SAMPLE-Exam AETF5952Questions for Final ExamINSTRUCTIONS TO STUDENTS1. Answer ALL questions and type all your answers. Any answers by image or handwriting will received zeromark.2. When certain decimal places for answers are stated, your answer should be rounded.3. When word limits for answers are stated, your answers should satisfy the limits. If your answer exceedsthe stated limit, you will receive zero mark for the question.1Question 1 (25 points=5+5+5+10)We consider a data set on weekly stock return. The variables in the data set are as follows:• Year: The year that the observation was recorded• Lag1: Percentage return for previous week• Lag2: Percentage return for 2 weeks previous• Lag3: Percentage return for 3 weeks previous• Lag4: Percentage return for 4 weeks previous• Lag5: Percentage return for 5 weeks previous• Volume: Volume of shares traded (average number of daily shares traded in billions)• Today: Percentage return for this week• Direction: A factor with levels Down and Up indicating whether the market had a positive or negativereturn on a given week1. We obtain the sample statistics. Answer the average weekly return.Year Lag1 Lag2 Lag3 Lag4 Lag51996 : 53 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950 Min. :-18.1950 Min. :-18.19502007 : 53 1st Qu.: -1.1540 1st Qu.: -1.1540 1st Qu.: -1.1580 1st Qu.: -1.1580 1st Qu.: -1.16601991 : 52 Median : 0.2410 Median : 0.2410 Median : 0.2410 Median : 0.2380 Median : 0.23401992 : 52 Mean : 0.1506 Mean : 0.1511 Mean : 0.1472 Mean : 0.1458 Mean : 0.13991993 : 52 3rd Qu.: 1.4050 3rd Qu.: 1.4090 3rd Qu.: 1.4090 3rd Qu.: 1.4090 3rd Qu.: 1.40501994 : 52 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260 Max. : 12.0260(Other):775Volume Today Direction timeMin. :0.08747 Min. :-18.1950 Down:484 Min. : 11st Qu.:0.33202 1st Qu.: -1.1540 Up :605 1st Qu.: 273Median :1.00268 Median : 0.2410 Median : 545Mean :1.57462 Mean : 0.1499 Mean : 5453rd Qu.:2.05373 3rd Qu.: 1.4050 3rd Qu.: 817Max. :9.32821 Max. : 12.0260 Max. :10892. We estimate an autoregressitve with lag order of 1 or AR(1) model as reported below. Explain the effectof Lag1 on Today.glm(formula = Today ~ Lag1, data = DATA)Deviance Residuals:Min 1Q Median 3Q Max-19.061 -1.271 0.113 1.280 11.235Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) 0.16120 0.07140 2.258 0.0242 *Lag1 -0.07503 0.03024 -2.481 0.0133 *---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 13. We estimate an autoregressitve with lag order of 5 or AR(5) model as reported below. Explain the effectof Volume on Today (no more than 20 words).2glm(formula = Today ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume,data = DATA)Deviance Residuals:Min 1Q Median 3Q Max-18.5410 -1.2622 0.0873 1.2579 11.9316Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) 0.25050 0.09964 2.514 0.0121 *Lag1 -0.07143 0.03046 -2.346 0.0192 *Lag2 0.04760 0.03058 1.556 0.1199Lag3 -0.06857 0.03045 -2.252 0.0245 *Lag4 -0.02215 0.03047 -0.727 0.4674Lag5 0.01406 0.03039 0.463 0.6437Volume -0.05441 0.04274 -1.273 0.2033---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 14. We estimate the model with more variables and also obtain the AIC for all three models, as reportedbelow (fit1, fit2 and fit3 are estimation results from Question1.2, Question1.3, Question1.4, respectively).In terms of R2, the model here is better. Explain why this model is better or worse for predicting stockreturn. (no more than 30 words).glm(formula = Today ~ Lag1 + Lag2 + Lag3 + Lag4 + Lag5 + Volume +Year, data = DATA)Deviance Residuals:Min 1Q Median 3Q Max-18.0478 -1.2246 0.0719 1.2435 12.4662Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) -0.05787 0.34205 -0.169 0.86569Lag1 -0.09849 0.03104 -3.173 0.00155 **Lag2 0.02080 0.03149 0.661 0.50907Lag3 -0.09284 0.03101 -2.994 0.00281 **Lag4 -0.04905 0.03100 -1.582 0.11394Lag5 -0.01007 0.03080 -0.327 0.74389Volume -0.06976 0.15874 -0.439 0.66043Year1991 0.68642 0.47174 1.455 0.14594Year1992 0.20143 0.47144 0.427 0.66926Year1993 0.24403 0.47139 0.518 0.60478Year1994 0.04760 0.47122 0.101 0.91956Year1995 0.78668 0.47503 1.656 0.09800 .Year1996 0.55581 0.47238 1.177 0.23961Year1997 0.74433 0.47768 1.558 0.11948Year1998 0.69017 0.48118 1.434 0.15177Year1999 0.58028 0.48512 1.196 0.23190Year2000 -0.04876 0.49021 -0.099 0.92079Year2001 -0.09390 0.49905 -0.188 0.85078Year2002 -0.39523 0.50943 -0.776 0.43803Year2003 0.64860 0.51383 1.262 0.20712Year2004 0.38829 0.51445 0.755 0.45056Year2005 0.28258 0.54832 0.515 0.60641Year2006 0.52796 0.59237 0.891 0.372993Year2007 0.31644 0.67343 0.470 0.63853Year2008 -0.48258 0.88789 -0.544 0.58689Year2009 0.97455 0.99170 0.983 0.32598Year2010 0.69020 0.84774 0.814 0.41573---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1> c(AIC(fit1), AIC(fit2), AIC(fit3) )[1] 4956.627 4956.663 4969.6274Question 2 (25 points=5+5+5+10)We consider a data set on credit card default. The variables in the data set are• default: A factor with levels No and Yes indicating whether the customer defaulted on their debt• student: A factor with levels No and Yes indicating whether the customer is a student• balance: The average balance that the customer has remaining on their credit card after making theirmonthly payment• income: Income of customer1. We obtain summary statistics and a frequency table as below. Obtain the probability of default = YESand Student = NO and the probability of default = YES condition on student = NO (2 decimal places).> summary(DATA)default student balance incomeNo :9667 No :7056 Min. : 0.0 Min. : 772Yes: 333 Yes:2944 1st Qu.: 481.7 1st Qu.:21340Median : 823.6 Median :34553Mean : 835.4 Mean :335173rd Qu.:1166.3 3rd Qu.:43808Max. :2654.3 Max. :73554> table(DATA$default, DATA$student)No YesNo 6850 2817Yes 206 1272. We obtain an estimate result as below. Interpret the estimated coefficient of income.glm(formula = default ~ balance + income + student, family = binomial, data = DATA)Deviance Residuals:Min 1Q Median 3Q Max-2.4691 -0.1418 -0.0557 -0.0203 3.7383Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) -1.087e+01 4.923e-01 -22.080 balance 5.737e-03 2.319e-04 24.738 income 3.033e-06 8.203e-06 0.370 0.71152studentYes -6.468e-01 2.363e-01 -2.738 0.00619 **---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 13. Using the previous estimation result, interpret the estimated coefficient of student.4. We obtain another estimate result as below. Interpret the effect of balance in the result (no more than 30words, 4 decimal places).glm(formula = default ~ balance * student + income, family = binomial, data = DATA)Deviance Residuals:Min 1Q Median 3Q Max-2.4949 -0.1417 -0.0555 -0.0202 3.75765Coefficients:Estimate Std. Error z value Pr(>|z|)(Intercept) -1.099e+01 5.667e-01 -19.399 balance 5.817e-03 2.938e-04 19.801 studentYes -2.856e-01 8.239e-01 -0.347 0.729income 3.016e-06 8.226e-06 0.367 0.714balance:studentYes -2.184e-04 4.781e-04 -0.457 0.648---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 16Question 3 (25 points=5+5+5+10)We consider a data set on cars. The data set includes the following variables:• mpg: miles per gallon• cylinders: Number of cylinders between 4 and 8• displacement: Engine displacement (cu. inches)• horsepower: Engine horsepower• weight: Vehicle weight (lbs.)• acceleration: Time to accelerate from 0 to 60 mph (sec.)• year: Model year (modulo 100)• origin: Origin of car (1. American, 2. European, 3. Japanese)• name: Vehicle nameThe summary statistics is as follows:mpg cylinders displacement horsepower weight acceleration yearMin. : 9.00 Min. :3.000 Min. : 68.0 Min. : 46.0 Min. :1613 Min. : 8.00 73 : 401st Qu.:17.00 1st Qu.:4.000 1st Qu.:105.0 1st Qu.: 75.0 1st Qu.:2225 1st Qu.:13.78 78 : 36Median :22.75 Median :4.000 Median :151.0 Median : 93.5 Median :2804 Median :15.50 76 : 34Mean :23.45 Mean :5.472 Mean :194.4 Mean :104.5 Mean :2978 Mean :15.54 75 : 303rd Qu.:29.00 3rd Qu.:8.000 3rd Qu.:275.8 3rd Qu.:126.0 3rd Qu.:3615 3rd Qu.:17.02 82 : 30Max. :46.60 Max. :8.000 Max. :455.0 Max. :230.0 Max. :5140 Max. :24.80 70 : 29(Other):193origin nameMin. :1.000 amc matador : 51st Qu.:1.000 ford pinto : 5Median :1.000 toyota corolla : 5Mean :1.577 amc gremlin : 43rd Qu.:2.000 amc hornet : 4Max. :3.000 chevrolet chevette: 4(Other) :3651. We obtain a regression result as below. Explain the effect of weight on miles per gallon (no more than 30words.)glm(formula = log(mpg) ~ log(weight) + cylinders + displacement +horsepower + weight + acceleration + year, data = DATA)Deviance Residuals:Min 1Q Median 3Q Max-0.34100 -0.05913 0.00180 0.05995 0.37563Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) 1.133e+01 1.097e+00 10.334 log(weight) -1.056e+00 1.560e-01 -6.772 4.96e-11 ***cylinders -2.045e-02 1.062e-02 -1.924 0.055082 .displacement 1.166e-04 2.351e-04 0.496 0.620209horsepower -1.656e-03 4.706e-04 -3.520 0.000485 ***weight 1.280e-04 5.935e-05 2.157 0.031676 *acceleration -3.262e-03 3.236e-03 -1.008 0.314127year71 1.542e-03 3.095e-02 0.050 0.960290year72 -2.846e-02 2.996e-02 -0.950 0.342860year73 -5.343e-02 2.721e-02 -1.964 0.050319 .year74 3.420e-02 3.222e-02 1.061 0.2892947year75 3.843e-02 3.131e-02 1.227 0.220567year76 6.273e-02 3.010e-02 2.084 0.037830 *year77 1.163e-01 3.067e-02 3.791 0.000175 ***year78 1.316e-01 2.903e-02 4.533 7.84e-06 ***year79 2.196e-01 3.072e-02 7.148 4.67e-12 ***year80 3.328e-01 3.238e-02 10.280 year81 2.518e-01 3.186e-02 7.904 3.07e-14 ***year82 2.926e-01 3.105e-02 9.423 ---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 12. We apply the lasso procedure, where the dependent variables is log of mpg and the rest of variables in thedata are used as regressors. Explain why we should not simply run standard regression.> x = sparse.model.matrix(log(mpg) ~ ., data=DATA)[,-1]> y = log(DATA$mpg)> dim(x)[1] 392 321> x = sparse.model.matrix(log(mpg) ~ ., data=DATA)[,-1]> x = sparse.model.matrix(log(mpg) ~ ., data=DATA)[,-1]> y = log(DATA$mpg)> dim(x)[1] 392 321> sclasso = gamlr(x, y, family = gaussian, nfold=10)> sum(coef(sclasso) !=0)[1] 583. In outputs from the lasso above, explain whether the lasso works properly (no more than 20 words).4. We obtain a plot after we implement the lasso procedure above. This plot shows trajectories of estimatedcoefficients. Explain why lines spread widely on the left side and does not on the right side (no more than40 words).8Question 4 (25 points=5+5+5+10)We have a data set collected from 506 geographical areas and the data set contains a list of variables:• price: median housing price in US$1,000,• crime: crimes committed per capita• nox: nitric oxide pollution per 100m• rooms: average number of rooms• dist: weighted distance to city center• radial: access index to radial highways (radial),• stratio: average student-teacher ratioWe estimate a regression tree by using price as the dependent variable and the other variables as regressors.The estimation result is reported in Figure 1.Figure 1: Regression Tree Result1. The far-right node has 45 and 6%. Explain those numbers (no more than 20 words.)2. Interpret prediction from the above estimation result for an area with rooms = 4, stratio = 15, dist = 2,crime = 8 and nox = 3 (no more than 20 words).3. Find the node with 18 and 8% and explain characters of observations in the node.4. Explain what possibly causes low housing prices at the far-left terminal node of the regression tree result.9转自：http://www.6daixie.com/contents/13/5216.html