Welcome to Econ 420 Applied Regression Analysis

1. Welcome to Econ 420 Applied Regression Analysis Study Guide Week Five Ending Wednesday, September 26 (Note: Exam 1 is on September 27)

2. Assignment 3Key • Remember • It had 40 points (10 points per question)

3. Question 5, Page 37 • = 182.44 + -4.62 PRICE • = 27.05 + 13.95 RAINFALL • The results for the multiple regression model are: • = 89.09 + 0.064 INCOME –3.18 PRICE + 6.05 RAINFALL • When PRICE is the only independent variable in the model, its slope estimate is • –4.62 instead of –3.18. • When RAINFALL is the only independent variable in the model, its slope estimate is 13.95 instead of 6.05. • The estimates are different because they have different meanings. The –3.18 slope estimate for PRICE in the multiple regression model means that if PRICE increases by one dollar, we can expect DVDEXP to decrease by an average of 3.18, keeping INCOME and RAINFALL constant. • The –4.62 slope estimate for PRICE in the simple regression has the same interpretation except that INCOME and RAINFALL are not being held constant, because they are not in the regression. • The same reasoning is true for the different slope estimates for RAINFALL. • In general, a slope coefficient is estimated keeping the other independent variables constant. In other words, the effects of the other independent variables are accounted for in the model and measured separately from the slope estimate we are considering. • If there is only one independent variable in the model, then no other possible independent variables are being taken into account.

4. Question 6, Page 37 • b. No, it is reasonable to use the predicted value in part a. The values of INCOME, PRICE, and RAINFALL used in part a to find the predicted value are all within the range of values for the three independent variables that exist in the data set (see Table 2.A). If any of the values used for INCOME, PRICE, or RAINFALL in making the prediction were from outside the range for each variable in the data set, then that would be a problem. (See the discussion of extrapolation in “Estimating and Interpreting the Multiple Regression Model: DVD Expenditures” in Section 2.1.)

5. Question 8, Page 37 • Multiply the slope estimate for each independent variable by the change in that independent variable and add the results together to find the expected change in Quinn’s DVD expenditures: (0.064 X100 ) + (-3.18 X 4) +(6.05 X 3) =11.83 b. Multiplying the estimated slope for PRICE by the change in price gives us the expected change in Quinn’s DVD expenditures from the price change: 2.00 X (-3.18) = -6.36. For every dollar increase in income, her expected DVD expenditures increase by 0.064. Therefore she needs a 6.36/0.064=99.38 increase in income to keep her expected DVD expenditures the same. c. Divide the change in expected expenditures by the INCOME slope estimate to find the answer. 30/0.064=468.75.

6. Question 13, Page 40 a. No. If several unbiased estimators are compared, only the estimator with the lowest variance is considered efficient. b. Yes. An estimator is consistent if its estimates approach the true value when sample size becomes very large. Since an unbiased estimator has a sampling distribution centered on the true value, a large sample size will give estimates that approach the true value. All unbiased estimators are consistent. c. No. Consistency is a weaker condition than unbiasedness. If an estimator is unbiased, that is better than if it is only consistent. However, if an estimator is biased but it’s consistent, that is still better than nothing; it’s still better than if the estimator is just biased.

7. Assignment 4 Key(30 points, 10 points each) • Set up the appropriate null and alternative hypotheses for our height- weight equation that we estimated before. Test your hypothesis at alpha = 10 percent. Don’t skip any steps. Evaluate your results. {Note: EViews output includes both the stand errors and the t-stats [for null hypotheses that have zero in them (= 0. ≥0 or ≤0 ]} • #3, Page 61 • # 8, Page 62

8. Question 1 • Our height-weight model: Wi = B^0 + B1^Hi, • The one sided hypothesis is like • H0: B1 ≤ 0 Higher height do not increase the same person’s weight. • HA: B1 > 0 Higher height increases the same person’s weight. • Degrees of freedom = n–k–1 = 8–1–1 = 6. Alpha = 10 percent, one sided, so tc = 1.440. • From EView, we get the estimate regression equation: Wi = – 231 + 5.9 Hi, and t-statistics is 4.47. Therefore, │t│> tc, we reject the null hypothesis, and conclude that with 10% significance level, the higher height will increase the person’s weight.

9. Question 2: #3, Page 61 a. It means that given the estimate of B2 that was found, there is only a 1% chance that the true value of B2 is zero. b. It means that given the estimate of B2 that was found, there is only a 5% chance that the true value of B2 is zero. c. You are more likely to prove statistical significance at the 5% error level. Proving statistical significance at a 1% error level is more convincing, because there is less chance that you are wrong to reject the null hypothesis. You are more likely to reject the null hypothesis when you use a 5% error level instead of a 1% error level because you allowing more room for a Type I error. Using a 1% error level provides a tougher standard for rejecting the null hypothesis, so it is more convincing.

10. Question 3: # 8, Page 62 a. False. There is a 1% change that the null hypothesis is true. We can not say for certain that the null hypothesis is false. b. False. There is a 25% chance that the null hypothesis is true, so it is more likely that the null hypothesis is false. However, the null hypothesis is not rejected in this case because a 25% change of a Type I error is too big of a chance of error. c. False. The null hypothesis is not rejected, but you cannot tell the chance that the null hypothesis is false from the information given. The 5% figure is the probability of Type I error, not Type II error. (It would be a Type II error if the null hypothesis is not rejected and it is actually false.) d. True. (See Table 3.A).

11. Remember the EViews results of height /weight problem • Dependent Variable: W • Method: Least Squares • Date: 09/09/07 Time: 08:15 • Sample: 1 8 • Included observations: 8 • Variable Coefficient Std. Error t-Statistic Prob. • C -231.1391 92.54581 -2.497565 0.0467 • H 5.879205 1.314659 4.472039 0.0042 • R-squared 0.769223 Mean dependent var 181.8750 • Adjusted R-squared 0.730760 S.D. dependent var 32.39681 • S.E. of regression 16.81016 Akaike info criterion 8.694162 • Sum squared resid 1695.489 Schwarz criterion 8.714023 • Log likelihood -32.77665 F-statistic 19.99913 • Durbin-Watson stat 1.128086 Prob(F-statistic) 0.004228 Prob= Pvalue described on Page 53. For example, 0.0042 means that we can reject the null hypothesis that the coefficient of H is zero (two sided test) at a minimum of 4.2%

12. Confidence Interval • The range within which B is likely to fall a specified percentage of the time • 95% of the time B = (B hat ± critical t for two sided test at 5% level of significance * SE of B hat)

13. Limitation of Statistical Significance • It does not provide a theoretical validity • It does not test the importance • The importance is determined based on the absolute value of coefficients

14. Exam 1 • Will cover everything up to Page 56. • Will have two parts: • Part One: will be given in class (Thomas 223) on WebCT at 8:00 AM on Thursday, September 27. it is a closed book closed notes exam. This part will have 60 points. • Part Two: will be an EViews exam. It will be available on Thursday after in class exam and it will be due Friday, September 28 before 10 PM. This part carries 40 points.