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MGEC11 Assignment 3, Suggested Solutions Yue Yu Fall 2023 Question 1 (a) ∂voteA = β 1 ∂log ( expendA ) = β 1 ∂expendA expendA . ∂expendA expendA = 0 . 1 represents a 10% increase in A’s campaign expenditures. β 1 tells us that all else equal, if A’s campaign expenditures increase by 10%, she is expected to receive 0 . 1 β 1 more percentage of the vote. (b) library ( wooldridge ) data ( vote1 ) q1.ols1 <- lm ( voteA ~ lexpendA + lexpendB, data= vote1 ) summary (q1.ols1) ## ## Call: ## lm(formula = voteA ~ lexpendA + lexpendB, data = vote1) ## ## Residuals: ## Min 1Q Median 3Q Max ## -20.4388 -5.1743 -0.6518 4.9251 25.4123 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 52.0389 2.7501 18.92 <2e-16 *** ## lexpendA 6.3420 0.3726 17.02 <2e-16 *** ## lexpendB -6.7568 0.3799 -17.79 <2e-16 *** ## --- ## Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 ## ## Residual standard error: 7.825 on 170 degrees of freedom ## Multiple R-squared: 0.7852, Adjusted R-squared: 0.7827 ## F-statistic: 310.7 on 2 and 170 DF, p-value: < 2.2e-16 (c) MLR1 to MLR4. You need to explain each assumption with one sentence. 1

(d) This is an open question. Any variables that both affect candidate A’s vote share and correlate with any one explanatory variable in the regression are acceptable answers. (e) Given that there are multiple explanatory variables in the regression model in a., the direction of the bias is unclear. (f) Null hypothesis: candidate A’s campaign expenditures do not affect her voting outcome, or β 1 = 0 . Alternative hypothesis: candidate A’s campaign expenditures affect her voting outcome, or β 1 = 0 . (g) The t-statistic is 17.02 and the critical value associated with 1% significance level is 2.576. We reject the null hypothesis in f. (h) The t-statistic is 17.02 and the critical value associated with 10% significance level is 1.645. We reject the null hypothesis in f. Question 2 (a) data ( ceosal2 ) q2.ols1 <- lm (lsalary ~ lsales + lmktval, data = ceosal2) summary (q2.ols1) ## ## Call: ## lm(formula = lsalary ~ lsales + lmktval, data = ceosal2) ## ## Residuals: ## Min 1Q Median 3Q Max ## -2.28060 -0.31137 -0.01269 0.30645 1.91210 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 4.62092 0.25441 18.163 < 2e-16 *** ## lsales 0.16213 0.03967 4.087 6.67e-05 *** ## lmktval 0.10671 0.05012 2.129 0.0347 * ## --- ## Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 ## ## Residual standard error: 0.5103 on 174 degrees of freedom ## Multiple R-squared: 0.2991, Adjusted R-squared: 0.2911 ## F-statistic: 37.13 on 2 and 174 DF, p-value: 3.727e-14 2

## Model 2: lsalary ~ lsales + lmktval ## ## Res.Df RSS Df Sum of Sq F Pr(>F) ## 1 176 64.646 ## 2 174 45.310 2 19.337 37.129 3.727e-14 *** ## --- ## Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Null hypothesis: θ 1 = 0 and θ 2 = 0 Alternative hypothesis: Null is not true (hence at least one of them is non-zero) Given the significance level at 5% and the degree of freedom at 169, for this F-test, the critical value would be: qf ( p= 0.05 , df1 = 2 , df2 = 174 , lower.tail= FALSE ) ## [1] 3.047906 Since the F-statistic is 37.129, which is greater than the critical value, we reject the null hypothesis in favor of the alternative hypothesis. Note that you can also compare the p-value (3.727e-14) with the significance level (5%). Since the p-value is less than the significance level, we reject the null hypothesis in favor of the alternative hypothesis. 4