2 Linear Model [Final]

8/13/2019 2 Linear Model [Final]

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DrMudassirUddin

DrMudassirUddin

Review of Linear Model Basics


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Linear Regression Lecture [1]

Gauss-Markov Assumptions for Classical Linear Regression

Functional Form: Y

(n1)= X

(nk)(k1)+

n1(Xis full rank and has leading column of 1s)

Mean Zero Errors: E[|X] =0Homoscedasticity: Var[] =2I

Non-Correlated Errors: Cov[i, j] = 0, i =j

Exogeneity of Explanatory VariablesCov[i|X] = 0, i


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Other Considerations

Requirements: conformability, Xis rankk

Freebee: eventual normality. . . |X N(0, 2I)


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Least Squares Estimation (OLS)

Define the following function:

S() == (Y X)(Y X)= YY

(1n)(n1) 2YX

(1n)(nk)(k1)+ XX

(1k)(kn)(nk)(k1)

Take the derivative ofS() with respect to :

S() = 0 2 XY

(kn)(n1)+ 2XX

(kn)(nk)(k1) 0

So there exists a solution at some value of: XXb= XYwhich is theNormal Equation

Premultiplying the Normal Equation by (XX)1, gives: = (XX)1XY, where we can call as bfor notational convenience (this is where the requirement for XXto be nonsingular comesfrom)


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Implications

Normal Equation: XXbXY= X(Y Xb) = Xe 0(by assumption)

ei 0

the regression hyperplane pass through means: Y= Xb

mean(Y) =mean(Y)

the hat matrix (H, P, or (IM)):e= Y Xb

=Y X((XX)1XY)=Y (X(XX)1X)Y=Y HY= (IH)Y=MY

whereMis symmetric and idempotent.


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The HAT Matrix

the name is because Y = Xb= X((XX)

1

XY) = (X(XX)1

X)Y = HY, but projectionmatrix is better for geometric reasons.

properties of interest:

IM= P, IP= M PX= X

PM= MP = 0 (orthogonality)

ee= Y MMY= Y MY= Ye(sum of squares)

Using the Normal Equation, XXb= XY,

ee= (Y Xb)(Y Xb) =Y Y YXb bXY+ bXXb

=Y Y (XXb)b b(XXb)+ bXXb=Y Y (bX)(Xb)=Y Y YY


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Fit & Decomposition

Definitions:

SST =n

i=1

(Yi Y)2

SSR =n

i=1(Yi Y)2

SSE =n

i=1

(Yi Yi)2


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Fit & Decomposition

Interesting manipulations of the sum of squares total:

SST =n

i=1

(Y2i 2YiY+ Y2)

=n

i=1

Y2i 2n

i=1

YiY+nY2

=n

i=1

Y2i 2nY2 +nY2

=n

i=1

Y2i nY2 (scalar description)

=n

i=1 Y2i

n

1

n

n

i=1 Yi2

=Y Y 1nYJY (matrix algebra description)

whereJis an nmatrix of all 1s.


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More Decomposition

Sum of Squares Regression:

SSR =n

i=1

(Y2i 2YiY+ Y2)

= YY 2Yn

i=1Yi+nY

2

= (bX)Xb 2nY2 +nY2

= (bX)Xb nY2

=bXY 1nYJY


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Sum of Squares Error:

SSE =

n

i=1

(Yi Yi)2

=ee

= (Y X)(Y X)=Y Y YXb bXY+ bXXb=Y Y

bXY+ (bX

Y)Xb

=Y Y bXY+ (0)Xb=Y Y bXY


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Magic

SSR + SSE:SS R+SS T = (bXY 1

nYJY) + (YY bXY)

=Y Y 1nYJY

= SST


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A Measure of Fit

The R-Square or R-Squared measure:

R2 =SSR

SS T =

SST SSESST

= 1 ee

YMoY=bXMoXbYMoY

whereMo =I 1nii, i= c(1, 1, . . . , 1).Note: Mo is idempotent and transforms means to deviances for the explanatory variables.

Warning: R2 is not a statistic and does not have quite the meaning that some expect.

There is another version that accounts for sample size and the number of explanatory variables:

R2adj = ee/(n k)YMoY/(n 1)


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Properties of OLS

bis unbiased for :b= (XX)1XY

= (XX)1X(X+)

= (XX)1XX+ (XX)1X

= + (XX)1X

SoE[b] =.


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Properties of OLS

The variance ofbis:Var[b] =E[(b )](b )] (E[b ])(E[b ])

=E[(b )](b )]=E[((XX)1X)((XX)1X)]

= (XX)1XE[]X(XX)1

= (XX)1X(2I)X(XX)1

=2(XX)1


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Properties of OLS

Given Gauss-Markov assumptions, bis BLUE for

if calculated from OLS. b|X N(, 2(XX)1).Est.Var[b] =s2(XX)1, wheres2 =ee/(n k), the standard error of the regression.Most econometric texts make a distinction between non-stochastic explanatory variables (those

set in experimental settings) and stochastic explanatory variables which are set by nature/survey/others.

Perspective here: Xis fixed once observed and is the random variable:

since Var[] =I, then no single term dominates and we get the Lindberg-Feller CLT result.

so (the sample quantity) is IID normal and we write the joint PDF as:

f() =

n

i=1 f(i) = (22)n2 exp[ee/22](estimation fromE[ee] =2).


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Estimating From Sample Quantitities

Population derived variance/covariance matrix: Var[b

] =

2

(XX

)1

.We also know: E[ei] =i.

And: E[e2i ] = Var[i] + (E[i])2 =tr(2I) =n2.

So why not use: 2 1n

e2i .

But:

ei= yi Xib (now insert population values)= (Xi+i) Xib=i Xi(b )

whereE[ei] =i.


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Recall that:M

=IH

=IX

(XX

)1X

. So that:My= (IX(XX)1X)y

=y X(XX)1Xy=y Xb=e

So e= Me since e= y y.And therefore:

My= M[X+]

=MX+ M

= (I

X(XX)1)X+ M

=X X(XX)1)XX+ M=M

So e= Me and ee= (M)M= MM= M.


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So we can use this:

E[ee|X] =E[eMe|X]=E[tr(eMe)|X] (Gauss-Markov assumption)=E[tr(Mee)|X] (properties of traces: tr(ABC) = tr(CBA)= tr(ME[ee|X]) (M is fixed for observedX= tr(M)2

= [tr(I) tr(X(XX)1X)]2= [n k]2

Tricks:

tr[X(XX)1X] = tr[(XX)1XX] =k

rank[A] = tr[A] for symmetric idempotentArank[ABC] = rank[B] ifA, B nonsingular

so tr[H] = rank[X] = rank[X] =k


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From the biased estimator of2,E[ee|X] = (n

k)2, we get:

2 = een k =s

2,

so that an estimator of Var[b] is:

Var[b] =s2(XX)1.

This sets up Wald-style traditional linear inference:

zk= bk nullk

2(XX)1asym. N(0, 1),

provided that we know2 (which we usually do not).

But we know that:

(n k)s22

2nk=X2

zk/

X2/df t(nk)(0)if the random variableszk andX

2 are independent.


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Making the obvious substitution gives:

t(nk)= bk nullk

2(XX)1 1

(nk)s22(nk)

= bk nullk

s2(XX)1

Typical (Wald) regression test:

H0 : k= 0 H1 : k= 0

making:

t(nk)= bk nullk

s2(XX)1= bk

SE(k)

Alternatives usually look like:

H0 : k


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Summary Statistics

(1

) Confidence Interval for bk:bk SE(b)t/2,df:bk+SE(b)t/2,df

(1 ) Confidence Interval for2:

(n k)s22/2

:(n k)s2

21

/2

F-statistic test for all but b0zero:

F = SSR/(k 1)SSE/(n k) Fk1,nkunder the null


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Multicollinearity Issues

If one explanatory variable is a linear combination of another then rank(X) =k

1.

Therefore rank(XX) =k 1 (matrix sizek k), and it is singular and non-invertible.Now no parameter estimates are possible, and model is now unidentified.

More typically: 2 explanatory variables are highly but not perfectly correlated.

Symptoms: small changes in data give large changes in parameter estimates.

coefficients have large standard errors and poort-statistics even if F-statistics andR2 are okay.

coefficients seem illogical (wrong sign, huge magnitude)


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Multicollinearity Remedies

respecify model (if reasonable)

center explanatory variables, or standardize

ridge regression (add a little bias):

b= [XX+RI]1Xy

such that the [ ] part barely inverts.


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Is bUnbiased?

Starting with b:

b= (XX)1Xy= (XX)1X(X+) = + (XX)1X,

and take expected values:

E[b] =E[+ (XX)1X] =E[] +E[(XX)1X] = +E[K] =


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Testing Linear Restrictions

A theory has testable implicationsif it implies some testable restrictions on the model definition:

H0 : k= 0 versus H1 : k= 0for example.

Most restrictions involve nestedparameter space:

unrestricted:[0, 1, 2, 3]

restricted:[0, 0, 2, 3].

Note that non-nested comparisons cause problems for non-Bayesians.

Li R i L t [26]


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Testing Linear Restrictions

Linear restrictions for regression are clear when:

r111+r122+. . .+r1kk=q1

r211+r222+. . .+r2kk=q2...

rj11+rj22+. . .+rjkk=qj

or in more succinct matrix algebra form: R(Jk)

= q.

Notes:

Each row ofR is one restriction.

J < k for R to be full rank.

This setup imposes J restrictions onk parameters, so there arek J free parameters left. We are still assuming that i N(0, 2)

General test:

H0 : R q= 0 H1 : R q =0



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Testing Linear Restrictions, Examples

One of the coefficients is zero, j = 0,J= 1:

R= [0, 0, 0, 1j

, . . . , 0, 0, 0], q= 0

Two coefficients are equal,j =k,J= 2:

R= [0, 0, 0, 1j

, . . . ,1k

, . . . , 0, 0, 0], q= 0

Three coefficients are zero, J= 3: 1 0 0 0 . . . 00 1 0 0 . . . 00 0 1 0 . . . 0

= [I:0], q= 00

0



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A set of coefficients sum to one, 2+3+4= 1:

R= [0, 1, 1, 1, 0, . . . , 0], q= 1

Several restrictions,2+3= 4,4+6= 0,5= 9:

R=

0 1 1 0 0 0

0 0 0 1 0 1

0 0 0 0 1 0

, q=

4

0

9

All coefficients except the constant are zero:

R= [0:I], q= [0]



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Define now the discrepency vectordictated by the null hypothesis:

Rb q= m 0which asks whether mis sufficiently different from zero. Note that mis linear function ofband

therefore also normally distributed.

This makes it easy to think about:

E[m|X] =RE[b|X] q= R q= 0Var[m|X] = Var[Rb q|X] =R[Var[b|X]R= 2R(XX)1R

Wald Test:

W =m[Var[m|X]]1m= (Rb q)[2R(XX)1R]1(Rb q) 2JwhereJis the number of rows ofR, i.e. the number of restrictions.



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Unfortunately we do not have 2, so we use the test:

F =2(Rb q)(R(XX)1R)(Rb q) 12J

2

s2

n kn k

=

(Rb q)(2R(XX)1R)1(Rb q)/J[(n k)s2/2]/(n k)

= Xn

2J/J

Xd 2nk/(n k) Fj,nk =thats the distributional interpretation, now simplify:

= 1

J(Rb q)(s2R(XX)1R)1(Rb q)



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g [ ]


Example with 1 linear restriction, start with the definition:

H0 : r11+r22+. . .+rkk=r= q

F1,nk=

j(rjbj q)2

j

krjrkEst.Cov.[bj, bk]

and be more specific:

= 0, so R= [0, 0, . . . , 0, 1, 0, . . . , 0, 0],q= [0]

so that R(XX)1Rsimplifies to thej th diagonal of (XX)1. Now:

Rb q= bj q, F1,nk = bj qEst.V ar.[bj]

12

.



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Non-Normal Errors

Withouti

N(0, 2) we do not have F,t,2 results (bummer).

Despite that, we knowthat the asymptotic disribution ofbis:

b N,

2

nQ1

whereQ= plim

XX

n and plims2 =2 wheres2 = e

enk .

So the test statistic is:

tk=

n(bk nullk )s2(XX/n)1

tnk

provided that ei N(0, 2).Since the denominator converges to

2Q1, then:

k=n(bk nullk )

2Q1=

n(bk nullk )Asym.V ar.[bk]

1

2

thus asymptotically justifying at-test without the assumption of normality.



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Non-Normal Errors Summary

Major theorem, if:

n(b ) dN(0, 2Q1)

and if

H0 : R q= 0is true, then

W = (Rbq)[Rs2(XX)1R]1(Rb

q) =J F d

2

J



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Testing Nonlinear Restriction

H0 :c() =qwherec() is some nonlinear function.

Simple 1-restriction case:

z= c()

est.SE tnk

(or equivalentlyz2 F1,nk).But getting est.SE is hard, so use the first two terms of a Taylor series expansion to get an

estimate:

f(b) =f(a) +f(a)(b 1)1

1! +f(a)

(b a)22!

+f(a)(b a)3

3! +. . .

meaning:

c() c() +

c()

( )

so plim= justifies usingc() instead ofc().



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Testing Nonlinear Restriction

Now we can calculate the needed variance term:

Var(c() =E

c()2 (E[c()])2

=E

c() +

c()

( )

(E[c()])2

=Ec()2 2c()c()

(

) + c()

(

)2c()

(E[c()])2=q2 2q

c()

(0) +E

c()

( )2

c()

q2

=E

c()

Var()

c()

sinceE( )2 =E()2 2 = Var().All this means that we can use sample estimates forc()/and plug ins2(XX)1 for Var()

and then test with a normal distribution.



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Linear Model Predictions

We want the predicted value for x0 not in the sample:

y0 =x0+0 y0 =x0b

since y0 is the LMVUE ofE[y0|x0].The prediction error is:

e0 =y0 y0 =x0( b) +0.The Prediction variance is:

Var[e0|X,x0] =2 + Var[x0( b)|X,x0] =2 + x02(XX)1x0

and if we have a constant term in the regression, this is equivalent to:

21 +1n+K1

j=1K1

k=1 (x0j xj)(x0k xk)(X1M0X1)jk ,where X1 is Xomitting the first column, Kis the number of explanatory variables (includingthe constant), and M0 =I 1nii.



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Linear Model Predictions

This shows that the variance is greater the further away x0 is from x. Classic diagram: where

prediction intervals are modeled y0 t/2Var(e0)



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Running Lowess

postscript("Class.Multilevel/trends1.ps")par(bg="lightgrey")

x


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Running Lowess

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1.0



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Example: Poverty Among the Elderly, Europe

Governments often worry about the economic condition of senior citizens for political and social

reasons.

Typically in a large industrialized society, a substantial portion of these people obtain the bulk of

their income from government pensions.

An important question is whether there is enough support through these payments to provide

subsistence above the poverty rate.

To see if this is a concern, the European Union (EU) looked at this question in 1998 for the (then)

15 member countries with two variables:

1. the median (EU standardized) income of individuals age 65 and older as a percentage of the

population age 064,

2. the percentage with income below 60% of the median (EU standardized) income of the national

population.



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The data from the European Household Community Panel Survey are:

Relative PovertyNation Income RateNetherlands 93.00 7.00Luxembourg 99.00 8.00Sweden 83.00 8.00Germany 97.00 11.00

Italy 96.00 14.00Spain 91.00 16.00Finland 78.00 17.00France 90.00 19.00United.Kingdom 78.00 21.00Belgium 76.00 22.00Austria 84.00 24.00Denmark 68.00 31.00Portugal 76.00 33.00Greece 74.00 33.00Ireland 69.00 34.00



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eu.pov


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60 70 80 90 100

5

10

15

20

25

30

35

NetherlandsLuxembourgSweden

Germany

Italy

SpainFinland

France

United.Kingdom

Belgium Austria

Denmark

Portugal

GreeceIreland

Relative Income, Over 65

Povertyrate

,O

ver65



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postscript("Class.Multilevel/trends3.ps")

par(mfrow=c(1,1),mar=c(4,4,2,2),lwd=5,bg="lightgrey")plot(eu.pov,pch=15,xlab="",ylab="",ylim=c(2,37),xlim=c(61,104))

x.y.fit


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60 70 80 90 100

5

10

15

20

25

30

35

NetherlandsLuxembourgSweden

Germany

Italy

SpainFinland

France

United.Kingdom

Belgium Austria

Denmark

Portugal

GreeceIreland

Relative Income, Over 65

Povertyrate

,O

ver65



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summary(x.y.fit)Call:

lm(formula = eu.pov[, 2] ~ eu.pov[, 1])

Residuals:

Min 1Q Median 3Q Max

-12.224 -3.312 1.482 3.923 7.424

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 83.6928 12.2526 6.831 1.21e-05

eu.pov[, 1] -0.7647 0.1458 -5.246 0.000158

Residual standard error: 5.611 on 13 degrees of freedom

Multiple R-Squared: 0.6792, Adjusted R-squared: 0.6545

F-statistic: 27.52 on 1 and 13 DF, p-value: 0.0001580



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More on Linear Models

In this example the slope of the line is

0.7647, which tells us how much the poverty rate changes

for a one unit increase in income.

The intercept is 83.6928, which is what poverty would be for zero income.

The standard error is a measure of how reliable these estimates are. One common rule of thumb

is to see if the standard error is half the coefficient estimates or less.

R-Squared tells us how much of the variance of the outcome variable can be explained by theexplanatory variable.

F-statistic tells us how well significant the model fit is.



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OECD Example

The data are from the Organization for Economic Cooperation and Development (OECD) and

highlight the relationship between

commitment to employment protection measured on an interval scale (0-4) indicating the

quantity and extent of national legislation to protect jobs,

the total factor productivity difference in growth rates between 1980-1990 and 1990-1998.

for 19 countries.

For details, see The Economist, September 23, 2000.



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OECD Example

Prot. Prod.

United States 0.2 0.5Canada 0.6 0.6

Australia 1.1 1.3

New Zealand 1.0 0.4

Ireland 1.0 0.1

Denmark 2.0 0.9

Finland 2.2 0.7Austria 2.4 -0.1

Belgium 2.5 -0.4

Japan 2.6 -0.4

Sweden 2.9 0.5

Netherlands 2.8 -0.5

France 2.9 -0.9

Germany 3.2 -0.2

Greece 3.6 -0.3

Portugal 3.9 0.3

Italy 3.8 -0.3

Spain 3.5 -1.5



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OECD Example

oecd


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OECD Example

0 1 2 3 4

1

.5

1

.0

0

.5

0.0

0.5

1.0

1.5

UnitedStatesCanada

Australia

NewZealand

Ireland

Denmark

Finland

Austria

Belgium Japan

Sweden

Netherlands

France

GermanyGreece

Portugal

Italy

Spain

Employment Protection Scale

Tota

lFactor

Pro

ductiv

ity

Difference



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OECD Example

oecd.fit |t|)

(Intercept) 0.8591 0.3174 2.706 0.0156

oecd$Prot -0.3496 0.1215 -2.878 0.0109

---Residual standard error: 0.5761 on 16 degrees of freedom

Multiple R-Squared: 0.3411, Adjusted R-squared: 0.2999

F-statistic: 8.284 on 1 and 16 degrees of freedom, p-value: 0.01093



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So What Does a BadRegression Model Look Like?

Run the following....

time = c(0.25, 0.5, 0.75, 1, 1.25, 2, 3, 4, 5, 6, 8, 0.25, 0.5,

0.75, 1, 1.25, 2, 3, 4, 5, 6, 8, 0.25, 0.5, 0.75, 1, 1.25, 2, 3,

4, 5, 6, 8, 0.25, 0.5, 0.75, 1, 1.25, 2, 3, 4, 5, 6, 8, 0.25,

0.5, 0.75, 1, 1.25, 2, 3, 4, 5, 6, 8, 0.25, 0.5, 0.75, 1, 1.25,

2, 3, 4, 5, 6, 8)

conc = c(1.5, 0.94, 0.78, 0.48, 0.37, 0.19, 0.12, 0.11,

0.08, 0.07, 0.05, 2.03, 1.63, 0.71, 0.7, 0.64, 0.36, 0.32, 0.2,0.25, 0.12, 0.08, 2.72, 1.49, 1.16, 0.8, 0.8, 0.39, 0.22, 0.12,

0.11, 0.08, 0.08, 1.85, 1.39, 1.02, 0.89, 0.59, 0.4, 0.16, 0.11,

0.1, 0.07, 0.07, 2.05, 1.04, 0.81, 0.39, 0.3, 0.23, 0.13, 0.11,

0.08, 0.1, 0.06, 2.31, 1.44, 1.03, 0.84, 0.64, 0.42, 0.24, 0.17,

0.13, 0.1, 0.09)

#postscript("Class.Multilevel/conc.ps"); par(bg="lightgrey")

conc.fit


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Extended Policy Example

# SIMPLE REGRESSION EXAMPLE USING THE DETROIT MURDER DATASET, YEARS 1961-1974# R CODE HERE RUNS A BASIC MODEL AND VARIOUS DIAGNOSTICS, EMAIL QUESTIONS

#FTP - Full-time police per 100,000 population

#UEMP - % unemployed in the population

#MAN - number of manufacturing workers in thousands#NMAN - Number of non-manufacturing workers in thousands

#GOV - Number of government workers in thousands

#LIC - Number of handgun licences per 100,000 population (you can buy one)

#GR - Number of handgun registrations per 100,000 population (you own one)

#CLEAR - % homicides cleared by arrests

#WM - Number of white males in the population

#HE - Average hourly earnings

#WE - Average weekly earnings



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#HOM - Number of homicides per 100,000 of population

#ACC - Death rate in accidents per 100,000 population

#ASR - Number of assaults per 100,000 population

# LOAD DATA FILE, CREATE DATA FRAME

count.fields("public_html/data/detroit.data")

count.fields("http://jgill.wustl.edu/data/etroit.data")

detroit.df


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# RUN A MODEL, SUMMARIZE THE RESULTS

detroit.ols


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rstudent(detroit.ols)

dfbetas(detroit.ols)

dffits(detroit.ols)

covratio(detroit.ols)cooks.distance(detroit.ols)

X


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for (i in 1:length(cooks.vals))

segments(i,0,i,cooks.vals[i])

mtext(side=3,cex=1.3,"Leverage and Influence: Detroit Murder Data",outer=T,line=2)

# dev.off()

# JACKKNIFE OUT THE 8TH CASE AND RERUN MODEL

detroit2.df

2 Linear Model [Final]

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