Download Econ 299 Chapter 05

Chapter 5 – Statistical Review
• Chapter 5 is a brief review of statistical
concepts. It is NOT a replacement for a
statistics course.
• By the end of the chapter, you will be able to:
1) Identify population and sample data and
perform population and sample statistical
calculations.
2) Define, interpret and evaluate statistics.
3) Demonstrate the use of statistical tables.
4) Construct confidence intervals and hypothesis
tests from sample data.
1
5) Begin to calculate OLS estimations
5.1 Simple Economic Models
and Random Components
• Consider the linear economic model:
Yi = β1 + β2Xi + єi
• The variable Y is related to another variable X
– Utility is related to hours of TV watched
• Єi (or epsilon) represents error; everything
included in Y that is not explained by X
– Ie: Quality of TV show, Quality of Popcorn,
Other Facts of Life
2
5.1 Observed or Random Components
• Єi (or epsilon) is the RANDOM ERROR TERM; it
takes on values according to chance
• Since Yi depends on Єi, it is also random
• β1 + β2Xi is assumed to be fixed in most simple
models (which simplifies everything)
– Referred to as the deterministic part of the
model
– X, β1 and β2 are Non-Random
• β1 and β2 are unknown, and must be estimated
3
5.1 Example
• Consider the function:
Utilityi = β1 + β2Sistersi + єi
• Happiness depends on the number of sisters
• єi captures: number of brothers, income, and
other factors (ie: bad data collection and
shocks)
• Utility and Sisters are Observable
• Utility and єi are random
• β1 and β2 must be estimated (< or > 0?) 4
5.2 Random Variables and Probabilities
• Random Variable
–A variable whose value is determined by
the outcome of a chance experiment
• Ie: Sum of a dice roll, card taken out of a deck,
performance of a stock, oil discovered in a
province, gender of a new baby, etc.
• Some outcomes can be more likely than others
(ie: greater chance to discover oil in Alberta,
more likely to roll an 8 than a 5)
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5.2 Random Variables
• Discrete Variable
–Can take on a finite # of values
–Ie: Dice roll, card picked
• Continuous Variable
–Can take on any value within a range
–Ie: Height, weight, time
• Variables are often assumed discrete to aid in
calculations and economic assumptions (ie:
Money in increments of 1 cent)
6
5.2 Probability Terminology
Experiment –Any action that creates at least one
possible outcome
-ie: Taking tests, speeding, flipping a coin
Trial – one occurrence of an experiment
-ie: Taking the Econ 299 midterm, speeding
tomorrow morning, flipping a coin once
Outcome – Any result of an experiment
-ie: Passing the test, getting a speeding ticket,
getting a “tails”
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5.2 Probability Terminology
• Probabilities are assigned to the various
outcomes
Sample Space – set of all possible outcomes from
a random experiment
-ie S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
-ie E = {Pass exam, Fail exam, Fail horribly}
Event – a subset of the sample space
-ie B = {3, 6, 9, 12} ε S
-ie F = {Fail exam, Fail horribly} ε E
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5.2 Quiz Example
Students do a 4-question quiz with each question
worth 2 marks (no part marks). Handing in the
quiz is worth 2 marks, and there is a 1 mark
bonus question.
Events:
-getting a zero (not handing in the quiz)
-getting at least 40% (at least 1 right)
-getting 100% or more (all right or all right plus
the bonus question)
-getting 110% (all right plus the bonus question)
*Events contain one or more
possible outcomes
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5.2 Probability
Probability = the likelihood of an event
occurring (between 0 and 1)
P(a) = Prob(a) = probability that event a
will occur
P(Y=y) = probability that the random
variable Y will take on value y
P(ylow < Y < yhigh) = probability that the
random variable Y takes on any value
between ylow and yhigh
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5.2 Probability Examples
P(true love) = probability that you will find
true love
P(Sleep=8 hours) = probability that the
random variable Sleep will take on the
value 8 hours
P($80 < Wedding Gift < $140) = probability
that the random variable Wedding Gift
takes on any value between $80 and $140
11
5.2 Probability Calculations
IF probability can be calculated, two
methods are: P(a)  # of outcomes including a
# of total outcomes
# of outcomes including a
P(a) 
# of trials
-Method 1 is used for full information (die
roll)
-Method 2 is used for a test (asking names
at a party)
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5.2 Probability of Small World’s Die
In the board game Small
World, the 6-sided die
has a blank (0) on 3
sides, and a 1, 2, and 3
on each of the remaining
sides
3
P ( 0) 
6
1
P (1) 
6
1
P ( 2) 
6
1
P (3) 
6
13
5.2 Probability Extremes
If Prob(a) = 0, the event will never occur
ie: Canada moves to Europe
ie: the price of cars drops below zero
ie: your instructor turns into a giant llama
If Prob(b) = 1, the event will always occur
ie: you will get a mark on your final exam
ie: you will either marry your true love or not
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ie: the sun will rise tomorrow
5.2 Probability Terminology
Mutually Exclusive Events – cannot occur at the
same time
-rolling both a 3 and an 11; being both dead
and alive;
Exhaustive Events – cover all possible outcomes
-a dice roll must lie within S ε [2,12]
-a person is either married or not married
Complement– everything other than the event
-being dead is the complement to being alive
-rolling an 11 or 12 is a complement to rolling a
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2, 3, 4, 5, 6, 7, 8, 9, or 10
5.2 Probability Rules
1) Probability Range
P(a) must be greater than or equal to 0 and less
than or equal to 1 :
0≤ P(a) ≤1
2) Something Must Happen
∑ P(a) =1
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5.2 Probability Rules
3) Exhaustive Rule
If any set of events (ie: {A,B,C}) are exhaustive,
then P(A or B or C) = 1
ex) Prob(winning, losing or tying)=1
4a) Addition Rule (Mutually Exclusive)
If any set of events (ie: {A,B,C}) are mutually
exclusive, then
P(A or B or C)=P(A)+P(B)+P(C)
ex) Prob. of marrying the person to the right or left 17
5.2 Probability Rules
4b) Addition Rule (General)
P(A or B)=P(A)+P(B)-P(A+B)
Example:
P(Marrying a fellow student OR someone named Joe)
= P(Marry a fellow student)
+P(Marry Joe)
-P(Marry a fellow student named Joe)
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5.2 Probability Rules
5) Complement Rule
P(not A)=1-P(A)
Example:
P(Not eating ice cream)=1-P(Eating Ice Cream)
5) Multiplication Rule (A and B are independent/unrelated)
P(A and B)=P(A)P(B)
Example:
P(Owning a cat AND having black hair)
=P(Owning a Cat)P(Having black hair)
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5.2 Probability of Small World
I have the last action in
Smallworld. Colin has 81 points,
and I have 72, plus 8 guaranteed
points this turn. (If our points
are tied, I win.)
I have 3 final attacks with the die.
I lose each attack on a roll of
zero, otherwise I win. Each
attack I win gets me one point.
What is the probability that I will
win and Colin will lose?
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5.2 Probability of Small World
3
6
1
P (1) 
6
P ( 0) 
1
6
1
P (3) 
6
P ( 2) 
P(Win )  1  P( Lose)
P(Win )  1  {P( Lose1st ) P( Lose 2nd ) PLose3rd )}
P(Win )  1  {( 0.5)0.5(0.5)}
P(Win )  1  0.125
P(Win )  0.875
There is an 87.5% chance that I will win.
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5.2 Probability Examples
1)
2)
3)
4)
5)
6)
7)
8)
P(coin flip=heads) = ½
P(2 coin flips=2 heads) = ¼
Probability of tossing 6 heads in a row = 1/64
Probability of rolling less than 4 with 1 sixsided die = 3/6
Probability of throwing a 13 with 2 dice= 0
Probability of winning rock, paper, scissors =
1/3 (or 3/9)
Probability of being in love or not in love=1
Probability of passing the course = ?
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5.2.1 Probability Density Functions
•
The probability density function (pdf)
summarizes probabilities associated with
possible outcomes
Discrete Random Variables – pdf
f(y) = Prob (Y=y)
Σf(y) = 1
-(the sum of the probabilities of all possible
outcomes is one)
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5.2.1 Dice Example
•
The probabilities of
rolling a number with
the sum of two sixsided die
• Each number has
different die
combinations:
7={1+6, 2+5, 3+4, 4+3,
5+2, 6+1}
y
f(y)
2
1/36 8
5/36
3
2/36 9
4/36
4
3/36 10
3/36
5
4/36 11
2/36
6
5/36 12
1/36
Exercise: Construct a
table with one 4-sided
and one 8-sided die
7
6/36
y
f(y)
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5.2.1 Probability Density Functions
Continuous Random Variables – pdf
f(y) = pdf for continuous random variable Y
∫f(y)dy = 1 (sum/integral of all probabilities of
all possibilities is one)
-probabilities are measured as areas under the
pdf, which must be non-negative
-technically, the probability of any ONE event is
zero
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5.2.1 Continuous Headache
Continuous Random Variables – pdf
f(y) = 0.2 for 2<y<7
= 0 for y <2 or y >7
7
f(y)
P(3  Y  7) 
7
0
.
2
dy

3 [0.2 y ]  0.2(7)  0.2(3)  0.8

y 3
Continuous
probabilities
are the area
under the
pdf curve.
0.2
Y
0
3
7
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5.3 Expected Values
Expected Value – measure of central tendency;
center of the distribution; population mean
-If the variable is collected an infinite number of
times, what average/mean would we expect?
Discrete Variable:
μY=E(Y) = Σyf(y)
Continuous Variable:
μ(Y)=E(Y) = ∫yf(y)dy
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5.3 Expected Example
What is the expected value from a dice roll?
E(W) = Σyf(y)
=2(1/36)+3(2/36)+…+11(2/36)+12(1/36)
=7
Exercise: What is the expected value of rolling a
4-sided and an 8-sided die? A 6-sided and a 10sided die?
28
5.3 Expected Application – Pascal’s
Wager
Pascal’s Wager, from Philosopher, Mathematician,
and Physicist Blaise Pascal (1623-62) argued that
belief in God could be justified through expected
value:
-If you live as if God exists, you get huge
rewards if you are right, and wasted some time
and effort if you’re wrong
-If you live as if God does not exist, you save
some time and effort if you’re right, and suffer
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huge penalties if you’re wrong
5.3 Expected Application – Pascal’s
Wager
Mathematically:
E(belief) = Σutility * f(utility)
=(Utility if God exists)*p(God exists)+
+(Utility if no God)*p(no God)
=1,000,000(0.01)+(50)(0.99)
=10,000+49.52
=10,049.52
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5.3 Expected Application – Pascal’s
Wager
Mathematically:
E(no belief)= Σutility * f(utility)
=(Utility if God exists)*p(God exists)+
+(Utility if no God)*p(no God)
=-1,000,000(0.01)+(150)(0.99)
=-10,000+148.5
=-9,851.5
Since -9,851.5 is less than 10,049.52, Pascal
argued that belief in God is rational.
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5.3.1 Properties of Expected Values
a) Constant Property
E(a) = a if a is a constant or non-random variable
Ie: E(14)=14
Ie: E(β1+ β2Xi) = β1+ β2Xi
b) Constants and random variables
E(a+bW) = a+bE(W)
If a and b are non-random and W is random
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5.3.1 Properties of Expected Values
Applications:
If E(єi) =0, then
E(Yi) = E(β1 + β2Xi + єi)
= β1 + β2Xi + E(єi)
= β1 + β2Xi
E(6sided+10sided)
=E (6-sided) + E (10 sided)
= 3.5 + 5.5
=9
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5.3.1 Properties of Expected Values
c) “Not so Fast” Property
E(WV) ≠ E(W)E(V)
E(W/V) ≠ E(W)/E(V)
d) Non-Linear Functions
E(Wk) = Σwkf(w)
E(2 six-sided die2) =22(1/36)+32(2/36)+
…+112(2/36)+122(1/36)
=54.83
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5.4 Variance
Consider the following 3 midterm distributions:
1) Average = 70%; everyone in the class
received 70%
2) Average = 70%; half the class received
50% and half received 90%
3) Average = 70%; most of the class was in
the 70’s, with a few 100’s and a few 40’s
who got a Bachelor in Pottery
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5.4 Variance
Although these midterm results share the
same average, their distributions differ
greatly.
While the first results are clustered together,
the other two results are quite dispersed
Variance – a measure of dispersion (how far a
distribution is spread out) for a random
variable
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5.4 Variance Formula
σY2= Var(Y) = E(Y-E(Y))2
= E(Y2) – [E(Y)]2
Discrete Random Variable:
σY2= Var(Y)= Σ(y-E(Y))2f(y)
Continuous Random Variable:
σY2= Var(Y)= ∫(y-E(Y))2f(y)dy
37
5.4 Variances
Example 1:
E(Y)=70
Yi =70 for all i
Var(Y)
= Σ(y-E(Y))2f(y)
= (70-70)2 (1)
= (0)(1)
=0
If all outcomes are the same, there is no
variance.
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5.4 Variances
Example 2:
E(Y)=70
f(50)=0.5, f(90)=0.5
Var(Y)
= Σ(y-E(Y))2f(y)
= (50-70)2(0.5)+ (90-70)2(0.5)
=200+200
=400
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5.4 Variances
Example 3:
E(Y)=70
f(40)=1/5, f(70)=3/5, f(100)=1/5
Var(Y)
= Σ(y-E(Y))2f(y)
= (40-70)2(1/5)+ (70-70)2(3/5)+
(100-70)2(1/5)
=900/5+0+900/5
=1800/5
=360
40
5.4 Standard Deviation
While Variance is a good tool for measuring
dispersion, it is difficult to represent
graphically (ie: Bell Curve)
Standard Deviation is more useful for a visual
view of dispersion
Standard Deviation = Variance1/2
sd(W)=[var(W)]1/2
σ= (σ2)1/2
41
5.4 SD Examples
In our first example, σ =01/2=0
No dispersion exists
In our second example, σ =4001/2≈20
In our third example, σ =3601/2=19.0
Results where most dispersed in the second
42
example.
5.4.1 Properties of Variance
a) Constant Property
Var(a) = 0 if a is a constant or non-random
variable
Ie: Var(14)=0
Ie: Var(β1+ β2Xi) = 0
b) Constants and random variables
Var(a+bW) = b2 Var(W)
If a and b are non-random and W is random 43
5.4.1 Properties of Variance
Applications:
If Var(єi) =k, then
Var(Yi)
= Var(β1 + β2Xi + єi)
= 0 + Var(єi)
=k
Exercise: Calculate the variance from:
a) A coin flip
b) A 4-sided die roll
c) Both a and b, where the coin flip represents 0 or 1.
44
5.4.1 Properties of Variance
c) Covariance Property
If W and V are random variables, and a, b, and c
are non-random, then
Var(a+bW+cV)= Var(bW+cV)
= b2 Var(W) + c2 Var (V)
+2bcCov(W,V)
Where Covariance will be examined in 5.6
45
5.4.1 Properties of Variance
Application:
If Var(Cost of Gas)=10 cents
And Var(Cost of a Slurpee)=5 cents
And Cov(Cost of Gas, Cost of Slurpee)=-1 cent
Var(Cost of Gas+Cost of 2 Slurpees)
= b2 Var(G) + c2 Var (Sl)+2bcCov(G,Sl)
=12(10)+22(5)+2(2)(-1)
=10+20-4
=26 cents
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5.5 Joint Probability Density Functions
Sometimes we are interested in the isolated
occurrence or effects of one variable. In this
case, a simple pdf is appropriate.
Often we are interested in more than one
variable or effect. In this case it is useful to use:
 Joint Probability Density Functions
 Conditional Probability Density Functions
47
5.5 Joint Probability Density Functions
Joint Probability Density Function-summarizes the probabilities associated
with the outcomes of pairs of random variables
f(w,z) = Prob(W=w and Z=z)
∑ f(w,z) = 1
Similar statements are valid for continuous
random variables.
48
5.5 Joint PDF and You
Love and Econ Example:
On Valentine’s Day, Jonny both wrote an Econ
299 midterm and sent a dozen roses to his love
interest.
He can either pass or fail the midterm, and his
beloved can either embrace or spurn him.
E = {Pass, Fail}; L = {Embrace, Spurn}
49
5.5 Joint PDF and You
Love and Econ Example:
Joint pdf’s are expressed as follows:
P(pass and embrace) = 0.32
P(pass and spurn) = 0.08
P(fail and embrace) = 0.48
P(fail and spurn) = 0.12
(Notice that:
∑f(E,L) = 0.32+0.08+0.48+0.12 = 1)
50
5.5 Joint and Marginal Pdf’s
Marginal (individual) pdf’s can be determined
from joint pdf’s. Simply add all of the joint
probabilities containing the desired outcome of
one of the variables.
Ie: f(Y=7)=∑f(Y=7,Z=zi)
Probability that Y=7 = sum of ALL joint
probabilities where Y=7
51
5.5 Love and Economics
f(pass) = f(pass and embrace)+f(pass and spurn)
= 0.32 + 0.08
= 0.40
f(fail) = f(fail and embrace)+f(fail and spurn)
= 0.48 + 0.12
= 0.60
Exercise: Find f(embrace) and f(spurn)
52
5.5 Love and Economics
Notice:
Since passing or failing are exhaustive outcomes,
Prob (pass or fail) = 1
Also, since they are mutually exclusive,
Prob (pass or fail)
= Prob (pass) + Prob (fail)
53
= 0.4 + 0.6
=1
5.5 Conditional Probability
Density Functions
Conditional Probability Density Function-summarizes the probabilities associated
with the possible outcomes of one random
variable conditional on the occurrence of a
specific value of another random variable
Conditional pdf = joint pdf/marginal pdf
Or
Prob(a|b) = Prob(a&b) / Prob(b)
(Probability of “a” GIVEN “b”) 54
5.5 Conditional Love and Economics
From our previous example:
Prob(pass|embrace) = Prob(pass and embrace)/
Prob (embrace)
= 0.32/0.80
= 0.4
Prob(fail|embrace) = Prob(fail and embrace)/
Prob (embrace)
= 0.48/0.80
= 0.6
55
5.5 Conditional Love and Economics
From our previous example:
Prob(spurn|pass) = Prob(pass and spurn)/
Prob (pass)
= 0.08/0.40
= 0.2
Prob(spurn|fail)
= Prob(fail and spurn)/
Prob (fail)
= 0.12/0.60
= 0.2
56
5.5 Conditional Love and Economics
Exercise: Calculate the other conditional pdf’s:
Prob(pass|spurn)
Prob(fail|spurn)
Prob(embrace|pass)
Prob(embrace|fail)
57
5.5 Statistical Independence
If two random variables (W and V) are
statistically independent (one’s outcome
doesn’t affect the other at all), then
f(w,v)=f(w)f(v)
And:
1) f(w)=f(w|any v)
2) f(v)=f(v|any w)
As seen in the Love and Economics example.58
5.5 Statistically Dependent Example
Bob can either watch Game of Thrones or
Yodeling with the Stars: W={T, Y}. He can
either be happy or sad V={H,S}. Joint pdf’s are
as follows:
Prob(Thrones and Happy) = 0.7
Prob(Thrones and Sad)=0.05
Prob(Yodeling and Happy)=0.10
Prob(Yodeling and Sad)=0.15
59
5.5 Statistically Dependent Example
Calculate Marginal pdf’s:
Prob(H)=Prob(T and H) + Prob(Y and H)
=0.7+0.10
=0.8
Prob(S)=Prob(T and S) + Prob(Y and S)
=0.05+0.15
=0.2
Prob(H)+Prob(S)=1
60
5.5 Statistically Dependent Example
Calculate Marginal pdf’s:
Prob(T)=Prob(T and H) + Prob(T and S)
=0.7+0.05
=0.75
Prob(Y)
=Prob(Y and H) + Prob(Y and S)
=0.10+0.15
=0.25
Prob(T)+Prob(Y)=1
61
5.5 Statistically Dependent Example
Calculate Conditional pdf’s:
Prob(H|T)=Prob(T and H)/Prob(T)
=0.7/0.75
=0.93
Prob(H|Y)=Prob(Y and H)/Prob(Y)
=0.10/0.25
=0.4
Exercise: Calculate the other conditional pdf’s62
5.5 Statistically Depressant Example
Notice that since these two variables are NOT
statistically independent – Game of Thrones is
utility enhancing – our above property does
not hold.
P(Happy) ≠ P(Happy given Thrones)
0.8 ≠ 0.93
P(Sad) ≠ P(Sad given Yodeling)
0.2 ≠ 0.6 (1-0.4)
63
5.5 Conditional Expectations and Variance
Assuming that our variables take numerical
values (or can be interpreted numerically),
conditional expectations and variances can be
taken:
E(P|Q=500)=Σpf(p|Q=500)
Var(P|Q=500)=Σ[p-E(P|Q=500)]2f(p|Q=500)
Ie) money spent on a car and resulting utility
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(both random variables expressed numerically).
5.5 Conditional Expectations and Variance
Example: A consumer can spend $5000 or
$10,000 on a car, yielding utility of 10 or 20. The
conditional probabilities are :
f(10|$5,000)=0.7
f(20|$5,000)=0.3
E(U|P=$5000) =ΣUf(U|P=$5000)
=10(0.7) +20(0.3) =13
Var(U|P=$5K) =Σ[U-E(U|P=$5K)]2f(U|P=$5K)
=(10-13)2(0.7)+(20-13)2(0.3)
65
= 21
5.6 Covariance and Correlation
If two random variables are NOT statistically
independent, it is important to measure the
amount of their interconnectedness.
Covariance and Correlation are useful for this.
Covariance and Correlation are also useful in
model testing, as you will learn in Econ 399.
66
5.6 Covariance
Covariance – a measure of the degree of linear
dependence between two random variables. A
positive covariance indicates some degree of
positive linear association between the two
variables (the opposite likewise applies)
Cov(V,W)=E{[W-E(W)][V-E(V)]}
67
5.6 Discrete and Continuous
Covariance
Discrete Random Variable:
Cov(V ,W )   (v  E (v))( w  E ( w)) f (v, w)
v
w
Continuous Random Variable:
Cov(V ,W )    (v  E (v))( w  E ( w)) f (v, w)vw
v w
68
5.6 Covariance Example
Joe can buy either a burger ($2) or ice cream
($1) and experience utility of 1 or zero.
C={$1, $2}, U={0,1}
Prob($1
Prob($1
Prob($2
Prob($2
and
and
and
and
0)=0.2
1)=0.6
0)=0.1
1)=0.1
69
5.6 Covariance Example
Prob($1
Prob($1
Prob($2
Prob($2
and
and
and
and
0)=0.2
1)=0.6
0)=0.1
1)=0.1
Prob($1)=0.2+0.6=0.8
Prob($2)=0.1+0.1=0.2
Prob(0)=0.2+0.1=0.3
Prob(1)=0.6+0.1=0.7
70
5.6 Covariance Example
E(C) =∑cf(c)
=$1(0.8)+$2(0.2)
=$1.20
E(U) = ∑uf(u)
= 0(0.3)+1(0.7)
=0.7
71
5.6 Covariance Example
E(C) =$1.20
E(U) =0.7
Cov(C,U)=∑∑(c-E(C))(u-E(U))f(c,u)
=(1-1.20)(0-0.7)(0.2)
+(1-1.20)(1-0.7)(0.6)+(2-1.20)(0-0.7)(0.1)
+(2-1.20)(1-0.7))0.1)
=(-0.2)(-0.7)(0.2)+(-0.2)(0.3)(0.6)
+(0.8)(-0.7)(0.1)+(0.8)(0.3)(0.1)
=0.028-0.036-0.056+0.032
=-0.032 (Negative Relationship)
72
5.6 Correlation
Covariance is an unbounded measure of
interdependence between two variables.
Often, it is useful to obtain a BOUNDED measure
of interdependence between two variables, as
this opens the door for comparison.
Correlation is such a bounded variable, as it lies
between -1 and 1.
73
5.6 Correlation
Correlation Formulas:
 W V  Corr (W ,V ) 
Corr (V ,W ) 
Cov(V , W )
WV
 (v  E (v))( w  E (w)) f (v, w)
v
w
Var (v) Var ( w)
74
5.6 Correlation Example
From the Data above:
Var(C)
=∑ (c-E(C)2f(v)
=(1-1.20)2(0.8)+(2-1.20)2(0.2)
=0.032 + 0.128
=0.16
Var(W) =∑ (u-E(U)2f(w)
=(0-0.7)2(0.3)+(1-0.7)2(0.7)
=0.147 + 0.063
=0.21
75
5.6 Correlation Example
From the Data above:
Corr(C,U) =Cov(C,U)/[sd(C)sd(U)]
=-0.032 / [0.16(0.21)]1/2
=-0.175 (significant, but not too strong)
Still represents a negative relationship.
76
5.6 Graphical Correlation
If Correlation = 1, observations of the two variables lie
upon an upward sloping line
If Correlation = -1, observations of the two variables lie
on a downward sloping line
If Correlation is between 0 and 1, observations of the
two variables will be scattered along an upward sloping
line.
If Correlation is between 0 and -1, observations of the
two variables will be scattered along a downward
77
sloping line.
5.6 Correlation, Covariance and
Independence
Covariance, correlation and independence have
the following relationship:
If two random variables are independent, their
covariance (correlation) is zero.
INDEPENDENCE => ZERO COVARIANCE
If two variables have zero covariance
(correlation), they may or may not be
independent.
ZERO COVARIANCE ≠> INDEPENDENCE
78
5.6 Correlation, Covariance and
Independence
INDEPENDENCE => ZERO COVARIANCE
ZERO COVARIANCE ≠> INDEPENDENCE
From these relationships, we know that
Non-zero Covariance => Dependence
but
Dependence ≠> Non-zero Covariance
79
5.7 POPULATION VS. SAMPLE DATA
Population Data – Full information on the ENTIRE
population.
-Includes population probability (pdf)
-Uses the previous formulas
-ex) data on an ENTIRE class
Sample Data
– Partial information from a RANDOM
SAMPLE (smaller selection) of the population
-Individual data points (no pdf)
-Uses the following formulas
-ex) Study of 2,000 random students
80
5.7 Estimators
Population Expected Value:
μ = E(Y) = Σ y f(y)
Sample Mean:
Y

Y
i
N
__
Note: From this point on, Y may be expressed as
Ybar (or any other variable - ie:Xbar). For example,
via email no equation editor is available, so answers
81
may be in this format.
5.7 Sample Mean Example
Tom and Rodney both go to a 4-day gaming
convention. How much they spend (S)each day is
listed below:
Day
1
2
3
4
Tom
110
85
90
135
Rodney
200
20
10
220
S
T
R
S


S 
T
i
N
R
S i
N
110  85  90  135

 105
4
200  20  10  220

 105
4
82
5.7 Estimators
Population Variance:
σY2 = Var(Y) = Σ [y-E(y)]2 f(y)
Sample Variance:
S y2 
2
(
Y

Y
)
 i
N 1
83
5.7 Sample Variance Example
Day
1
2
3
4
Tom
110
85
90
135
Rodney
200
20
10
220
S y2 
2
(
Y

Y
)
 i
N 1
Tom : S S2 
T
T 2
(
S

S
)
i

N 1
2
2
2
2
(
110

105
)

(
85

105
)

(
90

105
)

(
135

105
)
2
SS 
4 1
25  400  225  900
2
SS 
 517
3
84
5.7 Sample Variance Example
Day
1
2
3
4
Tom
110
85
90
135
Rodney
200
20
10
220
S y2 
2
(
Y

Y
)
 i
N 1
Rodney : S S2 
R
R 2
(
S

S
)
i

N 1
2
2
2
2
(
200

105
)

(
20

105
)

(
10

105
)

(
220

105
)
2
SS 
4 1
9025  7225  9025  13225
2
SS 
 12,833
3
85
5.7 Sample Variance Example
Day
1
2
3
4
Tom
110
85
90
135
Rodney
200
20
10
220
T
S  105
R
S  105
S y2 
2
(
Y

Y
)
 i
N 1
Tom : S S2  517
Rodney : S S2  12,833
Even though Tom and Rodney spent on average the
same average amount every day ($105), Rodney’s
spending was MUCH more spread out (changed more
86
from day to day), as seen by the variance.
5.7 Estimators
Population Standard Deviation:
σY = (σ2)1/2
Sample Standard Deviation:
Sy = (Sy2)1/2
87
5.7 Sample Standard Deviation Example
Tom : S S  517
2
S S  S S  517  22.7
2
Rodney : S S  12,833
2
S S  S S  12,833  113
2
88
5.7 Estimators
Population Covariance:
Cov(V,W)=∑∑(v-E(v))(w-E(w))f(v,w)
Sample Covariance:
(V  V )(W  W )

Cov(V ,W ) 
i
i
N 1
89
5.7 Covariance Example
Day
1
2
3
4
Tom
110
85
90
135
Rodney
200
20
10
220
T
Cov( S , S
R
 (S
)
T
i
 S T )( S R i  S R )
N 1
(110  105)( 200  105)  (85  105)( 20  105)  (90  105)(10  105)  (135  105)( 220  105)
Cov( S T , S R ) 
4 1
475  1700  1425  3450
Cov( S T , S R ) 
 2350
3
There is a POSITIVE covariance between Tom
and Rodney’s spending. They seem to go up and
90
down at the same time.
5.7 Estimators
Population Correlation:
σvw = corr(V,W)= Cov(V,W)/ σv σw
Sample Correlation:
rvw = corr(V,W)= Cov(V,W)/ Sv Sw
91
5.7 Correlation Example
Day
1
2
3
4
Tom
110
85
90
135
Rodney
200
20
10
220
T
R
Cov
(
S
,
S
)
T
R
Cor ( S , S ) 
ST S R
2350
T
R
Cor ( S , S ) 
 0.916
(22.7)(113)
There is a VERY STRONG positive correlation
between Tom and Rodney’s spending.
92
5.7 Estimators
Population Regression Function:
Yi = β1 + β2Xi + єi
Estimated Regression Function:
ˆ
ˆ
ˆ
Yi  1  2 X i
93
5.7 Estimators
OLS Estimation:
ˆ
( X  X )(Y  Y )


(X  X )
i
2
i
2
i
Cov( X , Y )
ˆ
2 
S X2
ˆ  Y  ˆ X
1
2
^
Note: B2 may be expressed as b2hat
94
5.7 OLS Estimation
Tom
Rodney
ˆ
ˆ
ˆ
St  1  1St
T
R
Cov( S , S ) 2350
ˆ
2 

 20.8
2
SR
113
ˆ 1 S T  ˆ2 S R  105  20.8(105)  2079
Tom
Rodney
ˆ
St  2079  20.8St
Note: Although mathematically there is a relationship
between Tom and Rodney’s spending, in reality there is
no connection – both are influenced by a 3rd variable –
the fact that most people spend the most on the first
95
and last day of the convention.
5.7 Estimators Example 2
Given the data set:
Price
4
3
3
6
Quantity
10
15
20
15
Find sample means, variance, covariance, correlation,
and ols estimation
96
5.7 Estimators Example 2
Price
4
3
3
6
Quantity
10
15
20
15
Sample Means:
Pbar = (4+3+3+6)/4 = 4
Qbar = (10+15+20+15)/4 = 15
97
5.7 Estimators Example 2
Price
4
3
3
6
Pbar = 4
Quantity
10
15
20
15
Qbar=15
Sample Variance:
Sp2
Sq2
= [(4-4)2+(3-4)2+(3-4)2+(6-4)2]/(N-1)
=(0+1+1+4)/3
=2
= [(10-15)2+(15-15)2+(20-15)2+(15-15)2]/(N-1)
=(25+0+25+0)/3
=50/3
98
5.7 Estimators Example 2
Price
4
3
3
6
Pbar = 4
Quantity
10
15
20
15
Qbar=15
Sample Covariance:
Cov(p,q)= [(4-4)(10-15)+(3-4)(15-15)
+(3-4)(20-15)+(6-4)(15-15)]/(N-1)
=[ 0 + 0 -5 +0] /3
= -5/3
99
5.7 Estimators Example 2
Price
4
3
3
6
Pbar = 4
Quantity
10
15
20
15
Qbar=15
Sample Correlation
Corr(p,q) =
=
=
=
Cov(p,q)/SpSq
5/3 / [2(50/3)]1/2
-5/3 / (10/31/2)
-0.2886
100
5.7 Estimators Example 2
Price
4
3
3
6
Pbar = 4
Quantity
10
15
20
15
Qbar=15
Ols Estimation
B2hat = ∑(Xi-Xbar)(Yi-Ybar)
---------------------∑(Xi-Xbar)2
= [(4-4)(10-15)+(3-4)(15-15)+(3-4)(20-15)+(6-4)(15-15)
-----------------------------------------------------------------------------
(4-4)2+(3-4)2+(3-4)2+(6-4)2
=-5/6
101
5.7 Estimators Example 2
Price
4
Quantity 10
3
3
6
Pbar = 4
15
20
15
Qbar=15
Ols Estimation
B1hat = Ybar – (B2hat)(Xbar)
= 15- (-5/6)4
 110 5
= 90/6 + 20/6
Yi 
 Xi
= 110/6
6 6
110 5
ˆ
Qi 
 Pi
6 6
102
5.7.1 Estimators as random variables
Each of these estimators will give us a result based upon
the data available.
Therefore, two different data sets can yield two
different point estimates.
Therefore the value of the point estimate can be seen as
being the result of a chance experiment – obtaining a
data set.
Therefore each point estimate is a random variable,
with a probability distribution that can be analyzed using
the expectation and variance operator.
103
5.7.1 Estimators Distribution
Since the same mean is a variable, we can easily apply
expectation and summation rules to find the expected
value of the sample mean:
Y

Y
i
N
  Yi  1
  E  Yi 
E Y  E 
 N
N


1
1
E Y   E (Yi )   Y
N
N
1
E Y  NY  Y
N



104
5.7.1 Estimators Distribution
If we make the simplifying assumption that there is no
covariance between data points (ie: one person’s
consumption is unaffected by the next person’s
consumption), we can easily calculate variance for the
sample mean:
2
  Yi   1 
    Var  Yi 
Var Y  Var 
 N
N




2

2
1
Var Y   
N
1
Var (Yi )   N 
2
1

 
2
Var Y    N Y  Y
N
N
2
2

 Y
105
5.7.1 Estimators Distribution
Although we can’t observe the population variance of
Ybar, we can calculate its sample variance, therefore,

Var Y 

2
Y
N

2
Y
S
SampleVar Y 
N
106
5.8 Common Economic Distributions
In order to test assumptions and models,
economists need be familiar with the
following distributions:
 Normal
 t
 Chi-square
 F
For full examples and explanations of these
tables, please refer to a statistics text.
107
5.8 Normal Distribution
The Normal (Z) Distribution produces a
symmetric bell-shaped curve with a mean of
zero and a standard deviation of one.
 The probability that z>0 is always 0.5
 The probability that z<0 is always 0.5
 Z-tables generally (but not always) measure area
from the centre
 Probabilities decrease as you move from the center
108
5.8 Normal Example
Weekly weight gain can be argued to have a
normal distribution:
 On average, no weight is gained or lost
 A few pounds may be gained or lost
 It is very unlikely to lose or gain many pounds
Find Prob(Gain between 0 and 1 pound)
Prob(0<z<1) = 0.3413
= 34.13%
109
5.8 Normal Example
Find Prob(Lose more than 2 pounds)
Prob(z<-2) = 0.5 - 0.4772
= 0.0228 (2.28%)
Find Prob (Do not gain more than 2 pounds)
Prob(z<2) = 0.5+0.4772
= 0.9772(97.72%)
110
5.8 Converting to a normal distribution
Z distributions assume that the mean is zero and
the standard deviation is one.
If this is not the case, the distribution needs to
be converted to a normal distribution using
the following formula:
Z
x  x
x
111
5.8 Assignment Example
The average for the Fall 2005 Assignment #2
was 82%. Standard deviation was aprox. 6.
What is the probability of a random student
getting above 90%?
Prob(Y>90)
= Prob[{(Y-82)/6}>{(90-82)/6}]
= Prob(Z>1.33)
= 0.5 - Prob (0<Z<1.33)
= 0.5 - 0.4082
=9.18%
112
5.8 Assignment Example
What is the probability of getting a mark in the
80’s?
Prob(79<Y<90)
= Prob[{(79-82)/6}<{(Y-82)/6}<{(90-82)/6}]
= Prob(-0.5<Z<1.33)
= Prob(0<Z<0.5) + Prob(0<Z<1.33)
=0.1915 + 0.4082
=0.5997
=59.97%
113
5.8 Assignment Example
Find the mark (Y*) wherein there is a 15%
probability that Y<Y* (Bottom 15% of the class)
(Since 0.15<50, Z*<0)
Prob(Z<Z*)
= 0.5-Prob(0<Z<-Z*)
0.15
= 0.5-Prob(0<Z<-Z*)
Prob (0<Z<-Z*)=0.35
From tables, -Z*= 1.04
Therefore
Z* = -1.04
114
5.8 Example Continued
We know that
Z = (x-μ)/σ
So
X= μ+z(σ)
X= 82+(-1.04)6
X= 82-6.24
X= 75.76
There is a 15% chance that a student scored less
than 75.76%
115
5.8 Other Distributions
All other distributions depend on DEGREES OF
FREEDOM
Degrees of Freedom are generally dependant on
two things:
 Sample size (as sample rise rises, so does
degrees of freedom)
 Complication of test (more complicated
statistical tests reduce degrees of freedom)
 Simple conclusions are easier to make
than complicated ones
116
5.8 t-distribution
t-distributions can involve 1-tail or 2-tail tests
Interpolation is often needed within the table
Example 1:
Find the critical t-values (t*) that cuts of 1% of
both tails with 27df
(Note: 1% off both tails = 0.5% off each tail)
For p=0.495, df 27 gives t*=2.77, -2.77
117
5.8 t-distribution
Example 2:
Find the critical t-value (t*) that cuts of 1% of
the right tail with 35df
For 1T=0.01, df 30 gives t*=2.46
df 40 gives t*=2.42
Since 35 is halfway between 30 and 40, a good
approximation of df 35 would be:
t*=(2.46+2.42)/2 = 2.44
118
5.8 t-distribution
Typically, the following variable (similar to the
normal Z variable seen earlier) will have a tdistribution: (we will see examples later)
Estimator  E ( Estimator)
t
Sample sd ( Estimator)
119
5.8 chi-square distribution
Chi-square distributions are 1-tail tests
Interpolation is often needed within the table
Example:
Find the critical chi-squared value that cuts off
5% of the right tail with 2df
For Right Tail = 0.05, df=2
Critical Chi-Squared Value = 5.99
120
5.8 F-distribution
F-distributions are 1-tail tests
Interpolation is often needed within the table
Example:
Find the critical F value (F*) that cuts of 1% of
the right tail with 3df in the numerator and
80df in the denominator
For Right Tail = 0.01, df1=3, df2=80,
df2=60 gives F*=4.13 df2=120 gives F*=3.95
121
5.8 Interpolation
df2=60 gives F*=4.13 df2=120 gives F*=3.95
Since 80 is 1/3rd of the way between 60 and 120:
60
80
100
120
Our F-value should be 1/3 of the way between
4.13 and 3.95:
4.13
?
3.95
Approximization:
F*=4.13-(4.13-3.95)/3=4.07
122
5.8 Distribution Usage
Different testing of models will use different
tables, as we will see later in the course.
In general:
1) Normal tables do distribution estimations
2) t-tables do simple tests
3) F-tables do simultaneous tests –Prob(a & b)
4) Chi-squared tables do complicated tests
devised by mathematicians smarter than you
or I (they invented them, we use them) 123
5.9 Confidence Intervals
Thus far, all our estimates have been POINT estimates;
a single number emerges as our estimate for an
unknown parameter.
Ie)
X  3.74
Even if we have good data and have an estimator with
a small variance, the chances that our estimate will
equal our actual value are very low.
Ie) If a coin is expected to turn heads half the time.
The chance that it actually does that in an
124
experiment is very low
5.9.1 Constructing Confidence Intervals
Confidence intervals or interval estimators
acknowledge underlying uncertainties and are
an alternative to point estimators
Confidence intervals propose a range of values
in which the true parameter could lie, given a
range of probability.
Confidence intervals can be constructed since our
125
point estimates are RANDOM VARIABLES.
5.9.1 Degrees of Freedom
When given actual population data, we converted
into a z-score:
Z = (x-μ)/σ
With random samples, we convert into a t-score:
t = (x – E(x)) / sample sd(x)
with n-1 degrees of freedom
This is proven by various complicated central
limit theorems
126
5.9.1 CI’s and Alpha
Probabilities of confidence intervals are denoted
by α (alpha).
Given α, we construct a 100(1- α)% confidence
interval. If α=5%, we construct a 95%
confidence interval.
P(Lower limit<true parameter<Upper limit)=1- α
127
5.9.1 Formula
Given a repeated sample, we want to construct
confidence intervals for the mean such that:
P{t*  ( X   X ) / s X  t*}  1  
(1-α)%
-t*
t*
t
Where t has n-1 degrees of freedom, and ±t*
128
cuts α/2 off both tails.
5.9.1 Formula
Rearranging we get:
P{ X  t * s X   X  X  t * s X }  1  
(1-α)%
X  t * sX
X  t * sX
μX
129
5.9.1 Formula
Our final formula becomes:
CI  X  X  t * s X
Or in general:
CItruevalue  estimate  t * sestimate
Which gives us an upper and lower bound for
our CI.
130
5.9.1 Example
Flipping a coin has given us 25 heads with a value of
1, and 15 tails with a value of zero. Find the
95% CI if n=40.
We therefore have:
25(1)  15(0) 25
C

 0.625
40
40
131
5.9.1 IMPORTANT - Estimated Standard
Deviation of a Sample Mean
We have already seen that sample standard deviation
is found through the formula:
SY 
 (Y  Y )
2
i
N 1
Standard deviation of a sample mean is found
through:
sY  sY / N
132
5.9.1 Example
SC 
 (C  C )
2
i
N 1
25(1  0.625)  15(0  0.625)
SC 
40  1
2
2
9.375
SC 
 0.49
39
sC  sC / N  0.49 / 40  0.077
133
5.9.1 Example
A 95% CI has 2.5% off each tail. If n=40,
t* = 2.02
CI C  C  t * sC
CI C  0.625  2.02(0.077)
CI C  [0.47,0.78]
134
5.9.1 Interpretation:
In this example, we have a confidence interval of
[0.47, 0.78].
In other words, in repeated samples, 95% of
these intervals will include the probability of
getting a “heads” when flipping a coin.
135
5.9.1 Confidence Requirements
In order to construct a confidence interval, one
needs:
a) A point estimate of the parameter
b) Estimated standard deviation of the
parameter
c) A critical value from a probability distribution
(or α and the sample size, n)
136
5.10 Hypothesis Testing
After a model has been derived, it is often useful to test
various hypotheses:
 Are a pair of dice weighted towards another number
(say 11)?
 Does a player get blackjack more often than he
should?
 Will raising tuition increase graduation rates?
 Will soaring gas costs decrease car sales?
 Will the recession affect Xbox sales?
 Does fancy wrapping increase the appeal of
Christmas presents?
 Does communication between rivals affect price?
137
5.10 Hypothesis Testing
Question: Is our data CONSISTENT with a
particular parameter having a specific value?
Although we may observe an outcome (ie: a
Blackjack player has 150% of his starting
chips) (assume the average outcome should
be 80%),
We need to test if this outcome is:
1) Consistent with typical chance or
2) Inconsistent – perhaps showing cheating
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5.10 Hypothesis Testing
Testing Consistency of a Hypothesized Parameter:
1) Form a null and an alternate hypothesis.
H0 = null hypothesis = variable is equal to a
number
Ha = alternate hypothesis = variable is not equal
to a number
EX)
H0: Outcome=0.8
Ha: Outcome≠0.8
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5.10 Hypothesis Testing
Testing Consistency of a Hypothesized Parameter:
2) Collect appropriate sample data
3) Select an acceptable probability (α) of rejecting
a null hypothesis when it is true
-Type one error
-Lower α, more unlikely to find a sample that
rejects the null hypothesis
- α is often 10%, 5%, or 1%
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5.10 Hypothesis Testing
Testing Consistency of a Hypothesized Parameter:
4) Construct an appropriate test statistic
-ensure the test statistic can be calculated from
the sample data
-ensure its distribution is appropriate to that being
tested (ie: t-statistic for test for mean)
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5.10 Hypothesis Testing
Testing Consistency of a Hypothesized Parameter:
5) Establish (do not) reject regions
-Construct bell curve
-Tails are Reject H0 regions
-Centre is Do not Reject H0 regions
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5.10 Hypothesis Testing
Testing Consistency of a Hypothesized Parameter:
6) Compare the test statistic to the critical statistic
-If the test statistic lies in the tails, reject
-If the test statistic doesn’t lie in the tails, do not
reject
-Never Accept
7) Interpret Results
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5.10 Hypothetical Example
Johnny is a poker player who has an average of 8 times
out of ten (from 120 games and the standard deviation
is 0.5). Test the hypothesis that Johnny never wins.
H0: W=0
Ha: W ≠0
2) We have estimated W=8. The standard
deviation was 0.5
3) We let α=1%; we want a strong result.
4) t= (estimate-hypothesis)/sd = (8-0)/0.5=16
5) t* for n-1=119, α=1%: t*=2.62
6) t*<t; Reject H0
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1)
5.10 Hypothetical Example
7) Allowing for a 1% chance of a Type 1 error, we
reject the null hypothesis that Johnny never
wins at Poker.
According to our data, it is consistent that
Johnny sometimes wins at Poker.
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5.10 Hypothesis Example 2
Tom spends the following on 4 days of gaming
conventions: $110, $85, $95, $130. Test the hypothesis
that Tom spends an average of $100 a day (α=5%)
H0: μS =0
Ha: μS ≠0
2) From before:
T
S 
1)
2
SS
 (S

T
T
S
 i
N
i
110  85  90  135

 105
4
 S T )2
N 1
(110  105) 2  (85  105) 2  (90  105) 2  (135  105) 2
2
SS 
4 1
25  400  225  900
2
SS 
 517
3
sS
517
sd S  sS 

 11.4
N
4
146
5.10 Hypothesis Example 2
3) We let α=5%
4) t= (estimate-hypothesis)/sd
S   S 105  100
t

 0.439
sd S
11.4
5)
t *df 3, 0.05  3.18
6) -t*<t<t*; Do Not reject H0.
At a 5% level of significance, according to our
data, Tom could spend an average of $100 a day.
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