Simple Linear Regression

1
Simple Linear Regression

• Chapter 17

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17.1 Introduction

• In Chapters 17 to 19 we examine the relationship
  between interval variables via a mathematical
  equation.
• The motivation for using the technique
• Forecast the value of a dependent variable (y)
  from the value of independent variables (x1,
  x2,xk.).
• Analyze the specific relationships between the
  independent variables and the dependent variable.

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17.2 The Model
The model has a deterministic and a probabilistic
components
House Cost
Building a house costs about 75 per square
foot.
House cost 25000 75(Size)
Most lots sell for 25,000
House size
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17.2 The Model
However, house cost vary even among same size
houses!
Since cost behave unpredictably, we add a random
component.
House Cost
Most lots sell for 25,000
House cost 25000 75(Size)
e
House size
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17.2 The Model

• The first order linear model
• y dependent variable
• x independent variable
• b0 y-intercept
• b1 slope of the line
• e error variable

b0 and b1 are unknown populationparameters,
therefore are estimated from the data.
y
Rise
b1 Rise/Run
Run
b0
x
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17.3 Estimating the Coefficients

• The estimates are determined by
• drawing a sample from the population of interest,
• calculating sample statistics.
• producing a straight line that cuts into the data.

y
w
Question What should be considered a good line?
w
w
w
w
w w w w
w
w w
w w
w
x
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The Least Squares (Regression) Line
A good line is one that minimizes the sum of
squared differences between the points and the
line.
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The Least Squares (Regression) Line
Sum of squared differences
(2 - 1)2
(4 - 2)2
(1.5 - 3)2
(3.2 - 4)2 6.89
Let us compare two lines
(2,4)
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The second line is horizontal
w
(4,3.2)
w
3
2.5
2
w
(1,2)
(3,1.5)
w
The smaller the sum of squared differences the
better the fit of the line to the data.
3
4
2
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The Estimated Coefficients
The regression equation that estimates the
equation of the first order linear model is
To calculate the estimates of the line
coefficients, that minimize the differences
between the data points and the line, use the
formulas
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Example 17.2 (Xm17-02)
The Simple Linear Regression Line

• A car dealer wants to find the relationship
  between the odometer reading and the selling
  price of used cars.
• A random sample of 100 cars is selected, and the
  data recorded.
• Find the regression line.

Independent variable x
Dependent variable y
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The Simple Linear Regression Line

• Solution
• Solving by hand Calculate a number of statistics

where n 100.
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The Simple Linear Regression Line

• Solution continued
• Using the computer (Xm17-02)

Tools gt Data Analysis gt Regression gt Shade the
y range and the x range gt OK
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The Simple Linear Regression Line
Xm17-02
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Interpreting the Linear Regression -Equation
17067
No data
0
This is the slope of the line. For each
additional mile on the odometer, the price
decreases by an average of 0.0623
The intercept is b0 17067.
Do not interpret the intercept as the Price of
cars that have not been driven
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17.4 Error Variable Required Conditions

• The error e is a critical part of the regression
  model.
• Four requirements involving the distribution of e
  must be satisfied.
• The probability distribution of e is normal.
• The mean of e is zero E(e) 0.
• The standard deviation of e is se for all values
  of x.
• The set of errors associated with different
  values of y are all independent.

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The Normality of e
The standard deviation remains constant,
m3
m2
but the mean value changes with x
m1
From the first three assumptions we have y is
normally distributed with mean E(y) b0 b1x,
and a constant standard deviation se
x1
x2
x3
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17.5 Assessing the Model

• The least squares method will produces a
  regression line whether or not there are linear
  relationship between x and y.
• Consequently, it is important to assess how well
  the linear model fits the data.
• Several methods are used to assess the model. All
  are based on the sum of squares for errors, SSE.

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Sum of Squares for Errors

• This is the sum of differences between the points
  and the regression line.
• It can serve as a measure of how well the line
  fits the data. SSE is defined by

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Standard Error of Estimate

• The mean error is equal to zero.
• If se is small the errors tend to be close to
  zero (close to the mean error). Then, the model
  fits the data well.
• Therefore, we can, use se as a measure of the
  suitability of using a linear model.
• An estimator of se is given by se

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Standard Error of Estimate,Example

• Example 17.3
• Calculate the standard error of estimate for
  Example 17.2, and describe what does it tell you
  about the model fit?
• Solution

Calculated before
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Testing the slope

• When no linear relationship exists between two
  variables, the regression line should be
  horizontal.

q
q
Linear relationship.
Linear relationship.
Linear relationship.
Linear relationship.
No linear relationship. Different inputs (x)
yield the same output (y).
Different inputs (x) yield different outputs (y).
The slope is not equal to zero
The slope is equal to zero
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Testing the Slope

• We can draw inference about b1 from b1 by testing
• H0 b1 0
• H1 b1 0 (or lt 0,or gt 0)
• The test statistic is
• If the error variable is normally distributed,
  the statistic is Student t distribution with d.f.
  n-2.

where
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Testing the Slope,Example

• Example 17.4
• Test to determine whether there is enough
  evidence to infer that there is a linear
  relationship between the car auction price and
  the odometer reading for all three-year-old
  Tauruses, in Example 17.2. Use a 5.

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Testing the Slope,Example

• Solving by hand
• To compute t we need the values of b1 and
  sb1.
• The rejection region is t gt t.025 or t lt -t.025
  with n n-2 98.Approximately, t.025 1.984

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Testing the Slope,Example
Xm17-02

• Using the computer

There is overwhelming evidence to infer that the
odometer reading affects the auction selling
price.
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Coefficient of determination

• To measure the strength of the linear
  relationship we use the coefficient of
  determination.

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Coefficient of determination

• To understand the significance of this
  coefficient note

The regression model
Overall variability in y
The error
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Coefficient of determination
y2
Two data points (x1,y1) and (x2,y2) of a certain
sample are shown.
Variation in y SSR SSE
y1
x1
x2
Total variation in y
Variation explained by the regression line
Unexplained variation (error)
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Coefficient of determination

• R2 measures the proportion of the variation in y
  that is explained by the variation in x.

• 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.

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Coefficient of determination,Example

• Example 17.5
• Find the coefficient of determination for Example
  17.2 what does this statistic tell you about the
  model?
• Solution
• Solving by hand

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Coefficient of determination

• Using the computer From the regression
  output we have

65 of the variation in the auction selling price
is explained by the variation in odometer
reading. The rest (35) remains unexplained
by this model.
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17.6 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 b0 b1Rm e

Rate of return on a particular stock
Rate of return on some major stock index
The beta coefficient measures how sensitive the
stocks rate of return is to changes in the
level of the overall market.
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The Market Model, Example
Example 17.6 (Xm17-06)

• Estimate the market model for Nortel, a stock
  traded in the Toronto Stock Exchange (TSE).
• Data consisted of monthly percentage return for
  Nortel and monthly percentage return for all the
  stocks.

This is a measure of the stocks market related
risk. In this sample, for each 1 increase in
the TSE return, the average increase in Nortels
return is .8877.
This is a measure of the total market-related
risk embedded in the Nortel stock. Specifically,
31.37 of the variation in Nortels return are
explained by the variation in the TSEs returns.
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17.7 Using the Regression Equation

• Before using the regression model, we need to
  assess how well it fits the data.

• If we are satisfied with how well the model fits
  the data, we can use it to predict the values of
  y.
• To make a prediction we use
• Point prediction, and
• Interval prediction

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Point Prediction

• Example 17.7
• Predict the selling price of a three-year-old
  Taurus with 40,000 miles on the odometer (Example
  17.2).

• It is predicted that a 40,000 miles car would
  sell for 14,575.
• How close is this prediction to the real price?

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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.

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Interval Estimates,Example

• Example 17.7 - 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

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Interval Estimates,Example

• Solution
• A prediction interval provides the price estimate
  for a single car

t.025,98 Approximately
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Interval Estimates,Example

• 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)

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

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

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

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

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

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

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17.8 Coefficient of Correlation

• The coefficient of correlation is used to measure
  the strength of association between two
  variables.
• The coefficient values range between -1 and 1.
• If r -1 (negative association) or r 1
  (positive association) every point falls on the
  regression line.
• If r 0 there is no linear pattern.
• The coefficient can be used to test for linear
  relationship between two variables.

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Testing the coefficient of correlation

• To test the coefficient of correlation for linear
  relationship between X and Y
• X and Y must be observational
• X and Y are bivariate normally distributed

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Testing the coefficient of correlation

• When no linear relationship exist between the two
  variables, r 0.
• The hypotheses are
• H0 r 0H1 r ¹ 0
• The test statistic is

The statistic is Student t distributed with d.f.
n - 2, provided the variables are bivariate
normally distributed.
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Testing the Coefficient of correlation

• Foreign Index Funds (Index)
• A certain investor prefers the investment in an
  index mutual funds constructed by buying a wide
  assortment of stocks.
• The investor decides to avoid the investment in a
  Japanese index fund if it is strongly correlated
  with an American index fund that he owns.
• From the data shown in Index.xls should he avoid
  the investment in the Japanese index fund?

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Testing the Coefficient of correlation

• Foreign Index Funds
• A certain investor prefers the investment in an
  index mutual funds constructed by buying a wide
  assortment of stocks.
• The investor decides to avoid the investment in a
  Japanese index fund if it is strongly correlated
  with an American index fund that he owns.
• From the data shown in Index.xls should he avoid
  the investment in the Japanese index fund?

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Testing the Coefficient of Correlation,Example

• Solution
• Problem objective Analyze relationship between
  two interval variables.
• The two variables are observational (the return
  for each fund was not controlled).
• We are interested in whether there is a linear
  relationship between the two variables, thus, we
  need to test the coefficient of correlation

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Testing the Coefficient of Correlation,Example

• Solution continued
• The hypothesesH0 r 0H1 r ¹ 0.
• Solving by hand
• The rejection regiont gt ta/2,n-2 t.025,59-2
  2.000.
• The sample coefficient of correlation Cov(x,y)
  .001279 sx .0509 sy 0512 r
  cov(x,y)/sxsy.491

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Testing the Coefficient of Correlation,Example

• Excel solution (Index)

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Spearman Rank Correlation Coefficient

• The Spearman rank test is a nonparametric
  procedure.
• The procedure is used to test linear
  relationships between two variables when the
  bivariate distribution is nonnormal.
• Bivariate nonnormal distribution may occur when
• at least one variable is ordinal, or
• both variables are interval but at least one
  variable is not normal.

52
Spearman Rank Correlation Coefficient

• The hypotheses are
• H0 rs 0
• H1 rs ¹ 0
• The test statistic iswhere a and b are the
  ranks of x and y respectively.

• For a large sample (n gt 30) rs is approximately
  normally distributed

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Spearman Rank Correlation Coefficient,Example

• Example 17.8 (Xm17-08)
• A production manager wants to examine the
  relationship between
• Aptitude test score given prior to hiring, and
• Performance rating three months after starting
  work.
• A random sample of 20 production workers was
  selected. The test scores as well as performance
  rating was recorded.

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Spearman Rank Correlation Coefficient,Example
Scores range from 0 to 100
Scores range from 1 to 5
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Spearman Rank Correlation Coefficient,Example

• Solution
• The problem objective is to analyze the
  relationship between two variables.(Note
  Performance rating is ordinal.)

• The hypotheses are
• H0 rs 0
• H1 rs 0

• The test statistic is rs, and the rejection
  region is rs gt rcritical (taken from the
  Spearman rank correlation table).

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Spearman Rank Correlation Coefficient,Example
Ties are broken by averaging the ranks.

• Solving by hand
• Rank each variable separately.

• Calculate sa 5.92 sb 5.50 cov(a,b) 12.34
• Thus rs cov(a,b)/sasb .379.
• The critical value for a .05 and n 20 is .450.

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Spearman Rank Correlation Coefficient,Example
Conclusion Do not reject the null
hypothesis. At 5 significance level there is
insufficient evidence to infer that the two
variables are related to one another.
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Spearman Rank Correlation Coefficient,Example

• Excel Solution (Data Analysis Plus Xm17-08)

gt 0.05
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17.9 Regression Diagnostics - I

• The three conditions required for the validity of
  the regression analysis are
• 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?

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Residual Analysis

• Examining the residuals (or standardized
  residuals), help detect violations of the
  required conditions.
• Example 17.2 continued
• Nonnormality.
• Use Excel to obtain the standardized residual
  histogram.
• Examine the histogram and look for a bell shaped.
  diagram with a mean close to zero.

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Residual Analysis
A Partial list of Standard residuals
For each residual we calculate the standard
deviation as follows
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Residual Analysis
It seems the residual are normally distributed
with mean zero
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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.

Residual

















y








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Homoscedasticity

• When the requirement of a constant variance is
  not violated we have a condition of
  homoscedasticity.
• Example 18.2 - continued

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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 autocorrelated.
• Autocorrelation can be detected by graphing the
  residuals against time.

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Non Independence of Error Variables
Patterns in the appearance of the residuals over
time indicates that autocorrelation exists.
Residual
Residual














0
0

Time
Time













Note the runs of positive residuals, replaced by
runs of negative residuals
Note the oscillating behavior of the residuals
around zero.
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Outliers

• 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 gt 2

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An influential observation
An outlier


but, some outliers may be very influential














The outlier causes a shift in the regression line
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Procedure for Regression Diagnostics

• Develop a model that has a theoretical basis.
• Gather data for the two variables in the model.
• Draw the scatter diagram to determine whether a
  linear model appears to be appropriate.
• Determine the regression equation.
• Check the required conditions for the errors.
• Check the existence of outliers and influential
  observations
• Assess the model fit.
• If the model fits the data, use the regression
  equation.