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*BASIC ECONOMETRICS

*THE NATURE OF LINEAR REGRESSION

Hypothesis testing , and

Estimation



2

INTRODUCTION

What is Econometrics?

Econometrics consists of the application of

mathematical statistics to economic data to lend

empirical support to the models constructed bymathematical economics and to obtain numerical

results.

Econometrics may be defined as the quantitativeanalysis of actual economic phenomena based on

the concurrent development of theory and

observation, related by appropriate methods of

inference.



3

WHAT IS ECONOMETRICS?

Statistics

Economics

Econometrics

Mathematics



4

PURPOSE OF ECONOMETRICS

Structural Analysis

Policy Evaluation

Economic Prediction

Empirical Analysis



5

METHODOLOGY OF ECONOMETRICS

1. Statement of theory or hypothesis.

2. Specification of the mathematical model of the theory.

3. Specification of the statistical, or econometric model.

4. Obtaining the data.

5. Estimation of the parameters of the econometric model.

6. Hypothesis testing.

7. Forecasting or prediction.

8. Using the model for control or policy purposes.



6

EXAMPLE：KYNESIAN THEORY OF

CONSUMPTION

1. Statement of theory or hypothesis.

Keynes stated: The fundamental psychological law is

that men/women are disposed, as a rule and onaverage, to increase their consumption as their income increases, but not as much as the increasein their income.

In short, Keynes postulated that the marginalpropensity to consume (MPC), the rate of change of consumption for a unit change in income, is greater than zero but less than 1



2.SPECIFICATION OF THE MATHEMATICAL

MODEL OF THE THEORY

A mathematical economist might suggest the

following form of the Keynesian consumption

function:

10 110

X Y

Consumption

expenditure

Income



3. SPECIFICATION OF THE STATISTICAL,

OR ECONOMETRIC MODEL.

To allow for the inexact relationships between

economic variables, the econometrician would modify

the deterministic consumption function as follows:

This is called an econometric model.

u X Y 10

U, known as disturbance, or error term



4. OBTAINING THE DATA.

ye ar Y X

1 9 8 2 3 0 8 1 .5 4 6 2 0 .3

1 9 8 3 3 2 4 0 .6 4 8 0 3 .7

1 9 8 4 3 4 0 7 .6 5 1 4 0 .1

1 9 8 5 3 5 6 6 .5 5 3 2 3 .5

1 9 8 6 3 7 0 8 .7 5 4 8 7 .7

1 9 8 7 3 8 2 2 .3 5 6 4 9 .5

1 9 8 8 3 9 7 2 .7 5 8 6 5 .2

1 9 8 9 4 0 6 4 .6 6 0 6 2

1 9 9 0 4 1 3 2 .2 6 1 3 6 .3

1 9 9 1 4 1 0 5 .8 6 0 7 9 .4

1 9 9 2 4 2 1 9 .8 6 2 4 4 .4

1 9 9 3 4 3 4 3 .6 6 3 8 9 .61 9 9 4 4 4 8 6 6 6 1 0 .7

1 9 9 5 4 5 9 5 .3 6 7 4 2 .1

1 9 9 6 4 7 1 4 .1 6 9 2 8 .4

Sourse: Data on Y (Personal Consumption Expenditure) and X (Gross

Domestic Product),1982-1996) all in 1992 billions of dollars



10

5. ESTIMATION OF THE PARAMETERS OF

THE ECONOMETRIC MODEL.

reg y x

Source | SS df MS Number of obs = 15

-------------+------------------------------ F( 1, 13) = 8144.59

Model | 3351406.23 1 3351406.23 Prob > F = 0.0000

Residual | 5349.35306 13 411.488697 R-squared = 0.9984 -------------+------------------------------ Adj R-squared = 0.9983

Total | 3356755.58 14 239768.256 Root MSE = 20.285

------------------------------------------------------------------------------

y | Coef. Std. Err. t P>|t| [95% Conf. Interval]

-------------+---------------------------------------------------------------- x | .706408 .0078275 90.25 0.000 .6894978 .7233182

_cons | -184.0779 46.26183 -3.98 0.002 -284.0205 -84.13525

------------------------------------------------------------------------------



11

6. HYPOTHESIS TESTING.

Such confirmation or refutation of

econometric theories on the basis of

sample evidence is based on a branch of

statistical theory know as statistical

As noted earlier, Keynes expected the

MPC to be positive but less than 1. In

our example we found it is about 0.70. Then, is 0.70 statistically less than 1?

If it is, it may support keynes’s theory.



12

7.FORECASTING OR PREDICTION.

To illustrate, suppose we want to predict the mean

consumption expenditure for 1997. The GDP value

for 1997 was 7269.8 billion dollars. Putting this

value on the right-hand of the model, we obtain4951.3 billion dollars.

But the actual value of the consumption expenditure

reported in 1997 was 4913.5 billion dollars. The

estimated model thus overpredicted.

The forecast error is about 37.82 billion dollars.



13

TYPES OF DATA SETS



Assume that we have collected data on

two variables X and Y. Let

( x 1

, y 1

) ( x 2

, y 2

) ( x 3

, y 3

) … ( x n

, y n

)

denote the pairs of measurements on the

on two variables X and Y for n cases in a

sample (or population)



THE STATISTICAL MODEL



Each y i is assumed to be randomlygenerated from a normal distribution with

mean m i = a + x i and

standard deviation s .

(a , and s are unknown)

yi

a + xi

s

xi

Y = a + X

slope =

a



THE DATATHE LINEAR REGRESSION MODEL

The data falls roughly about a straight line.

0

20

40

60

80

100

120

140

160

40 60 80 100 120 140

Y = a + X

unseen



THE LEAST SQUARES LINE

Fitting the best straight line

to “linear” data



Let

Y = a + b X denote an arbitrary equation of a straight line.

a and b are known values.

This equation can be used to predict for each

value of X , the value of Y .For example, if X = x i (as for the ith case) thenthe predicted value of Y is:

ii bxa y ˆ



The residual

can be computed for each case in the sample,

The residual sum of squares (RSS) is

a measure of the “goodness of fit of the lineY = a + bX to the data

iiiii bxa y y yr ˆ

,ˆ,,ˆ,ˆ222111 nnn y yr y yr y yr

n

i

ii

n

i

ii

n

i

i bxa y y yr RSS 1

2

1

2

1

2ˆ



The equation for the least squares line

Let

n

i

i xx x xS 1

2

n

i

i yy y yS 1

2

n

i

ii xy y y x xS 1



LINEAR REGRESSION

Hypothesis testing and Estimation



THE LEAST SQUARES LINE

Fitting the best straight line

to “linear” data



n

x x x xS

n

i

in

i

i

n

i

i xx

2

1

1

2

1

2

n

y x

y x

n

i

i

n

i

in

i

ii

11

1

n

y y y yS

n

i

in

i

i

n

i

i yy

2

1

1

2

1

2

n

i

ii xy y y x xS 1

Computing Formulae:



Then the slope of the least squares line

can be shown to be:

n

i

i

n

i

ii

xx

xy

x x

y y x x

S

S b

1

2

1



and the intercept of the least squares line

can be shown to be:

x

S

S y xb ya

xx

xy



The residual sum of Squares

22

1 1

ˆ

n n

i i i i

i i

RSS y y y a bx

2

xy

yy

xx

S S

S

Computing

formula



Estimating s , the standard deviation in the

regression model :

22

ˆ

1

2

1

2

n

bxa y

n

y y

s

n

i

ii

n

i

ii

xx

xy

yy S

S

S n

2

2

1

This estimate of s is said to be based on n – 2

degrees of freedom

Computing

formula



SAMPLING DISTRIBUTIONS OF THE

ESTIMATORS



The sampling distribution s lope of the

least squares line :

n

i

i

n

i

ii

xx

xy

x x

y y x x

S

S b

1

2

1

It can be shown that b has a normal

distribution with mean and standard deviation

n

i i

xx

bb

x xS

1

2

and s s

s m



Thus

has a standard normal distribution, and

b

b

xx

b b z

S

m s s

b

b xx

b bt

s s S

m

has a t distribution with df = n - 2



(1 – a )100% Confidence Limits for slope

:

t a /2 critical value for the t-distribution with n – 2

degrees of freedom

xxS

st ˆ

2/a



Testing the slope

The test statistic is:

0 0 0: vs : A H H

0

xx

bt

s

S

- has a t distribution with df = n – 2 if H 0 is true.



The Critical Region

Reject0 0 0: vs : A H H

0/ 2 / 2if or

xx

bt t t t s

S

a a

df = n – 2

This is a two tailed tests. One tailed tests are

also possible



The sampling distribution intercept of the

least squares line :

It can be shown that a has a normal

distribution with mean and standard deviation

n

i

i

aa

x x x

n

1

2

2

1 and s s a m

xS

S y xb ya

xx

xy



Thus

has a standard normal distribution and

2

2

1

1

a

a

n

i

i

a a z

x

n x x

m a

s

s

2

2

1

1

a

a

n

i

i

a at

s x s n

x x

m a

has a t distribution with df = n - 2



(1 – a )100% Confidence Limits for intercept

a :


degrees of freedom

1ˆ

2

2/

xxS

x

n

st a a



Testing the intercept


0 0 0: vs : A H H a a a a


0

2

2

1

1

n

i

i

at

x s

n x x

a



The Critical Region

Reject0 0 0: vs : A H H a a a a

0/ 2 / 2if or

a

at t t t

sa a

a

df = n – 2



EXAMPLE



THE FOLLOWING DATA SHOWED THE PER CAPITA CONSUMPTION OF

CIGARETTES PER MONTH (X) IN VARIOUS COUNTRIES IN 1930, AND THE

DEATH RATES FROM LUNG CANCER FOR MEN IN 1950.

TABLE : PER CAPITA CONSUMPTION OF CIGARETTES PER MONTH (XI) IN N

= 11 COUNTRIES IN 1930, AND THE DEATH RATES, Y I (PER 100,000),

FROM LUNG CANCER FOR MEN IN 1950.

COUNTRY (I) XI YI

AUSTRALIA 48 18CANADA 50 15

DENMARK 38 17

FINLAND 110 35

GREAT BRITAIN 110 46

HOLLAND 49 24

ICELAND 23 6

NORWAY 25 9

SWEDEN 30 11

SWITZERLAND 51 25

USA 130 20



Australia

CanadaDenmark

Finland

Great Britain

Holland

Iceland

NorwaySweden

Switzerland

USA

0

5

10

15

20

25

30

35

40

45

50

0 20 40 60 80 100 120 140

d e a t h r a t e s f r o m l u

n g c a n c e r ( 1 9 5 0 )

Per capita consumption of cigarettes



404,541

2

n

i

i x

914,16

1

n

i

ii y x

018,61

2

n

ii y

Fitting the Least Squares Line

6641

n

i

i x

2261

n

ii y



55.14322

11

66454404

2

xxS

73.1374

11

2266018

2

yyS

82.327111

22666416914 xyS

Fitting the Least Squares Line

First compute the following three quantities:



Computing Estimate of Slope (), Intercept (a)

and standard deviation (s),

288.055.14322

82.3271

xx

xy

S

S b

756.611

664288.0

11

226

xb ya

35.82

1 2

xx

xy

yyS

S S

n s



95% Confidence Limits for slope :

t .025 = 2.262 critical value for the t-distribution with 9

degrees of freedom

xxS st ˆ

2/a

0.0706 to 0.3862

8.350.288 2.2621432255



95% Confidence Limits for intercept a :

1ˆ

2

2/

xxS

x

n st a a

-4.34 to 17.85

t .025 = 2.262 critical value for the t-distribution with 9

degrees of freedom

2664 111

6.756 2.262 8.3511 1432255

50



Iceland

NorwaySweden

Denmark Canada

Australia

HollandSwitzerland

Great Britain

Finland

USA

0

5

10

15

20

25

30

35

40

45

50

0 20 40 60 80 100 120 140

Per capita consumption of cigarettes

d e a t h r a t e

s f r o m l u

n g c a n c e

r ( 1 9 5 0

Y = 6.756 + (0.228) X

95% confidence Limits for slope 0.0706 to 0.3862

95% confidence Limits for intercept -4.34 to 17.85



Testing the positive slope


0 : 0 vs : 0 A H H

0

xx

bt

s

S



The Critical Region

Reject0 : 0 in favour of : 0 A H H

0.050if =1.833

xx

bt t s

S

df = 11 – 2 = 9

A one tailed test

b



and conclude

0 : 0 H

0Since

xx

bt

s

S

0.28841.3 1.833

8.35

1432255

we reject

: 0 A H



CONFIDENCE LIMITS FOR POINTS ON THE

REGRESSION LINE

The intercept a is a specific point on the regressionline.

It is the y – coordinate of the point on theregression line when x = 0.

It is the predicted value of y when x = 0.

We may also be interested in other points on theregression line. e.g. when x = x 0

In this case the y – coordinate of the point on theregression line when x = x 0 is a + x 0



x0

a + x0

y = a + x



(1- a )100% Confidence Limits for a + x 0 :

12

02/0

xxS

x x

n

st bxa

a

t a /2 is the a /2 critical value for the t-distribution with

n - 2 degrees of freedom

PREDICTION LIMITS FOR NEW VALUES



PREDICTION LIMITS FOR NEW VALUES

OF THE DEPENDENT VARIABLE Y

An important application of the regression line

is prediction.

Knowing the value of x ( x 0) what is the value

of y ? The predicted value of y when x = x 0 is:

This in turn can be estimated by:.

ˆ0 x y a

00 ˆ

ˆˆ bxa x y a



The predictor

Gives only a single value for y .

A more appropriate piece of information

would be a range of values.

A range of values that has a fixed

probability of capturing the value for y.

A (1- a )100% predict ion interval for y.

00 ˆˆˆ

bxa x y a



(1- a )100% Prediction Limits for y when x =

x 0:

11

2

02/0

xxS

x x

n

st bxa

a





EXAMPLEIn this example we are studying bu i ld ing f i res in a city and interested in the relationship

between:

1. X = the distance of the closest fire hall

and the building that puts out the alarm

and

2. Y = cost of the damage (1000$)

The data was collected on n = 15 fires .



THE DATA

Fire Distance Damage1 3.4 26.2

2 1.8 17.8

3 4.6 31.3

4 2.3 23.1

5 3.1 27.5

6 5.5 36.0

7 0.7 14.1

8 3.0 22.3

9 2.6 19.6

10 4.3 31.3

11 2.1 24.012 1.1 17.3

13 6.1 43.2

14 4.8 36.4

15 3.8 26.1



SCATTER PLOT

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0.0 2.0 4.0 6.0 8.0

Distance (miles)

D a m a g e

( 1 0 0 0 $ )



COMPUTATIONS

Fire Distance Damage

1 3.4 26.2

2 1.8 17.8

3 4.6 31.3

4 2.3 23.1

5 3.1 27.5

6 5.5 36.0

7 0.7 14.1

8 3.0 22.3

9 2.6 19.6

10 4.3 31.3

11 2.1 24.0

12 1.1 17.3

13 6.1 43.2

14 4.8 36.4

15 3.8 26.1

2.491

n

ii x

2.3961

n

i

i y

65.14701

n

i

ii y x

16.1961

2

n

i

i x

5.113761

2

n

i i

y



COMPUTATIONS CONTINUED

28.315

2.491

n

x

x

n

i

i

4133.2615

2.3961

n

y

y

n

i

i





784.3415

2.4916.1962

2

1

1

2

n

x

xS

n

iin

i

i xx

517.911152.3965.113762

2

1

1

2

n

y

yS

n

i

in

i

i yy

n

y x

y xS

n

i

i

n

i

in

i

ii xy

11

1

114.171

152.3962.49

65.1470





92.4784.34114.171ˆ

xx

xy

S S b

28.1028.3919.44133.26ˆ xb ya a

2

2

n

S

S S

s xx

xy yy

316.213

784.34114.171517.911

2



95% Confidence Limits for slope :

t .025 = 2.160 critical value for the t-distribution with

13 degrees of freedom

xxS st ˆ

2/a

4.07 to 5.77



95% Confidence Limits for intercept a :

1ˆ

2

2/

xxS

x

n st a a

7.21 to 13.35

t .025 = 2.160 critical value for the t-distribution with

13 degrees of freedom



LEAST SQUARES LINE

0.0

10.0

20.0

30.0

40.0

50.0

60.0

0.0 2.0 4.0 6.0 8.0

Distance (miles)

D a m a g e

( 1 0 0 0 $ )

y=4.92 x+10.28




12

02/0

xxS

x x

n

st bxa

a



95% CONFIDENCE LIMITS FOR A + B X



95% CONFIDENCE LIMITS FOR A + B X 0

:

x 0 lower upper

1 12.87 17.52

2 18.43 21.803 23.72 26.35

4 28.53 31.38

5 32.93 36.826 37.15 42.44

95% CONFIDENCE LIMITS FOR A + B



95% CONFIDENCE LIMITS FOR A B

X 0

0.0

10.0

20.0

30.0

40.0

50.0

60.0

0.0 2.0 4.0 6.0 8.0

Distance (miles)

D a m a g e ( 1 0 0 0 $ )

Confidence limits




x 0:

11

2

02/0

xxS

x x

n

st bxa

a



95% PREDICTION LIMITS FOR Y WHEN X



95% PREDICTION LIMITS FOR Y WHEN X =

X 0

x 0 lower upper

1 9.68 20.71

2 14.84 25.403 19.86 30.21

4 24.75 35.16

5 29.51 40.246 34.13 45.45

95% PREDICTION LIMITS FOR Y



95% PREDICTION LIMITS FOR Y

WHEN X = X 0

0.0

10.0

20.0

30.0

40.0

50.0

60.0

0.0 2.0 4.0 6.0 8.0

Distance (miles)

D a m a g e ( 1

0 0 0 $ )

Prediction limits



LINEAR REGRESSION

SUMMARY

Hypothesis testing and Estimation



(1 – a )100% Confidence Limits for slope

:


degrees of freedom

xxS

st ˆ

2/a



Testing the slope


0 0 0: vs : A H H

0

xx

bt

sS




(1 – a )100% Confidence Limits for intercept

a :


degrees of freedom

1ˆ

2

2/

xxS

x

n

st a a



Testing the intercept


0 0 0: vs : A H H a a a a

- has a t distribution with df = n – 2 if H 0is true.

0

2

2

1

1

n

i

i

at x

s

n x x

a




12

02/0

xxS

x x

n

st bxa

a






x 0:

11

2

02/0

xxS

x x

n

st bxa

a





CORRELATION

Definition



The statistic:

n

i

i

n

i

i

n

i

ii

yy xx

xy

y y x x

y y x x

S S

S r

1

2

1

2

1

is called Pearsons correlation coeff icient



The test for independence (zero correlation)




The test statistic:

22

1r t n

r

Reject H 0 if |t | > t a/2 (df = n – 2)

H 0: X and Y are independent

H A: X and Y are correlated

The Critical region

This is a two-tailed critical region, the critical

region could also be one-tailed



EXAMPLEIn this example we are studying bu i ld ing f i res

in a city and interested in the relationship

between:

1. X = the distance of the closest fire hall

and the building that puts out the alarm

and

2. Y = cost of the damage (1000$)

The data was collected on n = 15 fires .

THE DATA



THE DATA


1 3.4 26.2

2 1.8 17.8

3 4.6 31.3

4 2.3 23.1

5 3.1 27.5

6 5.5 36.07 0.7 14.1

8 3.0 22.3

9 2.6 19.6

10 4.3 31.3

11 2.1 24.012 1.1 17.3

13 6.1 43.2

14 4.8 36.4

15 3.8 26.1

SCATTER PLOT



SCATTER PLOT

0.05.0

10.0

15.0

20.025.0

30.0

35.0

40.045.0

50.0

0.0 2.0 4.0 6.0 8.0

Distance (miles)

D a m a g e

( 1 0 0 0 $ )

COMPUTATIONS



COMPUTATIONS


1 3.4 26.2

2 1.8 17.8

3 4.6 31.3

4 2.3 23.1

5 3.1 27.5

6 5.5 36.0

7 0.7 14.1

8 3.0 22.3

9 2.6 19.6

10 4.3 31.3

11 2.1 24.0

12 1.1 17.3

13 6.1 43.2

14 4.8 36.4

15 3.8 26.1

2.491

n

ii x

2.3961

n

ii y

65.14701

n

i

ii y x

16.1961

2

n

i

i x

5.113761

2

n

i

i y



784.3415

2.4916.1962

2

1

1

2

n

x

xS

n

i

in

i

i xx

517.911152.3965.11376

2

2

1

1

2

n

y

yS

n

i

in

i

i yy

n

y x

y xS

n

i

i

n

i

in

iii xy

11

1

114.171

152.3962.49

65.1470

THE CORRELATION COEFFICIENT



171.114

0.96134.784 911.517

xy

xx yy

S

r S S


The test statistic:

2 2

0.9612 13 12.525

1 1 0.961

r t n

r

We reject H 0: independence, if |t | > t 0.025 = 2.160

H 0: independence, is rejected



RELATIONSHIP BETWEEN REGRESSION

AND CORRELATION

Recall xyS r



Recall

xx yy

r S S

Also

ˆxy yy xy yy y

xx xx xx x xx yy

S S S S sr r

S S S sS S

since and

1 1

yy xx x y

S S s s

n n

Thus the slope of the least squares line is simply the ratio

of the standard deviations × the correlation coefficient





Uses the test statistic:

22

1r t n

r

H 0: X and Y are independent


Note: andˆ yy

xx

S r S

ˆ xx

yy

S r S

The two tests



1. The test for independence (zero correlation) H 0: X and Y are independent


are equivalent

2. The test for zero slope H 0: = 0.

H A: ≠ 0

1. the test statistic for independence:



22

1

r t n

r

2 22 2

1 1

xy xy

xx yy xx

xy xy yy

xx yy xx yy

S S

S S S t n n

S S S S S S S

Thus

2

ˆ

12

the same statistic for testing for slope.

xy

xx

xy

yy xx

xx xx

S

S

sS S n S

S S

zero



REGRESSION (IN GENERAL)

In many experiments we would have collected data on a



In many experiments we would have collected data on asingle variable Y (the dependent variable ) and on p (say) other variables X 1, X 2, X 3, ... , X p (the independent

variables).

One is interested in determining a model thatdescribes the relationship between Y (the response(dependent) variable) and X

1

, X 2

, …, X p

(the predictor (independent) variables.

This model can be used for

Prediction Controlling Y by manipulating X 1, X 2, …, X p



The Model:

is an equation of the form

Y = f ( X 1, X 2,... , X p | q1, q2, ... , qq) + e

where q1, q2, ... , qq are unknownparameters of the function f and e is arandom disturbance (usually assumed to

have a normal distribution with mean 0and standard deviation s.



2. Y = average of five best times for running

the 100m X the ear



8

8.5

9

9.5

10

10.5

11

11.5

12

12.5

1930 1940 1950 1960 1970 1980 1990 2000 2010

the 100m, X = the year

The model

Y = a e- X + g e, thus q1 = a, q2 = and q2 =

g .

This model is called:

the exponential Regression Model

Y = a e- X + g



THE MULTIPLE LINEAR

REGRESSION MODEL

In Multiple Linear Regression we assume the



In Multiple Linear Regression we assume thefollowing model

Y = 0 + 1 X1 + 2 X2 + ... + p Xp + e

This model is called the Multiple Linear Regression Model.

Again are unknown parameters of the modeland where 0, 1, 2, ... , p are unknownparameters and e is a random disturbanceassumed to have a normal distribution withmean 0 and standard deviation s.

THE IMPORTANCE OF THE LINEAR MODEL



THE IMPORTANCE OF THE LINEAR MODEL

1. It is the simplest form of a model in whicheach dependent variable has some effect onthe independent variable Y.

When fitting models to data one tries to find thesimplest form of a model that still adequatelydescribes the relationship between thedependent variable and the independentvariables.

The linear model is sometimes the first model tobe fitted and only abandoned if it turns out to beinadequate.

I i t li d l i th



2. In many instance a linear model is the

most appropriate model to describe

the dependence relationship betweenthe dependent variable and the

independent variables.

This will be true if the dependent variableincreases at a constant rate as any or the

independent variables is increased while

holding the other independent variables

constant.

3 Man non Linear models can be



3. Many non-Linear models can be

Linearized (put into the form of a

Linear model by appropriatelytransformation the dependent variables

and/or any or all of the independent

variables.) This important fact ensures the wide utility

of the Linear model. (i.e. the fact the many

non-linear models are linearizable.)



AN EXAMPLE

The following data comes from an experimentthat was interested in investigating the sourcefrom which corn plants in various soils obtaintheir phosphorous.

The concentration of inorganic phosphorous (X1)and the concentration of organic phosphorous (X2)was measured in the soil of n = 18 test plots.

In addition the phosphorous content (Y) of corngrown in the soil was also measured. The data isdisplayed below:



Inorganic

Phosphorous

X1

Organic

Phosphorous

X2

Plant

Available

Phosphorous Y

Inorganic

Phosphorous

X1

Organic

Phosphorous

X2

Plant

Available

Phosphorous Y

0.4 53 64 12.6 58 51

0.4 23 60 10.9 37 76

3.1 19 71 23.1 46 96 0.6 34 61 23.1 50 77

4.7 24 54 21.6 44 93

1.7 65 77 23.1 56 95

9.4 44 81 1.9 36 54

10.1 31 93 26.8 58 168

11.6 29 93 29.9 51 99



Coefficients

Intercept 56.2510241 (0)

X1 1.78977412 (1)

X2 0.08664925 (2)

Equation:Y = 56.2510241 + 1.78977412 X1 + 0.08664925 X2



THE MULTIPLE LINEAR

REGRESSION MODEL

In Multiple Linear Regression we assume the



In Multiple Linear Regression we assume thefollowing model

Y = 0 + 1 X1 + 2 X2 + ... + p Xp + e

This model is called the Multiple Linear Regression Model.

Again are unknown parameters of the modeland where 0, 1, 2, ... , p are unknownparameters and e is a random disturbanceassumed to have a normal distribution withmean 0 and standard deviation s.

SUMMARY OF THE STATISTICS



USED IN

MULTIPLE REGRESSION

The Least Squares Estimates:



q

0 1 2, , , , , p

2

1

ˆ

n

i i

i

RSS y y

2

0 1 1 2 2

1

n

i i i p pi

i

y x x x

- the values that minimize

The Analysis of Variance Table Entries

a) Adjusted Total Sum of Squares (SSTotal)n



b) Residual Sum of Squares (SSError )

c) Regression Sum of Squares (SSReg)

Note:

i.e. SSTotal = SSReg +SSError

SSTotal n

i1

yi y _

2. d.f. n 1

RSS SSError n

i1

yi yî2. d.f. n p 1

SSReg SS1,2, ... , p n

i1

yˆ i y _

2. d.f. p

n

i1

yi y _

2

n

i1

yî y _

2

n

i1

yi yî 2

.

THE ANALYSIS OF VARIANCE TABLE



Source Sum of Squares d.f. Mean Square F

Regression SSReg p SSReg /p = MSReg MSReg /s2

Error SSError n-p-1 SSError /(n-p-1) =MSError = s2

Total SSTotal n-1

USES:



1. To estimate s2 (the error variance).

- Use s2 = MSError to estimate s2.

2. To test the Hypothesis

H0: 1 = 2= ... = p = 0.

Use the test statistic

2

Reg Reg Error F MS MS MS s

Reg 1 Error SS p SS n p

- Reject H 0 if F > F a ( p,n-p-1).

3. To compute other statistics that are useful indescribing the relationship between Y (the dependent



describing the relationship between Y (the dependent

variable) and X1, X2, ... ,Xp (the independent variables).

a) R2

= the coefficient of determination= SSReg /SSTotal

=

= the proportion of variance in Y explained by

X1, X2, ... ,Xp

1 - R2 = the proportion of variance in Y that is left unexplained by X1, X2, ... , Xp

= SSError /SSTotal.

y i y 2i1

n

y i y 2

i1

n



b) Ra2 = "R2 adjusted" for degrees of freedom.

= 1 -[the proportion of variance in Y that is leftunexplained by X1, X2,... , Xp adjusted for d.f.]

1 Error Total MS MS

11

1

Error

Total

SS n p

SS n

11

1

Error

Total

n SS

n p SS

2

11 1

1

n R

n p

c) R= R2 = the Multiple correlation coefficient of



Y with X 1, X 2, ... , X p

=

= the maximum correlation between Y and a

linear combination of X 1, X 2, ... , X p

Comment: The statistics F, R 2, R a2 and R are

equivalent statistics.

SSRe g

SSTotal



USING STATISTICAL PACKAGES

To perform Multiple Regression



USING SPSS

Note: The use of another statistical package

such as Minitab is similar to using SPSS

AFTER STARTING THE SSPS PROGRAM THE FOLLOWING

DIALOGUE BOX APPEARS:



DIALOGUE BOX APPEARS:

IF YOU SELECT OPENING AN EXISTING FILE AND PRESS OK

THE FOLLOWING DIALOGUE BOX APPEARS



THE FOLLOWING DIALOGUE BOX APPEARS:



IF THE VARIABLE NAMES ARE IN THE FILE ASK IT TO

READ THE NAMES IF YOU DO NOT SPECIFY THE



READ THE NAMES. IF YOU DO NOT SPECIFY THE

RANGE THE PROGRAM WILL IDENTIFY THE RANGE:

Once you “click OK”, two windows will appear

ONE THAT WILL CONTAIN THE OUTPUT:



THE OTHER CONTAINING THE DATA:



TO PERFORM ANY STATISTICAL ANALYSIS SELECT

THE MENU:



THE ANALYZE MENU:

THEN SELECT REGRESSION AND LINEAR.



THE FOLLOWING REGRESSION DIALOGUE BOX

APPEARS



SELECT THE DEPENDENT VARIABLE Y .



SELECT THE INDEPENDENT VARIABLES X 1, X 2, ETC.



IF YOU SELECT THE METHOD - ENTER.



All variables will be put into the equation.

There are also several other methods that can be

used :

1. Forward selection

2. Backward Elimination

3. Stepwise Regression



Forward selection

1. This method starts with no variables in the




equation

2. Carries out statistical tests on variables not in

the equation to see which have a significant

effect on the dependent variable.

3. Adds the most significant.

4. Continues until all variables not in the

equation have no significant effect on the

dependent variable.

Backward Elimination

1. This method starts with all variables in the



1. This method starts with all variables in the

equation

2. Carries out statistical tests on variables in the

equation to see which have no significant


3. Deletes the least significant.

4. Continues until all variables in the equation

have a significant effect on the dependent

variable.

epw se egress on uses o orwar anbackward techniques)

1 This method starts with no variables in the




equation2. Carries out statistical tests on variables not in

the equation to see which have a significant


3. It then adds the most significant.

4. After a variable is added it checks to see if any

variables added earlier can now be deleted.5. Continues until all variables not in the

equation have no significant effect on the

dependent variable.

All of these methods are procedures for



All of these methods are procedures for attempting to find the best equation

The best equation is the equation that is the

simplest (not containing variables that are notimportant) yet adequate (containing variablesthat are important)

ONCE THE DEPENDENT VARIABLE, THE INDEPENDENT VARIABLES

AND THE METHOD HAVE BEEN SELECTED IF YOU PRESS OK, THE

ANALYSIS WILL BE PERFORMED.



THE OUTPUT WILL CONTAIN THE FOLLOWING TABLE



Model Summary

.822a .676 .673 4.46

Model

1

R R Square

Adjusted

R Square

Std. Error

of the

Estimate

Predictors: (Constant), WEIGHT, HORSE, ENGINEa.

R 2 and R 2 adjusted measures the proportion of variance

in Y that is explained by X 1, X 2, X 3, etc (67.6% and

67.3%)

R is the Multiple correlation coefficient (the maximum

correlation between Y and a linear combination of X 1,

X 2, X 3, etc)

THE NEXT TABLE IS THE ANALYSIS OF VARIANCE

TABLE



The F test is testing if the regression coefficients of

the predictor variables are all zero. Namely none of the independent variables X 1, X 2, X 3,

etc have any effect on Y

ANOVAb

16098.158 3 5366.053 269.664 .000a

7720.836 388 19.899

23818.993 391

Regression

Residual

Total

Model

1

Sum of Squares df MeanSquare F Sig.

Predictors: (Constant), WEIGHT, HORSE, ENGINEa.

Dependent Variable: MPGb.

THE FINAL TABLE IN THE OUTPUT

Coefficientsa



Gives the estimates of the regression coefficients,

there standard error and the t test for testing if they arezero

Note: Engine size has no significant effect on

Mileage

44.015 1.272 34.597 .000

-5.53E-03 .007 -.074 -.786 .432

-5.56E-02 .013 -.273 -4.153 .000

-4.62E-03 .001 -.504 -6.186 .000

(Constant)

ENGINE

HORSEWEIGHT

Model1

B Std. Error

Unstandardized

Coefficients

Beta

Standardi

zedCoefficien

ts

t Sig.

Dependent Variable: MPGa.

THE ESTIMATED EQUATION FROM THE TABLE BELOW:C ffi i t a



5.53 5.56 4.6244.0

1000 100 1000 Mileage Engine Horse Weight Error

Is:

Coefficientsa

44.015 1.272 34.597 .000

-5.53E-03 .007 -.074 -.786 .432

-5.56E-02 .013 -.273 -4.153 .000

-4.62E-03 .001 -.504 -6.186 .000

(Constant)

ENGINE

HORSE

WEIGHT

Model1

B Std. Error

Unstandardized

Coefficients

Beta

Standardi

zed

Coefficien

ts

t Sig.

Dependent Variable: MPGa.

NOTE THE EQUATION IS:



5.53 5.56 4.6244.0

1000 100 1000

Mileage Engine Horse Weight Error

Mileage decreases with:

1. With increases in Engine Size (notsignificant, p = 0.432)

With increases in Horsepower (significant,

p = 0.000)

With increases in Weight (significant, p =0.000)



LOGISTIC REGRESSION

Recall the simple linear regression model:

y = 0 + 1x + e



y = 0 + 1 x + e

where we are trying to predict a continuousdependent variable y from a continuous

independent variable x.

This model can be extended to Multiple linear

regression model:

y = 0 + 1 x1 + 2 x2 + … + + p x p + e

Here we are trying to predict a continuous

dependent variable y from a several continuous

dependent variables x1 , x2 , … , x p .

Now suppose the dependent variable y is

binary



binary.

It takes on two values “Success” (1) or “Failure” (0)

This is the situation in which Logistic

Regression is used

We are interested in predicting a y from a

continuous dependent variable x.

EXAMPLE



We are interested how the success (y ) of anew antibiotic cream is curing “acne problems”

and how it depends on the amount ( x ) that is

applied daily.The values of y are 1 (Success) or 0 (Failure).

The values of x range over a continuum

THE LOGISITIC REGRESSION MODEL



Let p denote P [y = 1] = P [Success].

This quantity will increase with the value of

x.

1

p

p

The ratio: is called the odds ratio

This quantity will also increase with the value of

x, ranging from zero to infinity.

The quantity: ln1

p p

is called the log odds ratio

EXAMPLE: ODDS RATIO, LOG ODDS

RATIO



Suppose a die is rolled:

Success = “roll a six”, p = 1/6

1 16 6

516 6

1

1 1 5

p

p

The odds ratio

1

ln ln ln 0.2 1.690441 5

p

p

The log odds ratio




0 1

1

x p

e p

i. e. :

In terms of the odds ratio

0 1ln

1

p x

p

Assumes the log odds ratio is linearlyrelated to x.




0 1

1

x pe

p

or

Solving for p in terms x.

0 1 1 x p e p

0 1 0 1 x x p pe e

0 1

0 11

x

x

e p

e

INTERPRETATION OF THE PARAMETER B 0(DETERMINES THE INTERCEPT)



0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

p

0

0

1

e

e

x

INTERPRETATION OF THE PARAMETER B 1(DETERMINES WHEN P IS 0.50 (ALONG WITH

B0))



B 0))

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

p0 1

0 1

1 1

1 1 1 2

x

x

e p

e

x

00 1

1

0 or x x

when

ALSO0 1

0 11

x

x

dp d e

dx dx e



1dx dx e

0

1

x

when

0 1 0 1 0 1 0 1

0 1

1 1

2

1

1

x x x x

x

e e e e

e

0 1

0 1

1 1

241

x

x

e

e

1

4

is the rate of increase in p with respect to x

when p = 0.50

INTERPRETATION OF THE PARAMETER B 1(DETERMINES SLOPE WHEN P IS 0.50 )



0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

p

x

1slope4

THE DATA



The data will for each case consist of

1. a value for x, the continuous independent

variable

2. a value for y (1 or 0) (Success or Failure)

Total of n = 250 cases

case x y230 4.7 1

231 0.3 0

232 1.4 0

case x y

1 0.8 0

2 2.3 1

3 2 5 0



233 4.5 1

234 1.4 1235 4.5 1

236 3.9 0

237 0.0 0

238 4.3 1

239 1.0 0

240 3.9 1

241 1.1 0

242 3.4 1

243 0.6 0

244 1.6 0

245 3.9 0246 0.2 0

247 2.5 0

248 4.1 1

249 4.2 1

250 4.9 1

3 2.5 0

4 2.8 1

5 3.5 16 4.4 1

7 0.5 0

8 4.5 1

9 4.4 1

10 0.9 011 3.3 1

12 1.1 0

13 2.5 1

14 0.3 1

15 4.5 1

16 1.8 0

17 2.4 1

18 1.6 0

19 1.9 1

20 4.6 1

ESTIMATION OF THE PARAMETERS



The parameters are estimated by Maximum

Likelihood estimation and require a

statistical package such as SPSS

USING SPSS TO PERFORM LOGISTIC REGRESSION

O th d t fil



Open the data file:

Choose from the menu:

Analyze -> Regression -> Binary Logistic



The following dialogue box appears



Select the dependent variable ( y) and the independent

variable ( x) (covariate).

Press OK .

Here is the output



The Estimates and their S.E.

THE PARAMETER ESTIMATES



SEX 1.0309 0.1334

Constant -2.0475 0.332

1 1.0309

0 -2.0475



Another interpretation of the parameter 1

1

4

is the rate of increase in p with

respect to x when p = 0.50

1 1.03090.258

4 4

The Logistic Regression Model



The dependent variable y is binary.

It takes on two values “Success” (1) or

“Failure” (0)

We are interested in predicting a y from a

continuous dependent variable x.




Let p denote P [y = 1] = P [Success].

This quantity will increase with the value of

x.

1

p

p

The ratio: is called the odds ratio

This quantity will also increase with the value of

x, ranging from zero to infinity.

The quantity: ln1

p p

is called the log odds ratio




0 1

1

x p

e p

i. e. :

In terms of the odds ratio

0 1ln

1

p x

p

Assumes the log odds ratio is linearlyrelated to x.




In terms of p

0 1

0 11

x

x

e p

e

THE GRAPH OF P VS X



0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

p0 1

0 11

x

x

e

p e

x



THE MULTIPLE LOGISTIC REGRESSIONMODEL

Here we attempt to predict the outcome of

bi i bl Y f l



a binary response variable Y from several

independent variables X 1, X 2 , … etc

0 1 1ln 1 p p

p

X X p

0 1 1

0 1 1or 1

p p

p p

X X

X X

e

p e



For n = 223 infants in prenatal ward thefollowing measurements were determined



following measurements were determined

1. X 1 = gestational Age (weeks),

2. X 2 = Birth weight (grams) and

3. Y = presence of BPD

THE DATAcase Gestational Age Birthweight presence of BMD

1 28.6 1119 1

2 31.5 1222 0

3 30.3 1311 1

4 28.9 1082 0

5 30 3 1269 0



5 30.3 1269 0

6 30.5 1289 0

7 28.5 1147 08 27.9 1136 1

9 30 972 0

10 31 1252 0

11 27.4 818 0

12 29.4 1275 0

13 30.8 1231 0

14 30.4 1112 0

15 31.1 1353 1

16 26.7 1067 1

17 27.4 846 1

18 28 1013 0

19 29.3 1055 0

20 30.4 1226 0

21 30.2 1237 0

22 30.2 1287 0

23 30.1 1215 0

24 27 929 1

25 30.3 1159 0

26 27.4 1046 1

THE RESULTS

Variables in the Equation

003 001 4 885 1 027 998BirthweightStep

B S.E. Wald df Sig. Exp(B)



ln 16.858 .003 .5051

p

BW GA p

-.003 .001 4.885 1 .027 .998

-.505 .133 14.458 1 .000 .604

16.858 3.642 21.422 1 .000 2.1E+07

Birthweight

GestationalAge

Constant

Step

1a

Variable(s) entered on step 1 : Birthweight, GestationalAge.a.

16.858 .003 .505

1

BW GA pe

p

16.858 .003 .505

16.858 .003 .5051

BW GA

BW GA

e p

e

GRAPH: SHOWING RISK OF BPD VS GA ANDBRTHWT



0

0.2

0.4

0.6

0.8

1

700 900 1100 1300 1500 1700

GA = 27

GA = 28

GA = 29

GA = 30

GA = 31

GA = 32



NON PARAMETRIC STATISTICS

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