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Simple Linear Regression Chapter 18

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Title: Simple Linear Regression Chapter 18


1
Lecture 17-18
  • Simple Linear Regression (Chapter 18)
  • Homework 5 due Tue Apr 1st 3pm.

2
Coefficient of determination
  • To measure the strength of the linear
    relationship we use the coefficient of
    determination.

3
Coefficient of determination
  • To understand the significance of this
    coefficient note

The regression model
Overall variability in y
The error
4
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)
5
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.

6
Coefficient of determination,Example
  • Example 18.5
  • Find the coefficient of determination for Example
    18.2 what does this statistic tell you about the
    model?
  • Solution
  • Solving by hand

7
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.
8
18.9 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?

9
Residual Analysis
  • Examining the residuals (or standardized
    residuals), help detect violations of the
    required conditions.
  • Example 18.2 continued
  • Nonnormality.
  • Use JMP to save the residuals and to obtain the
    Normal Quantile Plot along with the Historgram.
  • Examine the histogram and look for a bell shaped.
    diagram with a mean close to zero.
  • The points should fall along the straight line in
    the Normal Quantile plot.

10
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11
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 or by plotting
    the residual against X variable.

Residual

















y








12
Homoscedasticity
  • When the requirement of a constant variance is
    not violated we have a condition of
    homoscedasticity.
  • Example 18.2 - continued

13
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.

14
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.
15
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

16
An influential observation
An outlier


but, some outliers may be very influential














The outlier causes a shift in the regression line
17
Outlier, Leverage, Influential Points
  • Residual The vertical distance of a point from
    the fitted regression line.
  • Outlier -- An atypical observation, either in the
    horizontal (X) or vertical (Y) direction.
  • Leveraged Observations that are unusual in the
    X direction.
  • Influential Leveraged points which also possess
    substantial residual value in the vertical
    direction (Y) are influential.

18
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.

19
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 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.
20
The Market Model, Example
Example 18.6 (Xm18-06)
  • 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.
21
18.7 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

22
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?

23
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.

24
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

25
Interval Estimates,Example
  • Solution
  • A prediction interval provides the price estimate
    for a single car

t.025,98 Approximately
26
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)

27
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.

28
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.

29
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.

30
18.8 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.

31
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

32
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.
33
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?

34
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?

35
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

36
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

37
Testing the Coefficient of Correlation,Example
  • Excel solution (Index)

38
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.

39
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

40
Spearman Rank Correlation Coefficient,Example
  • Example 18.8 (Xm18-08)
  • 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.

41
Spearman Rank Correlation Coefficient,Example
Scores range from 0 to 100
Scores range from 1 to 5
42
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).

43
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.

44
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.
45
Spearman Rank Correlation Coefficient,Example
  • Excel Solution (Data Analysis Plus Xm18-08)

gt 0.05
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