Economics 173Business Statistics
Lecture 14
Fall, 2001
Professor J. Petry
http://www.cba.uiuc.edu/jpetry/Econ_173_fa01/
• Hypothesis Test to Evaluate Model: Testing the slope
– When no linear relationship exists between two variables, the regression line should be horizontal.
Linear relationship.Different inputs (x) yielddifferent outputs (y).
No linear relationship.Different inputs (x) yieldthe same output (y).
The slope is not equal to zero The slope is equal to zero
• We can draw inference about 1 from b1 by testingH0: 1 = 0
H1: 1 = 0 (or < 0,or > 0)– The test statistic is
– If the error variable is normally distributed, the statistic is Student t distribution with d.f. = n-2.
1b
11
sb
t
1b
11
sb
t
The standard error of b1.
2x
bs)1n(
ss
1
2x
bs)1n(
ss
1
where
• Solution– Solving by hand– To compute “t” we need the values of b1 and sb1.
49.1300231
0312
00231.688,528,43)(99(
6.151
)1(
312.
1
1
11
2
1
..
s
bt
sn
ss
b
b
x
b
– Using the computerCoefficients Standard Error t Stat P-value
Intercept 6533.383035 84.51232199 77.30687 1.22E-89Odometer -0.031157739 0.002308896 -13.4947 4.44E-24
There is overwhelming evidence to inferthat the odometer reading affects the auction selling price.
• Example– Evaluate the model used in the Armani’s Pizza example by
testing the value of the slope. You are given:
0722.1)1(
89425.10
2
1
1
x
bsn
ss
b
• Armani Pizza Example -- Output
Regression StatisticsMultiple R 0.963366334R Square 0.928074693Adjusted R Square 0.91908403Standard Error 28.13339035
SUMMARY OUTPUT Observations 10
ANOVAdf SS MS F Significance F
Regression 1 81702.49878 81702.49878 103.2264983 7.5384E-06Residual 8 6331.90122 791.4876525Total 9 88034.4
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%Intercept -23.49273678 28.85565265 -0.814146783 0.439120548 -90.0340341 43.04856X Variable 1 10.89424753 1.072263792 10.16004421 7.53839E-06 8.42160119 13.36689
• R2 (Coefficient of Determination) to Evaluate Model
– When we want to measure the strength of the linear relationship, we use the coefficient of determination.
SST
SSR
SST
SSERor
ss
YXR
yx
1)],[cov( 2
22
22
SST
SSR
SST
SSERor
ss
YXR
yx
1)],[cov( 2
22
22
– To understand the significance of this coefficient note:
Overall variability in y
The regression model
Remains, in part, unexplained The error
Explained in part by
x1 x2
y1
y2
y
Two data points (x1,y1) and (x2,y2) of a certain sample are shown.
22
21 )yy()yy( 2
22
1 )yy()yy( 222
211 )yy()yy(
Total variation in y = Variation explained by the regression line)
+ Unexplained variation (error)
• R2 measures the proportion of the variation in y that is explained by the variation in x.
SST
SSR
SST
SSESST
SST
SSER
12
Variation in y (SST) = SSR + SSE
• 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.
Regression StatisticsMultiple R 0.8063R Square 0.6501
Adjusted R Square 0.6466Standard Error 151.57Observations 100
• Example 17.4– Find the coefficient of determination for example 17.1;
what does this statistic tell you about the model?• Solution
– Solving by hand;– Using the computer
• From the regression output we have
6501.ss
)]Y,X[cov(R )999,64)(688,528,43(
]256,356,1[2y
2x
22 2
65% of the variation in the auctionselling price is explained by the variation in odometer reading. Therest (35%) remains unexplained bythis model.
• Example– Find the coefficient of determination for the Armani’s
Pizza example; what does this statistic tell you about the model? You are given values of SSR = 81702.499, and SSE = 6331.901.
• Armani Pizza Example -- Output
Regression StatisticsMultiple R 0.963366334R Square 0.928074693Adjusted R Square 0.91908403Standard Error 28.13339035
SUMMARY OUTPUT Observations 10
ANOVAdf SS MS F Significance F
Regression 1 81702.49878 81702.49878 103.2264983 7.5384E-06Residual 8 6331.90122 791.4876525Total 9 88034.4
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%Intercept -23.49273678 28.85565265 -0.814146783 0.439120548 -90.0340341 43.04856X Variable 1 10.89424753 1.072263792 10.16004421 7.53839E-06 8.42160119 13.36689
Project I: Finance Application• Project simulates the job of professional portfolio managers.• Your job is to advise high net-worth clients on investment decisions. This often involves “educating” the client.• Client is Medical Doctor, with little financial expertise, but serious cash ($1,000,000). • The stock market has declined significantly over the last year or so, and she believes now is the time to get in.• She gives you five stocks and asks which one she should invest in. How do you respond?
Project I: Finance ApplicationThe project is divided into three parts
1. Calculate mean, standard deviation and beta for each asset.– gather 5 years of historical monthly returns for your team’s five companies, the S&P 500 total return index and the 3-month constant maturity Treasury bill. Put in table and explain to client.
2. Illustrate benefits of diversification with concrete example. – You are provided with average annual returns and standard deviations for the S&P 500 and the 3-month T-bill. You create a portfolio with these two instruments weighted differently to illustrate the impact on risk (standard deviation).
3. Analyze impact of correlations on diversification.– Create table of correlation coefficients between all assets, and explain. Select asset pair with r closest to –1. Explain. – Adjust graph in 2 with r = 1, and –1 instead of 0. Explain.
Part 1. Finding Beta (17.6 in text)
• 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 17.5 The market modelSUMMARY OUTPUT
Regression StatisticsMultiple R 0.560079R Square 0.313688Adjusted R Square0.301855Standard Error0.063123Observations 60
ANOVAdf SS MS F Significance F
Regression 1 0.10563 0.10563 26.50969 3.27E-06Residual 58 0.231105 0.003985Total 59 0.336734
CoefficientsStandard Error t Stat P-valueIntercept 0.012818 0.008223 1.558903 0.12446TSE 0.887691 0.172409 5.148756 3.27E-06
• Estimate the market model for Nortel, a stock traded in the Toronto Stock Exchange.
• Data consisted of monthly percentage return for Nortel and monthly percentage returnfor all the stocks.
This is a measure of the stock’smarket related risk. In this sample, for each 1% increase in the TSE return, the average increase in Nortel’s return is .8877%.
This is a measure of the total risk embeddedin the Nortel stock, that is market-related.Specifically, 31.37% of the variation in Nortel’sreturn are explained by the variation in the TSE’s returns.
Part 2. Diversification (Ex 6.8, sect 6.7)Investment portfolio diversification– An investor has decided to invest equal amounts of money
in two investments.
– Find the expected return on the portfolio– If = 1, .5, 0 find the standard deviation of the portfolio.
Mean returnStandard dev.Investment 1 15% 25%Investment 2 27% 40%
Mean returnStandard dev.Investment 1 15% 25%Investment 2 27% 40%
– The return on the portfolio can be represented by Rp = w1R1 + w2R2 = .5R1 + .5R2
The relative weights are proportional to the amounts invested.
– The variance of the portfolio return isV(Rp) = w1
2V(R1) + w22V(R1) +2w1w212
– Calculate the portfolio return and risk for = 0, = 1 and = -1
– Substituting the required coefficient of correlationwe have:
• For = 1 : V(Rp) = .1056 = .3250
• For = .5: V(Rp) = .0806 = .2839
• For = 0: V(Rp) = .0556 = .2358
Larger diversification is expressed bysmaller correlation.As the correlation coefficient decreases,the standard deviation decreases too.
pp
p
Benefits of DiversificationInvestment Proportion Portfolio
Bonds Stocks Expected Return Standard Deviation0% 100% 17.0% 25.0%
20% 80% 15.6% 20.1%40% 60% 14.2% 15.7%60% 40% 12.8% 12.3%80% 20% 11.4% 10.8%
100% 0% 10.0% 12.0%
Investment Opportunity Set
0.0%
2.0%
4.0%
6.0%
8.0%
10.0%
12.0%
14.0%
16.0%
18.0%
0.0% 5.0% 10.0% 15.0% 20.0% 25.0% 30.0%
Risk
Re
turn
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