Multiple Regression

Multiple RegressionChapter 18

18.1 Introduction

• In this chapter we extend the simple linear regression model, and allow for any number of independent variables.

• We expect to build a model that fits the data better than the simple linear regression model.

• We will use computer printout to – Assess the model

• How well it fits the data• Is it useful• Are any required conditions violated?

– Employ the model• Interpreting the coefficients• Predictions using the prediction equation• Estimating the expected value of the dependent variable

Coefficients

Dependent variable Independent variables

Random error variable

18.2 Model and Required Conditions

• We allow for k independent variables to potentially be related to the dependent variable

y = 0 + 1x1+ 2x2 + …+ kxk +

y = 0 + 1xy = 0 + 1xy = 0 + 1xy = 0 + 1x

X

y

X2

1

The simple linear regression modelallows for one independent variable, “x”

y =0 + 1x +

The multiple linear regression modelallows for more than one independent variable.Y = 0 + 1x1 + 2x2 +

Note how the straight line becomes a plain, and...

y = 0 + 1x1 + 2x2

y = 0 + 1x1 + 2x2

y = 0 + 1x1 + 2x2

y = 0 + 1x1 + 2x2y = 0 + 1x1 + 2x2

y = 0 + 1x1 + 2x2

y = 0 + 1x1 + 2x2

X

y

X2

1

… a parabola becomes a parabolic surface

y= b0+ b1x2

y = b0 + b1x12 + b2x2

b0

• Required conditions for the error variable – The error is normally distributed with mean equal

to zero and a constant standard deviation

(independent of the value of y). is unknown.– The errors are independent.

• These conditions are required in order to – estimate the model coefficients,– assess the resulting model.

– If the model passes the assessment tests, use it to interpret the coefficients and generate predictions.

– Assess the model fit and usefulness using the model statistics.

– Diagnose violations of required conditions. Try to remedy problems when identified.

18.3 Estimating the Coefficients and Assessing the Model

• The procedure– Obtain the model coefficients and statistics using a

statistical computer software.

– La Quinta Motor Inns is planning an expansion.– Management wishes to predict which sites are likely to be

profitable.– Several areas where predictors of profitability can be

identified are:• Competition• Market awareness• Demand generators• Demographics• Physical quality

Example 18.1 Where to locate a new motor inn?

Profitability

Competition Market awareness Customers Community Physical

Margin

Rooms Nearest Officespace

Collegeenrollment

Income Disttwn

Distance to downtown.

Medianhouseholdincome.

Distance tothe nearestLa Quinta inn.

Number of hotels/motelsrooms within 3 miles from the site.

– Data was collected from randomly selected 100 inns that belong to La Quinta, and ran for the following suggested model:

Margin =Rooms NearestOfficeCollege

+ 5Income + 6Disttwn + INN MARGIN ROOMS NEAREST OFFICE COLLEGE INCOME DISTTWN1 55.5 3203 0.1 549 8 37 12.12 33.8 2810 1.5 496 17.5 39 0.43 49 2890 1.9 254 20 39 12.24 31.9 3422 1 434 15.5 36 2.75 57.4 2687 3.4 678 15.5 32 7.96 49 3759 1.4 635 19 41 4

SUMMARY OUTPUT

Regression StatisticsMultiple R 0.724611R Square 0.525062Adjusted R Square0.49442Standard Error5.512084Observations 100

ANOVAdf SS MS F Significance F

Regression 6 3123.832 520.6387 17.13581 3.03E-13Residual 93 2825.626 30.38307Total 99 5949.458

CoefficientsStandard Error t Stat P-value Lower 95%Upper 95%Intercept 72.45461 7.893104 9.179483 1.11E-14 56.78049 88.12874ROOMS -0.00762 0.001255 -6.06871 2.77E-08 -0.01011 -0.00513NEAREST -1.64624 0.632837 -2.60136 0.010803 -2.90292 -0.38955OFFICE 0.019766 0.00341 5.795594 9.24E-08 0.012993 0.026538COLLEGE 0.211783 0.133428 1.587246 0.115851 -0.05318 0.476744INCOME -0.41312 0.139552 -2.96034 0.003899 -0.69025 -0.136DISTTWN 0.225258 0.178709 1.260475 0.210651 -0.12962 0.580138

• Excel outputThis is the sample regression equation (sometimes called the prediction equation)

MARGIN = 72.455 - 0.008ROOMS -1.646NEAREST + 0.02OFFICE +0.212COLLEGE - 0.413INCOME + 0.225DISTTWN

Let us assess this equation

• Standard error of estimate– We need to estimate the standard error of estimate

– Compare s to the mean value of y• From the printout, Standard Error = 5.5121 • Calculating the mean value of y we have

– It seems s is not particularly small. – Can we conclude the model does not fit the data well?

1knSSE

s

739.45y

• Coefficient of determination– The definition is

– From the printout, R2 = 0.5251– 52.51% of the variation in the measure of profitability is

explained by the linear regression model formulated above.– When adjusted for degrees of freedom,

Adjusted R2 = 1-[SSE/(n-k-1)] / [SS(Total)/(n-1)] = = 49.44%

2i

2

)yy(

SSE1R

• Testing the validity of the model– We pose the question:

Is there at least one independent variable linearly related to the dependent variable?

– To answer the question we test the hypothesis

H0: 1 = 2 = … = k = 0

H1: At least one i is not equal to zero.

– If at least one i is not equal to zero, the model is valid.

• To test these hypotheses we perform an analysis of variance procedure.

• The F test – Construct the F statistic

– Rejection regionF>F,k,n-k-1

MSEMSR

F

MSR=SSR/k

MSE=SSE/(n-k-1)

[Variation in y] = SSR + SSE. Large F results from a large SSR. Then, much of the variation in y is explained by the regression model. The null hypothesis shouldbe rejected; thus, the model is valid.

Required conditions mustbe satisfied.



• Excel provides the following ANOVA results

Example 18.1 - continued

SSESSR

MSEMSR

MSR/MSE



• Excel provides the following ANOVA results

Example 18.1 - continued

F,k,n-k-1 = F0.05,6,100-6-1=2.17F = 17.14 > 2.17

Also, the p-value (Significance F) = 3.03382(10)-13

Clearly, = 0.05>3.03382(10)-13, and the null hypothesisis rejected.

Conclusion: There is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. At least one of the i is not equal to zero. Thus, at least one independent variable is linearly related to y. This linear regression model is valid

• Let us interpret the coefficients

– This is the intercept, the value of y when all the variables take the value zero. Since the data range of all the independent variables do not cover the value zero, do not interpret the intercept.

– In this model, for each additional 1000 rooms within 3 mile of the La Quinta inn, the operating margin decreases on the average by 7.6% (assuming the other variables are held constant).

5.72b

0076.b1

– In this model, for each additional mile that the nearest competitor is to La Quinta inn, the average operating margin decreases by 1.65%

– For each additional 1000 sq-ft of office space, the average increase in operating margin will be .02%.

– For additional thousand students MARGIN

increases by .21%.

– For additional $1000 increase in median

household income, MARGIN decreases by .41%

– For each additional mile to the downtown

center, MARGIN increases by .23% on the average

65.1b2

02.b3

21.b4

41.b5

23.b6

CoefficientsStandard Error t Stat P-value Lower 95% Upper 95%Intercept 72.45461 7.893104 9.179483 1.11E-14 56.78048735 88.12874ROOMS -0.00762 0.001255 -6.06871 2.77E-08 -0.010110582 -0.00513NEAREST -1.64624 0.632837 -2.60136 0.010803 -2.902924523 -0.38955OFFICE 0.019766 0.00341 5.795594 9.24E-08 0.012993085 0.026538COLLEGE 0.211783 0.133428 1.587246 0.115851 -0.053178229 0.476744INCOME -0.41312 0.139552 -2.96034 0.003899 -0.690245235 -0.136DISTTWN 0.225258 0.178709 1.260475 0.210651 -0.12962198 0.580138

• Testing the coefficients– The hypothesis for each i

– Excel printout

H0: i = 0H1: i = 0

Test statistic

d.f. = n - k -1ib

iis

bt

• Using the linear regression equation

– The model can be used by• Producing a prediction interval for the particular value of y,

for a given set of values of xi.• Producing an interval estimate for the expected value of y,

for a given set of values of xi.

– The model can be used to learn about relationships between the independent variables xi, and the dependent variable y, by interpreting the coefficients i

• Example 18.1 - continued. Produce predictions

– Predict the MARGIN of an inn at a site with the following characteristics:• 3815 rooms within 3 miles,• Closet competitor 3.4 miles away,• 476,000 sq-ft of office space,• 24,500 college students,• $39,000 median household income,• 3.6 miles distance to downtown center.

MARGIN = 72.455 - 0.008(3815) -1.646(3.4) + 0.02(476) +0.212(24.5) - 0.413(39) + 0.225(3.6) = 37.1%

• The required conditions for the model assessment to apply must be checked.

– Is the error variable normally distributed?

– Is the error variance constant?– Are the errors independent?– Can we identify outliers?– Is multicollinearity a problem?

18.4 Regression Diagnostics - II

Draw a histogram of the residuals

Plot the residuals versus y

Plot the residuals versus the time periods

• Example 18.2 House price and multicollinearity

– A real estate agent believes that a house selling price can be predicted using the house size, number of bedrooms, and lot size.

– A random sample of 100 houses was drawn and data recorded.

– Analyze the relationship among the four variables

Price Bedrooms H Size Lot Size124100 3 1290 3900218300 4 2080 6600117800 3 1250 3750

. . . .

. . . .



Regression 3 7.65E+10 2.55E+10 40.7269 4.57E-17Residual 96 6.01E+10 6.26E+08Total 99 1.37E+11

CoefficientsStandard Error t Stat P-value Lower 95%Upper 95%Intercept 37717.59 14176.74 2.660526 0.009145 9576.963 65858.23Bedrooms 2306.081 6994.192 0.329714 0.742335 -11577.3 16189.45H Size 74.29681 52.97858 1.402393 0.164023 -30.8649 179.4585Lot Size -4.36378 17.024 -0.25633 0.798244 -38.1562 29.42862

• Solution• The proposed model is

PRICE = 0 + 1BEDROOMS + 2H-SIZE +3LOTSIZE + – Excel solution

The model is valid, but no variable is significantly relatedto the selling price !!

– when regressing the price on each independent variable alone, it is found that each variable is strongly related to the selling price.

– Multicollinearity is the source of this problem.Price Bedrooms H Size Lot Size

Price 1Bedrooms 0.645411 1H Size 0.747762 0.846454 1Lot Size 0.740874 0.83743 0.993615 1

• Multicollinearity causes two kinds of difficulties:– The t statistics appear to be too small.– The coefficients cannot be interpreted as “slopes”.

• However,

• Remedying violations of the required conditions

– Nonnormality or heteroscedasticity can be remedied using transformations on the y variable.

– The transformations can improve the linear relationship between the dependent variable and the independent variables.

– Many computer software systems allow us to make the transformations easily.

• A brief list of transformations» y’ = log y (for y > 0)

• Use when the s increases with y, or• Use when the error distribution is positively skewed

» y’ = y2

• Use when the s2 is proportional to E(y), or

• Use when the error distribution is negatively skewed» y’ = y1/2 (for y > 0)

• Use when the s2 is proportional to E(y)

» y’ = 1/y• Use when s2

increases significantly when y increases beyond some value.

• Example 18.3: Analysis, diagnostics, transformations.

– A statistics professor wanted to know whether time limit affect the marks on a quiz?

– A random sample of 100 students was split into 5 groups.– Each student wrote a quiz, but each group was given a

different time limit. See data below.Time 40 45 50 55 60

20 24 26 30 3223 26 25 32 31. . . . .. . . . .

Marks Analyze these results, and include diagnostics

0

10

20

30

40

50

-2.5 -1.5 -0.5 0.5 1.5 2.5 MoreSUMMARY OUTPUT

Regression StatisticsMultiple R 0.86254R Square 0.743974Adjusted R Square 0.741362Standard Error 2.304609Observations 100


Regression 1 1512.5 1512.5 284.7743 9.42E-31Residual 98 520.5 5.311224Total 99 2033

CoefficientsStandard Error t Stat P-value Lower 95%Upper 95%Intercept -2.2 1.64582 -1.33672 0.184409 -5.46608 1.066077Time 0.55 0.032592 16.87526 9.42E-31 0.485322 0.614678

This model is useful andprovides a good fit.

The errors seem to benormally distributed

The model tested:MARK = 0 + 1TIME +

-3

-2-1

0

1

23

4

20 22 24 26 28 30 32

Standardized errors vs. predicted mark.

The standard error of estimate seems to increase with the predicted value of y.

Two transformations are used to remedy this problem:1. y’ = logey2. y’ = 1/y

Let us see what happens when a transformation is applied

Mark

15

20

25

30

35

40

0 20 40 60 80

LogMark

2

3

4

0 20 40 60 80

40,18

40,2340, 3.135

40, 2.89

Loge23 = 3.135

Loge18 = 2.89

The original data, where “Mark” is a function of “Time”

The modified data, where LogMark is a function of “Time"

The new regression analysis and the diagnostics are:

The model tested:LOGMARK = ’0 + ’1TIME + ’

SUMMARY OUTPUT




CoefficientsStandard Error t Stat P-value Lower 95%Upper 95%Intercept 2.129582 0.0603 35.31632 1.51E-57 2.009918 2.249246Time 0.021716 0.001194 18.18566 3.58E-33 0.019346 0.024086

Predicted LogMark = 2.1295 + .0217Time

This model is useful andprovides a good fit.

The errors seem to benormally distributed

Standard Residuals

-4

-2

0

2

4

2.9 3 3.1 3.2 3.3 3.4 3.5

0

10

20

30

40

-2.5 -1.5 -0.5 0.5 1.5 2.5 More

The standard errors still changes with the predicted y, but the change is smaller than before.

Let TIME = 55 minutes

LogMark = 2.1295 + .0217Time = 2.1295 + .0217(55) = 3.323

To find the predicted mark, take the antilog:

antiloge3.323 = e3.323 = 27.743

How do we use the modified model to predict?

18.5 Regression Diagnostics - III

• The Durbin - Watson Test– This test detects first order auto-correlation between

consecutive residuals in a time series– If autocorrelation exists the error variables are not

independent

4d0isdofrangeThe

r

)rr(

dn

1i

2i

n

2i

21ii

Residual at time i

+

+

+

+

++

+

++

+ Residuals

Time

Positive first order autocorrelation occurs when consecutive residuals tend to be similar. Then,the value of d is small (less than 2).

Positive first order autocorrelation

Negative first order autocorrelation

+

+

+

+

0

0

Residuals

Time

+

Negative first order autocorrelation occurs when consecutive residuals tend to markedly differ. Then, the value of d is large (greater than 2).

• One tail test for positive first order auto-correlation– If d<dL there is enough evidence to show that positive first-order

correlation exists– If d>dU there is not enough evidence to show that positive first-

order correlation exists– If d is between dL and dU the test is inconclusive.

• One tail test for negative first order auto-correlation– If d>4-dL, negative first order correlation exists

– If d<4-dU, negative first order correlation does not exists

– if d falls between 4-dU and 4-dL the test is inconclusive.

• Two-tail test for first order auto-correlation– If d<dL or d>4-dL first order auto-correlation exists

– If d falls between dL and dU or between 4-dU and 4-dL the test is inconclusive

– If d falls between dU and 4-dU there is no evidence for first order auto-correlation

dL dU 20 44-dU 4-dL

First ordercorrelationexists

First ordercorrelationexists

Inconclusivetest

Inconclusivetest

First ordercorrelationdoes notexist

First ordercorrelationdoes notexist

– How does the weather affect the sales of lift tickets in a ski resort?

– Data of the past 20 years sales of tickets, along with the total snowfall and the average temperature during Christmas week in each year, was collected.

– The model hypothesized was

TICKETS=0+1SNOWFALL+2TEMPERATURE+ – Regression analysis yielded the following results:

• Example 18.4

SUMMARY OUTPUT

Regression StatisticsMultiple R 0.3464529R Square 0.1200296Adjusted R Square 0.0165037Standard Error 1711.6764Observations 20

ANOVAdf SS MS F Signif. F

Regression 2 6793798.2 3396899.1 1.1594 0.3372706Residual 17 49807214 2929836.1Total 19 56601012

CoefficientsStandard Error t Stat P-value Lower 95% Upper 95%Intercept 8308.0114 903.7285 9.1930391 5E-08 6401.3083 10214.715Snowfall 74.593249 51.574829 1.4463111 0.1663 -34.22028 183.40678Tempture -8.753738 19.704359 -0.444254 0.6625 -50.32636 32.818884

The model seems to be very poor:

• The fit is very low (R-square=0.12),• It is not valid (Signif. F =0.33)• No variable is linearly related to Sales

Diagnosis of the required conditions resulted with the following findings

-4000-3000-2000-1000

0100020003000

7500 8500 9500 10500 11500 12500

-4000-3000-2000-1000

0100020003000

0 5 10 15 20 25

Residual over time

Residual vs. predicted y

The errors are not independent

The error variance is constant

01234567

-2.5 -1.5 -0.5 0.5 1.5 2.5 More

The errors may benormally distributed

The error distribution

Residuals

-4000

-2000

0

2000

4000

0 5 10 15 20 25

Test for positive first order auto-correlation:n=20, k=2. From the Durbin-Watson table we have:dL=1.10, dU=1.54. The statistic d=0.59

Conclusion: Because d<dL , there is sufficient evidence to infer that positive first order auto-correlation exists.

Durbin-Watson Statistic-2793.99-1723.23 d = 0.5931-2342.03-956.955-1963.73

.

.

Using the computer - ExcelTools > data Analysis > Regression (check the residual option and then OK)Tools > Data Analysis Plus > Durbin Watson Statistic > Highlight the range of the residualsfrom the regression run > OK

The residuals

The modified regression model

TICKETS=0+ 1SNOWFALL+ 2TEMPERATURE+ 3YEARS+

The autocorrelation has occurred over time.Therefore, a time dependent variable added to the model may correct the problem

• All the required conditions are met for this model.

• The fit of this model is high R2 = 0.74.

• The model is useful. Significance F = 5.93 E-5. • SNOWFALL and YEARS are linearly related to ticket sales.

• TEMPERATURE is not linearly related to ticket sales.

Multiple Regression

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