1 chapter 5 – statistical review chapter 5 is a brief review of statistical concepts. it is not a...
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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.5) Begin to calculate OLS estimations
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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
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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
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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?)
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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)
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5.2 Probability Terminology
• Probabilities are assigned to the various outcomes of random variables
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 Probability Terminology
Mutually Exclusive Events – cannot occur at the same time-rolling both a 3 and an 11; being both dead and alive; having both a son and a daughter (and only one child)
Exhaustive Events – cover all possible outcomes-a dice roll must lie within S ε [2,12]-a person is either married or not married
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5.2 Quiz ExampleStudents 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 ProbabilityProbability = 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
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5.2 Probability Extremes
If Prob(a) = 0, the event will never occur ie: Canada moves to Europeie: the price of cars drops below zeroie: 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 examie: you will either marry your true love or
notie: the sun will rise tomorrow
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5.2 Probability Rules
1) P(a) must be greater than or equal to 0 and less than or equal to 1 : 0≤ P(a) ≤1
2) If any set of events (ie: {A,B,C}) are exhaustive, then
P(A or B or C) = 1ex) Prob. of winning, losing or tying
3) 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
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5.2 Probability Examples
1) P(coin flip=heads) = ½ 2) P(2 coin flips=2 heads) = ¼ 3) Probability of tossing 6 heads in a row =
1/644) Probability of rolling less than 4 with 1
six-sided die = 3/6 5) Probability of throwing a 13 with 2 dice=
06) Probability of winning rock, paper,
scissors = 1/3 (or 3/9)7) Probability of being in love or not in
love=18) 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 six-sided die
• Each number has different die combinations:
7={1+6, 2+5, 3+4, 4+3, 5+2, 6+1}
Exercise: Construct a table with one 4-sided and one 8-sided die
y f(y) 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
7 6/36
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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 HeadacheContinuous Random Variables – pdf
f(y) = 0.2 for 2<y<7= 0 for y <2 or y >7
730
0.2
f(y)
Continuous probabilities are the area under the pdf curve.
Y
7
3
73 8.0)3(2.0)7(2.0]2.0[2.0)73(
y
ydyYP
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5.3 Expected ValuesExpected 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 10-sided die?
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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 huge penalties if you’re wrong
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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)=14Ie: 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, thenE(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(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:σY
2= Var(Y)= Σ(y-E(Y))2f(y)
Continuous Random Variable:σY
2= Var(Y)= ∫(y-E(Y))2f(y)dy
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5.4 VariancesExample 1:E(Y)=70Yi =70 for all i
Var(Y) = Σ(y-E(Y))2f(y)= Σ(70-70)2 (1)= Σ(0)(1)=0
If all outcomes are the same, there is no variance.
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5.4 Variances
Example 2:E(Y)=70f(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 VariancesExample 3:E(Y)=70f(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
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5.4 Standard DeviationWhile 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
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5.4 SD Examples
In our first example, σ =01/2=0No 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 example.
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5.4.1 Properties of Variance
a) Constant PropertyVar(a) = 0 if a is a constant or non-random
variableIe: Var(14)=0Ie: Var(β1+ β2Xi) = 0
b) Constants and random variablesVar(a+bW) = b2 Var(W)If a and b are non-random and W is random
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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 flipb) A 4-sided die rollc) Both a and b, where the coin flip represents 0
or 1.
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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
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5.4.1 Properties of VarianceApplication:
If Var(Cost of Gas)=10 centsAnd Var(Cost of a Slurpee)=5 centsAnd 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
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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.
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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}
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5.5 Joint PDF and You
Love and Econ Example:
Joint pdf’s are expressed as follows:P(pass and embrace) = 0.32P(pass and spurn) = 0.08P(fail and embrace) = 0.48P(fail and spurn) = 0.12
(Notice that:∑f(E,L) = 0.32+0.08+0.48+0.12 = 1)
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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
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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)
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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)
= 0.4 + 0.6 = 1
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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 pdfOr
Prob(a|b) = Prob(a&b) / Prob(b)(Probability of “a” GIVEN “b”)
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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
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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
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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)
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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.
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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) = 0.7Prob(Thrones and Sad)=0.05Prob(Yodeling and Happy)=0.10Prob(Yodeling and Sad)=0.15
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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
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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
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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’s
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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 ≠ 0.4 (1-0.6)
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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 (both random variables expressed numerically).
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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)=0.7f(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)= 21
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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.
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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)]}
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5.6 Discrete and Continuous Covariance
Discrete Random Variable:
Continuous Random Variable:
v w
wvfwEwvEvWVCov ),())())(((),(
v w
wvwvfwEwvEvWVCov ),())())(((),(
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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 and 0)=0.2Prob($1 and 1)=0.6Prob($2 and 0)=0.1Prob($2 and 1)=0.1
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5.6 Covariance Example
Prob($1 and 0)=0.2Prob($1 and 1)=0.6Prob($2 and 0)=0.1Prob($2 and 1)=0.1
Prob($1)=0.2+0.6=0.8Prob($2)=0.1+0.1=0.2Prob(0)=0.2+0.1=0.3Prob(1)=0.6+0.1=0.7
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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
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5.6 Covariance ExampleE(C) =$1.20E(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)
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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.
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5.6 Correlation
Correlation Formulas:
VWWV
WVCovVWCorr
),(
),(
)()(
),())())(((),(
wVarvVar
wvfwEwvEvWVCorr v w
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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
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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
Still represents a negative relationship.
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5.6 Graphical CorrelationIf 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 sloping line.
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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
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5.6 Correlation, Covariance and Independence
INDEPENDENCE => ZERO COVARIANCE
ZERO COVARIANCE ≠> INDEPENDENCE
From these relationships, we know that
Non-zero Covariance => Dependencebut
Dependence ≠> Non-zero Covariance
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5.7 POPULATION VS. SAMPLE DATAPopulation 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
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5.7 EstimatorsPopulation Expected Value:
μ = E(Y) = Σ y f(y)
Sample Mean:
__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 may be in this format.
N
YY i
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5.7 Estimators
Population Variance:
σY2 = Var(Y) = Σ [y-E(y)]2 f(y)
Sample Variance:
1
)( 22
N
YYS iy
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5.7 Estimators
Population Standard Deviation:
σY = (σ2)1/2
Sample Standard Deviation:
Sy = (Sy2)1/2
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5.7 Estimators
Population Covariance:
Cov(V,W)=∑∑(v-E(v))(w-E(w))f(v,w)
Sample Covariance:
1
))((),(
N
WWVVWVCov ii
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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
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5.7 Estimators
Population Regression Function:
Yi = β1 + β2Xi + єiEstimated Regression Function:
ii XY 21
ˆˆˆ
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5.7 Estimators
OLS Estimation:
B2hat = ∑(Xi-Xbar)(Yi-Ybar)
---------------------- ∑(Xi-Xbar)2
B1hat = Ybar – B2hatXbar ^
Note: B2 may be expressed as b2hat
XY
S
YXCov
XX
YYXX
X
i
ii
21
22
22
ˆˆ
),(ˆ
)(
))((ˆ
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5.7 Estimators Example
Given the data set:
Find sample means, variance, covariance, correlation, and ols estimation
Price 4 3 3 6
Quantity 10 15 20 15
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5.7 Estimators Example
Sample Means:
Pbar = (4+3+3+6)/4 = 4
Qbar = (10+15+20+15)/4 = 15
Price 4 3 3 6
Quantity 10 15 20 15
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5.7 Estimators Example
Sample Variance:
Sp2 = [(4-4)2+(3-4)2+(3-4)2+(6-4)2]/(N-1)
=(0+1+1+4)/3=2
Sq2 =
[(10-15)2+(15-15)2+(20-15)2+(15-15)2]/(N-1)=(25+0+25+0)/3=50/3
Price 4 3 3 6 Pbar = 4
Quantity 10 15 20 15 Qbar=15
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5.7 Estimators Example
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
Price 4 3 3 6 Pbar = 4
Quantity 10 15 20 15 Qbar=15
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5.7 Estimators Example
Sample Correlation
Corr(p,q)= Cov(p,q)/SpSq
= 5/3 / [2(50/3)]1/2
= -5/3 / (10/31/2)= -0.2886
Price 4 3 3 6 Pbar = 4
Quantity 10 15 20 15 Qbar=15
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5.7 Estimators Example
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
Price 4 3 3 6 Pbar = 4
Quantity 10 15 20 15 Qbar=15
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5.7 Estimators Example
Ols Estimation
B1hat = Ybar – (B2hat)(Xbar)= 15- (-5/6)4= 90/6 + 20/6= 110/6
Yhat = 110/6 –(5/6)XQhat = 110/6 –(5/6)P
Price 4 3 3 6 Pbar = 4
Quantity
10 15 20 15 Qbar=15
ii
ii
PQ
XY
6
5
6
110ˆ
6
5
6
110
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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.
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5.7.1 Estimators DistributionSince the same mean is a variable, we can easily apply expectation and summation rules to find the expected value of the sample mean:
YY
Yi
ii
i
NN
YE
NYE
NYE
YENN
YEYE
N
YY
1
1)(
1
1
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5.7.1 Estimators DistributionIf 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:
N
NN
YVar
NYVar
NYVar
YVarNN
YVarYVar
YY
Yi
ii
22
2
222
2
1
1)(
1
1
90
5.7.1 Estimators DistributionAlthough we can’t observe the population variance of Ybar, we can calculate its sample variance, therefore,
N
SYSampleVar
NYVar
Y
Y
2
2
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5.8 Common Economic Distributions
In order to test assumptions and models, economists need be familiar with the following distributions:
Normal t Chi-square FFor full examples and explanations of these
tables, please refer to a statistics text.
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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
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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%
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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%)
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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:
x
xxZ
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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%
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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%
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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 *= 1.04Therefore Z* = -1.04
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5.8 Example Continued
We know that
Z = (x-μ)/σSoX= μ+z(σ)X= 82+(-1.04)6X= 82-6.24X= 75.76There is a 15% chance that a student
scored less than 75.76%
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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
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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
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5.8 t-distribution
Example 2: Find the critical t-value (t*) that cuts of 1%
of the right tail with 35dfFor 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
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5.8 t-distribution
Typically, the following variable (similar to the normal Z variable seen earlier) will have a t-distribution: (we will see examples later)
)(
)(
EstimatorsdSample
EstimatorEEstimatort
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5.8 chi-square distribution
Chi-square distributions are 1-tail testsInterpolation is often needed within the
tableExample: Find the critical chi-squared value that cuts
off 5% of the right tail with 2dfFor Right Tail = 0.05, df=2Critical Chi-Squared Value = 5.99
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5.8 F-distribution
F-distributions are 1-tail testsInterpolation is often needed within the
tableExample: 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
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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 120Our F-value should be 1/3 of the way
between 4.13 and 3.95:4.13 ? 3.95Approximization:
F*=4.13-(4.13-3.95)/3=4.07
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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
estimations2) t-tables do simple tests3) 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)
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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)
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 experiment is very low
74.3X
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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 point estimates are RANDOM VARIABLES.
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5.9.1 Degrees of FreedomWhen 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
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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
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5.9.1 FormulaGiven a repeated sample, we want to construct confidence intervals for the mean such that: 1*}/)(*{ tsXtP
XX
-t* t*
(1-α)%
Where t has n-1 degrees of freedom, and ±t* cuts α/2 off both tails.
t
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5.9.1 Formula
1}**{XXXstXstXP
Rearranging we get:
(1-α)%
XstX * X
stX * μX
114
5.9.1 Formula
Our final formula becomes:
XstXCI
X*
Or in general:
estimatetruevalue stestimateCI *Which gives us an upper and lower
bound for our CI.
115
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:
625.040
25
40
)0(15)1(25
C
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5.9.1 IMPORTANT - Estimated Standard Deviation of a Sample
MeanWe have already seen that sample standard deviation is found through the formula:
Standard deviation of a sample mean is found through:
1
)( 2
N
YYS iY
Nss YY/
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5.9.1 Example
49.039
375.9
140
)625.00(15)625.01(25
1
)(
22
2
C
C
iC
S
S
N
CCS
077.040/49.0/ Nss CC
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5.9.1 Example
A 95% CI has 2.5% off each tail. If n=40,t* = 2.02
]78.0,47.0[
)077.0(02.2625.0
*
C
C
C
CI
CI
stCCIC
119
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.
120
5.9.1 Confidence Requirements
In order to construct a confidence interval, one needs:
a) A point estimate of the parameterb) Estimated standard deviation of the
parameterc) A critical value from a probability
distribution (or α and the sample size, n)
121
5.10 Hypothesis TestingAfter 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?
122
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
123
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
numberHa = alternate hypothesis = variable is not
equal to a numberEX)H0: Outcome=0.8Ha: Outcome≠0.8
124
5.10 Hypothesis Testing
Testing Consistency of a Hypothesized Parameter:
2) Collect appropriate sample data3) 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%
125
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)
126
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
127
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
128
5.10 Hypothetical ExampleJohnny is a poker player who wins an average of 8 times out of ten (the standard deviation is 0.5). Test the hypothesis that Johnny never wins.
1) H0: W=0Ha: W ≠0
2) We have estimated W=8. The standard deviation was 0.53) We let α=1%; we want a strong result.4) t= (estimate-hypothesis)/sd = (8-0)/0.5=165) t* for n-1=119, α=1%: t*=2.62
6) t*<t; Reject H0
129
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.