enemployment Jate Lhange and
∆( >  ! X =  ) ∆( ≤  ! X = 1 ) Total

! ,nc(mb*nt − (*) >  ! Y =  ) .14 .'( .%

! ,nc(mb*nt − (*) ≤  ! Y = 1 ) .+4 .41 .%8

Total .' .4% 1.

1) The accompanying table lists the @oint distribution of unemployment in the Enited 2tates in 1 by demographic characteristics !race and gender). Soint istribution of Enemployment by emographic Lharacteristics, Enited 2tates, 1

Age 131+ ! X =  ) Age  and above ! X = 1 ) Total

5hite ! Y =  ) .1%

Dlac& and ther ! Y = 1 ) .(

Total

.

.

.8

.4%

.4

1.

.18

!a) 5hat is the percentage of unemployed white teenagersI Answer: "r!Y

= , X = ) = .1%.

!b) Lalculate the conditional distribution for the categories 6white7 and 6blac& and other.7 +

Ans)er* Londitional istribution of Enemployment by emographic

Lharacteristics, Enited 2tates, 1

Age 131+ ! X =  ) Age  and above ! X = 1 ) Total

5hite ! Y =  ) .18

Dlac& and ther ! Y = 1 ) .1+

.8

.81

1.

1.

!c) Given your answer in the previous question, how do you reconcile this fact with the probability to be * of finding an unemployed adult white person, and only * for the category 6blac& and other.7 Answer: The original table showed the @oint probability distribution, while the table in !b) presented the conditional probability distribution.

%

Mathematical and Graphical Problems 1)

Thin& of an e$ample involving five possible quantitative outcomes of a discrete random variable and attach a probability to each one of these outcomes. isplay the outcomes, probability distribution, and cumulative probability distribution in a table. 2&etch both the probability distribution and the cumulative probability distribution. Answer: Answers will vary by student. The generated table should be similar to Table .1 in the te$t, and figures should resemble Oigures .1 and . in the te$t.

)

The height of male students at your college9university is normally distributed with a mean of 4 inches and a standard deviation of %.( inches.
!a) !b) !c) !d) !e)

taller than UVI between (U%V and U(VI shorter than (U4V, the mean height of female studentsI shorter than (UVI taller than 2haq RPeal, the center of the .A. a&ers, who is 4U1V tallI Lompare this to the probability of a woman being pregnant for 1 months !% days), where days of pregnancy is normally distributed with a mean of  days and a standard deviation of 1 days. Answer: !a) "r!W = .(41') 0 .8%+M !b) "r! B ? W ? ) 0 .+('( or appro$imately .+(M !c) "r!W ? 3.8(41) 0 .1+(4M !d) "r!W ? 3.8(41) 0 .1M !e) "r!W = '.8(4) 0 .+ !the te$t does not show values above .++ standard deviations, "r!W=.++ 0 .1') and "r!W = .1() 0 .18.

%)

Lalculate the following probabilities using the standard normal distribution. 2&etch the probability distribution in each table case, shading in the area of the calculated probability.

!a) !b) !c) !d) !e) !f) !g) !h) !i) !@)

"r! Z ? .) "r! Z ≤ 1.) "r! Z = 1.+) "r! Z ? B.) "r! Z = 1.'() "r! Z = B1.'() "r!B1.+ ? Z ? 1.+) "r! Z ? .(4 or Z = .(4) "r! Z = -) 0 .1M find -. "r! Z ? 3 - or Z = -) 0 .(M find -.

%1

Answer: !a) .(M !b) .8'1%M !c) .(M !d) .8M !e) .(M !f) .+(M !g) .(M !h) .1M !i) 1.81M !@) 1.+. ') Esing the fact that the standardized variable Z is a linear transformation of the normally distributed random variable Y , derive the e$pected value and variance of Z .

Y − µY

1 µ Y and b = . Given !.+) σY σ Y σ Y σ Y 1 µ Y 1 + µ Y =  , and σ Z =  σ Z  = 1 . and !.%) in the te$t, E ! Z ) = − σ Y σ Y σ Z

Answer: Z

=

=−

µ Y σY

+

1

Y = a + bY , with a = −

() 2how in a scatterplot what the relationship between two variables X and Y would loo& li&e if there was !a)

a strong negative correlation. Answer:

!b)

%

a strong positive correlation. Answer:

!c) no correlation.

!d) 5hat would the correlation coefficient be if all observations for the two variables were on a curve described by Y = X  I Answer: The correlation coefficient would be zero in this case, since the relationship is non3 linear.

)

Oind the following probabilities:

!)a

Y is distributed χ ' . Oind "r!Y = +.'+). Answer: .(.

!)b

Y is distributed t ∞ . Oind "r!; = B.(). Answer: .+1(.

!c)

Y is distributed  ',∞ . Oind "r!Y ? %.%). Answer: .++.

!d)

Y is distributed N !(,1) . Oind "r!Y = + or Y ? %'). Answer: .(. %%

4)

"robability

#1(

.

#(

.%



.'

3(

.(

31(

.(

5hat is the e$pected value, the variance, and the standard deviationI 5hich is the most li&ely outcomeI 2&etch the cumulative distribution function. .  Answer: F!;) 0 %.(M σ Y

= 8.'+M σ Y 0 .+1M most li&ely: .

Stock Price Change During Next !elve Months 1 0.9 0.8 y 0.7 t i l i 0.6 b a 0.5 b o 0.4 r P

0.3 0.2 0.1 0 !15

!5

0

%5

Percentage Change &toc' Price ( C)an*e

%'

%15

8)

;ou consider visiting Hontreal during the brea& between terms in Sanuary. ;ou go to the relevant 5eb site of the official tourist office to figure out the type of clothes you should ta&e on the trip. The site lists that the average high during Sanuary is B4 L, with a standard deviation of ' L. Enfortunately you are more familiar with Oahrenheit than with Lelsius, but find that the two are related by the following linear function: ( +

. = −% +  . Oind the mean and standard deviation for the Sanuary temperature in Hontreal in Oahrenheit. Answer: Esing equations !.+) and !.%) from the te$tboo&, the result is 1+.' and 4.. +)

Two random variables are independently distributed if their @oint distribution is the product of their marginal distributions.
= y / X = 1) = "r!Y = y / X = ) = ... = "r!Y = y / X = l) .

Dut if all conditional distributions are equal, then they must also equal the marginal distribution, i.e. "r!Y

= y / X = x) = "r!Y = y) .

Given the definition of the conditional distribution of Y given X 0 x, you then get "r!Y

= y / X = x) =

= y, X = x ) = "r!Y = y) , "r! X = x)

"r!Y

which gives you the condition "r!Y 1)

= y, X = x) = "r!Y = y) "r! X = x) .

There are frequently situations where you have information on the conditional distribution of Y given X , but are interested in the conditional distribution of X given Y .

%(

Jecalling "r!Y "r! X

= x /Y =

"r! X = x,Y = y ) , derive a relationship between "r! X = x ) y ) and "r!Y = y / X = x) . This is called DayesR theorem.

= y / X = x) =

"r! X = x, Y = y ) , "r! X = x ) "r!Y = y / X = x) × "r! X = x) = "r! X = x, Y = y) M "r! X = x, Y = y) similarly "r! X = x / Y = y ) = and "r!Y = y) "r! X = x / Y = y ) × "r!Y = y) = "r! X = x, Y = y) . Fquating the two and solving for "r! X = x / Y = y ) then results in "r!Y = y / X = x) × "r! X = x) "r! X = x / Y = y ) 0 . "r!Y = y )

Answer: Given "r!Y

11)

= y / X = x) =

;ou are at a college of roughly 1, students and obtain data from the entire freshman class !( students) on height and weight during orientation. ;ou 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. 5eight !Y ) is measured in pounds and height ! X ) is measured in inches. ;ou calculate the following sums: n

∑ y

 i

i =1

n

0 +',8.8,

∑ x

 i

n

0 1,'8.+,

i =1

∑ x y 0 4,(.+ i

i

i =1

!small letters refer to deviations from means as in -i

= Zi − Z ).

!a) Given your general &nowledge about human height and weight of a given age, what can you say about the shape of the two distributionsI Answer: Doth distributions are bound to be normal. !b) 5hat is the correlation coefficient between height and weight hereI Answer: .4%. 1)

Ese the definition for the conditional distribution of Y given X = x and the marginal distribution of X to derive the formula for "r! X = x, Y = y ) . This is called the multiplication rule. Ese it to derive the probability for drawing two aces randomly from a dec& of cards !no @o&er), where you do not replace the card after the first draw. Pe$t, generalizing the multiplication rule and assuming independence, find the probability of having four girls in a family with four children.

%

Answer: 1%)

× = .'( M .( or  1 ( (1   '

%

'

= 1  1   

.

The systolic blood pressure of females in their s is normally distributed with a mean of 1 with a standard deviation of +. 5hat is the probability of finding a female with a blood pressure of less than 1I Hore than 1%(I Detween 1( and 1%I ;ou visit the womenRs soccer team on campus, and find that the average blood pressure of the ( members is 11'.
Answer: σ

 Y

k

k

= ∑ ! yi − µ Y ) pi = ∑ ! y + µ −  µY yi ) pi = ∑ ! yi pi + µ Y p i −  µ Y yi p i) . 

 i

i =1

 Y

i =1

i =1

Hoving the summation sign through results in k

σ

 Y

= ∑ y pi + µ  i

i =1

k

 Y

k

k

k

∑ p − µ ∑ y p . Dut ∑ p = 1 and µ = ∑ y p , giving i

i =1

Y

i

i =1

i

i

i =1

you the second e$pression after simplification.

%4

Y

i

i =1

i

1)

The Economic %*port o+ t* /r*"id*nt gives the following age distribution of the Enited 2tates population for the year : +nited tates -opulation ./ Age 0roup, 2'''

utcome !age category) "ercentage

Ender (

(31(

131+

3'

(3''

'(3'

.

.1

.

.4

.%

.

( and over .1%

() 0 .%(M

utcome !age category) Lumulative probability distribution

Ender (

(31(

131+

3'

(3''

'(3'

.

.

.8

.%(

.(

.84

( and over 1.

"r!Y ≤ ') = .%(. 14)

The accompanying table gives the outcomes and probability distribution of the number of times a student chec&s her e3mail daily: -robabilit/ of Checing &#ail

utcome !number of e3 mail chec&s) "robability distribution



1



%

'

(



.(

.1(

.%

.(

.1(

.8

.

2&etch the probability distribution. Pe$t, calculate the c.d.f. for the above table. 5hat is the probability of her chec&ing her e3mail between 1 and % times a dayI f chec&ing it more than % times a dayI %8

Answer: utcome !number of e3 mail chec&s) Lumulative probability distribution



1



%

'

(



.(

.

.(

.4(

.+

.+8

1.

"r!1 ≤ Y ≤ %) = .4 M "r!Y > .() . Cumulative Distribution +unction 1 y t i l i b a b o r P

0.8 0.6 0.4 0.2 0 0

1

2

3

4

5

6

Number of "#mail Checks

Cumulative Distribution +unction

18)

The accompanying table lists the outcomes and the cumulative probability distribution for a student renting videos during the wee& while on campus. ideo !entals per ee during emester

utcome !number of wee&ly video rentals) "robability distribution



1



%

'

(



.(

.((

.(

.(

.4

.

.1

%+

2&etch the probability distribution. Pe$t, calculate the cumulative probability distribution for the above table. 5hat is the probability of the student renting between  and ' a wee&I f less than % a wee&I Answer: The cumulative probability distribution is given below. The probability of renting between two and four videos a wee& is .%4. The probability of renting less than three a wee& is .8(. utcome !number of wee&ly video rentals) Lumulative probability distribution 1+)



1



%

'

(



.(

.

.8(

.+

.+4

.++

1.

The te$tboo& mentioned that the mean of Y , E !Y ) is called the first moment of Y , and  that the e$pected value of the square of Y , E !Y ) is called the second moment of Y , and so on. These are also referred to as moments about the origin. A related concept is r moments about the mean, which are defined as E-!Y − µ Y )  . 5hat do you call the second moment about the meanI 5hat do you thin& the third moment, referred to as 6s&ewness,7 measuresI o you believe that it would be positive or negative for an earnings distributionI 5hat measure of the third moment around the mean do you get for a normal distributionI Answer: The second moment about the mean is the variance. 2&ewness measures the departure from symmetry. Oor the typical earnings distribution, it will be positive. Oor the normal distribution, it will be zero.

,umber o- ee'ly /i#eo entals 0.6 0.5 y t i l i b a b o r P

0.4 0.3 0.2 0.1 0 1

2

3

4

5

Number of $entals

,umber o- ee'ly /i#eo entals

'

6

7

)

F$plain why the two probabilities are identical for the standard normal distribution: "r! −1.+ ≤ X ≤ 1.+) and "r! −1.+ < X < 1.+) . Answer: Oor a continuous distribution, the probability of a point is zero.

Chapter 3

Thin& of at least nine e$amples, three of each, that display a positive, negative, or no correlation between two economic variables.
Adult males are taller, on average, than adult females. isiting two recent American ;outh 2occer rganization !A;2) under313year3old !E1) soccer matches on a 2aturday, you do not observe an obvious difference in the height of boys and girls of that age. ;ou suggest to your little sister that she collect data on height and gen der of children in 'th to th grade as part of her science pro@ect. The accompanying table shows her findings. eight of 5oung .o/s and 0irls, 0rades 4&%, in inches

Doys

Y Boy" (4.8

!e)

" Boy" %.+

Girls

nBoy"

Y 0irl"

((

(8.'

"0irl" '.

n0irl" (4

et your null hypothesis be that there is no difference in the height of females and males at this age level. 2pecify the alternative hypothesis.

'1

Answer: #  : µ Boy" − µ 0irl" !f)

=  vs. # 1 : µ Boy" − µ 0irl" ≠ 

Oind the difference in height and the standard error of the difference. Answer: Y Boy" − Y 0irl" 0 3., 2F! YBoy" − Y 0irl" ) 0

!g)

%.+ ((

+

'. (4

0 .44.

Generate a +(* confidence interval for the difference in height. Answer: 3. ± 1.+ × .44 0 !3.11, .+1).

!h)

Lalculate the t 3statistic for comparing the two means.
)'

Hath 2AT scores !Y ) are normally distributed with a mean of ( and a standard deviation of 1. An evening school advertises that it can improve studentsR scores by roughly a third of a standard deviation, or % points, if they attend a course which runs over several wee&s. !A similar claim is made for attending a verbal 2AT course.) The statistician for a consumer protection agency suspects that the co urses are not effective. 2he views the situation as follows: #  : µ Y = ( vs. # 1 : µ Y = (% .

!e)

2&etch the two distributions under the null hypothesis and the alternative hypothesis. Answer:

!f)

The consumer protection agency wants to evaluate this claim by sending ( students to attend classes. ne of the students becomes sic& during the course and drops out. 5hat is '

the distribution of the average score of the remaining '+ students under the null, and under the alternative hypothesisI Answer: Y of the '+ participants is normally distributed, with a mean of ( and a standard deviation of 1'.8 under the null hypothesis. Ender the alternative hypothesis, it is normally distributed with a mean of (% and a standard deviation of 1'.8.

'%

!g)

Assume that after graduating from the course, the '+ participants ta&e the 2AT test and score an average of (.
!h)

5hat would be the critical value under the null hypothesis if the size of your test were (*I Answer: The critical value would be (%.

!i)

Given this critical value, what is the power of the testI 5hat options does the statistician have for increasing the power in this situationI Answer: "r!Y < (% / # 1 is true) 0 .%1. Cence the power of the test is .88. 2he could increase the power by decreasing the size of the test. Alternatively, she could try to convince the agency to hire more test sub@ects, i.e., she could increase the sample size.

)(

;our pac&aging company fills various types of flour into bags. Jecently there have been complaints from one chain of stores: a customer returned one opened ( pound bag which weighed significantly less than the label indicated. ;ou view the weight of the bag as a random variable which is normally distributed with a mean of ( pounds, and, after studying the machine specifications, a standard deviation of .( pounds.

.a

;ou ta&e a sample of  bags and weigh them. 2&etch below what the average pattern of individual weights might loo& li&e. et the horizontal a$is indicate the sampled bag number !1, , Q, ). n the vertical a$is, mar& the e$pected value of the weight under the null hypothesis, and two ! ≈ 1.+) standard deviations above and below the e$pected value. raw a line through the graph for E !Y ) #  σ Y , E !Y ), and E !Y ) B  σ Y . Cow many of the bags in a sample of  will you e$pect to weigh either less than '.+ pounds or more than (.1 poundsI

''

Answer: n average, there should be one bag in every sample of  which weighs less than '.+ pounds or more than (.1 pounds.

.b

;ou sample ( bags of flour and calculate the average weight. 5hat is the distribution of the average weight of these ( bagsI Jepeating the same e$ercise  times, s&etch what the distribution of the average weights would loo& li&e in a graph similar to the one you drew in !b), where you have ad@usted the standard error of Y accordingly. Answer: The average weight of ( bags will be normally distributed, with a mean of ( pounds and a standard deviation of .1 pounds.

'(

.c

Oor each of the twenty observations in !b), a +(* confidence interval is constructed. raw these confidence intervals, using the same graph as in !b). Cow many of these  confidence intervals would you e$pect to weigh ( pounds under the null hypothesisI Answer: ;ou would e$pect 1+ of the  confidence intervals to contain ( pounds.

()

Assume that two presidential candidates, call them Dush and Gore, receive (* of the votes in the population. ;ou can model this situation as a Dernoulli trial, where Y is a random variable with success probability "r!Y = 1) = p , and where Y 0 1 if a person votes for Dush and Y 0  otherwise. Ourthermore, let p> be the fraction of successes !1s) p!1 − p) in a sample, which is distributed N ! p, ) in reasonably large samples, say for n ≥

n

'. !a) Given your &nowledge about the population, find the probability that in a random sample of ', Dush would receive a share of '* or less.

Answer:

!a)

µ < .') = "r! Z "r! p

<

.' − .(

) = "r! Z < −1.) ≈ .1'.

Cow would this situation change with a random sample of 1I '

Answer:

!b)

µ < .') = "r! Z "r! p

<

.' − .(

) = "r! Z < −.) ≈ .%. 5ith 5ith this sample .( 1 size, you would e$pect this to happen only every (th sample.

Given Given your your answ answers ers in !a) !a) and and !b), !b), woul would d you you be comf comfort ortabl ablee to pred predict ict what what the the voting voting intentions for the entire population are if you did not &now p but had polled 1, individuals at random and calculated p> I F$plain. Answer: The answers in !a) !a) and !b) suggest that for even moderate increases in the sample size, the estimator does not vary too much from the population mean. µ of .'8, for e$ample, "olling 1, individuals, the probability of finding a p would be .%. Enless the election was e$tremely close, which the  election was, polls are quite accurate even for sample sizes of ,(.

!c)

This This resul resultt seem seemss to hold hold wheth whether er you you poll poll 1, 1, peopl peoplee at random random in the the Peth Petherl erland andss or the Enited 2tates, where the former has a population of less than  million people, while the Enited 2tates is 1( times as populous. 5hy does the population size not come into playI Answer: The distribution distribution of sample means shrin&s very quic&ly depending on the sample size, not the population size. Although at first this does not seem intuitive, the standard error of an estimator is a value which indicates by how much the estimator varies around the population value. Oor large sample sizes, the sample mean typically is very close to the population p opulation mean. ) ;ou ;ou have collected collected wee&ly wee&ly earnings earnings and age data from a sub3sam sub3sample ple of 1,4'' individu individuals als using the Lurrent "opulation 2urvey in a given year.

!a)

. Given Given the the over overall all mean mean of N'%'.'+ N'%'.'+ and a standa standard rd deviati deviation on of N+ N+'.4 '.4,, const construc ructt a ++* ++* confidence interval for average earnings in the entire population. 2tate the meaning of this interval in words, rather than @ust in numbers.
+'.4

14'' 0 '%'.'+ ± 18.1% 0 !'1.%, '(.). Dased on the sample at hand, the best guess for the population mean is N'%'.'+. N'% '.'+. Cowever, because of random '4

sampling error, this guess is li&ely to be wrong.
5hen 5hen div divid idin ing g your your samp sample le int into o peopl peoplee '( year yearss and and olde older, r, and and youn younge gerr than than '(, '(, the the information shown in the table is found.

Age Lategory Age ≥ '( '( Age < '( ' (

Average Farnings

2tandard eviation

Y

"Y

N'88.84 N'1.

N%8.' N4.%

N ( 4 1 % 4

Test Test whether or not the difference in average earnings is statistically significant. Given your &nowledge of age3earning profiles, p rofiles, does this result ma&e senseI

t =

!'8 !'88.84 − '1.) )

%8.' 4.% 0 '., + ( 4 1% 4 which is statistically significant at conventional levels whether you use a two3 sided or one3sided alternative. Cence the null hypothesis of equal average earnings in the two groups is re@ected. Age3earning profiles typically ta&e on an inverted E3shape. Ha$imum earnings occur in the 's, depending on some other factors such as years of education, which are not considered here. Cence it is not clear if the alternative hypothesis should be one3sided o ne3sided or two3sided.
Answer: Assuming unequal population variances,

4)

!)c

A ma manuf nufac actu ture rerr clai claims ms that that a cert certai ain n brand brand of of LJ LJ pla playe yerr has an ave avera rage ge lif lifee e$pe e$pect ctan ancy cy of ( years and  months with a standard deviation of 1 year and  months. Assume that the life e$pectancy is normally distributed. 2elect 2electing ing one one LJ LJ play player er from from this this brand brand at at random random,, calcul calculate ate the the probab probabili ility ty of its its life life e$pectancy e$ceeding 4 years. Answer: "r!Y

!)d

> 4) = "r! Z > 1) 0 .1(84.

The .ritical .on"(m*r magazine magazine decides to test fifty LJs of this brand. The average life in this sample is  years and the sample standard deviation is  years. Lalculate a ++* confidence interval for the average life. '8

Answer:  ± .(4 ×

 (

0  ± .4% 0 !(.4, .4%).

'+

!c)

Cow many more LJs would the magazine have to test in order to halve the width of the confidence intervalI Answer :

1 

× !  .( 4 ×

 (

) = .(4×

1 

×

 (

= .(4×

 ' × (

, or n 0 .

8) '!' N*" and $orld %*port ran&s ran&s colleges and universities annually. ;ou ;ou randomly sample 1 of the national universities u niversities and liberal arts colleges from the year  issue. The average cost, which includes tuition, fees, and room and board, is N%,(41.'+ with a standard deviation of N4,1(.(. !a) Dased on this sample, construct a +(* confidence interval interval of the average cost of attending a university9college in the Enited 2tates. Answer: %,(41.'+ ± 1.+ ×

4,1(.( 1

0 %,(41.'+ ± 41.(( 0 !,8+.+', ',4%.').

!b) Lost varies by quite a bit. ne of the reasons may be that some universities9colleges universities9colleges have a better reputation than others. '!' N*" and $orld %*port tries tries to measure this factor by as&ing university presidents and chief academic officers about the reputation of institutions. The ran&ing is from 1 !6marginal7) to ( !6distinguished7). ;ou ;ou decide to split the sample according to whether the academic aca demic institution has a reputation of greater than %.( or not. Oor comparison, in , Laltech had a reputation ran&ing of '.4, 2mith Lollege had '.(, and Auburn Eniversity had %.1. This gives you the statistics shown in the accompanying table.

Jeputation La Lategory

Average Lo Lost

Y Jan&ing = %.( Jan&ing ≤ %. %.(

N+,%11.%1 N1,4.

2tandard eviation of Lost ! "Y ) N(,'+.1 N,1%%.%8

P + 41

Test Test the hypothesis that the average cost co st for all universities9colleges is the same independent of the reputation. 5hat alternative hypothesis did you useI

t =

!+,%1 !+,%11. 1.%1 %1− 1, 1, 4. 4. ))

(, ' '+.1 ,1%%.%8 0 .%%, + + 41 which is statistically significant whether or not you use a one3sided or two3sided hypothesis test. ;our ;our prior e$pectation is that academic institutions with a


(

higher reputation will charge more for attending, and hence a one3sided alternative would have been appropriate here. !c) 5hat other factors should you consider before ma&ing a decision based on the data in !b)I Answer: There may be other variables which potentially have an effect on the cost of attending the academic institution. 2ome of these factors might be whether or not the college9university is private or public, its size, whether or not it has a religious affiliation, etc.
The development office and the registrar have provided you with anonymous matches of starting salaries and G"As for 18 graduating economics ma@ors. ;our sample contains a variety of @obs, from church pastor to stoc&bro&er. !a) The average starting salary for the 18 students was N%8,''.8 with a standard deviation of N4,('1.'. Lonstruct a +(* confidence interval for the starting salary of all econo mics ma@ors at your university9college.

Answer: %8,''.8 ± 1.+ ×

4,('1.' 18

0 %8,''.8 ± 1,'.% 0 !%4,.(', ',4.18).

!b) A similar sample for psychology ma@ors indicates a significantly lower starting salary. Given that these students had the same number of years of education, does this indicate discrimination in the @ob mar&et against psychology ma@orsI Answer:
Average Farnings

2tandard eviation

Y

"Y

N%+,+1(.( N%4,8%.%%

N8,%%.1 N,14'.8 (1

n (+ '+

Londuct a t 3test for the hypothesis that the two starting salaries are the same in the population. Given that this data was collected in 1+++, do you thin& that your results will hold for other years, such as I

t =

!%+,+1(.( − %4,8%.%%)

8, %%.1 ,14'.8 0 .%. + (+ '+ The critical value for a one3sided test is 1.', for a two3sided test 1.+, both at the (* level. Cence you can re@ect the null hypothesis that the two starting salaries are equal. "resumably you would have chosen as an alternative that better students receive better starting salaries, so that this becomes your new wor&ing hypothesis. 1+++ was a boom year.

1%)

uring the last few days before a presidential election, there is a frenzy of voting intention surveys. n a given day, quite often there are conflicting results from three ma@or polls.

!a) Thin& of each of these polls as reporting the fraction of successes !1s) of a Dernoulli random variable Y , where the probability of success is "r!Y = 1) = p . et p> be the fraction of successes in the sample and assume that this estimator is normally distributed with a mean of p!1 − p) . 5hy are the results for all polls different, even though they are p and a variance of

n

ta&en on the same dayI Answer: 2ince all polls are only samples, there is random sampling error. As a result, p> will differ from sample to sample, and most li&ely also from p. !b) Given the estimator of the variance of p> ,

pµ !1 − µp ) , construct a +(* confidence interval n

for p> . Oor which value of p> is the standard deviation the largestI 5hat value does it ta&e in the case of a ma$imum p> I

(

µ ± 1.+ × Answer: p

µp !1 − µp)

n

. A bit of thought or calculus will show that the standard

deviation will be largest for p> 0 .(, in which case it becomes

!)e

n

.

5hen the results from the polls are reported, you are told, typically in the small print, that the 6margin of error7 is plus or minus two percentage points. Esing the appro$imation of 1.+ ≈  , and assuming, 6conservatively,7 the ma$imum standard deviation derived in !b), what sample size is required to add and subtract !6margin of error7) two percentage points from the point estimateI Answer: n 0 ,(.

!)f

.(

5hat sample size would you need to halve the margin of errorI Answer: n 0 1,.

(%

Mathematical and Graphical Problems a.

;our te$tboo& defined the covariance between X and Y as follows: n

1

∑ ! X − X )!Y − Y )

n − 1 i =1

i

i

"rove that this is identical to the following alternative specification:

n X Y − XY ∑ n −1 n −1 n

1

i i

i =1

Answer: n

1

∑ ! X − X )!Y − Y ) = n − 1 ∑ ! X Y − XY − YX + YX )

n − 1 i =1

= 0

n

1

1

n −1

i

i

i i

n

!

i

i

i =1

n

n

1

n

∑ X Y − X ∑ Y − Y ∑ X + nYX ) = n − 1 !∑ X Y − nXY − nYX + nYX ) i i

i

i =1

i

i =1

i i

i =1

i =1

n X Y − XY . ∑ n −1 n −1 1

n

i i

i =1

b.

Oor each of the accompanying scatterplots for several pairs of variables, indicate whether you e$pect a positive or negative correlation coefficient between the two variables, and the li&ely magnitude of it !you can use a small range).

!a) 10

8

2

6

4

2 2

4

6

8

10

12

14



Answer: "ositive correlation. The actual correlation coefficient is .'.

('

!b) 0.08

0.06

0.04

2

0.02

0.00

!0.02

!0.04 0.0

0.2

0.4

0.6

0.8

1.0



Answer: Po relationship. The actual correlation coefficient is .4. !c) 0.65

0.60

0.55

2 0.50

0.45

0.40

0.35 3.5

4.0

4.5

5.0

5.5

6.0

6.5

1

Answer: Pegative relationship. The actual correlation coefficient is B.4.

((

!d) 2000

1500

2 1000

500

0 0

20

40

60

80

100



Answer: Ponlinear !inverted E) relationship. The actual correlation coefficient is .%. c.

;our te$tboo& defines the correlation coefficient as follows: n

1

∑ !Yi − Y ) n −1



! X i − X ) 

i =1

r =

n

1

∑ !Yi − Y ) n −1

1

n

∑! Xi − X ) n −1



i =1



i =1

Another te$tboo& gives an alternative formula:

n r =

n

n

∑ Y X − !∑ Y )!∑ X ) i

i

i =1

n

n

∑

Yi  − !

i =1

n

i

i =1

n

∑

Yi ) n

i =1

n

∑ i =1

"rove that the two are the same. Answer:

(

i

i=1

X i − !

n

∑X ) 

i

i=1

n

1

∑ !Y − Y ) ! X − X ) 

n − 1 i =1

r =

i

!Y − Y ) n −1 ∑

!X − X ) n −1 ∑



i

∑

n

i =1

0

n

∑Y



i

− nY

∑X

i =1

n

 i

− nX n 

i =1

n

∑

Yi X i − !

i =1

n

0

n



n

n

n

− YYi + Y

∑Y

i

i

n



i =1

)

∑! X

 i

−  XX i + X  )

i =1

∑ Y X − nYnX

n



i

i

i

− nY



0

n

∑X

 i

−X 

i =1

n

i

i =1

n

n

 i

i

i =1

i

i

∑ ∑ X ) Yi )!



i

∑ !Y



i =1

∑ Y − ! ∑ Y ) n∑ X − ! ∑ X ) i =1

n

i

n

i =1

i =1



n −1

i =1

Yi X i − nYX

1



i

i =1

n

=

n

1

∑ !Y X − YX − XY + YX )

n − 1 i=1

i

n

1

n

1



i =1

.



i

i =1

')
n 0 (, find "r!Y < 1() . n 0 1, find "r!Y > +4) . n 0 1'', find "r!11 < Y < 1%) . Answer: a. .+'M b. .+4M c. .1.

()

Lonsider the following alternative estimator for the population mean: . 1 1 4 1 4 1 4 Y° = ! Y1 + Y + Y% + Y' + ... + Yn −1 + Yn ) ' ' ' ' ' n '

° is unbiased and consistent, but not efficient when compared to Y . "rove that Y 1 1 4 1 4 1 4 Answer: E !Y° ) = ! E !Y1 ) + E !Y ) + E!Y% ) + E!Y' ) + ... + E! Yn−1) + E! Y n )) ' ' ' ' ' n ' 1 1 4 n ° is unbiased. 0 µY ! +  + ... + + ) = µY = µ Y . Cence Y n n ' ' ° − µ )  0 E- 1 ! 1 Y1 + 4 Y + 1 Y% + 4 Y' + ... + 1 Yn −1 + 4 Y n ) − µ Y  var!Y° ) = E !Y Y ' ' ' ' ' n ' 1 1 4 1 4  0  E- !Y1 − µY ) + !Y − µY ) + ... + !Yn −1 − µY ) + !Y n − µY ) n ' ' ' '

(4

0

1

-

1

E !Y1 − µY )  +

n 1 

'+

1

E!Yn−1 − µY )  +

'+

E!Y n − µY )  

1 1  1 1  '+  1  '+  σ Y n 1 '+ σ Y  0  - σ Y + σ Y + ... + σ Y + σ Y  0  - ! + ) 0 1.(( . n 1 1 1 1 n  1 1 n ° ) →  as n → ∞, Y ° is consistent. Y ° has a larger variance than Y 2ince var!Y and is therefore not as efficient. )(

1

E !Y − µY )  + ... +

4)

et Y be a Dernoulli random variable with success probability "r!Y = 1) = p , and let Y1 ,..., Y n be i.i.d. draws from this distribution. et p> be the fraction of successes !1s) in this sample. will be appro$imately normal, i.e. p>

p !1 − p) ) . Pow let X be the n(mb*r o+ "(cc*""*" and n n the sample size. , the success proportion, or fraction is appro$imately distributed N ! p,

of successes. Pe$t, using your &nowledge of linear transformations, derive the distribution of X .

p!1 − p) ) , then, given that X is a n µ , X is distributed N !np, np !1 − p)) . linear transformation of p

Answer: X

= n × µp . Cence if pµ is distributed N ! p,

(8

8) 5hen you perform hypothesis tests, you are faced with four possible outcomes described in the accompanying table.

6ecision based on sample

$ruth -opulation8

#  is true

# 1 is true

Je@ect # 

I

Z

o not re@ect # 

Z

II

6Z7 indicates a correct decision, and < and << indicate that an error has been made.
Assume that under the null hypothesis, Y has an e$pected value of ( and a standard deviation of . Ender the alternative hypothesis, the e$pected value is ((. 2&etch the probability density function for the null and the alternative hypothesis in the same figure. "ic& a critical value such that the p3value is appro$imately (*. Har& the areas, which show the size and the power of the test. 5hat happens to the power of the test if the alternative hypothesis moves closer to the null hypothesis, i.e., µ Y 0 (', (%, (, etc.I

Answer: Oor a given size of the test, the power of the test is lower.

(+



/ro+*""or "car 4ord5

1' B Endergraduate Fconometrics 2pring % 1)

The net weight of a bag of flour is guaranteed to be ( pounds with a standard deviation of .( pounds. ;ou are concerned that the actual weight is less. To test for this, you sample ( bags. Larefully state the null and alternative hypothesis in this situation. etermine a critical value such that the size of the test does not e$ceed (*. Oinding the average weight of the ( bags to be '.4 pounds, can you re@ect the null hypothesisI 5hat is the power of the test hereI 5hy is it so lowI Answer: et Y be the net weight of the bag of flour. Then #  : E !Y ) = ( and #1 : E!Y ) < ( . Ender the null hypothesis, Y is distributed normally, with a mean of ( pounds and a standard deviation of .1 pounds. The critical value is appro$imately '.+8 pounds. 2ince '.4 pounds falls in the re@ection region, the null hypothesis is re@ected. The power of the test is low here, since there is no simple alternative.
11)

2ome policy advisors have argued that education should be subsidized in developing countries to reduce fertility rates. To investigate whether or not education and fertility are correlated, you collect data on population growth rates !Y ) and education ! X ) for 8 countries. Given the sums below, compute the sample correlation: n

n

n

n

n

∑ Y 0 1.(+'M ∑ X 0 ''+.M ∑ Y X 0 .'+4M ∑ Y 0 .%+8M ∑ X 0 

i

i

i =1

i =1

i

i =1

i

 i

i

i =1

i =1

%,.4 Answer: r 0 B.41. 1)

!Advanced) Enbiasedness and small variance are desirable properties of estimators. Cowever, you can imagine situations where a trade3off e$ists between the two: one estimator may be have a small bias but a much smaller variance than another, unbiased estimator. The concept of 6mean square error7 estimator > be an estimator of µ . Then the mean square combines the two concepts. et µ > ) 0 E ! µ> − µ )  . "rove that H2F! µ >)0 error !H2F) is defined as follows: H2F! µ > ). !Cint: subtract and add E ! µ > ) in E ! µ> − µ )  .) bias # var! µ Answer:

M!E ! µµ ) = E ! µ> − E ! µ> ) + E! µ> ) − µ)  = E-! µ> − E! µ> )) + ! E ! µ> ) − µ )  = E-! µ> − E ! µ> ))  + ! E ! µ> ) − µ)  + ! µ> − E! µ> ))! E ! µ> ) − µ ) Pe$t, moving through the e$pectation operator results in E- µ> − E ! µ> ) + E- E! µ> ) − µ) +  E-! µ> − E! µ> ))! E ! µ> ) − µ ) .

1


1' B Endergraduate Fconometrics 2pring %

The first term is the variance, and the second term is the squared bias,   > ) 0 bias # var! since E- E ! µ> ) − µ ) = - E ! µ> ) − µ ) . This proves H2F! µ > ) if the last term equals zero. Dut µ > ) E-! µ> − E ! µ> ))! E! µ> ) − µ) = E- E! µ> ) µ> − µµ> − ! E! µ> ))  + µE ! µ  > ) = . 0 E ! µ> ) E ! µ> ) − µ E ! µ> ) − ! E! µ> )) + µ E ! µ 1%)

;our te$tboo& states that when you test for differences in means and you assume that the two population variances are equal, then an estimator of the population variance is the following 6pooled7 estimator:

"

 pool*d

n n   Y Y Y Y ! ) ! ) = − + − ∑ ∑ i 2  nm + n2 −   i =1 i m i =1 

1

m

2

F$plain why this pooled estimator can be loo&ed at as the weighted average of the two variances. Answer: n n  1  ! nm − 1) "m + ! n2 − 1) "2 !Yi − Ym ) + ∑ !Yi − Y2 )  = " = ∑   nm + n2 −   i =1 i =1  nm + n2 −  !n − 1)  ! n2 − 1)  "m + " . = m nm + n2 −  nm + n2 −  2  pool*d

i.

1

m

2

;our te$tboo& suggests using the first observation from a sample of n as an estimator of the population mean.




° ) →  as n → ∞, does not hold, Y ° is not consistent. 2ince var!Y var!Y° ) < var!Y ) , and is therefore more efficient than the estimator, 1

which only uses the first observation. 1() et p be the success probability of a Dernoulli random variable Y , i'*'6 p = "r!Y = 1) .
p!1 − p ) µ , ) . Esing the estimator of the variance of p n

pµ !1 − µp) , construct a +(* confidence interval for p. 2how that the margin for n sampling error simplifies to 19 n if you used  instead of 1.+ assuming, conservatively, that the standard error is at its ma$imum. Lonstruct a table indicating the sample size needed to generate a margin of sampling error of 1*, *, (* and 1*. 5hat do you notice about the increase in sample size needed to halve the margin of errorI !The margin of sampling error is 1.+ ×!E ! µp) .)

µ ± 1.+ × Answer: The +(* confidence interval for p is p

µp !1 − µp)

n

.

µ !1 − µp) p is n

µ 0 .(, in which case the confidence interval reduces at a ma$imum for p µ ± 1.+ × .( to p

n

≈ µp ±

1

n

, and the margin of sampling error is

1

n

.

n

1

n .1 . .( .1

1, ,( ' 1

To halve the margin of error, the sample size has to increase fourfold. 1)

et Y be a Dernoulli random variable with success probability "r!Y = 1) = p , and let Y1 ,..., Y n be i.i.d. draws from this distribution. et p> be the fraction of successes !1s) in this sample. Given the following statement "r!−1.+ < - < 1.+) = .+( and assuming that p> is appro$imately distributed N ! p,

p!1 − p) ) , derive the +(* n

confidence interval for p by solving the above inequalities.

%



µ −p p "r!−1.+ < < 1.+) = .+( Answer: . Hultiplying through by the p!1 − p) n standard deviation results in "r!−1.+ ×

p !1 − p) µ p!1 − p) < p − p < 1.+× ) = .+( . 2ubtraction of n n

p> then yields, after multiplying both sides by !31), µ − 1.+ × p !1 − p) < p < µp + 1.+× p!1 − p) ) = .+( . The +(* "r! p n n µ ± 1.+ × p !1 − p) . confidence interval for p then is p n 14)

;our te$tboo& mentions that dividing the sample variance by n 7 1 instead of n is called a degrees of freedom correction. The meaning of the term stems from the fact that one degree of freedom is used up when the mean is estimated. Cence degrees of freedom can be viewed as the number of independent observations remaining after estimating the sample mean. Lonsider an e$ample where initially you have  independent observations on the height of students. After calculating the average height, your instructor claims that you can figure out the height of the th student if she provides you with the height of the other 1+ students and the sample mean. Cence you have lost one degree of freedom, or there are only 1+ independent bits of information. F$plain how you can find the height of the th student. 



∑

1+

∑

∑

Y i ,  × Y = Yi = Y + Yi . Cence &nowledge of the  i =1 i =1 i =1 sample mean and the height of the other 1+ students is sufficient for finding the height of the th student.

Answer: 2ince Y

=

1

18) The accompanying table lists the height !!&8##& ) in inches and weight !$*it ) in pounds of five college students. Lalculate the correlation coefficient.

STUDHGHT

WEIGHT

74 73 72 68 66

165 165 145 155 140

Answer: r 0 .4.

'



1+) !Jequires calculus.) The variance of the success probability p !a Dernoulli p!1 − p) random variable) is . Ese calculus to show that this variance is minimized

n

for p 0 .(.

Answer:

∂  

p!1 − p ) 1 n   = 1 − p − p = . Cence 1 −  p =  or p =  . n n ∂ p

) Lonsider two estimators: one which is biased and has a smaller variance, the other which is unbiased and has a larger variance. 2&etch the sampling distributions and the location of the population parameter for this situation. iscuss conditions under which you may prefer to use the first estimator over the second one. Answer: The bias indicates 6how far away,7 on average, the estimator is from the population value. Although this average is zero for an unbiased estimator, there may be quite some variation around the population mean.
(



Chapter 4

1) ;ou have obtained measurements of height in inches of + female and 81 male students !!t(d*nt) at your university. A regression of the height on a constant and a binary variable ! B*mm*), which ta&es a value of one for females and is zero otherwise, yields the following result:

· 0 41. 3 '.8'[ B*mm* , % 0 .', !E% 0 . !t(d*nt !.%) !.(4) !a)

!b)

Test the hypothesis that females, on average, are shorter than males, at the 1* level. Answer:

!c)

The t 3statistic for the difference in means is 38.'+. Oor a one3sided test, the critical value is B.%%. Cence the difference is statistically significant.



1' B Endergraduate Fconometrics 2pring % Answer:

)


;ou have obtained a sub3sample of 14'' individuals from the Lurrent "opulation 2urvey !L"2) and are interested in the relationship between wee&ly earnings and age. The regression, using heteros&edasticity3robust standard errors, yielded the following result:

· 0 %+.1 # (.[ A* , % 0 .(, !E% 0 84.1., Earn !.') !.(4) where Earn and A* are measured in dollars and years respectively. !a)

!b)

!c)

5hy should age matter in the determination of earningsI o the results suggest that there is a guarantee for earnings to rise for everyone as they become olderI o you thin& that the relationship between age and earnings is linearI Answer:
!d)

The variance of the error term and the variance of the dependent variable are related. Given the distribution of earnings, do you thin& it is plausible that the distribution of errors is normalI

4



Answer: 2ince the earnings distribution is highly s&ewed, it is not reasonable to assume that the error distribution is normal. !e)

!Jequires Appendi$ Haterial) The average age in this sample is %4.( years. 5hat is annual income in the sampleI

¶ Answer: 2ince β 

the estimates for the = Y − βµ1 X ⇒ Y = β¶  + βµ1 X . 2ubstituting

slope and the intercept then results in average wee&ly earnings of N'%'.1 or annual average earnings of N,(4.%.

%)

The baseball team nearest to your home town is, once again, not doing well. Given that your &nowledge of what it ta&es to win in baseball is vastly superior to that of management, you want to find out what it ta&es to win in Ha@or eague Daseball !HD). ;ou therefore collect the winning percentage of all % baseball teams in HD for 1+++ and regress the winning percentage on what you consider the primary determinant for wins, which is quality pitching !team earned run average). ;ou find the following information on team performance:

ummar/ of the 6istribution of inning -ercentage and $eam arned !un A"erage for #9. in 1::: A"erage

Team FJA 5inning "ercentage !a)

tandard de"iation

-ercentile 1';

2(;

4';

'.41

.(%

%.8'

'.%(

'.4

('; median8 '.48

.(

.8

.'

.'%

.'

.'8

%';

7(;

:';

'.+1

(.

(.(

.'+

.(+

.

5hat is your e$pected sign for the regression slopeI 5ill it ma&e sense to interpret the interceptI
!b)

The authors of your te$tboo& have informed you that unless you have more than 1 observations, it may not be plausible to assume that the distribution of your 2 estimators is normal. 5hat are the implications here for testing the significance of your theoryI 8



Answer: 2ince there are only % observations, the distribution of the t 3statistic is un&nown. ;ou should therefore not conduct statistical inference. !c)

2 estimation of the relationship between the winning percentage and the team FJA yields the following:

· 0 .+' B .1[t*am*ra , %0.'+, !E% 0 ., $inpct !.8) !.) where inpct is measured as wins divided by games played, so for e$ample a team that won half of its games would have $inpct 0 .(.

!e)

5hat would be the effect on the slope, the intercept, and the regression % if you measured $inpct in percentage points, i.e. as !5ins9Games)[1I Answer: Llearly the regression % will not be affected by a change in scale, since a descriptive measure of the quality of the regression would depend on whim otherwise. The slope of the regression will compensate in such a way that the interpretation of the result is unaffected, i'*' it will become 1 in the above e$ample. The intercept will also change to reflect the fact that if K were , then the dependent variable would now be measured in percentage, i'*', it will become +'. in the above e$ample.

!f)

Are you impressed with the size of the regression %I Given that there is (1* of une$plained variation in the winning percentage, what might some of these factors beI

+



Answer:
;ou have learned in one of your economics courses that one of the determinants of per capita income !the 65ealth of Pations7) is the population growth rate. Ourthermore you also found out that the "enn 5orld Tables contain income and population data for 1' countries of the world. To test this theory, you regress the G" per wor&er !relative to the Enited 2tates) in 1++ ! %*l/*r",nc) on the difference between the average population growth rate of that country !n) to the E.2. average population growth rate !n(" ) for the years 1+8 to 1++. This results in the following regression output:

· l/*r",nc 0 .(18 B 18.8%1[!n 7 n(") , %0.(, !E% 0 .1+4 Je !.() !%.144) !a)

!b)

5hat would happen to the slope, intercept, and regression % if you ran another regression where the above e$planatory variable was replaced by n only, i'*', the average population growth rate of the countryI !The population growth rate of the Enited 2tates from 1+8 to 1++ was .+.) 2hould this have any affect on the t 3 statistic of the slopeI Answer: The interpretation of the partial derivative is unaffected, in that the slope still indicates the effect of a one percentage point increase in the population growth rate. The regression % and t 3statistic will remain the same since only a constant was removed from the e$planatory variable. The intercept will change as a result of the change in X .

!c)

4



stages of economic and institutional development. ther factors vary as well.
%1 of the 1' countries have a dependent variable of less than .1. oes it therefore ma&e sense to interpret the interceptI Answer: To interpret the intercept, you must observe values of X close to zero, not Y .

()

The neoclassical growth model predicts that for identical savings rates and population growth rates, countries should converge to the per capita income level. This is referred to as the convergence hypothesis. ne way to test for the presence of convergence is to compare the growth rates over time to the initial starting level.

!a)

!b)

The results of the regression for 1' countries were as follows:

· + 0 .1+ B .[ %*l/rod :0 , % 0 .4, !E% 0 .1,  !.') !.4%) where :090 is the average annual growth rate of G" per wor&er for the 1+3 1++ sample period, and %*l/rod :0 is G" per wor&er relative to the Enited 2tates in 1+.
41



by  percent per year over the sample period. The regression %  indicates that the regression has virtually no e$planatory power. This is confirmed by the very low t 3statistic, indicating that the slope is not statistically significant. The result is not surprising given that there are not many theories that predict unconditional convergence between the countries of the world. !c)

Esing the 2 estimator with homos&edasticity3only standard errors, the results changed as follows:

· + 0 .1+ B .[ %*l/rod :0 , % 0 .4, !E% 0 .1  !.) !.8) 5hy didnRt the estimated coefficients changeI Given that the standard error of the slope is now smaller, can you re@ect the null hypothesis of no beta convergenceI Are the results in !c) more reliable than the results in !b)I F$plain. Answer: Esing homos&edasticity3only standard errors has no effect on the 2 estimator. The t 3 statistic remains small and is certainly below the critical value. The results are less reliable since there is no reason to believe that the error variance is homos&edastic. !d)

;ou decide to restrict yourself to the ' FL countries in the sample. This changes your regression output as follows:

· + 0 .'8 B .'' %*l/rod :0 , % 0 .8 , !E% 0 .'  !.') !.%) Cow does this result affect your conclusions from aboveI 5hen you test for convergence, should you worry about the relatively small sample sizeI Answer: Sudging by the size of the slope coefficient, there is strong evidence of unconditional convergence for the FL countries. The regression % is quite high, given that there is only a single e$planatory variable in the regression. Cowever, since we do not &now the sampling distribution of the estimator in this case, we cannot conduc t inference.

)
4



· 0 B.%'4 # .+ AZ!*a"a; , %0.11, !E% 0 .1'(, AZ$"a; !.') !.1()

· 0 .1%' # .1% NY!*a"a; , %0.1, !E% 0 .+, NY$"a; !.%41) !1.%'4) where $"a; and !*a"a; indicate the batting average during the 5orld 2eries and the regular season respectively. !a)

Oocusing on the coefficients first, what is your interpretationI Answer: The two regressions are quite different. Oor the iamondbac&s, players who had a 1 point higher batting average during the regular season had roughly a % point higher batting average during the 5orld 2eries. Cence top performers did relatively better. The opposite holds for the ;an&ees.

!b)

5hat can you say about the e$planatory power of your equationI 5hat do you conclude from thisI Answer: Doth regressions have little e$planatory power as seen from the regression %. Cence performance during the season is a poor forecast of 5orld 2eries performance.

!c)

Lalculate the t 3statistics for the various regression coefficients. Are any of these significant at the (* levelI 5hen using statistical inference in this case, should you be concerned about the number of observationsI Answer: The respective t 3statistics are B.(4(, 1., .%1, and .11. Pone of these are statistically significant at the (* level. Cowever, given that there are only 11 observations, you should not conduct inference, since the sampling distribution is un&nown.

Mat*matical <(*"tion" 1)

"rove that the regression % is identical to the square of the correlation coefficient between two variables Y and X . Jegression functions are written in a form that suggests causation running from X to Y . Given your proof, does a high regression % present supportive evidence of a causal relationshipI Lan you thin& of some regression e$amples where the direction of causality is not clearI
4%



E!! Answer: The regression % = , where F22 is given by &!! 

n

∑ !Yµ

i

− Y )  . Dut

i =1



µ X and Y = β µ + β µ X . Cence µ µ ! X − X )  , Yµ i = βµ  + β !Y i − Y )  = β 1 i  1 i 1

and therefore E!!

=

 n

µ β

1

∑

! Xi

− X ) . Esing small letters to indicate

i =1

deviations from mean, i'*', -i  n

∑ x ∑ y

µ β

%  =

= Zi − Z , we get that the regression

 i

1

i =1

n

. The square of the correlation coefficient is

 i

i =1

n

n

n

∑ ∑ ∑ r = = ∑ x ∑ y !∑ x ) ∑ y 

! yi xi ) 

i =1 n

i =1

n

 i

i =1

! yi xi ) 

i =1 n

n

 i

i =1

  i

i =1

xi

i=1

 i

n

∑x ∑y 

=

µ β 1

i =1

n

 i

. Cence the two are the

 i

i=1

same. Lorrelation does not imply causation.
Oor the following estimated slope coefficients and their standard errors, find the t 3 statistics for the null hypothesis # : β 1 0 .
> ) = .' β>1 = '., !E ! β 1

4'


!c)

> ) = .%4 β>1 = .(, !E ! β 1 > ) = . β> = .%, !E ! β

!d)

> ) = % β>1 = %, !E ! β 1

!b)

1


1

Answer: a) t 0 1.4(M re@ect null 1* level of two3sided test, and (* of one3sided test. b) t 0 1.%(M cannot re@ect null at 1* of two3sided test, re@ect null at 1* of one3sided test. c) t 0 1.(M cannot re@ect null at 1* of two3sided test, re@ect null at 1* of one3sided test. d) t 0 1.M cannot re@ect null at 1* of both two3sided and one3sided test. ')

F$plain carefully the relationship between a confidence interval, a one3sided hypothesis test, and a two3sided hypothesis test. 5hat is the unit of measurement of the t 3statisticI Answer:
()

;ou have analyzed the relationship between the weight and height of individuals. Although you are quite confident about the accuracy of your measurements, you feel that some of the observations are e$treme, say, two standard deviations above and below the mean. ;our therefore decide to disregard these individuals. 5hat consequence will this have on the standard deviation of the 2 estimator of the slopeI Answer: ther things being equal, the standard error of the slope coefficient will decrease the larger the variation in X . Cence you prefer more variation rather than less. This is easier to see in the case of homos&edasticity3

4(



only standard errors, but carries over to the heteros&edasticity3robust standard errors.

∑ !Yµ

Answer: The E!! is given by

i

i =1

µ X and − Y )  . Dut Yµ i = βµ  + β 1 i 

µ X . Cence µ µ ! X − X )  , and therefore Y = βµ  + β !Y i − Y )  = β 1 i 1  n

µ E!! = β 1

∑

! X i − X )  . The right3hand side contains the estimated

i =1

slope squared and the denominator of the slope, i'*', all values that have already been calculated. 4)

!Jequires Appendi$ Haterial) and β > . The sum of squared residuals with respect to the two parameters β  1 resulting two equations imply two restrictions that 2 places on the data, n

namely that

n

∑ (> 0  and ∑ (> X 0 . 2how that you get the same formula for i

i

i =1

i

i =1

the regression slope and the intercept if you impose these two conditions on the sample regression function. Answer: The sample regression function is Yi n

sides results in

∑Y =

n βµ

i

o

+

µ β

i =1

1

µ X +( $i . 2umming both = βµ o + β i 1

n

n

∑ X + ∑ (µ .
i

i =1

i =1

restriction, namely that the sum of the residuals is zero, dividing both µ gives the 2 formula sides of the equation by n, and solving for β o

for the intercept. Oor the second restriction, multiply both sides of the sample regression function by X i and then sum both sides to get n

∑Y X i

i

i =1

=

µ β

n

o

n

n

imposing the restriction ∑ X + ∑ X + ∑ (µ X . After i

i =1

µ β

 i

1

i =1

i

i

i =1

n

∑ (> X 0 and substituting the formula for the intercept, you get i

i

i =1

4


1' B Endergraduate Fconometrics 2pring % n

∑

µ Yi X i = !Y − βµ 1 X ) nX + β 1

i =1

n

∑

X i or

i =1

n

∑

Yi X i − nYX = βµ 1

i =1

n

∑X

 i

µ X − β , 1

i =1

µ and dividing by the variation in ,X results in which, after isolating β 1

the 2 estimator for the slope. 8)

!Jequires Appendi$ Haterial) 2how that the two alternative formulae for the slope given in your te$tboo& are identical. n

1

∑ 1 X n∑

n

n

X iYi − XY

i =1 n

∑ ! X i − X )!Yi − Y ) i =1

0  i

n

− X

∑ ! X i − X )



i =1



i =1

n

n

n∑ X i Y i − !∑ X i )! ∑ Y i ) i =1

i =1

i =1

n

n

i =1

i =1

n ∑ X i − !∑ X i )  "rove that this formula is also the same as those given above. Answer: etRs start with the first equality. The numerator of the right3hand side e$pression can be written as follows: n

n

n

n

n

∑ ! X − X )!Y − Y ) =∑ ! X Y − XY − YX + XY ) = ∑ X Y − X ∑ Y − Y ∑ X + nYX i

i

i =1

i i

i

i

i =1

i i

i =1

n

n

i =1

i =1

= ∑ Yi X i − nXY − nXY + nXY = ∑ Yi X i − nXY .

i

i

i =1

i =1

n

!Pote that

∑ X

i

= nX .)

i =1

Hultiplying out the terms in the denominator and moving the summation n

∑ X

 i

− nX 

sign into the e$pression in parentheses similarly yields i =1 . ividing both of these e$pressions by n then results in the left3hand side fraction.

44



n

n

n

n

n

n

∑ X Y − !∑ X )!∑ Y ) n∑ X Y − nXnY ∑ X Y − nXY = = n ∑ X − !∑ X ) n∑ X − ! nX ) ∑ X − nX i i

i

i =1

i

i =1

i i

i =1

n

n

 i

i i

i =1 n



i =1 n

 i

i



 i



i =1 i =1 i =1 i =1 Oinally, ividing both numerator and denominator by n then gives you the desired result.

+)

.

!Jequires Lalculus) Lonsider the following model: Y i = β  + ( i .

erive the 2 estimator for = . Answer: To derive the 2 estimator, minimize the sum of squared prediction n

mista&es

∑ !Y − b )





i

. Ta&ing the derivative with respect to b results in

i =1

n n ∂ n ∂   Y b Y b − = − = ! ) ! ) !Yi − b )!−1) ∑ ∑ i  i  b ∂b ∑ ∂ i =1 i =1 i =1  n

n

i =1

i =1

= !−)∑ !Yi − b ) = !−) ∑ Yi − nb . 2etting the derivative to zero then results in the 2 estimator: !−)

n

∑ Y − nβµ i



µ = Y =  ⇒ β . o

i =1

1)

!Jequires Lalculus) Lonsider the following model: Y i = β 1 X i + ( i .

erive the 2 estimator for = 1. Answer: To derive the 2 estimator, minimize the sum of squared prediction n

mista&es

∑ !Y − b X ) i

1

i



. Ta&ing the derivative with respect to b1 results

i =1

in n n ∂ n ∂   ! ) ! ) !Yi − b1 X i )!− X i ) Y b X Y b X − = − = ∑ ∑ i 1 i i 1 i b ∂b1 ∑ ∂ i =1 i =1 i =1 1 n

n

i =1

i =1

= !−)∑ !Yi − b1 X i )! X i ) = !−)! ∑ Yi X i − b1 X i ) . 2etting the derivative to zero then results in the 2 estimator:

48


1' B Endergraduate Fconometrics 2pring % n

Y X ∑ µ !−)! ∑ Y X − β . ∑ X =  ⇒ β µ = ∑ X n

i

n

i

i

 i

1

i =1

i

i =1 n

1

 i

i =1

i =1

11)

2how first that the regression % is the square of the sample correlation coefficient. Pe$t, show that the slope of a simple regression of Y on X is only identical to the inverse of the regression slope of X on Y if the regression % equals one.

E!! Answer: The regression % = , where F22 is given by &!! 

n

∑ !Yµ

i

− Y )  . Dut

i =1



µ X and Y = β µ + β µ X . Cence µ µ ! X − X )  , Yµ i = βµ  + β !Y i − Y )  = β 1 i  1 i 1  n

µ and therefore E!! = β 1

∑

! Xi

− X ) . Esing small letters to indicate

i =1

= Zi − Z , we get that the regression

deviations from mean, i'*', -i  n

∑ x ∑ y

µ β

 i

1

%  =

i =1

n

. The square of the correlation coefficient is

 i

i =1

n

n

n

∑ ! y x ) ∑ ! y x ) ∑ x r = = ∑ x ∑ y !∑ x ) ∑ y 



i i



i =1 n

i

i =1

n

 i

i =1

i

 i

i =1 n

n

 i

  i

i =1

i =1

n

∑x ∑y 

µ β

1

=

i =1

n

 i

 i

. Cence the two are the

 i

i=1

i=1

same.  n

1 = r 

=

∑ ∑ y

µ β 1

xi

i =1

n

 i

n



µ ⇒ β 1

∑ = ∑x

yi

i =1 n

 i

n



µ β 1

∑ µ = β ∑ x 1

y ∑ µ = β ∑ x y  i

i =1 n

i =1 i =1 i =1 Pow . Dut which is the inverse of the regression slope of K on ;.

1)

n

xi y i

1

 i

i =1 n

i

and therefore

i

i =1

,

Lonsider the sample regression function

Yi = β> + β >1 X i + (>i . Oirst, ta&e averages on both sides of the equation. 2econd, subtract the resulting equation from the above equation to write the sample regression function in deviations from means. !Oor simplicity, you may want to use small letters to indicate deviations from the mean, i.e., -i = Z i − Z .) Oinally, illustrate in a two3 4+



dimensional diagram with !!% on the vertical a$is and the regression slope on the horizontal a$is how you could find the least squares estimator for the slope by varying its values through trial and error. > X , and subtracting this equation Answer: Ta&ing averages results in Y = β> + β 1 > x + (> . from the above one, we get y = β i

!!% =

n

∑ i =1



($i =

∑ ! y − β µ x ) i

1 i



1 i

i

is a quadratic which ta&es on different

µ !the y and x are given in this case, values for different choices of β 1

i'*', different from the usual calculus problems, they cannot vary here). ;ou could choose a starting value of the slope and calculate !!%. Pe$t you could choose a different value for the slope and calculate the new !!%. There are two choices for the new slope value for you to ma&e: first, in which direction you want to move, and second, how large a distance you want to choose the new slope value from the old one. !
Larefully discuss the advantages of using heteros&edasticity3robust standard errors over standard errors calculated under the assumption of homos&edasticity. Give at least five e$amples where it is very plausible to assume that the errors display heteros&edasticity. Answer:

There are virtually no e$amples where economic theory suggests that the errors are homos&edastic. Cence the maintained hypothesis should be that they are heteros&edastic. Esing homos&edasticity3only standard errors when in truth heteros&edasticity3robust standard errors should 8



be used, results in false inference. 5hat ma&es this worse is that homos&edasticity3only standard errors are typically smaller than heteros&edasticity3robust standard errors, resulting in t 3statistics that are too large, and hence re@ection of the null hypothesis too often. There is an alternative G2 estimator, weighted least squares, which is DEF, but requires &nowledge of how the error variance depends on X , e.g. X or K. Answers will vary by student regarding the e$amples, but earnings functions, cross country beta3convergence regressions, consumption functions, sports regressions involving teams from mar&ets with varying population size, weight3height relationships for children, etc., are all good candidates. 1')

The effect of decreasing the student3teacher ratio by one is estimated to result in an improvement of the districtwide score by .8 with a standard error of .(. Lonstruct a +* and ++* confidence interval for the size of the slope coefficient and the corresponding predicted effect of changing the student3teacher ratio by one. 5hat is the intuition on why the ++* confidence interval is wider than the +* confidence intervalI Answer: The +* confidence interval for the slope is calculated as follows: !.8 B 1.'( × .(, .8 # 1.'( × .() 0 !1.', %.1'). The corresponding predicted effect of a unit change in the student3 teacher ratio is the same, since the change in X is 1. The ++* confidence interval for the slope coefficient and the unit change in the student3teacher ratio is: !.8 B .(8 × .(, .8 # .(8× .() 0 !.+', %.). The ++* confidence interval corresponds to a smaller size of the test. This means that you want to be 6more certain7 that the population parameter is contained in the interval, and that requires a larger interval.

1()

Given the amount of money and effort that you have spent on your education, you wonder if it was !is) all worth it. ;ou therefore collect data from the Lurrent "opulation 2urvey !L"2) and estimate a linear relationship between earnings and the years of education of individuals. 5hat would be the effect on your regression slope and intercept if you measured earnings in thousands of dollars rather than in dollarsI 5ould the regression % be affectedI 2hould statistical inference be dependent on the scale of variablesI iscuss. Answer:
81



is zero. The change in the units of measurement have no effect on this, µ . The slope since the change in X is cancelled by the change in β 1

coefficient will change to compensate for the change in the units of measurement of K.
!Jequires Appendi$ Haterial) Lonsider the sample regression function >1 X iX + (>i , Yi X = γ> + γ where 6X7 indicates that the variable has been standardized. 5hat are the units of measurement for the dependent and e$planatory variableI 5hy would you want to transform both variables in this wayI 2how that the 2 estimator for the intercept equals zero. Pe$t prove that the 2 estimator for the slope in this case is identical to the formula for the least squares estimator where the variables hav e not been standardized, times the ratio of the sample standard deviation of X and Y , i.e., γ>1

" X . "Y

= β >1 X

Answer: The units of measurement are in standard deviations. 2tandardizing the variables allows conversion into common units and allows comparison of the size of coefficients. The mean of standardized variables is zero, and hence the 2 intercept must also be zero. The n

x y ∑ $ = slope coefficient is given by the formula γ ∑ x X i

1

X i

i =1 n

, where small

X i

i =1

letters indicate deviations from mean, i'*', - = Z − Z . n

∑ X i Y i X

$ = Pote that means of standardized variables are zero, and hence we get γ 1

i =1 n

X

.

∑ X

X i

i =1

5riting this e$pression in terms of originally observed variables results in 1 1 n

$ = γ 1

∑ x y 1 x ! ∑

! X ! Y

i

i =1 n

 X i =1

14)

i

, which is the same as the sought after e$pression after simplification.

 i

The 2 slope estimator is not defined if there is no variation in the data for the e$planatory variable. ;ou are interested in estimating a regression relating earnings to years of schooling.
8



for different individuals, but that all these individuals had completed a college education !1 years of education). 2&etch what the data would loo& li&e and e$plain intuitively why the 2 coefficient does not e$ist in this situation. Answer: There is no variation in X in this case, and it is therefore unreasonable to as& by how much Y would change if X changed by one unit. Jegression analysis cannot figure out the answer to this question, because a change in X never happens in the sample.

Farnings

K K K K K K K K

1

;ears of Fducation

8%

Practice Problems for Midterm 1

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