# CS代考 BUSANA 7001 – Predictive and Visual Analytics for Business – cscodehelp代写

Basics Simple regression Regression assumptions Multiple regression
BUSANA 7001 – Predictive and Visual Analytics for Business
Week 4: Predictive analytics using multiple regressions
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Basics Simple regression Regression assumptions Multiple regression
Simple regression
Regression assumptions
Multiple regression
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Introduction
The purpose of quantitative analysis is to 􏰂nd or test certain relations.
Correlation coe􏰆cients shed some light on the direction of the linear relation:
• positive
• negative or
• no linear relation.
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Let’s investigate the relation between car price and car length using SAS provided data set.
/* Creating data file: */
DATA work.car_data;
SET SAShelp.Cars;
/* Correlation coefficient: */
proc corr data=work.car_data;
var invoice length;
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Example II
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Example III
Correlation coe􏰆cient between car price and length is 0.16659 (p-val.=0.0005).
=⇒ The relation is positive and statistically signi􏰂cant.
However, correlation coe􏰆cient does not let us estimate or predict
the car price given certain car length:
• e.g., what is approximately price of 180-inch length car?
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Introduction II
One needs to use regression analysis in order to answer this question!
Regressions:
Simple regression: y = β0 + β1x
Multiple linear regression: y = β0 + β1×1 + β2×2 + · · · + βnxn
y is the dependent variable
x, x1, x2, xn β0
β1, β2, βn
are independent variables is intercept
are slopes.
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Simple regression
Let’s regress using SAS car price on car length:
• INVOICE = f(LENGTH) = intercept + slope × LENGTH.
/* OLS regression: */
PROC REG DATA=work.car_data;
MODEL invoice=length;
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Simple regression II
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Simple regression III
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Simple regression IV
No obvious trends or patterns in the residuals.
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Simple regression V
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95% con􏰂dence interval vs. 95% prediction interval
Con􏰂dence intervals tell you how well you have determined the mean. Assume that the data are randomly sampled from a normal distribution and you are interested in
determining the mean. If you sample many times, and calculate a con􏰂dence interval of the mean from each sample, you’d expect 95% of those intervals to include the true value of the population mean.
Prediction intervals tell you where you can expect to see the next data point sampled. Assume that the data are
randomly sampled from a normal distribution. Collect a sample of data and calculate a prediction interval. Then sample one more value from the population. If you repeat this process many times, you’d expect the prediction interval to capture the individual value 95% of the time.
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Simple regression: Interpretation of the results
We got that:
• intercept = 􏰅8131.76
• slope = 204.69
• INVOICE = 􏰅8131.76 + 204.69 × LENGTH.
Car price increases by \$204.69 for each additional inch of car length.
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Simple regression: Predictions
Suppose we would like to estimate the price of 180-inch length car: • INVOICE = 􏰅8131.76 + 204.69 × LENGTH
• INVOICE = 􏰅8131.76 + 204.69 × 180 = \$28,712.4
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Simple regression: Predictions II
Let’s check the actual prices of 180-inch length cars:
PROC PRINT DATA=work.car_data (obs=20);
var make model length invoice;
where length = 180;
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Simple regression: Predictions III
=⇒ most of the cars are more expensive than our predicted value.
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R-squared is a goodness of 􏰂t or accuracy measure.
The higher the R-squared, the better the model.
R-squared is the ratio of the variation explained to the total variation (of the dependent variable).
0 ≤ R-squared ≤ 1.
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R-squared and correlation
• R-squared of the regression model = 0.0278 • Correlation coe􏰆cient = 0.16659.
Correlation2 = R-squared 0.166592 = 0.0278
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R-squared and correlation II
R-squared of the regression model = 0.0278
=⇒ Car length can explain only 2.78% of variation in car prices: • explanatory power of the model is weak
• there should be other factors that better explain car prices.
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Predicted values and residuals
Predicted values of car prices (INVOICEpred) can be computed from:
• INVOICEpred = 􏰅8131.76 + 204.69 × LENGTH, where LENGTH is the actual car length.
Residuals (INVOICEres) are then computed as: • INVOICEres = INVOICE 􏰅 INVOICEpred,
where INVOICE is the actual car price.
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Predicted values and residuals II
Consider the following 180-inch length cars: • LX: \$18,630
• Chevrolet Corvette 2dr: \$39,068.
The predicted car prices are the same: \$28,712.4.
The residuals are:
• LX: 􏰅\$10,082.4
• Chevrolet Corvette 2dr: \$10,355.6.
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Predicted values and residuals III
We can compute predicted values and residuals manually or we can ask SAS to do this:
PROC REG DATA=work.car_data;
MODEL invoice=length;
OUTPUT OUT=work.reg_results r=Res p=Pred;
`Pred’ are predicted values `Res’ description are residuals.
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Regression analysis using SAS Visual Analytics
One should use `Linear regression’ object (from `SAS Visual Statistics’ list).
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Assumptions of linear regression
1. Linear relationship 2. Normality
3. Independence
4. Homoscedasticity
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1. Linear relationship
The relation between the independent variable(x) and the dependent variable (y) is linear.
Detecting a linear relationship is fairly simple.
In most cases, linearity is clear from the scatterplot.
Relevant SAS code:
/* Scatterplot: */
proc gplot data=work.car_data;
title ‘Scatter plot of Invoice and length’;
plot Invoice * Length=1;
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1. Linear relationship II
=⇒ no obvious non-linear relation found.
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2. Normality
The dependent variable y is distributed normally for each value of the independent variable x.
Outliers are the main reason for the violation of this assumption.
To check for the normality, one can use:
• scatterplots (y vs x)
• histogram of standardized residuals
• A normal probability versus residual probability distribution plot (P-P plot).
=⇒ there are a few outliers.
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3. Independence
The values of y should depend on independent variables but not on its own previous values.
The violation of this assumption is observed mostly in time series data (e.g., gross domestic product (GDP))
Autocorrelation coe􏰆cients for di􏰁erent lags can help detect dependencies:
• correlation between GDP and GDP lagged by 1 period • correlation between GDP and GDP lagged by 2 periods • correlation between GDP and GDP lagged by 3 periods • correlation between GDP and GDP lagged by 4 periods.
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4. Homoscedasticity
The variance in y is the same at each stage of x.
There is no special segment or an interval in x where the dispersion in y is distinct.
Scatterplots (y vs x) can be used to detect heteroscedasticity (which is opposite of homoscedasticity).
In our example, the variance is higher when car is 175-205 inch long but statistical tests imply that the model does not su􏰁er from heteroscedasticity
Plots with the residual versus predicted values can also be used to detect heteroscedasticity.
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4. Homoscedasticity II
The simple way to deal with heteroscedasticity is to segment out the data and build di􏰁erent regression lines for di􏰁erent intervals.
In general, if the 􏰂rst three assumptions are satis􏰂ed, then heteroscedasticity might not even exist.
As a rule of thumb, 􏰂rst three assumptions need to be 􏰂xed before attempting to 􏰂x heteroscedasticity.
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4. Homoscedasticity III
To detect heteroscedasticity, one can use White’s and Breusch-Pagan tests.
Procedure MODEL (rather than REG) includes them.
Relevant SAS code (both procedures give the same results):
PROC REG DATA=work.car_data;
MODEL invoice=EngineSize;
PROC MODEL DATA=work.car_data;
PARMS a1 b1;
invoice = a1 + b1 * EngineSize;
FIT invoice / WHITE PAGAN=(1 EngineSize);
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Tests’ results:
4. Homoscedasticity IV
=⇒ H0 of no heteroscedasticity is rejected.
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Solutions:
4. Homoscedasticity V
• adjust standard errors (a.k.a., (heteroskedasticity) robust standard errors, White-Huber standard errors etc.)
• transform non-normally distributed variables (e.g., using natural logarithm).
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PROC REG DATA=work.car_data;
MODEL invoice=EngineSize / ACOV;
This option can be used in SAS procedure REG only.
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The robust standard errors can be used even under homoskedasticity.
Then the robust standard errors will become just 􏰃regular􏰄 standard errors.
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Log-transformed variables
Variables with positive values can be log-transformed.
DATA work.car_data;
SET work.car_data;
log_MSRP=log(MSRP);
log_length=log(length);
BUSANA 7001, Week 4
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Log-transformed variables II
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Log-transformed variables III
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Log-transformed variables IV
Let’s estimate 3 regression models and predict MSRP of a 180 inches long car.
PROC REG DATA=work.car_data;
MODEL log_MSRP= length;
PROC REG DATA=work.car_data;
MODEL MSRP= log_length;
PROC REG DATA=work.car_data;
MODEL log_MSRP= log_length;
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Basics Simple regression Regression assumptions Multiple regression
Presentation of results
SAS does not present regression results properly.
We should manually make tables.
We should also make table descriptions that include:
• variable de􏰂nitions
• that t-statistics based on standard errors are reported in brackets.
• that ***, **, and * indicate signi􏰂cance at 1%, 5%, and 10% levels, respectively.
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We get the following results:
Log-transformed variables V
Dependent variable:
Independent variables LENGTH
ln(LENGTH) Intercept
ln(MSRP) Model 1
0.0095*** [6.05]
8.4922*** [28.83]
0.079 0.077 428
MSRP Model 2
44,467*** [3.70] 􏰅199,551*** [3.18]
0.031 0.029 428
ln(MSRP) Model 3
1.8039*** [6.16] 0.8446 [0.55]
0.082 0.080 428
Number of observations
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Description for the previous table
Table 1: Determinants of car prices
This table presents the results of OLS regressions where the dependent variable is the manufacturer suggested retail price (MSRP) or its natural logarithm. LENGTH is a car length in inches. The absolute values of t-statistics based on standard errors are reported in brackets. ***, **, and * indicate signi􏰂cance at 1%, 5%, and 10% levels, respectively.
The description above should be above the table.
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Log-transformed variables VI
Let’s compute predicted values of MRSP for 180 inches long car.
ln(180) ≈ 5.1930
Model 1: ln(MSRP) = 8.4922 + 0.0095 × 180 ≈ 10.2022. MSRP = exp(10.2022) ≈ 26,962.44
Model 2: MSRP = 􏰅199551 + 44,467 × 5.1930 ≈ 31,364.21 Model 3: ln(MSRP) = 0.8446 + 1.8039 × 5.1930 ≈ 10.2122.
MSRP = exp(10.2122) ≈ 27,232.73
Models 1 and 3 are preferred (one of the reasons is higher R2). The di􏰁erence between the predictions of Models 1 and 3 is 1%. However, the di􏰁erence between the predictions of Models 2 and 3 is 15%.
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When linear regression can’t be used
If any of the 4 assumptions are not satis􏰂ed, the linear regressions should not be used.
Linear regression can’t be used when:
• the relation between y and x is nonlinear
• the errors are not normally distributed
• there is a dependency within the values of the dependent variable
• the variance pattern of y is not the same for the entire range of x.
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Multiple regression
Let’s investigate the relation between the discount and car length.
It is likely that the discount is impacted by many other factors besides car length:
• car manufacturer (`make’) • car model (`model’)
• car type (`type’)
• drivetrain (`drivetrain’)
• production place (`origin’)
• engine (`enginesize’, `cylinders’, `horsepower’)
• fuel e􏰆ciency (`MPG_City’, `MPG_Highway’)
• and maybe car weight (`weight’) and wheel base (`wheelbase’).
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Multiple regression II
If our regression model does not include any of the important independent variables, then the model su􏰁ers from the omitted variable bias.
OLS estimator is likely to be biased and inconsistent due to omitted variable bias.
=⇒ coe􏰆cient estimates might become unreliable.
One should include as many variables as possible in the regression
model if the data is available.
I found that some variables in the previous slide cannot be included in the model together with car length (due to multicollinearity).
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is a phenomenon due to a high interdependency between the independent variables.
If we include highly correlated independent variables in the same regression model, then this could cause multicollinearity.
Implications of multicollinearity:
• it increases the variance of the coe􏰆cient estimates and make the estimates very sensitive to minor changes in the model
• coe􏰆cient estimates might become unstable and di􏰆cult to interpret
• T-test results are not trustworthy etc.
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Multiple regression III
First, let’s check the summary statistics of the discount.
/* Creating discount variable: */
DATA work.car_data;
SET work.car_data;
discount=msrp/invoice-1;
proc univariate data=work.car_data plots;
var discount;
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Properties of DISCOUNT
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Properties of DISCOUNT II
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Properties of DISCOUNT III
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Properties of DISCOUNT IV
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Correlation matrix
Let’s look at the correlation matrix of discount and other numerical variables:
proc corr data=work.car_data;
var discount invoice enginesize cylinders horsepower
MPG_City MPG_Highway Length weight wheelbase;
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Correlation matrix II
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Determinants of DISCOUNT
Let’s consider engine size as potential determinant for discount.
SAS code for scatterplot:
proc gplot data=work.car_data;
title ‘Scatter plot of discount and engine size’;
plot Discount * Enginesize=1;
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Determinants of DISCOUNT II
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