The Multiple Regression Model Onon Khanoi Berhanu Anbesa

2. Estimating a Multiple Regression Model Content 1. Introduction 2. Estimating a Multiple Regression Model 3. Forecasting from a Multiple Regression Model 4. Interval Estimation 5. Hypothesis Testing 6. Polynomial Equations 7. Interaction Variables 8. Goodness-of-fit 9. Summary

Introduction 1 Multiple regression model – econometric model with more than one explanatory variable. Our model is: SALES = ß1+ß2PRICE + ß3 ADVERT+e , where SALES – monthly sales revenue in a given city (1000USD) PRICE – price in that city ADVERT – monthly advertising expenditure in that city (1000USD) ß2 - the change in monthly SALES (1000USD) when price index PRICE is increased by 1 unit ($1) and advertising expenditure ADVERT is held constant ß3 - the change in monthly SALES (1000USD) when advertising expenditure ADVERT is increased by 1 unit ($1000) and price index PRICE is held constant. e – random error term

Opening Eviews work file 1 1. Open Eviews work file - Andy Number of observation Data series Page name 2. Create objects as a group: Sales, Price and Advert 3. Select the name table5_1

Spreadsheet 1 1. Click View → Descriptive Stats → Common Sample

Saving 1 1. Click Freeze to save 2. Click Name 2. Write name: Andy_statistics Table object

Estimating a Multiple Regression Model 2 Two ways: through Quick menu, Object menu and equation Command 1. Using the Quick menu write: sales c price advert Explanatory variables Constant term Dependent variable Click Quick → Estimation Equation Our equation is: SALES = ß1+ß2PRICE + ß3 ADVERT+e

Regression output 2 Name it burger_eqn Coefficient: The negative coefficient on PRICE suggests that demand is price elastic. With advertising held constant, an increase in price of $1 will lead to a fall in monthly revenue of $7,0908; or reduction in price of $1 will lead to an increase in revenue of $7,908. The coefficient on advertising is positive. With price held constant, an increase in advertising expenditure of $1000 will lead to an increase in sales revenue of $1,863. C is not directly interpretable. It is not possible that if both proce and advertising expenditure were zero the sales revenue would be $118,914.

Forecasting from a Multiple Regression Model 3 In addition to providing information about how sales change when price or advertising change, the estimated equation can be used for prediction. We will now forecast hamburger sales revenue for PRICE=5.5 and ADVERT=1.2 b1=C(1) b2=C(2) b3=C(3) Objective is to get Eviews to perform the calculation: SALES=b1 + b2x5.5+b3x1.2

Forecasting from a Multiple Regression Model: A Simple Forecasting Procedure 3 Eviews command is: scalar sales_f=c(1)+c(2)*5.5+c(3)*1.2 You will see: Double click sales_f in work file Forecast result:

Forecasting from a Multiple Regression Model: Using the Forecast option 3 We need to extend the size of work file to accommodate the observations for which we want forecast. Click Proc → Structure/Resize Current Page Change range of data to 76 Range and sample changed to 76 Double click Click Yes

Forecasting from a Multiple Regression Model: Using the Forecast option 3 Open PRICE and change 76 to 5.5 Same steps for ADVERT change to 1.2 Click to open for editing Change NA to 5.5 Click to open for editing Change NA to 1.2 Click to close for editing New value Click to close for editing New value

Forecasting from a Multiple Regression Model: Using the Forecast option 3 Open burger_eqn equation and click Forecast Open salesf and sef as a group Change: Forecast value Add name for standard error Forecast for observation 76 SALES= 77.6555 and se(f)=4.942 Can be used to compute a forecast Interval as SALES ±t(1-α/2,72) x se(f). Click OK

The least squares covariance matrix Interval Estimation: The least squares covariance matrix 4 The variances and covariances of the least squares estimators give us information about the reliability of the estimators. The smaller their variances, the higher the probability that they will produce estimates “near” the true parameter values. Open burger_eqn and click View → Covariance Matrix Var(b1) Cov(b1, b2) Cov(b1, b3) Var(b2) Cov(b2, b3) Var(b3)

Computing interval estimates Interval Estimation: Computing interval estimates 4 To view upper and lower bounds of the interval estimates Type the command one by one : scalar tc=@qtdist(0.975,72) scalar beta2_low=c(2)-tc*@stderrs(2) scalar beta2_up=c(2)+tc*@stderrs(2) scalar beta3_low=c(3)-tc*@stderrs(3) scalar beta3_up=c(3)+tc*@stderrs(3) Value BETA2_LOW -10.09268 BETA2_UP -5.723032 Easy way to get a same result: Open burger_eqn Click View → Coefficient Diagnostics → Confidence Intervals → choose .95 → OK

Interval estimates for linear combinations of coefficients Interval Estimation: Interval estimates for linear combinations of coefficients 4 Big Andy plans to increase advertising expenditure by $800 and drop the price by 40 cents. λ=E(SALES1)-E(SALES0)= -0.4ß2+0.8ß3 1. Open Burger_eqn → click Coefficient Diagnostics → Wald Test – Coefficient Restrictions… 2. Type -0.4*c(2)+0.8*c(3)=0 Estimate

Two-tail tests of significance Hypothesis Testing : Two-tail tests of significance 5 Ho =ß2=0 (no price effect) H1 =ß2≠0 (there is a price effect ) Ho =ß3=0 (no advertising effect) H1 =ß3≠0 (there is an advertising effect ) T-values and p-values for two-tail tests of significance

Two-tail tests of significance Hypothesis Testing : Two-tail tests of significance 5 P-value for ADVERT is 0.0080. We can confirm the p-value giving the command: scalar pval_advert=2*@ctdist(-2.726,72) Open pval_advert In the case of advertising expenditure we reject Ho =ß3=0 at a 5% significance level because p-value of 0.0080 is less than 0.05. If we want to make a decision about Ho = by comparing the calculated value t=2.726 to a 5% critical value. Give command: scalar tc=@qtdist(0.975,72) The answer is tc=1.993, a value that leads us to reject Ho =ß3=0 because 2.726 > 1.993

One-tail tests of significance Hypothesis Testing : One-tail tests of significance 5 Ho =ß2≥0 H1 =ß2<0 Eview command: scalar tc1=@qtdist(0.05,72) Thus, making the test decision by reference to the critical value, we reject Ho =ß2≥0 in favor of H1 =ß2<0 because -7.215 < - 1.666

Testing nonzero values Hypothesis Testing : Testing nonzero values 5 One tail: For advertising to be effective, ß2 must be greater than 1. Thus we test: Ho =ß3≤1 H1 =ß3>1. Eview command: scalar t3=(c(3)-1)/@stderrs(3) result is t=1.263 scalar pval3=1-@ctdist(t3,72) resultis p=0.105 Choosing α=0.05 as our significance level, relevant critical value is t(o.95,72)=1.666. Since 1.263<1.666, we don’t reject Ho. There is insufficient evidence in our sample conclude that advertising will be cost effective.

Testing nonzero values Hypothesis Testing : Testing nonzero values 5 We will now test two-tails test: Ho =ß3=1 H1 =ß3≠1 Open burger_eqn → Coefficient Diagnostics → Wald Test – Coefficient Restrictions Type null hypothesis c(3)=1 Or you can use this command: burger_eqn.wald c(3)

Testing nonzero values Hypothesis Testing : Testing nonzero values 5 T-value denominator T-value numerator t=b3-1/se(b3)

Testing linear combinations of coefficients Hypothesis Testing : Testing linear combinations of coefficients 5 Big Andy’s marketing advisor claims that dropping the price by 20 cents will be more effective for increasing sales revenue than increasing advertising expenditure by $500. -0.2ß2>0.5ß3. Hypothesis: Ho =-0.2ß2-o.5ß3≤ 0 H1 =-0.2ß2-o.5ß3>0 Open burger_eqn → Coefficient Diagnostics → Wald Test – Coefficient Restrictions Write -0.2*c(2)-0.5*c(3)=0 Two-tail p-value As 1.263<1.666, we do not reject Ho. At a 5% significance level, there is not enough evidence to support marketing advisor. You can also use command: burger_eqn.wald -0.2*c(2)-0.5*c(3)=0

Polynomial Equations 6 To take into account the fact that the marginal effect of advertising is most likely a diminishing function of advertising, the squared value of advertising is added to Andy’s SALES equation. SALES = ß1+ß2PRICE + ß3 ADVERT+ß4 ADVERT2+e Click Quick →Equation Estimation → write sales c price advert advert^2 Name it andy_quad

Polynomial Equations 6 Example: the marginal effect of advertising on SALES when ADVERT=2 ($2000 per week). Open ANDY_QUAD → View →Coefficient Diagnostics → Wald Test-Coefficient Restrictions Or use command → andy_quad.wald c(3)+2*c(4)*2=0 Marginal effect when ADVERT=2 If ADVERT=2, the marginal effect of an additional $1000 of advertising expenditure is $1,079

Polynomial Equations 6 The Wald test also reports the t –statistics for the hypothesis that the estimated marginal effect is zero: We cannot reject the null hypothesis that the marginal effect is zero at the 5% level of significance. From the figure you can see that 2 is near to the maximum point where the slope of the fitted quadratic function is zero.

The optimal level of advertising Polynomial Equations The optimal level of advertising 6 Andy’s objective is maximize profit, but not SALES. Thus, advertising should be increased to the point where ß3+2ß4ADVERT0=1 Enter command: andy_quad.wald (1-c(3))/(2*c(4))=0

Interaction Variables 7 Open new work file – pizza4 Our equation is PIZZA = ß1+ß2AGE + ß3 INCOME+e As age might moderate the income effect: PIZZA = ß1+ß2AGE + ß3 INCOME+ß4(AGExINCOME)+e Command: equation pizza_eq.Is pizza c age income age*income

Interaction Variables 7 For individuals of ages 20 and 50, the marginal effect are given by pizza_eq.wald c(3)+c(4)*20=0 pizza_eq.wald c(3)+c(4)*50=0 The results are: The marginal effect of income for the 50-year-old is much smaller than the marginal effect of income for the 20-year-old.

Goodness-of-fit 8 Below the parameter estimates are various summary measures including the R2=0.448258 Omitted the intercept:

Summary 9 Multiple regression model – econometric model with more than one explanatory variable. We did regression SALES = ß1+ß2PRICE + ß3 ADVERT+e We forecasted hamburger sales revenue for PRICE=5.5 and ADVERT=1.2 with SALES=b1 + b2x5.5+b3x1.2 Interval estimation: Computed Confidence Intervals with .95 significance level. Checked how Sales will change when advertising expenditure increased by $800 and the price dropped by 40 cents. Checked hypothesis: 1. Ho =ß2=0 H1 =ß2≠0 Ho =ß3=0 H1 =ß3≠0 3. Ho =ß3≤1 H1 =ß3>1. Polynominal equation with SALES = ß1+ß2PRICE + ß3 ADVERT+ß4 ADVERT2+e Interaction variables with PIZZA = ß1+ß2AGE + ß3INCOME+ß4(AGExINCOME)+e 2. Ho =ß2≥0 H1 =ß2<0 4. Ho =ß3=1 H1 =ß3≠1 5. Ho =-0.2ß2-o.5ß3≤ 0 H1 =-0.2ß2-o.5ß3>0