Lecture 20

• Simple linear regression (18.6, 18.9)• Homework 5 is posted and is due next

Tuesday at 3 p.m. (Note correction on question 4(e)).

• Regular office hours: Tuesday, 9-10, 12-1.• Extra office hours: Today (after class),

Monday, 10-11.• Midterm 2: Wednesday, April 2nd, 6-8 p.m.

Point Prediction

• Example 18.7– Predict the selling price of a three-year-old Taurus

with 40,000 miles on the odometer (Example 18.2).

– It is predicted that a 40,000 miles car would sell for $14,575.

– How close is this prediction to the real price?

575,14)000,40(0623.17067x0623.17067y A point prediction

Interval Estimates• Two intervals can be used to discover how closely the

predicted value will match the true value of y.– Prediction interval – predicts y for a given value of x,– Confidence interval – estimates the average y for a given x.

– The confidence interval

– The confidence interval

2x

2g

2 s)1n()xx(

n1

sty

2x

2g

2 s)1n()xx(

n1

sty

– The prediction interval– The prediction interval

2x

2g

2 s)1n()xx(

n1

1sty

2x

2g

2 s)1n()xx(

n1

1sty

Interval Estimates,Example

• Example 18.7 - continued – Provide an interval estimate for the bidding

price on a Ford Taurus with 40,000 miles on the odometer.

– Two types of predictions are required:• A prediction for a specific car

• An estimate for the average price per car

Interval Estimates,Example

• Solution– A prediction interval provides the price estimate for a

single car:

2x

2g

2 s)1n()xx(

n1

1sty

605575,14690,528,43)1100(

)009,36000,40(

100

11)1.303(984.1)]40000(0623.067,17[

2

t.025,98

Approximately

• Solution – continued– A confidence interval provides the estimate of

the mean price per car for a Ford Taurus with 40,000 miles reading on the odometer.

• The confidence interval (95%) =

2

i

2g

2)xx(

)xx(

n1

sty

70575,14690,528,43)1100(

)009,36000,40(

100

1)1.303(984.1)]40000(0623.067,17[

2

Interval Estimates,Example

– As xg moves away from x the interval becomes longer. That is, the shortest interval is found at

2x

2g

2 s)1n(

)xx(

n1

sty

x

g10 xbby

The effect of the given xg on the length of the interval

x

x1x)1x( 1x)1x(

g10 xbby

)1xx(y g )1xx(y g

1x 1x

– As xg moves away from the interval becomes longer. That is, the shortest interval is found at

The effect of the given xg on the length of the interval

2x

2g

2 s)1n(

)xx(

n1

sty

2x

2

2 s)1n(1

n1

sty

xx

x

– As xg moves away from the interval becomes longer. That is, the shortest interval is found at .

g10 xbby

2x)2x( 2x)2x(

2x 2x

2x

2g

2 s)1n()xx(

n1

sty

2x

2

2 s)1n(1

n1

sty

2x

2

2 s)1n(2

n1

sty

The effect of the given xg on the length of the interval

xx

Caveat about Prediction

• Remember that predicting y based on x from a regression is only reliable if x falls inside the range of the data observed.

• Extrapolation is dangerous.

Predicting Height Based on Age

7 6

7 8

8 0

8 2

8 4

Heigh

t

1 7 .5 2 0 2 2 .5 2 5 2 7 .5 3 0

A g e

H e i g h t i n c e n t i m e t e r s , A g e i n m o n t h s L in e a r F it

L i n e a r F i t H e i g h t = 6 4 . 9 2 8 3 2 2 + 0 . 6 3 4 9 6 5 A g e S u m m a r y o f F i t

R S q u a r e 0 . 9 8 8 7 6 4 R o o t M e a n S q u a r e E r r o r 0 . 2 5 5 9 6 4 P a r a m e t e r E s t i m a t e s T e r m E s t i m a t e S t d E r r o r t R a t i o P r o b > | t |

I n t e r c e p t 6 4 . 9 2 8 3 2 2 0 . 5 0 8 4 1 1 2 7 . 7 1 < . 0 0 0 1 A g e 0 . 6 3 4 9 6 5 0 . 0 2 1 4 0 5 2 9 . 6 6 < . 0 0 0 1

18.9 Regression Diagnostics - I

• The four conditions required for the validity of the simple linear regression analysis are:– the mean of the error variable conditional on x is zero

for each x

– the error variable is normally distributed.

– the error variance is constant for all values of x.

– the errors are independent of each other.

• How can we diagnose violations of these conditions?

Residual Analysis

• Examining the residuals helps detect violation of the required conditions

• A residual plot is a scatterplot of the regression residuals against another variable, usually the independent variable or time.

• If the simple linear regression model holds, there should be no pattern in the residual plots.

• Don’t read too much into these plots. You’re looking for gross departures from a random scatter.

Residual plot for utopia.jmp

• Utopia.jmp is a simulation from a simple linear regression model (all assumptions hold).

-2

0

2

Re

sid

ua

l

-10 -5 0 5 10

X

Residual Plot for Example 18.2

-800

-4000

400800

Re

sid

ua

l

15000 25000 35000 45000

Odometer

Detecting Curvature

• If the residual plot has a curved pattern, this indicates that the regression function is not a straight line.

• Transformations to deal with the problem of a curved regression function rather than a straight line regression function later in the lecture.

Heteroscedasticity• When the requirement of a constant variance is violated

we have a condition of heteroscedasticity.• Diagnose heteroscedasticity by plotting the residual

against the predicted y.

+ + ++

+ ++

++

+

+

+

+

+

+

+

+

+

+

++

+

+

+

The spread increases with y

y

Residualy

+

+++

+

++

+

++

+

+++

+

+

+

+

+

++

+

+

Homoscedasticity• When the requirement of a constant variance is not violated we

have a condition of homoscedasticity.• Example 18.2 - continued

-1000

-500

0

500

1000

13500 14000 14500 15000 15500 16000

Predicted Price

Re

sid

ua

ls

Residual plot for cleaning.jmp

-20

-100

1020

Re

sid

ua

l

0 5 10 15

NumberOfCrews

Non Independence of Error Variables

– 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 serially correlated (or autocorrelated)

– Serial correlation can be detected by graphing the residuals against time.

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

+

+++ +

++

++

+ +

++ + +

+

++ +

+

+

+

+

+

+Time

Residual Residual

Time+

+

+

Note the runs of positive residuals,replaced by runs of negative residuals. Positive serial correlation.

Note the oscillating behavior of the residuals around zero. Negative serial correlation.

0 0

Non Independence of Error Variables

Checking normality

• To check the normality of the error variable, draw a histogram of the residuals.

• Violation of normality only has a serious effect on confidence intervals and tests if the sample size is small (less than 30) and there is either strong skewness or outliers.

-800 -400 0 200 600

Outliers• An outlier is an observation that is unusually small or large.• Three types of outliers in scatterplots:

– Outlier in x direction– Outlier in y direction– Outlier in overall direction of scatterplot (residual has large

magnitude)• 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 scatterplot

Leverage and Influential Points

• An observation has high leverage if it is an outlier in the x direction.

• An observation is influential if removing it would markedly change the least squares line.

• Observations that have high leverage are influential if they do not fall very close to the least squares line for the other points.

+

+

+

+

+ +

+ + ++

+

+

+

+

+

+

+

The outlier causes a shift in the regression line

… but, some outliers may be very influential

++++++++++

An outlier An influential observation

Regression of Brain Weight on Body Weight for 96 Mammals

Bivariate Fit of Brain weight By Body Weight

0

1000

2000

3000

4000

Brain

weig

ht

0 5001000150020002500

Body Weight

Transformations

• Suppose that the residual plot indicates curvature in the regression function. What do we do?

• One possibility: Transform x or transform y.

• Tukey’s Bulging Rule (see Handout).

Transformation for display.jmp

• Y=Sales, X=Display Feet

• Y=Sales, X=Square Root of Display Feet/Log of Display Feet

-100

0

100

Re

sid

ua

l

0 1 2 3 4 5 6 7 8

DisplayFeet

-100

-500

50100

Re

sid

ua

l

1 1.5 2 2.5

Square Root DisplayFeet

-50

0

50

100

Re

sid

ua

l

0 .5 1 1.5 2

Log DisplayFeet

Predictions with Transformations

• Linear Fit

• Sales = -46.28718 + 154.90188 Square Root DisplayFeet

• For 5 display feet, the average amount of sales is

08.300590.15429.46

18.6 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.

R = 0 + 1Rm +

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.

Example 18.6B i v a r i a t e F i t o f N o r t e l B y T S E

-0 .2

-0 .1

0 .0

0 .1

0 .2

Nortel

-0 .2 5 -0 .1 5 -0 .0 5 .0 0 .0 5 .1 0

T S E

L in e a r F it

L i n e a r F i t N o r t e l = 0 . 0 1 2 8 1 8 1 + 0 . 8 8 7 6 9 1 2 T S E S u m m a r y o f F i t R S q u a r e 0 . 3 1 3 6 8 8 R S q u a r e A d j 0 . 3 0 1 8 5 5 R o o t M e a n S q u a r e E r r o r 0 . 0 6 3 1 2 3 P a r a m e t e r E s t i m a t e s T e r m E s t i m a t e S t d E r r o r t R a t i o P r o b > | t |

I n t e r c e p t 0 . 0 1 2 8 1 8 1 0 . 0 0 8 2 2 3 1 . 5 6 0 . 1 2 4 5 T S E 0 . 8 8 7 6 9 1 2 0 . 1 7 2 4 0 9 5 . 1 5 < . 0 0 0 1