Econ 371Problem Set #3Answer Sheet4.3 In this question, you are told that a OLS regression analysis of average weekly earnings yields the followingestimated model.dAW E = 696.7 + 9.6 × Age, R2= 0.023, SER = 624.1 (1)a. The first question asks you to explain what the coefficient values mean. The coefficient 9.6 shows themarginal effect of Age on AW E; that is, AW E is expected to increase by $9.6 for each additional yearof age. 696.7 is the intercept of the regression line. It determines the overall level of the line, indicatingthe the average weekly earnings for someone just born (i.e., Age = 0). Clearly, one would not want toput much emphasis on this prediction.b. The second question asks what the units of measurement are for the SER. SER is in the same units asthe dependent variable (Y , or AW E in this example). Thus SER is measures in dollars per week.c. This question asks for the units for R2. R2is unit-free.d. This question asks what the regression’s predicted earnings for a 25 year-old worker and a 45 year-oldworker. Our model implies thatdAW E = 696.7 + 9.6 × 25 = $936.7 (2)anddAW E = 696.7 + 9.6 × 45 = $1128.7 (3)e. Part e of the question asks if the regression will give a reliable prediction for a 99 year-old worker. Theanswer in this case is “no.” The oldest worker in the sample is 65 years old. 99 years is far outside therange of the sample data. It is usually inadvisable to use a regression model (particularly a linear one)to make predictions outside the range of the sample data.f. Here you are asked whether it is plausible that the distribution of the errors in the regression arenormal. It is unlikely that the underlying error terms are normal. Indeed, it is probably the case thatthe distribution of earning is positively skewed and has kurtosis larger than the normal. The incomelevels are bounded below by zero (which also will end up bounding the error terms). Also, there arelikely to large outliers on the right-hand side of the distribution (but not the left), due to extreme incomecases such as Bill Gates, Tiger Woods, etc., but no comparable extremes on the left-hand side (againbecause income is bounded below by zero).g. Finally, you are asked what the average value of AW E is in the sample. Sinceˆβ0=¯Y −ˆβ1¯X, then¯Y =ˆβ0+ˆβ1¯X. Thus the sample mean of AW E is 696.7 + 9.6 × 41.6 = $1, 096.06.4.9 This question has two parts.a. First, you are asked to show that ifˆβ1= 0, then R2= 0. But in this case,ˆβ0=¯Y , so thatˆYi=ˆβ0=¯Y ,yielding ESS = 0 and R2= 0.b. Second, you are asked if R2= 0 implies thatˆβ1= 0. If R2= 0 , then ESS = 0, so thatˆYi=¯Y for alli. ButˆYi=ˆβ0+ˆβ1Xiso that¯Y =ˆβ0+ˆβ1Xifor all i, which implies that eitherˆβ1= 0 or that Xiisconstant for all i. If Xiis constant for all i, thenPni=1(Xi−¯X)2= 0 andˆβ1is undefined (see equation(4.7)).5.1 In this question, you are told that an OLS regression analysis of test scores on class size (CS) yields.dT estScore =520. 4 − 5. 82CS, R2= 0.08, SER = 11.5(20.4) (2.21)a. The first part of the question asks you to construct a 95% confidence interval for β1. This is given by:ˆβ1± 1.96[SE(ˆβ1)] = −5.82 ± 1.96(2.21) = (−10.152, −1.4884) (4)1b. This second part of the question asks you to compute the p-vale associated with the hypothesis H0:β1= 0. The p-value is computed using:p − value = 2Φ¡−¯¯tact¯¯¢= 2ΦÃ−¯¯¯¯¯ˆβ1− 0SE(ˆβ1)¯¯¯¯¯!= 2Φµ−¯¯¯¯−5.822.21¯¯¯¯¶= 2Φ (−2.6335)= 2(0.0042)= 0.0084.Clearly we would reject the null hypothesis at both the 5% and 1% levels.c. In part c, you are asked to test the null hypothesis that H0: β1= −5.61 and to predict whether or not-5.6 would be contained in a 95% confidence interval for β1. The p-value is computed using:p − value = 2Φ¡−¯¯tact¯¯¢= 2ΦÃ−¯¯¯¯¯ˆβ1− (−5.61)SE(ˆβ1)¯¯¯¯¯!= 2Φµ−¯¯¯¯−0.212.21¯¯¯¯¶= 2Φ (−0.1)= 2(0.46)= 0.92.The p-value is larger than 0.10, so we cannot reject the null hypothesis at the 10%, 5% or 1% significancelevel. Because H0: β1= −5.61 is not rejected at the 5% level, this value is contained in the 95%confidence interval.5.7 This question considers a linear regression model with a sample size of n = 250. Specifically, the study findsbY =5.4 + 3.2X, R2= 0.26, SER = 6.2(3.1) (1.5)a. The first part of the question asks you to test the hypothesis H0: β1= 0 at the 5% level. The p-valueis computed using:p − value = 2Φ¡−¯¯tact¯¯¢= 2ΦÃ−¯¯¯¯¯ˆβ1− 0SE(ˆβ1)¯¯¯¯¯!= 2Φµ−¯¯¯¯3.21.5¯¯¯¯¶= 2Φ (−2.13)= 2(0.0166)= 0.0332.Clearly we would reject the null hypothesis at the 5% level since the p-value is less than 0.05.b. This part of the question asks you to construct a 95% confidence interval for β1. This is given by:ˆβ1± 1.96[SE(ˆβ1)] = 3.2 ± 1.96(1.5) = (0.26, 6.14) (5)c. Part c asks if you would be surprised to learn that Yiand Xiare independent. You should be. If Y andX are independent, then beta1= 0; but this null hypothesis was rejected at the 5% level in part (a).22 3 4 5course_eval−2 −1 0 1 2beautyFigure 1: Scatter Plot of CourseEval versus Beautyd. β1would be rejected at the 5% level in 5% of the samples; 95% of the confidence intervals would containthe value β1= 0.The two empirical exercises in this homework use the same dataset: TeachingRatings. The data can be down-loaded from the Web site listed in the assignment (which you can also reach from the class website). A programthat carries all of the tasks for problems E4.2 and E5.2 is appended to this answer sheet.E4.2 The specific questions you are asked to respond to are:a. From Figure 1, we can see that there appears to be a weak positive relationship between course evaluationand the beauty index.b. The regression results are as follows:dCourseEval =4.00 + 0.133 × BeautyThe variable Beauty has a mean that is equal to 0; the estimated intercept is the mean of the dependentvariable (CourseEval) minus the estimated slop e (0.133) times the mean of the regressor (Beauty).Thus, the estimated intercept is equal to the mean of CourseEval.c. Next, you are asked to predict the CoursEval of Watson (Beauty = 0) and Stock (Beauty = 0.789).Using our regression results we have:• Watson’s predicted CoursEval is 4.00 + 0.133 × 0 = 4.00.• Stock’s predicted CoursEval is 4.00 + 0.133 × 0.789 = 4.103.The program provides two different ways of computing these predicted values. One uses the