Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson10-1 Lesson 10: Regressions Part I.

Lesson10-1 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data

Lesson 10:

Regressions Part IRegressions Part I


Outline

Correlation Analysis

Regression Analysis

Standard error of estimate

Confidence interval and prediction interval

Inference about the regression slope

Cautions about the interpretation of significance

Evaluating the model

Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data

Correlation Analysis

Correlation Analysis is a group of statistical techniques used to measure the strength of the association between two variables.

A Scatter Diagram is a chart that portrays the relationship between the two variables.

The Dependent Variable is the variable being predicted or estimated.

The Independent Variable provides the basis for estimation. It is the predictor variable.


Example

Suppose a university administrator wishes to determine whether any relationship exists between a student’s score on an entrance examination and that student’s cumulative GPA. A sample of eight students is taken. The results are shown below

Student Exam Score GPA

A 74 2.6

B 69 2.2

C 85 3.4

D 63 2.3

E 82 3.1

F 60 2.1

G 79 3.2

H 91 3.8


Scatter Diagram: GPA vs. Exam Score

| | | | | | | | | |

50 55 60 65 70 75 80 85 90 95

Exam Score

4.00 -3.75 -3.50 -3.25 -3.00 -2.75 -2.50 -2.25 -2.00 -

Cu

mu

lati

ve G

PA

We would like to know whether there is a strong linear relationship between the two variables.


The Coefficient of Correlation, r

The Coefficient of Correlation (r) is a measure of the strength of the linear relationship between two variables. It can range from -1.00 to 1.00. Values of -1.00 or 1.00 indicate perfect and strong

correlation. Values close to 0.0 indicate weak correlation. Negative values indicate an inverse relationship

and positive values indicate a direct relationship.


Formula for r

n

1i

2i

n

1i

2i

n

1iii

n

1i

2i

n

1i

2i

n

1iii

yx

2xy

)yy()xx(

)yy)(xx(

)1n(

)yy(

)1n(

)xx(

)1n(

)yy)(xx(

ss

sr

We calculate the coefficient of correlation from the following formulas.

Sample covariance between x and y

Sample standard deviation of x

Sample standard deviation of y


Coefficient of Determination

The coefficient of determination (r2) is the proportion of the total variation in the dependent variable (Y) that is explained or accounted for by the variation in the independent variable (X). It is the square of the coefficient of correlation. It ranges from 0 to 1. It does not give any information on the direction of the

relationship between the variables.

Special cases: No correlation: r=0, r2=0. Perfect negative correlation: r=-1, r2=1. Perfect positive correlation: r=+1, r2=1.


EXAMPLE 1

Dan Ireland, the student body president at Toledo State University, is concerned about the cost to students of textbooks. He believes there is a relationship between the number of pages in the text and the selling price of the book. To provide insight into the problem he selects a sample of eight textbooks currently on sale in the bookstore. Draw a scatter diagram. Compute the correlation coefficient.

Book Page Price ($)

Intro to History 500 84

Basic Algebra 700 75

Intro to Psyc 800 99

Intro to Sociology 600 72

Bus. Mgt. 400 69

Intro to Biology 500 81

Fund. of Jazz 600 63

Princ. Of Nursing 800 93


Example 1 continued

400 500 600 700 800

60

70

80

90

100

Page

Scatter Diagram of Number of Pages and Selling Price of Text

Price ($)


Example 1 continued

Book Page Price ($)

X Y

Intro to History 500 84

Basic Algebra 700 75

Intro to Psyc 800 99

Intro to Sociology 600 72

Bus. Mgt. 400 69

Intro to Biology 500 81

Fund. of Jazz 600 63

Princ. Of Nursing 800 93

Total 4,900 636

n

1i

2i

n

1i

2i

n

1iii

)yy()xx(

)yy)(xx(r

The correlation between the number of pages and the selling price of the book is 0.614. This indicates a moderate association between the variable.


EXAMPLE 1 continued

Is there a linear relation between number of pages and price of books?

Test the hypothesis that there is no correlation in the population. Use a .02 significance level.

Under the null hypothesis that there is no correlation in the population. The statistic

)2/()1( 2

nr

rt

follows student t-distribution with (n-2) degree of freedom.


EXAMPLE 1 continued

Step 1: H0: The correlation in the population is zero. H1: The correlation in the population is not zero.

Step 2: H0 is rejected if t>3.143 or if t<-3.143. There are 6 degrees of freedom, found by n – 2 = 8 – 2 = 6.

Step 3: To find the value of the test statistic we use:

905.1)614(.1

28614.

)2/()1( 22

nr

rt

Step 4: H0 is not rejected. We cannot reject the hypothesis that there is no correlation in the population. The amount of association could be due to chance.


Regression Analysis

In regression analysis we use the independent variable (X) to estimate the dependent variable (Y).

The relationship between the variables is linear.


Simple Linear Regression Model

Relationship Between Variables Is a Linear Function

iii XY 10

Y intercept Slope Random Error

Dependent (Response) Variable

Independent (Explanatory) Variable x

y

0 Run

Rise

1 = Rise/Run

0 and 1 are unknown,therefore, are estimated from the data.


Finance Application: Market Model

One of the most important applications of linear regression is the market model.

It is assumed that rate of return on a stock (R) is linearly related to the rate of return on the overall market (Rm).

Rate of return on a particular stock

Rate of return on some major stock index

The beta coefficient measures how sensitive the stock’s rate of return is to changes in the level of the overall market.

R = 0 + 1Rm +


EstimationMethod of moments

Given a set of data (x1,y1),…,(xn,yn), how should we estimate the parameters 0 and 1?

Start from the simple case – suppose 1 is known to be 0. That is, we have only one parameter to estimate. yi=0 + i

Possible assumption #1 E() = 5. That is E(y) =0 + E() = 0 + 5

Possible assumption #2 E() = 0. That is E(y) =0 + E() = 0

0 has the interpretation of mean of y.

Better assumption!!


Estimation of 0

Method of moments

Given a set of data (x1,y1),…,(xn,yn), how should we estimate the parameters 0?

yi=0 + i

Assumption: E() = 0. That is E(y) =0 + E() = 0


Use the sample analog # 1: E() = 0 implies 0 = E(y)

Take the estimate as b0 = (y1 + y2 + … + yn)/n = ∑ yi /n


Estimation of 0

Method of moments

Given a set of data (x1,y1),…,(xn,yn), how should we estimate the parameters 0?

yi=0 + i

Assumption: E() = 0. That is E(y) =0 + E() = 0


Use the sample analog # 1: E() = 0 in population. Set the (e1 + e2 + … + en )/n = 0

(y1- b0) + (y2- b0) + … + (yn- b0) = 0

(y1 + y2 + … + yn) - nb0 = 0

b0 = (y1 + y2 + … + yn)/n = ∑ yi /n

(yn-b0): deviation of y from the assumed value of 0


Estimation of 1

Method of moments


Start from another simple case – suppose 0 is known to be 0. That is, we have only one parameter to estimate. yi=1xi + i

Possible assumption #1 E() = 0. That is E(y) =1E(x) + E() = 1E(x) and

1=E(y)/E(x) Possible assumption #2

E(x) = 0. That is E(yx) =1E(x2) + E(x) = 1E(x2) and 1=E(yx)/E(x2)


Estimation of 0 and 1

Method of moments


yi=0 + 1 xi+ i

Assumption #1 E() = 0.

If x is known and non-random: E(y) =0 + 1 x + E() = 0 + 1 x

If x is unknown and random: E(y) =0 + 1 E(x) + E() = 0 + 1 E(x)

This assumption gives us only one equation. Need an extra equation….

Assumption #2….



Method of moments


yi=0 + 1 xi+ i

Assumption #1 E() = 0 implies E(y) – 0 – 1 E(x) = 0

Assumption #2 How about E(x) = 0? Since Cov(x) = E(x) – E()E(x) = E(x), the

assumption really imply and x are uncorrelated. E(x) =0 implies E[(y – 0 – 1x)x]=0

E[yx – 0x – 1x2]= E(yx) – 0E(x) – 1E(x2) = 0Two equations are adequate to solve two unknowns.



Method of moments

Two equations are adequate to solve two unknowns. E(y) – 0 – 1 E(x) = 0

E(y) E(x) – 0 E(x) – 1 E(x)2 = 0

E(yx) – 0E(x) – 1E(x2) = 0

E(yx) – E(y) E(x) – 1E(x2) + 1 E(x)2 = 0

Cov(x,y) – 1Var(x) = 0

1= Cov(x,y)/Var(x)

E(y) – 0 – 1 E(x) = 0

0 = E(y)– 1 E(x)



Method of moments

The two assumptions: Assumption #1: E() = 0 Assumption #2: E(x) = 0

implies 1= Cov(x,y)/Var(x) and 0 = E(y)– 1 E(x)

To estimate the parameters, use the sample analog: Suffice to compute the

Sample covariance between x and y, Sx,y

Sample variance of x, Sxx

Sample mean of y, my

Sample mean of x, mx

and use them to replace the corresponding population quantities.



Method of moments

The two assumptions: Assumption #1: E() = 0 Assumption #2: E(x) = 0

To estimate the parameters, we can also use the sample analog of the two assumptions directly: Sample analog of E() = 0:

∑ (yi – b0 – b1 xi) =0∑yi– nb0 – b1 ∑xi =0

Sample analog of E(x) = 0?∑ (yi – b0 – b1 xi)xi =0∑(yi xi) – b0 ∑xi – b1 ∑xi

2 =0



Which way is better?

Matching the moments of the two assumptions directly: Assumption #1: E() = 0 Assumption #2: E(x) = 0

Or the implied parameters in terms of the moments 1= Cov(x,y)/Var(x) and 0 = E(y)– 1 E(x)

It is not surprise that both approaches yield the same estimator. b1 = Sxy / Sxx and b0 = my – b1 mx


E(x) = 0 one assumption implies two

By law of iterated expectations E(x) = 0 implies E [E(x) ] = E[ ] = 0

(assumption #1) E(x) = 0 implies E(xx) = 0,

implies E[ E(xx) ] = E[ x ] = 0 (assumption #2)

What does E(x) = 0 mean? E(x) =0 E[ (y – 0 – 1 x) | x] =

E (y|x) –0 – 1 x = E (y|x) = 0 + 1 x

That the expectation of y conditional on x is 0 + 1 x


What if the assumption E(x) = 0 fails?

Still have: E[ ] = 0 (assumption #1)

Need to find another moment conditions, say another variables z such that E[ z ] = 0 (assumption #2’)

instrumental variable


Maximum likelihood

Maximum likelihood estimation (MLE) is a popular statistical method used to make inferences about parameters of the underlying probability distribution from a given data set. That is to say, you have a sample of data x1, x2, …, xn

and you want to infer the distribution of the random variable x.

Commonly, one assumes the data are independent, identically distributed (iid) drawn from a particular distribution with unknown parameters and uses the MLE technique to create estimators for the unknown parameters, that define the distribution.


Maximum likelihood

Given a sample of data x1, x2, …, xn

We would like to find the distribution P0 such that for all feasible distribution P

Prob(P0 | x1,x2,…,xn) Prob(P | x1,x2,…,xn)

For example, the distribution P0 = N(0,02)


An example: Maximum likelihood

Consider tossing an unfair coin 80 times (i.e., we sample something like x1=H, x2=T, ..., x80=T, and count the number of HEADS "H" observed). Call the probability of tossing a HEAD p, and the probability of tossing TAILS 1-p (so here p is θ above).

Suppose we toss 49 HEADS and 31 TAILS, and suppose the coin was taken from a box containing three coins: one which gives HEADS with probability p=1/3, one which gives HEADS with probability p=1/2 and another which gives HEADS with probability p=2/3.

The coins have lost their labels, so we don't know which one it was. Using maximum likelihood estimation we can calculate which coin has the largest likelihood, given the data that we observed.



The likelihood function (defined below) takes one of three values: Pr(H=49 | p=1/3) = C(80,49) (1/3)49 (1-1/3)31≈0.000 Pr(H=49 | p=1/2) = C(80,49) (1/2)49 (1-1/2)31≈0.012 Pr(H=49 | p=2/3) = C(80,49) (2/3)49 (1-2/3)31≈0.054

Which coin is more likely? The one with p=2/3

because among the three possibilities, the observation of the sample data has the highest probability when p=2/3.



Now suppose we had only one coin but its p could have been any value 0 ≤ p ≤ 1. We must maximize the likelihood function:

L(p) = C(80,49) p49 (1-p)31

Over all possible values of 0 ≤ p ≤ 1.

To find the maximum, take the first derivative of L(p) with respect to p and set to zero.

p48(1-p)30(49-80p) = 0 Possible solutions of p are:

P=0, P=1 and p=49/80. L(0)=L(1)=0. L(49/80) >0. Hence the maximum likelihood

estimator for p is 49/80.


L(p) = C(80,49) p49 (1-p)31

over all possible values of 0 ≤ p ≤ 1.


To estimate 0 and 1 using ML

Assume i to be independent identically distributed with normal distribution of zero mean and variance 2. Denote the normal density for i be

f(i)=f(yi-0-1xi)

normal density (which has a rather ugly formula)

The joint likelihood of observing 1, 2, …, n:

L = f(1)*f(2)*…*f(n)

i = yi - 0 - 1xi


To estimate 0 and 1 using ML (Computer)

We do not know 0 and 1. Nor do we know i. In fact, our objective is estimate 0 and 1.

The procedure of ML:

1. Assume a combination of 0 and 1, call it b0 and b1. Compute the implied ei = yi-b0-b1xi and f(ei)=f(yi-b0-b1xi)

2. Compute the joint likelihood conditional on the assumed values of b0 and b1:

L(b0,b1) = f(e1)*f(e2)*…*f(en)

Assume many more combination of 0 and 1, and repeat the above two steps, using a computer program (such as Excel).

Choose the b0 and b1 that yield a largest joint likelihood.


To estimate 0 and 1 using ML (Calculus)

The procedure of ML:

1. Assume a combination of 0 and 1, call it b0 and b1. Compute the implied ei = yi-b0-b1xi and f(ei)=f(yi-b0-b1xi)

2. Compute the joint likelihood conditional on the assumed values of b0 and b1:

L(b0,b1) = f(e1)*f(e2)*…*f(en)

Choose b0 and b1 to maximize the likelihood function L(b0,b1) – using calculus. Take the first derivative of L(b0,b1) with respect to b0, set it

to zero. Take the first derivative of L(b0,b1) with respect to b1, set it

to zero. Solve b0 and b1 using the two equations.


MM versus ML in the estimation of 0 and 1

It turns out that in this special case, the estimators b0 and b1 turn out to the same as the one obtained by method of moments and OLS (to be discussed later).


Alternative assumptionX and Y are bivariate normal

Suppose (X1, X2) ~ BVN(1,2,12,2

2, 12),

The marginal distribution of X1 is normal: X1~N(1,1

2)

The conditional distribution of X2 given x1 is normal:

X2|x1 ~ N( + x1, 2)where = 2-1, = 1

2/12, 2 = 2

2(1-2)=22-21

2

Reference: Goldberger, Arthur S. (1991): A course in Econometrics, Harvard University Press. See page 75.

E(X2|X1)


Estimation Ordinary least squares

For each value of X, there is a group of Y values, and these Y values are normally distributed.

Yi~ N(E(Y|X),i2), i=1,2,…,n

The means of these normal distributions of Y values all lie on the straight line of regression.

E(Y|X) = 0+1X

The standard deviations of these normal distributions are equal.

i2= 2 i=1,2,…,n

i.e., homoskedasticity


Assumptions Underlying Linear Regression

yi and yk are independently drawn from the population, say, as in sampling with replacement. Cov(i,j) = 0 for all i ≠ j

Note that independence implies much more than zero covariance. For two discrete random variables, they are

independent if P(x,y) = P(x)P(y)


Choosing the line that fits best

The question is:Which straight line fits best?

x

y

Lesson10-43 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data3

3

41

1

4

(1,2)

2

2

(2,4)

(3,1.5)

Sum of squared differences = (2 - 1)2 + (4 - 2)2 + (1.5 - 3)2 +

(4,3.2)

(3.2 - 4)2 = 6.89Sum of squared differences = (2 -2.5)2 + (4 - 2.5)2 + (1.5 - 2.5)2 + (3.2 - 2.5)2 = 3.99

2.5

Let us compare two lines

The second line is horizontal

The smaller the sum of squared differences the better the fit of the line to the data. That is, the line with the least sum of squares (of differences) will fit the line best.

The best line is the one that minimizes the sum of squared vertical differences between the points and the line.

Choosing the line that fits best


Choosing the line that fits bestOrdinary Least Squares (OLS) Principle

Straight lines can be described generally by Y = b0 + b1X

Finding the best line with smallest sum of squared difference is the same as

Let b0* and b1

* be the solution of the above problem.

Y* = b0* + b1

*Xis known as the “average predicted value” (or simply “predicted value”) of y for any X.

Min S(b0,b1) = [yi – (b0 + b1xi)]2


Coefficient estimates from the ordinary least squares (OLS) principle

Solving the minimization problem implies the first order conditions:

S(b0,b1) = [yi – (b0 + b1xi)]2

∂S(b0,b1) /∂ b0= (2)[yi – (b0 + b1xi)](-1) = 0

yi – (b0 + b1xi)] = 0

∂S(b0,b1) /∂ b1= (2)[yi – (b0 + b1xi)](-xi) = 0

[yi – (b0 + b1xi)](xi) = 0

ei = 0

ei xi = 0

The same as what we had earlier in the discussion of method of moments.


The consistency of b0 and b1 as estimators of 0 and 1

Same equations implies same estimator: b1 = Sxy / Sxx and b0 = my – b1 mx

The estimators are consistent as long as Sxy, Sxx, my and mx

are consistent estimators of the corresponding population quantities. Sxy converges to Cov(x,y) as sample size increases.

Sxx converges to Var(x) as sample size increases.

my converges to E(y) as sample size increases.

mx converges to E(x) as sample size increases


The b0 and b1 are linear in yi

n

1i iii

n

1i n

1i

2i

i

n

1i

2i

n

1i ii

n

1i n

1i

2i

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1i i

n

1i

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1i

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1i ii1

ycyxx

xx

yxx

xxy

xx

xx

xx

yxx

xx

yxx

xx

yyxxb

)(

)(

)(

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))((

Similarly, it can be shown that b0 is also a linear combination of yi.


The unbiasedness of b0 and b1 as estimators of 0 and 1

The estimator.

b1 = ci yi

E(b1|x) = ci E(yi) = ci E(0+1xi+i) = 0 ci +1 ci xi+ ci E(i) = 0 +1 + = 1

n

1i

2i

ii

)xx(

)xx(c

0)xx(

)xx(

)xx(

)xx(c n

1i

2i

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1iin

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1i

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1ii

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x)xx(

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Best estimator

It can also be shown that the estimators b0 and b1 have the smallest variance among all unbiased estimators that are linear combination of yi.

That is, if there is another unbiased estimator b1’ that is linear combination of yi, say, b1’ = ci’ yi, we must have

Var(b1) Var(b1’)


BLUE

Best: smallest variance Linear: linear combination of yi

Unbiased: E(b0) = 0, E(b1) = 1

Estimator


yi = b0 + b1xi + ei

1. Slope (b1)

Estimated Y changes by b1 for each 1 unit increase in X

y* + y*= b0 + b1(x+1) y*= b1

More generally,y* + y*= b0 + b1(x+x) y*/x = b1

Y-Intercept (b0 )

1. Estimated value of Y when X = 0

Interpretation of Coefficients


EXAMPLE 2 continued from Example 1

Develop a regression equation for the information given in EXAMPLE 1. The information there can be used to estimate the selling price based on the number of pages.

The regression equation is:

Y* = 48.0 + .05143X

The equation crosses the Y-axis at $48. A book with no pages would cost $48.

The slope of the line is .05143. Each additional page costs about $0.05 or five cents.

Note: the sign of the b value and the sign of r will always be the same.


Result part 1

Coefficients Standard Error t Stat P-value

Intercept 48 16.94157037 2.833267456 0.029829

X 0.051428571 0.026998733 1.904851306 0.105458

=0.051428571/ 0.026998733=(b1-0)/Sb1

Pr(t>t Stat)

H0: 1 = 0

b1 Sb1


Example 2 continued from Example 1

We can use the regression equation to estimate values of Y.

The estimated selling price of an 800 page book is $89.14, found by

Y* = 48.0 + .05143X = 48.0 + .05143(800) = 89.14


Standard Error of Estimate (denoted se or Sy.x)

Additional assumption: Var(i) = Var(j) = Var() = 2 for all i,j

Recall yi=0 + 1 xi+ i

Var() = 0 implies y are perfectly linear with x.


More Accurate Estimatorof X, Y Relationship

Less Accurate Estimatorof X, Y Relationship

Scatter Around the Regression Line

Small Var() Large Var()


Standard Error of Estimate (denoted se or Sy.x)

Additional assumption: Var(i) = Var(j) = 2 for all i,j.

Se2 is a estimate of

2.

Hence, it may be interpreted as a measure of the reliability of the estimating equation

A measure of dispersion Measures the variability, or scatter of the observed values

around the regression line

2n

)xbby(

2n

)yy(ss

2i10

n

1ii

2n

1i

*ii

x.ye


Assumptions: Observed Y values are normally distributed

around each estimated value of Y*

Constant variance

se measures the dispersion of the points around the regression line If se = 0, equation is a “perfect” estimator

se may be used to compute confidence intervals of the estimated value

Interpreting the Standard Error of the Estimate


X1

X2

X

Y

f(e)y values are normally distributed around the regression line.

For each x value, the “spread” or variance around the regression line is the same.

Regression Line

Variation of Errors Around the Regression Line


X2

X

Y

Variation of Errors Around the Regression Line

E(y|x)=0+1x

y is distributed normal with mean E(y|x)=0+1x, and variance 2.Confidence intervals may be constructed in the usual fashion.


De

pen

de

nt

Va

ria

ble

( Y

)

Independent Variable (X)

Y = b0 + b1X + 2se

Y = b0 + b1X + 1se

Y = b0 + b1X - 1se

Y = b0 + b1X - 2se

Y = b0 + b1X regression line

2se (95.5% Lie in this Region)

Scatter around the Regression Line

1se (68% Lie in this Region)


Example 3 continued from Example 1 and 2.

Find the standard error of estimate for the problem involving the number of pages in a book and the selling price.

408.1028

)200,397(05143.0)636(48606,51

2n

)xbby(s

2i10

n

1ii

e


Equations for the Interval Estimates

htsy e*

2

1

2

)(

)(1

xx

xx

nh

n

ii

htsy e 1*

Confidence Interval for the Mean of y

Prediction Interval for y

y*=b0 + b1 x

y=b0 + b1 x + e

Minimized when

xx


Confidence Interval Estimate for Mean Response

X

The following factors influence the width of the interval: Std Error, Sample Size, X Value

y* = b0+b1xi


Confidence Interval continued from Example 1, 2 and 3.

For books of 800 pages long, what is that 95% confidence interval for the mean price? This calls for a confidence interval on the

average price of books of 800 pages long.

31.1514.898

)4900(000,150,3

)5.612800(

8

1)408.10(447.214.89

)(

)(1

2

2

2

1

2**

xx

xx

ntsyhtsy

n

ii

ee


Prediction Interval continued from Example 1, 2 and 3.

For a book of 800 pages long, what is the 95% prediction interval for its price? This calls for a prediction interval on the price of an

individual book of 800 pages long.

72.2914.898

)4900(000,150,3

)5.612800(

8

11)408.10(447.214.89

)(

)(111

2

2

2

1

2**

xx

xx

ntsyhtsy

n

ii

ee


1. Tests if there is a linear relationship between X & Y

2. Involves population slope

3. Hypotheses H0: = 0 (no linear relationship)

H1: 0 (linear relationship)

4. Theoretical basis is sampling distribution of slopes

Test of Slope Coefficient (b1)


Sampling Distribution of the Least Squares Coefficient Estimator

If the standard least squares assumptions hold, then b1 is an unbiased estimator of 1 and has a population variance

2

2

1

2

22

)1()(

1

xn

ii

b snxx

and an unbiased sample variance estimator

2

2

1

2

22

)1()(

1

x

en

ii

eb sn

s

xx

ss

n

1i iii

n

1i n

1i

2i

i1 ycy

)xx(

)xx(b


Basis for Inference About the Population Regression Slope

Let 1 be a population regression slope and b1 its least squares estimate based on n pairs of sample observations. Then, if the standard regression assumptions hold and it can also be assumed that the errors i are normally distributed, the random variable

1

11 -

bs

bt

is distributed as Student’s t with (n – 2) degrees of freedom. In addition the central limit theorem enables us to conclude that this result is approximately valid for a wide range of non-normal distributions and large sample sizes, n.


Tests of the Population Regression Slope

If the regression errors i are normally distributed and the standard least squares assumptions hold (or if the distribution of b1 is approximately normal), the following tests have significance value :

1. To test either null hypothesis H0: 1 = 1

* or H0:1 1*

against the alternative H1: 1 > 1

*

The decision rule is to reject if

2),(n-b

*11 t

sβ-b

t1




* or H0:1 > 1*

against the alternative H1: 1 1

*

the decision rule is to reject if

2),(nb

*11 t

sβ-b

t1

-≤




* against the alternative H1: 1 1

*

the decision rule is to reject if

/22),(nb

*11

/22),(nb

*11 t-

sβ-b

t or t s

β-bt

11

-- ≤≥

/22),(n-b

*11 t

sβb

t 1

≥ -

Equivalently


Confidence Intervals for the Population Regression Slope 1

If the regression errors i , are normally distributed and the standard regression assumptions hold, a 100(1 - )% confidence interval for the population regression slope 1 is given by

11 b/22),(n11b/22),(n-1 stbβst-b -


Some cautions about the interpretation of significance tests

Rejecting H0: b1 = 0 and concluding that the relationship between x and y is significant does not enable us to conclude that a cause-and-effect relationship is present between x and y.

Causation requires: Association Accurate time sequence Other explanation for correlation

Correlation Causation Correlation Causation


Some cautions about the interpretation of significance tests

Just because we are able to reject H0: 1 = 0 and demonstrate statistical significance does not enable us to conclude that there is a linear relationship between x and y.

Linear relationship is a very small subset of possible relationship among variables.

A test of linear versus nonlinear relationship requires another batch of analysis.


Variation Measures Coeff. Of Determination Standard Error of Estimate

Test Coefficients for Significance

yi* = b0 +b1xi

Evaluating the Model


Y

X

Y

Xi

Total Sum of Squares (Yi - Y)2

Unexplained Sum of Squares (Yi -Yi

*)2

Explained Sum of Squares (Yi

* - Y)2

Yi

SST

SSE

SSR

yi* = b0 +b1xi

Variation Measures


Total Sum of Squares (SST) Measures variation of observed Yi around the

mean,Y Explained Variation (SSR)

Variation due to relationship between X & Y

Unexplained Variation (SSE) Variation due to other factors

SST=SSR+SSE

Measures of Variation in Regression


Variation in y (SST) = SSR + SSE

n

1i

2i )yy(

n

1i

2**i )yyyy(

n

1i

**i

2*2*i )yy)(yy()yy()yy(

n

1i

**i

n

1i

2*n

1i

2*i )yy)(yy()yy()yy(

n

1i

2*n

1i

2*i )yy()yy(

SST:

SSE SSR

=0, as imposed in the estimation, E(x)=0.


R2 (=r2, the coefficient of determination) measures the proportion of the variation in y that is explained by the variation in x.

n

ii

n

ii

n

ii

n

ii yy

SSR

yy

SSEyy

yy

SSER

1

2

1

2

1

2

1

2

2

)()(

)(

)(1

R2 takes on any value between zero and one. R2 = 1: Perfect match between the line and the data

points. R2 = 0: There are no linear relationship between x and

y.

Variation in y (SST) = SSR + SSE


Summarizing the Example’s results (Example 1, 2 and 3)

The estimated selling price for a book with 800 pages is $89.14.

The standard error of estimate is $10.41.

The 95 percent confidence interval for all books with 800 pages is $89.14 ± $15.31. This means the limits are between $73.83 and $104.45.

The 95 percent prediction interval for a particular book with 800 pages is $89.14 ± $29.72. The means the limits are between $59.42 and $118.86.

These results appear in the following output.


Example 3 continued

Regression Analysis: Price versus Pages

The regression equation isPrice = 48.0 + 0.0514 Pages

Predictor Coef SE Coef T PConstant 48.00 16.94 2.83 0.030Pages 0.05143 0.02700 1.90 0.105

S = 10.41 R-Sq = 37.7% R-Sq(adj) = 27.3%

Analysis of VarianceSource DF SS MS F PRegression 1 393.4 393.4 3.63 0.105Residual Error 6 650.6 108.4Total 7 1044.0


Testing for Linearity

Key Argument: If the value of y does not change linearly with the

value of x, then using the mean value of y is the best predictor for the actual value of y. This implies is preferable.

If the value of y does change linearly with the value of x, then using the regression model gives a better prediction for the value of y than using the mean of y. This implies y=y* is preferable.

yy


Three Tests for Linearity

Testing the Coefficient of CorrelationH0: = 0 There is no linear relationship between x and y.

H1: 0 There is a linear relationship between x and y.

Test Statistic:

1b

1

sb

t

Testing the Slope of the Regression LineH0: = 0 There is no linear relationship between x and y.

H1: 0 There is a linear relationship between x and y.

Test Statistic:

)2/()1( 2

nr

rt


Three Tests for Linearity

The Global F-testH0: There is no linear relationship between x and y.

H1: There is a linear relationship between x and y.

Test Statistic:

)2/()(

)(

)2/(

1/

2

1

*

2

1

*

nyy

yy

nSSE

SSR

MSE

MSRF

n

iii

n

ii

[Variation in y] = SSR + SSE. Large F results from a large SSR. Then, much of the variation in y is explained by the regression model. The null hypothesis should be rejected; thus, the model is valid.

Note: At the level of simple linear regression, the global F-test is equivalent to the t-test on b1. When we conduct regression analysis of multiple variables, the global F-test will take on a unique function.


Purposes Examine Linearity Evaluate violations of assumptions

Graphical Analysis of Residuals Plot residuals versus Xi values

Difference between actual Yi & predicted Yi*

Studentized residuals:Allows consideration for the magnitude of the

residuals

Residual Analysis


Residual Analysis for Linearity

Not Linear LinearOK

X

e e

X

For example, if truth isy = 0 + 1 x + 2x2 + The estimated residuals are likely e=2x2 +


Heteroscedasticity OK Homoscedasticity

Using Standardized Residuals (e/se)

SR

X

SR

X

Residual Analysis for Homoscedasticity

When the requirement of a constant variance (homoscedasticity) is violated we have heteroscedasticity.

For example, for xi>xj

Var(i|xi)>var(j|xj)


Residual Analysis for Independence

Not Independent Independent

X

SR

X

SR

OK

Using Standardized Residuals (e/se)


A time series is constituted if data were collected over time.

Examining the residuals over time, no pattern should be observed if the errors are independent.

When a pattern is detected, the errors are said to be autocorrelated.

Autocorrelation can be detected by graphing the residuals against time.

Non-independence of error variables


+

+++ +

++

++

+ +

++ + +

+

++ +

+

+

+

+

+

+Time

Residual Residual

Time+

+

+

Note the runs of positive residuals,replaced by runs of negative residuals

Note the oscillating behavior of the residuals around zero.

0 0

Patterns in the appearance of the residuals over time indicates that autocorrelation exists.


n

ii

n

iii

e

eeD

1

2

2

21)( Should be close to 2.

If not, examine the model for autocorrelation.

Used when data is collected over time to detect autocorrelation (Residuals in one time period are related to residuals in another period)

Measures Violation of independence assumption

The Durbin-Watson Statistic

Intuition: If x and y are independent, Var(x-y)= Var(x) + Var(y)


An outlier is an observation that is unusually small or large.

Several possibilities need to be investigated when an outlier is observed: There was an error in recording the value. The point does not belong in the sample. The observation is valid.

Identify outliers from the scatter diagram. It is customary to suspect an observation is an

outlier if its |standard residual| > 2

Outliers


+

+

+

+

+ +

+ + ++

+

+

+

+

+

+

+

The outlier causes a shift in the regression line

… but, some outliers may be very influential

++++++++++

An outlier An influential observation


Nonnormality or heteroscedasticity can be remedied using transformations on the y variable.

The transformations can improve the linear relationship between the dependent variable and the independent variables.

Many computer software systems allow us to make the transformations easily.

Remedying violations of the required conditions


A brief list of transformations

Y’ = log y (for y > 0) Use when the se increases with y, or Use when the error distribution is positively skewed

y’ = y2

Use when the se2 is proportional to E(y), or

Use when the error distribution is negatively skewed y’ = y1/2 (for y > 0)

Use when the se2 is proportional to E(y)

y’ = 1/y Use when se

2 increases significantly when y increases beyond some value.


Example: Transformation to get linearity

Not Linear OK

X

e

Linear

e

X

Yi = b0 + b Xi + ei Yi = b0 + b1 Xi + b2 Xi2 +

ei


- END -

Lesson 10:Lesson 10: Regressions Part IRegressions Part I

Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson10-1 Lesson 10: Regressions Part I.

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Transcript of Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson10-1 Lesson 10: Regressions Part I.