16 Review of Part II

8/17/2019 16 Review of Part II

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REVIEW OF PART II

Topics Outline

Inference for Regression

Multiple Regression

Building Regression Models

Terms and Concepts

1. Equation of a regression model 18. ANOVA table

2. Equation of the true population line/plane/surface 19. Sums of Squares (SST, SSR, SSE)

3. Equation of the least squares regression

line/plane/surface

20. Mean squares (MSR, MSE)

4. Least squares criterion 21. Collinearity

5. Regression coefficients and their interpretation 22. Variance inflation factor (VIF )

6. Correlation r and its interpretation 23. Correlation matrix

7. Coefficient of determination 2r and its interpretation 24. Validation of the fit

8. Regression standard error e s and its interpretation 25. Dummy (indicator) variables

9. Confidence intervals for the intercept ( ) and for

the regression slopes ( ’s)26. Interaction (cross-product) terms

10. The overall F test27. Nonlinear transformations

(quadratic and logarithmic)

11. t test for a regression slope 28. The partial F test

12. Confidence interval for a mean response 29. Adjusted 2r

13. Prediction interval for an individual response 30. pC statistic

14. Residuals and residual plots 31. Include/exclude decisions

15. Normal probability plot (Q-Q plot) 32. Principle of parsimony

16. Regression assumptions and how to check them 33. Building regression models

17. Excel regression output (what each cell represents

and what the connections between the cells are)

34. Variable selection procedures

(forward selection, backward elimination,

stepwise regression, best subsets regression


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Example 1 Florida reappraises real estate every year, so the county appraiser’s Web site lists the current

“fair market value” of each piece of property. Property usually sells for somewhat more than theappraised market value. Data for the appraised market values and actual selling prices(in thousands of dollars) of 16 condominium units sold in a beachfront building over a 19-month

period are stored in the file Condominiums.xlsx.

Condominium Selling Price Appraised Value

1 850 758.02 900 812.7

15 1325 1031.816 845 586.7

Excel output for a linear regression of selling price on appraised value is shown on the next page.

(a) Write the equation for the model of the population regression line.

x y

(b) Write the equation of the true population regression line.

x y

(c) What is the equation of the least-squares regression line for predicting selling pricefrom appraised value?

x y 0466.127.127ˆ

(d) What is the correlation between appraised value and selling price?

The correlation r is the square root of .2

r 93.0861.0 r

(We take the positive square root because the sign of r must be the same as the sign ofthe slope, 1.0466.)

Reminder: For simple and multiple regression, r is the correlation between the observed values of y

and the predicted values ŷ . For simple linear regression, r is also the correlation between x and y.

(e) Explain why the pattern you see on the residual plot agrees with the conditions oflinear relationship and constant standard deviation needed for regression inference.

On the residual plot, as usual a horizontal line is added at residual zero, the mean of the residuals.This line corresponds to the regression line in the plot of selling price against appraised value.The residuals show a random scatter about the line, with roughly equal vertical spread acrosstheir range. This is what we expect when the conditions for regression inference hold.

(f) Does the histogram of the residuals suggest lack of normality?

The distribution of residuals has a bit of a cluster at the left, but there are no outliers orother strong deviations from normality that would prevent regression inference.


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Regression Statistics

Multiple R 0.9277

R Square 0.8606

Adjusted R Square 0.8506

Standard Error 69.7299

Observations 16

ANOVA

df SS MS F Significance F

Regression 1 420072.1418 420072.1418 86.3945 0.0000

Residual 14 68071.6082 4862.2577

Total 15 488143.7500

Coefficients Standard Error t Stat P-value Lower 95% Upper 95%

Intercept 127.2705 79.4892 1.6011 0.1317 -43.2168 297.7578

Appraised Value 1.0466 0.1126 9.2949 0.0000 0.8051 1.2881

y = 1.0466x + 127.27

R² = 0.8606

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P r i c e

Appraised Value

Fitted Line Plot

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Residuals versus Appraised Value

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Residual

Histogram of the Residuals


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(g) How many degrees of freedom does the t distribution used for statistical inference on these data have

There are n = 16 data pairs, so df = n – 1 – 1 = 16 – 2 = 14.

Reminder:

df = n – k – 1, where n is the number of observations, k is the number of explanatory variables.

(h) Explain what the slope of the true regression line means in this setting.

is the average rate of increase in selling price in a population of condominium units when

appraised value increases by $1,000.

(i) Find a 95% confidence interval for the population slope .

bSE t b * = 1.0466 ± (2.145)( 0.1126) = 1.0466 ± 0.2415 = 0.8051 to 1.2881

(j) Is there significant evidence that selling price increases as appraised value increases?

To test0:0 H

0: a H

we use the test statistic bSE

bt 9.2949 with df = 14. The P -value is 0.000.

We reject the null hypothesis and conclude that there is a positive linear relationship betweenselling prices and appraised values of beach houses in Florida.

(k) What is the selling price of a house appraised at $802,600?

2712.967)600.802(0466.127.127ˆ y

The selling price of a house appraised at $802,600 is $967,271.

(l) What is a 95% interval for the mean selling price of a unit appraised at $802,600?(Use SE mean = 21.6387.)

For a mean selling price, we should use the confidence interval :

415.462712.967)6387.21(145.22712.967**ˆ meanSE t y

$1,014,000 to000,921$

(m) Hamada owns a unit in this building appraised at $802,600.

What is a 95% interval for this appraised value? (Use SE ind = 73.0102.)

For a range of values for an individual unit, we should use the prediction interval .Hamada can be 95% confident that her unit would sell for

6069.1562712.967)0102.73(145.22712.967**ˆ ind SE t y

$1,124,000 to000,$811


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Example 2

Demand and Cost for Electricity

The Public Service Electric Company produces different quantities of electricity each month,depending on the demand. The file Cost_of_Power.xlsx lists the number of Units of electricity

produced and the total Cost (in dollars) of producing these units for a 36-month period.

Month Cost Units

1 45623 6012 46507 738

35 45218 70536 45357 637

(a) What does the scatterplot of Cost versus Units reveal about the relationship between Cost and Units?

The scatterplot indicates a definite positive relationship and one that is nearly linear.However, there is also some evidence of curvature in the plot. The points increase slightlyless rapidly as Units increases from left to right. In economic terms, there might beeconomies of scale, so that the marginal cost of electricity decreases as more units ofelectricity are produced.

(b) The output for a simple linear regression is shown on the next page.Does the residual plot suggest the need for a nonlinear transformation?

The residuals to the far left and the far right are all negative, whereas the majority of theresiduals in the middle are positive. This negative-positive-negative behavior of residualssuggests a parabola. Admittedly, the pattern is far from a perfect parabola because there areseveral negative residuals in the middle. However, this plot certainly suggests nonlinear behavior and exploring a quadratic relationship with the square of Units included in theequation is reasonable.

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Units

Scatterplot of Cost vs Units


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Regression output for the linear model

MultipleR-Square

Adjusted StErr of

Summary R R-Square Estimate

0.8579 0.7359 0.7282 2733.7

Degrees of Sum of Mean ofF-Ratio p-Value

ANOVA Table Freedom Squares Squares

Explained 1 708085273.8 708085273.8 94.7481 < 0.0001

Unexplained 34 254093815.2 7473347.506

CoefficientStandard

t-Value p-ValueConfidence Interval 95%

Regression Table Error Lower Upper

Constant 23651.5 1917.1 12.3369 < 0.0001 19755.4 27547.6

Units 30.533 3.137 9.7339 < 0.0001 24.158 36.908

(c) The regression output for estimating a quadratic relationship between Cost and Units isshown on the next page. What is the estimated regression equation? Does it provide a betterfit than the linear equation?

The estimated regression equation is

Predicted Cost = 5792.80 + 98.350Units – 0.0600(Units)2

The graph of the regression equation superimposed on the scatterplot of Cost versus Unitsshows a reasonably good fit, plus an obvious curvature.

The quadratic model provides a better fit as indicated by the coefficient of determination 2r

which has increased from 73.6% to 82.2% and the standard error of estimate e s which has

decreased from $2,734 to $2,281.

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Fit

Scatterplot of Residual vs Fit


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Regression output for the quadratic modelMultiple

R-SquareAdjusted StErr of


0.9064 0.8216 0.8108 2280.800

Degrees of Sum of Mean of F-Ratio p-Value


Explained 2 790511518.3 395255759.1 75.9808 < 0.0001

Unexplained 33 171667570.7 5202047.597

CoefficientStandard



Constant 5792.7983 4763.0585 1.2162 0.2325 -3897.7171 15483.3137

Units 98.3504 17.2369 5.7058 0.0000 63.2817 133.4191

(Units)^2 -0.0600 0.0151 -3.9806 0.0004 -0.0906 -0.0293

(d) Interpret the regression coefficients.

The interpretation of the y intercept is that the predicted cost for zero units of electricity produced is $5,792.80.

There is no easy way to interpret the slope coefficients in a quadratic equation. For example,you can't conclude from the 98.35 coefficient of Units that Cost increases by 98.35 dollarswhen Units increases by one. The reason is that when Units increases by one, (Units)2 doesn'tstay constant; it also increases. You can provide instead a qualitative description of the

relationship between x and y from the signs of the coefficients1

b and2

b . For example if1

b is

positive and2

b is negative (as it is in our example), then y increases as x increases, but the

marginal rate of change decreases as x increases.

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C o s t

Units

Quadratic Fit


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(e) Test the significance of the quadratic effect.

To test

0: 20 H (Including the quadratic term does not significantly improve the model.)

0: 20 H (Including the quadratic term significantly improves the model.)

we use the test statistic t = – 3.98 with df = n – k – 1 = 36 – 2 – 1 = 33 and P -value = 0.0004.The small P -value indicates that the quadratic effect is significant.

Notes:

1. The coefficient of (Units)2, 0.0600 is negative and it makes the parabola bend downward.This produces the decreasing marginal cost behavior, where every extra unit of electricityincurs a smaller cost. Actually, the curve described by the regression equation eventuallygoes downhill for large values of Units, but this part of the curve is irrelevant because the

company evidently never produces such large quantities.

2. You should not be fooled by the small magnitude of this coefficient.Remember that it is the coefficient of Units squared , which is a large quantity.Therefore, the effect of the product 0.0600(Units)2 is sizable.

(f) To examine the possibility for a logarithmic fit, a new variable – Log(Units), the naturallogarithm of Units – has been created. The output from a regression of Cost againstLog(Units) is shown on the next page. Interpret the slope of the regression line.

The estimated regression equation is

Predicted Cost = – 63993 + 16654 Log (Units)

Reminder: If b is the coefficient of the log of x, then the expected change in y when x increases by 1% is approximately 0.01 times b.

In the present case, you can interpret the slope coefficient as follows.Suppose that Units increases by 1%, for example, from 600 to 606.Then the regression equation implies that the expected Cost will increase by approximately(0.01)(16654) = 166.54 dollars.In words, every 1% increase in Units is accompanied by an expected $166.54 increase in Cost.

Note that for larger values of Units, a 1% increase represents a larger absolute increase(from 700 to 707 instead of from 600 to 606, say). But each such 1% increase entails the sameincrease in Cost. This is another way of describing the decreasing marginal cost property.


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Regression output for the logarithmic model

MultipleR-Square

Adjusted StErr of


0.8931 0.7977 0.7917 2392.8



Explained 1 767506900.9 767506900.9 134.0471 < 0.0001

Unexplained 34 194672188.1 5725652.59

CoefficientStandard



Constant -63993.3 9144.3 -6.9981 < 0.0001 -82576.8 -45409.8

Log(Units) 16653.6 1438.4 11.5779 < 0.0001 13730.4 19576.7

(g) Compare the quadratic and the logarithmic fits.

To the naked eye, the logarithmic curve appears to be similar, and about as good a fit,

as the quadratic curve. However, the values of 2r , adjusted 2r , and e s indicate that the

logarithmic fit is not quite as good as the quadratic fit:

Model 2r 2

ad jr e s

Quadratic 82.2% 81.1% $2,281

Logarithmic 79.8% 79.2% $2,393

The advantage of the logarithmic equation is that it is easier to interpret.

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Logarithmic Fit


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Example 3

Meddicorp

Meddicorp Company sells medical supplies to hospitals, clinics, and doctors’ offices. The company currently markets in three regions of the United States: the South, the West,and the Midwest. These regions are each divided into many smaller sales territories.

Meddicorp management is concerned with the effectiveness of a new bonus program.This program is overseen by regional sales managers and provides bonuses to salespeople basedon performance. Management wants to know if the bonuses paid in 2010 were related to sales.(Obviously, if there is a relationship here, the managers expect it to be a direct – positive – one.)In determining whether this relationship exists, they also want to take into account the effects ofadvertising, market share, and competitor’s sales. The variables to be used in the study include:

y = Sales – Meddicorp sales (in thousands of dollars) in each territory for 2010

1 x = Adv – the amount Meddicorp spent on advertising in each territory (in hundreds of dollars) in 201

2 x = Bonus – the total amount of bonuses paid in each territory (in hundreds of dollars) in 2010

3 x = MktShare – percentage of the market share currently held by Meddicorp in each territory

4 x = Compet – largest competitor’s sales (in thousands of dollars) in each territory

Data for a random sample of 25 of Meddicorp’s sales territories are contained in the file Meddicorp.xlsx

Territory Sales Adv Bonus MktShare Compet

1 963.50 374.27 230.98 33 202.22

2 893.00 408.50 236.28 29 252.77

24 1583.75 583.85 289.29 27 313.44

25 1124.75 499.15 272.55 26 374.11

(a) What is the hypothesized regression model?

44332211 x x x x y

(b) Interpret the equation of the true population surface.

The population regression equation

44332211 x x x x y

shows that the conditional mean of y given 4321 and,,, x x x x is a point on the four-dimensional

hyperplane described by 44332211 x x x x .


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(c) Below are the least squares regression results. Conduct the F test for overall fit of the regression.

Regression of y on1

x (Adv),2

x (Bonus),3

x (MktShare),4

x (Compet)


Multiple R 0.9269

R Square 0.8592



Observations 25

ANOVA


Regression 4 1073118.5420 268279.6355 30.5114 0.0000

Residual 20 175855.1980 8792.7599

Total 24 1248973.7400


Intercept -593.5375 259.1959 -2.2899 0.0330 -1134.2105 -52.8644

Adv 2.5131 0.3143 7.9966 0.0000 1.8576 3.1687

Bonus 1.9059 0.7424 2.5673 0.0184 0.3574 3.4545

MktShare 2.6510 4.6357 0.5719 0.5738 -7.0188 12.3208

Compet -0.1207 0.3718 -0.3247 0.7488 -0.8963 0.6549

The hypotheses are:

0: 43210 H

0 oneleastAt: j a H

Because of the small P -value (0) for the F statistic (= 30.51) we reject the null hypothesis

and conclude that at least one of the regression slopes ( 4321 ,,, ) is not equal to zero.

This means that at least one of the variables ( 4321 ,,, x x x x ) is important in explaining the

variation in y.

(d) At the 0.05 significance level, test the significance of the relationship between y and each ofthe explanatory variables.

The P -values for the four t tests are:

0.00 for Adv, 0.02 for Bonus, 0.57 for MktShare, 0.75 for Compet

Thus, the two explanatory variables1

x (amount spent on advertising) and2

x (amount of bonuses)

are related to y (sales). The variables3

x (market share) and4

x (competitor’s sales) are not

useful in explaining any of the variation in y (sales) and should be excluded from the model.


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(e) Below is the regression output for the model with1

x (amount spent on advertising) and

2 x (amount of bonuses). Interpret the estimated regression equation and its slope coefficients.

Regression of y on1

x (Adv),2

x (Bonus)


Multiple R 0.9246R Square 0.8549



Observations 25

ANOVA


Regression 2 1067797.3206 533898.6603 64.8306 0.0000Residual 22 181176.4194 8235.2918

Total 24 1248973.7400


Intercept -516.4443 189.8757 -2.7199 0.0125 -910.2224 -122.6662

Adv 2.4732 0.2753 8.9832 0.0000 1.9022 3.0441

Bonus 1.8562 0.7157 2.5934 0.0166 0.3719 3.3405

After rounding, the least squares regression equation describing the relationship between salesand the two explanatory variables may be written

Predicted Sales = – 516.4 + 2.47 Adv + 1.86 Bonus This equation can be interpreted as providing an estimate of mean sales for a given level ofadvertising and bonus payment.

If bonus payment is held fixed, the equation shows that mean sales tends to rise by $2,470(2.47 thousands of dollars) for each $100 spent on ads.

If advertising is held fixed, the equation shows that mean sales tends to rise by $1,860(1.86 thousands of dollars) for each $100 of bonus paid.

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Predicted Sales

Residual Plot

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Bonus


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(f) The best subsets regression procedure has been performed using all four explanatory variables.Below is a summary of the results. Which is the best model according to the best subsetsregression technique?

Variables in the Regression k + 1 pC 2r 2

ad jr e s

Adv 2 5.90 0.811 0.802 101.42Bonus 2 75.19 0.323 0.293 191.76Compet 2 100.85 0.142 0.105 215.83MktShare 2 120.97 0.001 0.000 232.97

Adv, Bonus 3 1.61 0.855 0.842 90.75Adv, MktShare 3 7.66 0.812 0.795 103.23Adv, Compet 3 7.74 0.812 0.795 103.38Bonus, Compet 3 68.03 0.387 0.332 186.51Bonus, MktShare 3 76.46 0.328 0.267 195.33MktShare, Compet 3 100.18 0.161 0.085 218.20

Adv, Bonus, MktShare 4 3.11 0.859 0.838 91.75

Adv, Bonus, Compet 4 3.33 0.857 0.836 92.26Adv, MktShare Compet 4 9.59 0.813 0.786 105.52Bonus, MktShare, Compet 4 66.95 0.409 0.325 187.48

Adv, Bonus, MktShare, Compet 5 5.00 0.859 0.831 93.71

Recall that small values of pC and values close to k + 1 are of interest in choosing good sets

of explanatory variables.

There are four competing models with relatively small pC values:

Variables in the Regression k + 1 pC 2r 2ad jr e s

Adv, Bonus 3 1.61 0.855 0.842 90.75

Adv, Bonus, MktShare 4 3.11 0.859 0.838 91.75

Adv, Bonus, Compet 4 3.33 0.857 0.836 92.26

Adv, Bonus, MktShare, Compet 5 5.00 0.859 0.831 93.71

The smallest pC value is for the regression with Adv and Bonus as explanatory variables.

It has a pC value of 1.61 and explains 85.5% of the variation in sales.

Note that only modest increases in 2r are achieved in the other three models.

The adjusted 2r is highest and the standard error of estimate is smallest for the regressionwith Adv and Bonus, again supporting this model as best.Therefore, the best subsets procedure suggests using this model.


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(g) Below is the StatTools output from forward selection, backward elimination, and stepwiseregression when applied to the Meddicorp data with P-value to enter = 0.05 and P-value to leave = 0.10. Which model appears to be the best?

Forward selectionMultiple



0.9246 0.8549 0.8418 90.7485



Explained 2 1067797.3206 533898.6603 64.8306 < 0.0001

Unexplained 22 181176.4194 8235.2918

CoefficientStandard



Constant -516.4443 189.8757 -2.7199 0.0125 -910.2224 -122.6662Adv 2.4732 0.2753 8.9832 0.0000 1.9022 3.0441

Bonus 1.8562 0.7157 2.5934 0.0166 0.3719 3.3405

MultipleR-Square

Adjusted StErr of Entry

Step Information R R-Square Estimate Number

Adv 0.9003 0.8106 0.8024 101.4173 1

Bonus 0.9246 0.8549 0.8418 90.7485 2

Backward eliminationMultiple



0.9246 0.8549 0.8418 90.7485



Explained 2 1067797.3206 533898.6603 64.8306 < 0.0001

Unexplained 22 181176.4194 8235.2918

CoefficientStandard



Constant -516.4443 189.8757 -2.7199 0.0125 -910.2224 -122.6662

Adv 2.4732 0.2753 8.9832 0.0000 1.9022 3.0441

Bonus 1.8562 0.7157 2.5934 0.0166 0.3719 3.3405

MultipleR-Square

Adjusted StErr of Exit

Step Information R R-Square Estimate Number

All Variables 0.9269 0.8592 0.8310 93.7697

Compet 0.9265 0.8585 0.8382 91.7508 1

MktShare 0.9246 0.8549 0.8418 90.7485 2


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Stepwise regressionMultiple



0.9246 0.8549 0.8418 90.7485

Degrees of Sum of Mean of F-Ratio p-Value ANOVA Table Freedom Squares Squares

Explained 2 1067797.3206 533898.6603 64.8306 < 0.0001

Unexplained 22 181176.4194 8235.2918

CoefficientStandard



Constant -516.4443 189.8757 -2.7199 0.0125 -910.2224 -122.6662

Adv 2.4732 0.2753 8.9832 0.0000 1.9022 3.0441

Bonus 1.8562 0.7157 2.5934 0.0166 0.3719 3.3405

Multiple

R-Square

Adjusted StErr of Enter or

Step Information R R-Square Estimate Exit

Adv 0.9003 0.8106 0.8024 101.4173 Enter

Bonus 0.9246 0.8549 0.8418 90.7485 Enter

Regardless of the procedure used, the result is the same. The equation chosen is

Predicted Sales = – 516.4 + 2.47 Adv + 1.86 Bonus

(h) Medicorp markets in three regions of the United States: the South, the West, and the Midwest.Management of Meddicorp believes that, in addition to advertising and bonus, the regionsit markets in may be important in explaining variation in sales.What is the equation of the regression model that includes the region information?

Since there are three regions, two indicator variables have to be included in the model.Let Midwest be the base category. Then the two dummy variables are:

otherwise0

Southin theisterritorytheif 1South3 x

otherwise0

Westin theisterritorytheif 1 West4 x

Region 3 x 4 x

South 1 0

West 0 1Midwest 0 0

The model is

44332211 x x x x y


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(i) The regression output for this model follows. Interpret the regression equation from the least squares

Regression of y on1

x (Adv),2

x (Bonus),3

x (South),4

x (West)


Multiple R 0.9730

R Square 0.9468Adjusted R Square 0.9362


Observations 25

ANOVA


Regression 4 1182559.8959 295639.9740 89.0296 0.0000

Residual 20 66413.8441 3320.6922Total 24 1248973.7400


Intercept 435.0989 206.2342 2.1097 0.0477 4.9020 865.2958

Adv 1.3678 0.2622 5.2165 0.0000 0.8208 1.9148

Bonus 0.9752 0.4808 2.0281 0.0561 -0.0278 1.9781

South -257.8916 48.4129 -5.3269 0.0000 -358.8792 -156.9040

West -209.7457 37.4203 -5.6051 0.0000 -287.8032 -131.6883

The regression equation is

Predicted Sales = 435.0989 + 1.3678 Adv + 0.9752 Bonus – 257.8916 South – 209.7457 West

The coefficient of each indicator variable represents the difference in the intercept betweenthe base-level (Midwest) group and the indicated group. This can be expressed through theuse of three separate equations:

South (3

x = 1,4

x = 0): 8916.2579752.03678.10989.435ˆ Bonus Adv y Bonus Adv 9752.03678.12073.177

West (3

x = 0,4

x = 1): 7457.2099752.03678.10989.435ˆ Bonus Adv y Bonus Adv 9752.03678.13532.225

Midwest ( 3 x = 0, 4 x = 0): Bonus Adv y 9752.03678.10989.435ˆ

For given amounts spent by Meddicorp on advertising and bonuses, the estimated mean salesin a territory that is in the South region will be $257,892 (257.8916 thousands of dollars) below the sales in a territory that is in the Midwest region.

For given amounts spent by Meddicorp on advertising and bonuses, the estimated mean salesin a territory that is in the West region will be $209,746 (209.7457 thousands of dollars) below the sales in a territory that is in the Midwest region.

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Residual Plot


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(j) Predict the average sales in each region when advertising expenditures equal 500 hundreds ofdollars and bonuses are 250 hundreds of dollars.

South: 9073.1104)250(9752.0)500(3678.12073.177ˆ y

West: 0532.1153)250(9752.0)500(3678.13532.225ˆ y

Midwest: 7989.1362)250(9752.0)500(3678.10989.435ˆ y

The mean sales figures ($1,104,907 for South, $1,153,053 for West, and $1,362,799 for Midwest)when advertising expenditures equal to $50,000 and bonus payments equal to $25,000 differ accordinto the coefficients of the dummy variables: the figure for South is $257,892 smaller than thefigure for Midwest; the figure for West is $209,746 smaller than the figure for Midwest.

(k) Determine whether there is a significant difference in sales for territories in different regions.

Because the location of territories is measured by two variables in a group – 3

x (South) and

4 x (West) – the appropriate test is the partial F test:

0: 430 H

0and/or0: 43 a H

The full model contains the dummy variables; the reduced model does not.

From the output for the regression of y on1

x (Adv),2

x (Bonus),3

x (South),4

x (West),

SSE(full) = 66413.8441 and MSE(full) = 3320.6922

From the output for the regression of y on1

x (Adv) and2

x (Bonus), SSE(reduced) = 181176.4194

The test statistic is:

2799.176922.3320

2877.57381

6922.3320

2

8441.664134194.181176

(full)

termsextraof number

(full)(reduced)

MSE

SSE SSE

F

n = 25, k = 4, j = 2; df1 = k –

j = 4 – 2 = 2, df2 = n –

k –

1 = 25 – 4 – 1 = 20 P -value = FDIST(17.2799,2,20) = 0.000044The P -value is very small and we reject the null hypothesis.Thus, at least one of the coefficients of the indicator variables is not zero.This means that there are statistically significant differences in average sales levels betweenthe three regions in which Meddicorp does business.

(l) How useful is the group of the dummy variables3

x (South) and4

x (West)?

Do they improve considerably the explanation of variation in sales?

Model 2r 2

ad jr e s

Adv, Bonus 85% 84% 91Adv, Bonus, South, West 95% 94% 58

A comparison of 22 , adjr r , and e s for the reduced and full models shows that the indicator variables

3 x (South) and

4 x (West) carry a lot of explanatory power. They help to explain about 10% more of

the variation in sales while reducing the standard error of estimate by about $33,000.

Therefore, the dummy variables3

x (South) and4

x (West) should be retained in the model.

Date post:	06-Jul-2018
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16 Review of Part II

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